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MP Board Class 11th Chemistry Important Questions Chapter 13 Hydrocarbons

Hydrocarbons Important Questions

Hydrocarbons Very Short Answer Type Questions

Question 1.
Full form of T.E.L?
Answer:
Tetra Ethyl Lead.

Question 2.
What is obtained when ethylene dibromide is heated with Zn dust?
Answer:
Ethylene.

Question 3.
The smelling substance in L.P.G is?
Answer:
Ethyl mercaptan.

Question 4.
What is the name of the method in which by electrolysis of potassium acetate, methane is formed?
Answer:
Kolbe’s reaction.

Question 5.
Write the name of the reaction

Answer:
Sabatier and Senderens reaction

Question 6.
Write the names of the products formed in following reaction:

Answer:
(A) CH E= CH
(B) CH3 – CHO
(C) CH3 – CH₂ – OH

Question 7.
Write the reaction in which alkane is formed by alkyl halide and sodium
Answer:
Wurtz – Fittig reaction

Question 8.
What is number of a and n bonds in ethene?
Answer:
5o and 1 n.

Question 9.
In benzene, carbon has which hybridization?
Answer:
sp² hybridization

Question 10.
In HC = CH, which hybridization is present in C – C?
Answer:
sp – sp² hybridization

Hydrocarbons Short Answer Type Questions – I

Addition to Alkyne and Alkene Reactions Cheat Sheet, Cheat Sheet for Organic Chemistry PDF Download.

Question 1.
Trans – alkene is formed by the reduction of alkyne by liquor ammonia. Is butene show geometrical isomerism obtained by reduction of 2 – butyne?

2 – butene shows geometrical isomerism.

Question 2.
Despite their I – effect, halogens are o – and p – directing in haloarenes. Explain?
Answer:
Halogen have (-1) and (+R) effect, these groups are deactivating due to their (-1) effect and they are ortho and para directing due to (+R) effect.

Question 3.
Alkenes are more reactive than alkanes. Why?
Answer:
In alkanes there is only σ – bond between C – C but in alkenes there is one and one σ – bond between C = C. Due to lateral overlapping the π – bond is weaker than the σ – bond. Hence, alkenes are more reactive than alkanes. (MPBoardSolutions.com) The bond energy of π – bond lower than the bond energy of π – bond. Due to this difference of bond energy alkenes are more reactive as compared to alkanes.

Question 4.
What are asymmetric carbon?
Answer:
The carbon atom in which four different groups are attached is called asymmetric carbon. Due to this property, compound shows optical activity.

Question 5.
What do you mean by Chirality?
Answer:
Those molecules which are not superimposable on their mirror images are chiral molecules and this property is called chirality. They are optically active. The Chirality due to presence of asymmetric carbon in the molecule.

Question 6.
What are Alkanes? Which type of bonds are present in it?
Answer:
Alkanes are saturated hydrocarbons because of their low reactivity, they are also called paraffin. The general formula is C_nH_2n+2, In alkanes each carbon atom is sp³ hybridized. Single σ – bonds are present between C – C and C – H.
Example: Methane CH₄, Ethane C₂H₆.

Question 7.
What are Alkenes? In Alkenes C is present in which hybridized state?
Answer:
A saturated hydrocarbon becomes unsaturated when two hydrogens are less in it. Such hydrocarbons are called olefins. In IUPAC system these olefins are called alkenes. Such alkenes are unsaturated and they contain C = C double bond. General formula is
The hybridization of carbon in C = C is sp².
Example: Ethene CH₂ = CH₂, Propylene CH₃ – CH = CH₂.

Question 8.
What are Alkynes? What type of bonds are present in them?
Answer:
Decrease in four hydrogens in saturated hydrocarbons result in the formation of triple bonds between two carbon atoms. The unsaturated hydrocarbon produced known as alkynes. (MPBoardSolutions.com) Their general formula is C_nH_2n-2. Carbon atom of alkyne is sp³ hybridized. Alkynes are also called acetylenes. They contain carbon – carbon triple bond.
Example: Acetylene CH₂ = CH₂, Propyne CH₃ – C = CH₂.

Question 9.
Why Cyclopropane is more reactive than cyclohexane?
Answer:
In cyclopropane the bond angle in C – C – C is 60° due to which ring feelstrain, and so it is reactive and less stable. Whereas in cyclohexane C – C – C have 109°28 and have less strain in the ring, therefore it is stable and less reactive.

Question 10.
Explain functional isomerism with example?
Answer:
The compound having same molecular formula but different functional groups in the molecule are called functional group isomers.
Example:
1. Alcohols and ethers (C₂H_2n+2 – O)

Question 11.
What is dissociation?
Answer:
The method of separation of dor l image isomers from racemic mixture is called dissociation. This is done through biochemical or chemical methods.

Question 12.
How will you obtain nitrobenzene from acetylene?
Answer:

Question 13.
What is Prototrophy?
Answer:
By hydration of Propyne, enol and keto forms are obtained. It shows tautomerism and this type of isomerism is present in that compounds which have at least one hydrogen. This isomerism is due to the transfer of proton from one place to another. This is called prototrophy.

Question 14.
What is conformational stereioisomerism?
Answer:
The phenomenon of easily interconvertibility at room temperature due to free rotation around the carbon – carbon single bond in alkanes and its derivatives is called conformational stereioisomerism.

Question 15.
Explain the process of polymerisation?
Answer:
In polymerisation many simple molecules of a substance combine together to form a big molecule. The simple molecule is called monomer and the big molecule as polymer or macromolecule. Rubber, nylon, bakelite, P.V.C. are examples of high polymers. (MPBoardSolutions.com) Polymerisation of alkenes takes place in presence of Lewis acid BF₃, AlCl₃ or organic and inorganic peroxides.

Polyethylene is used as electrical insulator and as packing materials.

Hydrocarbons Short Answer Type Questions – II

Question 1.
What is Cracking? Write a note on cracking and its uses?
Answer:
Cracking or Pyrolysis:
Higher hydrocarbons when heated to high temperature decomposes into smaller hydrocarbons (lower carbon atom containing molecule). This type of thermal decomposition is known as cracking or pyrolysis.

Pyrolysis is related to free radical reaction. Manufacture of oil gas or petrol gas is based on the concept of pyrolysis. For example, dodecane when heated to 973 K gives a mixture of heptane and pentene. Platinum, palladium or nickel is used as catalyst.

Steam phase cracking, catalytic cracking and liquid phase cracking are the different methods of cracking.

Uses of alkanes:

1. Alkanes are used for making carbon black which is used for making ink, black paints and polish.
2. As a gaseous fuel in industries and as L.P.G.
3. Higher alkanes like petrol, kerosene, lubricant, oil, paraffin, wax, etc.obtained from petroleum are useful.
4. Some halogen derivatives like chloroform and carbon tetrachloride are useful in laboratories and industries.
5. Catalytic oxidation of alkanes give important compounds like alcohol, aldehyde, acid, etc.

Question 2.
Explain Dehydrohalogenation and Dehalogenation with example?
Answer:
Dehydrohalogenation:
When alkyl halide is heated with alcoholic KOH, molecule of hydrogen halide is eliminated forming alkene. This reaction is known as dehydro – halogenation.
Example:
CH₃ – CH₂ – CH₂ – Cl + KOH CH₃ – CH = CH₂ + KCl + H₂O.

Dehalogenation:
Vicinal dihalogen derivatives of alkanes are compounds in which halogen atoms are present on adjacent carbon atoms. They are also called, 1,2 – dihalogen derivative. They form alkenes when heated with Zn dust.
This reaction is known as dehalogenation:

Question 3.
Write notes on Friedei – Crafts reaction?
Answer:
Alkylation:
When benzene or its higher homologous are heated with alkyl halide in presence of anhydrous AlCl₃ alkyl derivatives of benzene are formed.

Acetylation:
When benzene or its higher homologous are treated with acid chloride in presence of anhydrous AlCl₃, acyl benzene or aromatic ketones are formed.

Question 4.
What is Lindlar’s Catalyst? Write its uses?
Answer:
Lindlar catalyst is a palladium supported mixture of calcium carbonate and poisoned with sulphur or quinoline.

Uses:
Aikynes react with hydrogen in presence of nickel powder or platinum or palladium catalyst to form alkenes. This process is known as catalytic hydrogenation.

Question 5.
Geometrical isomerism is found in which type of compounds? Explain with example?
Answer:
Geometrical isomers and Geometrical isomerism or cis – trans isomerism:
This type of isomerism is shown by those alkene derivatives in which different groups are attached with carbons linked with double bond. (MPBoardSolutions.com) For example: compound like abc = cba. When similar atom or group is on same side of bond, it is called cis – isomer and when they are on opposite side of bond, it is called trans – isomer. This type of isomerism is called cis – trans isomerism.

Following compounds does not show geometrical isomerism.

Question 6.
What are Dienes, write their types? Explain with example?
Answer:
Diene is an unsaturated hydrocarbon. In this, two double bonds are present between carbon – carbon chain. On the basis of position of double bonds dienes are of three types:
1. Isolated dienes:
Dienes in which more than one single bonds are present between two double bonds in C – C chain.
CH₂ = CH – CH₂ – CH = CH₂

2. Conjugated dienes:
In these dienes the double bonds are present in alternate position in carbon – carbon chain.

3. Cumulative dienes:
In these dienes the double bonds are present continuously on two C atoms.
CH₃ = C = CH – CH₃
CH₃ – CH = C = CH₂.

Question 7.
Write Diel’s – Alder reaction with equation?
Answer:
When a conjugated diene is heated with ethene, then a cyclic compound is formed. This is called Diel’s – Alder reaction.

Question 8.
What is Huckel’s Rule?
Answer:
According to it, all those planar cyclic compounds exhibit aromatic character, whose ring contain (4n + 2) n electrons. Where n is an integer. Therefore, planar cyclic compounds in which there are 2 (n = 0), 6(n = 1), 10 (n = 2), 14 (n = 3) π electrons exhibit aromatic property.

Benzene and Naphthalene are aromatic compounds on the basis of Huckel’s rules several heterocyclic compounds should also show aromatic properties.

Question 9.
Write Kolbe’s method for manufacuture of alkanes?
Answer:
When sodium or potassium salt of carboxylic acids are electrolysed, alkanes are obtained at anode.
CH₃COONa ⇄ CH₃COO^– + Na⁺

At anode:
CH₃COO^– – e^– → CH₃COO\(\overset { \bullet }{ C } \)
CH₃ – COO^. → \(\overset { \bullet }{ C } \)H₃ + CO₂
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) H₃ → C₂H₆

At cathode:
Na⁺ + e^– → Na
2Na + 2H₂O → 2NaOH + H₂

Question 10.
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitution reactions with difficulty?
Answer:
The orbital structure of benzene shows that the π electrons cloud lying above and below the benzene ring is loosely held and therefore it is likely to be attacked by electrophiles which subsequently bring about substitution. The nucleophiles would be repelled by the n – electron ring and hence benzene ring reacts with nucleophiles with difficulty.

Question 11.
Why do alkenes prefer to undergo electrophilic addition reactions while arenes prefer electrophilic substitution reactions? Explain?
Answer:
Alkenes are source of loosely held π – bonds. Due to which they show electrophilic addition reactions. There occurs a tremendous change in the energy during the electrophilic addition of electrons to alkenes therefore they show electrophilic addition reactions.

In Arenes, during the electrophilic addition reactions the aromatic nature of benzene destroyed, but in electrophilic substitution reactions it remains constant. (MPBoardSolutions.com) So the electrophilic substitution reactions are more stable in order of energy.

Question 12.
What is tautomerism? Explain with example?
Answer:
Tautomerism:
This is a special type of functional isomerism in which the isomers differ in the arrangement of atoms but they exist in dynamic equilibrium with each other. For example, acetaldehyde and vinyl alcohol are tautomers which exist in equilibrium as:

It is of different types but the most common among them is the keto – enol tautomerism. This arises due to 1, 3 – migration of a hydrogen atom from one polyvalent atom to other within the same molecule.

Conditions for the molecule to show tautomerism:
1. Presence of electron withdrawing groups such as:

2. Presence pf α – hydrogen:

Question 13.
What is Newman projection formula?
Answer:
Newman projection:
Named after M.S. Newman, who first proposed this method of representing the three – dimensional structure on paper, this is easier to visualize than the one described before. In this projection, the molecule is viewed at the C – C bond head on. (MPBoardSolutions.com) In this formula, front carbon atom is shown by dot and rear carbon atom by a circle. Three hydrogen atoms bonded to the carbon atoms are shown by lines making an angle of 120° with each other.

Newman’s projection formula of ethane shown in fig. gives a planar representation for two – dimensional (2 – D) representation of the molecule. (MPBoardSolutions.com) Staggered form changes into eclipsed form when rotated through 60°. Similarly, eclipsed form also changes into staggered form.

Question 14.
What is Markownikoff s rule?
Answer:
During the addition across unsymmetrical double bond, the negative part of the adding molecule attaches itself to the carbon atom carrying less number of hydrogen atoms.

Markownikoffs rule can be explained on the basis of the mechanism of addition reaction. Stability of intermediate decides the yield of the product. Consider the attacks of H⁺ (an electrophile) on the propene molecule. The two inter – mediate carbocations are formed.

Since, a 2° carbocation (I) is more stable than 1° carbocation (II), therefore, carbocation (I) is predominantly formed. (MPBoardSolutions.com) This carbocation then rapidly undergoes nucleo – philic attack by the Br^– ion forming 2 – bromopropane as the major product. Thus, MarkownikofFs addition occurs through the more stable carbocation intermediate. .

Question 15.
Draw the cis and trans structures of hex – 2 – ene? Which isomer will have higher boiling point and why?
Answer:

As cis isomer is more polar than trans therefore magnitude of dipole – dipole interaction is more than trans. Hence boiling point of cis isomer is more than trans.

Question 16.
An alkene, ‘A’ on ozonolysis gives a mixture of ethanal and pentan – 3 – one. Write structure and IUPAC name of ‘A’? Write structure and IUPAC name of ‘A’?
Answer:
The product of ozonolysis are –

Remove the oxygen atom (= 0) and join the two ends by a double bond, the structure of the alkene ‘A’ is

Question 17.
Explain Racemic mixture with example?
Answer:
The mixture of equal amount of two optical isomers is called Racemic mixture. The rotation of these two isomers are in opposite directions. So it is represented by dl or ±.
Example:

Optical isomerism or Enantiomerism:
This type of isomerism is shown by unsymmetrical compounds. Structural, physical and chemical properties of optical isomers are nearly similar but optical properties are different.(MPBoardSolutions.com) Isomer which rotates plane of polarised light in clockwise direction is called dextrorotatory and one which rotates in anti – clockwise direction is called laevo – rotatory. These two forms are represented as d – and l – or (+) and (-) respectively.
Optical isomers have unsymmetrical i.e., chiral carbon in the molecule.

Question 18.
What is B.H.C.? What is its uses?
Answer:
It is prepared by the chlorination of benzene in the presence of ultraviolet light.

Benzene hexachloride is an addition compound and its y isomer is called Gammaxene. It is an important pesticide. It is also called Lindane or 666.

Question 19.
How will you do the following conversion?

1. Methane to Ethane
2. Ethane to Methane
3. Acetylene to Benzene.

Answer:
1. Methane to Ethane:

2. Ethane to Methane:

3. Acetylene to Benzene:

Question 20.
Write the name of the organic compounds which show geometrical isomerism in cyclic compounds?
Answer:
Some of the cyclic compounds also show geometrical isomerism.
Example:

Question 21.
What is Saytzeff’s rule? Explain with example?
Answer:
Saytzeff’s rule:
According to this:
“If an alkyl halide can eliminate the hydrogen in two different ways, that alkene will be formed in excess in which carbon atoms joined by double bond are more alkylated.

Question 22.
What are the essential conditions for a compound to show aromaticity?
Answer:
The property which gives extra stability to benzene and benzene like compounds is called aromaticity.
Aromaticity is decided by Huckel’s rule. According to Huckel’s rule, a compound will be aromatic if it fulfills the following four conditions:

1. Compound shall be cyclic.
2. Compound should be planar or nearly planar (sp² hybridisation)
3. Compound should be conjugated.
4. Compound should have (4n + 2) π electrons. Where n is a whole number and it may be n = 0, 1, 2, 3,4, 5, 6, …

Question 23.
What is cis and trans isomerism? Explain with example?
Answer:
cis and trans isomerism is also called geometrical isomerism. This isomerism is shown by that compounds which have carbon atoms attached to two different atoms. When the two same groups or hydrogen atoms are present at one side of double bond then the compound is called cis isomer. (MPBoardSolutions.com) But when the groups or H – atoms are present in opposite side of the C double bond then such isomers are called trans isomers.
Example:

Question 24.
Arrange benzene, n – hexane and ethyne in decreasing order of their acidic behaviour? What is the reason for this behaviour?
Answer:
The hybridised state of C in the given compounds are:

The acidic character increases with increase of s – character. So the decreasing order of acidty is:
Ethyne > Benzene > n – hexane.

Question 25.
How would you convert the following compounds into benzene:

1. Ethyne
2. Ethene
3. Hexane.

Answer:
Ethyne:

Ethene:

Hexane:

Question 26.
Explain why the following systems are not aromatic?

Answer:
(i)
 = CH₂ does not have six electrons i.e., (4n + 2) π – electrons in the ring. Therefore, it is not an aromatic compound.

(ii)

Reason: Due to sp³ hybridisation the molecule is not planar. It contains 4π – electrons, So, molecule is not aromatic.

(iii)

It is a conjugated system but does not have (4n + 2) π – electrons.

Hydrocarbons Long Answer Type Questions – I

Question 1.
Explain the conformation of n – butane?
Answer:
Conformations of n – butane:
n – Butane can be considered as dimethyl derivative ethane which is produced when terminal hydrogen of each carbon is replaced by methyl group. To assign conformations to n – butane is a difficult task because it has three carbon – carbon single bonds (one in the middle and two at the ends) which undergo free rotation.

Various conformations of n – butane can be obtained by rotating C₂ or C₃ through 360° in six steps (60° each time). Staggered and eclipsed forms are obtained alternately on rotating C₂ or C₃ by 60°. These conformations are shown below. (MPBoardSolutions.com) Fully eclipsed form is shown in (I) and other eclipsed forms are shown in II and III. On rotation of C₂ – C₃ bond by 120°. The completely staggered form is shown in (IV). It is called Antiform also. Other staggered forms shown in V and VI are called skew or gauche forms. The staggered form IV and gauche forms V and VI are termed conformational diastereoisomers.

Question 2.
Give any four points in favour of Kekule’s formula to show its resonance structures?
Answer:
Factors in favour of Kekule’s structure:
1. Benzene reacts with three molecules of hydrogen to form cyclohexane which proves presence of three double bonds.

2. Benzene reacts with three molecules of chlorine to form benzene hexachloride.

3. Three molecules of acetylene polymerize in a red hot tube to form benzene.
3CH ≡ CH → C₆H₆

4. Oxidation of benzene with air in presence of V₂O₅ gives maleic acid which loses a molecule of water to form maleic anhydride.

Question 3.
A hydrocarbon ‘A’ vapour density is 14, makes the Baeyer’s reagent colourless and perform following reaction:

Write the name and formula of A, B, C and D?
Answer:

(A) → Ethylene
(B) → 1, 2 – dibromoethane or ethylene bromide
(C) → Acetylene
(D) → Acetaldehyde.

Question 4.
Explain the laboratory method of formation of alkene with figure?
Answer:
Laboratory preparation of alkene (ethylene):
Ethylene is obtained in laboratory by heating ethyl alcohol with excess of cone. H₂SO₄ at 170°C.
1. Requirements:
Ethyl alcohol, cone. H₂SO₄, sand bath. Thistle funnel, potassiumm hydroxide, gas jar, stand etc.

2. Chemical equation:
C₂H₅OH + H₂SO₄ → C₂H₅HSO₄ + H₂O

3. Method:
50 cc of ethyl alcohol and 100 cc of conc.H₂SO₄ is taken in a flask, then 8 gm anhydrous Al₂ (SO₄)₃ and 50 gm of sand is added. The mixture of Al₂ (SO₄)₃ and sand checks up formation of foam and facilitates the reaction to take place at 140° C.

Now the flask is fitted with a thermometer, an exit tube and a dropping funnel. Flask is placed on a sand bath and fixed with a stand. The other end of exit tube dips in wash bottle containing NaOH. Another tube from wash bottle leads to a behive shelf placed in a trough of water. (MPBoardSolutions.com) A water jar is Inverted over behive shelf. Flask is heated at a temperature of 150° C and simultaneously mixture of alcohol and cone. H₂SO₄ is added dropwise into the flask. Along with ethylene, CO₂ (by oxidation of alcohol) and SO₂ (by reduction of H₂SO₄) are also present as impurities in flask. These impurities get adsorbed in NaOH solution and pure ethylene is collected in gas jars by downward displacement of water. Ethylene prepared by this method is pure.

4. Labelled diagram:

Question 5.
Explain the Laboratory method of preparation of acetylene? Or, Explain the Laboratory method of preparation of acetylene on following points:

1. Method and Chemical reaction
2. Labelled figure.

Answer:
Preparation of acetylene in laboratory:
Acetylene is prepared by dropping water on calcium carbide. Acetylene obtained by this method contains impurities of PH₃ and NH₃. These are removed by passing the gas through CuSO₄ solution.

Precautions:
Before experiment, air in flask is replaced by oil gas because acetylene forms explosive mixture with air.

Reaction of acetylene with water:
In presence of 1% HgSO₄ and 42% H₂SO₄, acetaldehyde is formed.

Reaction with ammoniacal silver nitrate solution:
When acetylene is passed through ammoniacal AgNO₃ solution, white precipitate of silver acetylide is obtained.

Question 6.
Out of benzene, m – dinitrobenzene and toluene which will undergo nitration most easily, why?
Answer:
CH₃ group is electron releasing while NO₂ group is electron withdrawing. Therefore, maximum electron density will be in Toluene followed by in benzene and least in m – nitrobenzene. Therefore, the ease of nitration decreases in the order:
Toluene > Benzene > m – dinitrobenzene.

Question 7.
How will you obtain:

1. B.H.C. from benzene
2. Acetophenone from benzene
3. P.V.C. from chloroethene
4. Teflon from tetrafluoroethene.

Answer:
1. B.H.C. from benzene:

2. Acetophenone from benzene:

3. P.V.C from chloroethene:

4. Teflon from tetrafluoroethene:

Question 8.
Write notes on following:

1. Sabatier and Senderens reaction
2. Wurtz reaction
3. Duma’s reaction,
4. Swart reaction.

Answer:
1. Sabatier and Senderens reaction:
The reaction of alkene with hydrogen in presence of Ni or Pt, after hydrogenation alkane is obtained.

2. Wurtz reaction:
Reaction of two molecules of alkyl halide with sodium in presence of dry ether, alkane is obtained.

3. Duma’s reaction:
Reaction of sodium salt of monocarboxylic acid with soda lime, after decarboxylation alkanes are formed.

4. Swart reaction:
Reaction of alkyl halide with mercuric chloride, chloroalkanes are obtained. During this reaction the substitution of halogen of alkyl halide occur by Cl.
2C₂H₅ – I + HgF₂ → 2C₂H₅ – F + HgI₂

Question 9.
An unsaturated hydrocarbon ‘A’ adds two molecules of H₂ and on reductive ozonolysis gives butane – 1, 4 – dial, ethanal and propanone. Give the structure of ‘A’. Write its IUPAC name and explain the reaction involved?
Answer:

Thus, the structure of compound ‘A’ may be written as:

Hydrocarbons Long Answer Type Questions – II

Question 1.
Write the equations for the following reactions:

1. Reaction of calcium carbide with water.
2. Reaction of bromine water on ethylene.
3. Heating of ethylene with alkaline KMnO₄.
4. Heating benzene with cone. HNO₃ and cone. H₂SO₄.
5. Heating benzene with methyl chloride in presence of anhydrous AlCl₃

Answer:
1. Reaction of calcium carbide with water:
CaC₂ + 2H.OH → CH = CH + Ca(OH)₂

2. Reaction of ethylene with bromine water:

3. Heating ethylene with alkaline KMnO₄:

4. Heating benzene with cone. HNO₃ and cone. H₂SO₄:

5. Heating benzene with methyl chloride in presence of anhydrous AlCl₃:

Question 2.
How will you obtain following:

1. Acetaldehyde from Acetylene.
2. Mustard gas from Ethylene.
3. Ethane from Grignard reagent
4. Cuprous acetylide from Acetylene.
5. Methane from Aluminium carbide.

Answer:
1. Acetaldehyde from Acetylene:

2. Mustard gas from Ethylene:

3. Ethane from Grignard reagent:

4. Cuprous Acetylide from Acetylene:
CH ≡ CH + Cu₂Cl₂ +2NH₄OH → Cu – C ≡ C – Cu + 2NH₄Cl + 2H₂O

5. Methane from Aluminium carbide:
Al₄C₃ + 12H₂O → 3CH₄ + 4Al(OH)₃.

Question 3.
What is conformation? Describe conformation found in ethane?
Answer:
In alkane the C – C bond is formed by the axial overlapping of sp³ hybrid orbitals of the adjacent carbon atoms. The electron distribution in molecular orbitals of sp³ – sp³ sigma bond is cylindrically symmetrical around the inter nuclear axis. This symmetry permits the free rotation about the bond axis without rupture of the molecule. (MPBoardSolutions.com) This result in a large number of different spatial arrangement of atom or group attached to the carbon atom. Thus, “The different spatial arrangement obtained by the free rotation around the bond axis of a C – C cr bond are called conformers and the molecular geometry corresponding to a conformer is known as conformation.

Conformation of ethane:
If the position of one carbon atom of ethane is fixed in space and the other carbon atom is rotated around the C – C bond, then various conformations of ethane are possible. Out of these the conformers which has the lowest energy is called staggered and the one having highest energy is called eclipsed conformation.

1. Staggered conformation:
In staggered conformation, the hydrogen atom of the two carbon atoms are oriented in such a way that they lie far apart from one another. In other words, they are staggered away with respect to one another.

2. Eclipsed conformation:
In eclipsed conformation, the hydrogen atoms of one carbon are lying directly behind the hydrogen atoms of the other. In other words, hydrogen atoms of one carbon are eclipsing the hydrogen atoms of the other.(MPBoardSolutions.com) The conformations of ethane do not have same stability. The staggered conformation is relatively more stable than the other conformation. The difference in the energy content of staggered and eclipsed conformation is 12.5 kJ mol^-1.

Question 4.
An alkane C₈H₁₈ is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and tertiary bromide?
Answer:
From Wurtz reaction of an alkyl halide gives an alkane with double the number of carbon atoms present in the alkyl halide. Here, Wurtz reaction of a primary alkyl gives an alkene (C₈H₁₆), therefore, the alkyl halide must contain four carbon atoms. Now the two possible primaiy alkyl halides having four carbon atoms each are, I and II.

Since, alkane C₈H₁₈ on monobromination yields a single isomer of tertiary alkyl bromide, therefore, the alkene must contain tertiary hydrogen. This is possible, only if primary alkyl halide (which undergoes Wurtz reaction) has a tertiary hydrogen.

Question 5.
Explain the free radical mechanism of halogenation of alkane?
Answer:
Mechanism of halogenation of alkane:
Halogenation of alkanes proceed through the formation of free radicals. Therefore, it is also called free radical substitution. The reaction proceeds in the following steps:

1. Chain initiation step:
The first step involves the homolytic fission of chlorine molecule to form chlorine free radicals. This fission takes place in the presence of light or heat.

Note that Cl – Cl bond being weaker than C – H bond and the C – C bonds. Therefore, undergo cleavage first.

2. Chain propagation:
Chlorine free radical is produced in the first step, attacks methane molecule forming methyl free radical and HCl. Methyl free radical reacts with other chlorine molecule forming methyl chloride and chlorine free radical.

When sufficient amount of methyl chloride has been formed, the Cl produced in reaction (ii) has a greater chance of colliding with a molecule of CH₃Cl rather than a molecule of CH₄. If such a collision occurs, a new free radical (CH₂Cl) is produced.
CH₃Cl + \(\overset { \bullet }{ C } \) l → CH₂Cl + HCl
This \(\overset { \bullet }{ C } \)H₂ Cl then reacts with Cl₂ to give CH₂Cl₂ and another \(\overset { \bullet }{ C } \) l free radical.
\(\overset { \bullet }{ C } \)H₂Cl + Cl₂ → CH₂Cl₂ + \(\overset { \bullet }{ C } \) l
This process continues till all the hydrogen is removed.

3. Chain termination:
The chain reaction steps if two or different free radicals combine amongst themselves without producing new free radicals. The possible termination steps are as follows:
\(\overset { \bullet }{ C } \) l + \(\overset { \bullet }{ C } \) l → Cl₂
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) H₃ → CH₃ – CH₃
\(\overset { \bullet }{ C } \) H₃ + \(\overset { \bullet }{ C } \) l → CH₃Cl

Question 6.
Explain the conformational isomerism in cyclohexane?
Answer:
Conformations of Cyclohexane:
Like alkanes cyclohexane (a cycloalkane) also exhibits conformational isomerism. Sachse (1890) suggested that if cyclohexane has a planar cyclic hexagonal structure, then it will be highly stable due to the presence of angular strain of the ring composed of sp² hybrid carbon atoms. (MPBoardSolutions.com) However, cyclohexane is quite stable. Sachse suggested that two non – planar models are possible which are free from angular strain. These are called chair and boat conformations as shown in Fig.

These are in a staggered way and eclipsed conformation of alkanes. The chain form is more stable that the boat form by an energy equal to about 7 kcal/mol. In the boat form, like the eclipsed form of ethane, the hydrogen atoms being very close repel each other and the system becomes unstable. (MPBoardSolutions.com) In the boat form there is considerable non – bonded interaction between the flagpole hydrogens and also between other eclipsed hydrogens. As a consequence, this form of cyclohexane can flex into what is known as twist boat form (flexible form) which is stable by about 1.8 kcal/mol than regular boat.



In the chain form, the hydrogen atoms are situated quite far apart from one another. As a result the force of repulsion between the nearest (v icinal) hydrogen atoms is minimum. Thus, out of two main conformations, chair and boat, the chair form is more stable and cyclohexane exists mainly in chair form.

Question 7.
Explain the electrophilic substitution reactions in aromatic hydrocarbons giving two examples?
Answer:
Mechanism of monosubstitution in benzene:
Study of many monosubstitution reactions of benzene show that these reactions follow mechanism of electrophilic substitution. In thesi 40% H₂SO₄tie reagent is an electrophile (E⁺). It can be understood in the following steps:

Step I.
Generation of electrophile:
Dissociation of attacking reagent results in the formation of electrophile (E⁺).
E – Nu → E⁺ + Nu^–

Step II.
Attack of the electrophile on the ring:

Step III.
Abstraction of a proton by the base:
Nucleophile displaces proton from the hybrid in fast step forming desired substituted products. In this way elimination takes place in this step.

In this way in electrophilic substitution reaction of benzene, electrophile is added in first step and then H⁺ ion is eliminated.

Example 1.
Mechanism of nitration of benzene:
Nitration of benzene is done in the ahead steps by treating with a nitrating mixture of concentrated nitric acid and concentrated sulphuric acid:

Step I.
Generation of electrophile (NO₂⁺):

Step II.
Attack of the electrophile on the ring:
Electrophile attacks on the benzene ring forming carbocation. It attains stability by ion resonance.

Step III.
Abstraction of H⁺ ion by the base:
Nucleophile HSO₄^– ion substitutes H⁺ ion of the ring in fast step forming nitrobenzene.

Example 2.
Mechanism of halogenation of benzene:
Mechanism ofhalogenation of benzene can be explained by the action of chlorine on benzene in presence of Lewis acid FeCl₃.

Step I.
Generation of electrophile (Cl⁺):

Step II.
Attack of the electrophile on the ring:
Electrophile attacks on the ring forming carbocation intermediate which is resonance stabilised.

Step III.
Abstraction of H⁺ ion by the base:
FeCl₄^– ion is a base here and it displaces H⁺ ion from hybrid forming desired products in the fast step.

MP Board Class 11 Chemistry Important Questions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 Integrals Ex 7.7 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.7

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 2 Inverse Trigonometric Functions Ex 2.2

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 10 Vector Algebra Ex 10.2 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 10 Vector Algebra Ex 10.2

MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry

Electrochemistry NCERT Intext Exercises

Electrochemistry Class 12 NCERT Solutions help to score more marks.

Question 1.
How would you determine the standard electrode potential of the system Mg²⁺ | Mg ?
Answer:
The standard electrode potential of the system Mg²⁺/ Mg can be determined as explained below. Prepare Mg²⁺/ Mg system by dipping Mg rod in a solution of Mg²⁺ ions and connect it to standard hydrogen electrode.
Mg | Mg²⁺ || H⁺_(aq) | H_2(g); Pt
When magnesium is connected with SHE, oxidation takes place at the Mg electrode. Hence, the potential of the magnesium electrode is taken as – ve The emf of the cell, determined potentiometrically, is equal to the potential of the magnesium electrode because the potential of SHE is taken as zero.

Question 2.
Can you store copper sulphate solutions in a zinc pot?
Answer:
No, it is not possible. The E° values of the copper and zinc electrodes are as follows :
Zn2+(aq) + 2e– → Zn(s) ; E° = – 0·76 V
Cu2+(aq) + 2e– → Cu(s) ; E° = + 0·34 V
This shows that zinc is a stronger reducing agent than copper. It will lose electrons to Cu2+ ions and a redox reaction will immediately set in.
Zn(s) + Cu2+ (aq) → Zn2+(aq) + Cu(s)
Thus, copper sulphate solution cannot be stored in zinc pot.

Question 3.
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Answer:
Substances having higher E^° values than Fe²⁺ ions can oxidise ferrous ions. Thus, Ag⁺, Hg²⁺, Br₂, Cl₂, F₂ etc. can oxidise Fe²⁺ ions to Fe³⁺ ions.

Question 4.
Calculate the potential of the hydrogen electrode in contact with a solution whose pH is 10.
Solution:
For hydrogen electrode, H+ + e– → 1/2H2
Applying Nernst equation,

Question 5.
Calculate the emf of the cell in which the following reaction takes place :
Ni(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag_(s) Given that E^∘_cell = 1.05 V.
Solution:

Question 6.
The cell in which the following reactions occurs :
2Fe³⁺(aq) + 2I^–(aq) → 2Fe²⁺(aq) + I₂(s) has E^∘_cell = 0-236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Solution:
The two half reactions are :
2Fe³⁺ + 2e^– → 2Fe²⁺ and 2I^– → I₂ + 2e^–
For the above reaction, n = 2

How to calculate emf of a cell PDF available in my website.

Question 7.
Why does the conductivity of a solution decrease with dilution?
Answer:
The number of ions per unit volume that carry the current in a solution decreases on dilution. Therefore the conductivity of a solution decreases with dilution.

Question 8.
Suggest a way to determine the \(\wedge_{m}^{0}\) value of water.
Answer:
Conductance of weak electrolytes can be determined by kohlraush’s law. Thus, molar conductance of water of infinite dilution can be determined by the molar conductances of NaOH, HCl and NaCl at infinite dilution.

Question 9.
The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1 Calculate its degree of dissociation and dissociation constant Given \(\lambda_{\left(\mathbf{H}^{+}\right)}^{\circ}\) = 349.6 S cm² mol^-1 and \(\lambda_{\left(\mathrm{H} \mathrm{COO}^{-}\right)}^{\circ}\) = 54.6 S cm² mol.
Solution:

Question 10.
If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ?
Solution:
Q (coulomb) = 1 (ampere) × t (sec)
Q = 0.5 ampere × 2 × 60 × 60
= 3600 C
A flow of IF, i.e. 96500 C is equivalent to flow of 1 mole of electrons
i. e., = 6.023 × 10²³ electrons
3600 C is equivalent to flow of electrons

Question 11.
Suggest a list of metals that are extracted electrolytically.
Answer:
Li, Na, Mg, Ba, Ca, Al

Question 12.
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O^-2₇ ? Consider the reaction :

Answer:
1 mole Cr₂O^-2₇ requires 6 moles electrons for reduction.
∴ Required charge = 6F
= 6 × 96500 coulomb.
= 579000 coulomb.

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer:
Lead storage battery : It is a secondary cell i.e., a cell which is rechargeable because the products of cell reaction sticks to the electrode. It is also called as lead storage cells. It consists of six cells connected in series. Each cell consist of spongy lead anode and a grid of lead packed with lead dioxide (PbO₂) acts as cathode. An aqueous solution of H₂SO₄ (38% by mass) acts as electrolyte. The reactions which takes place at electrodes can be represented as:

Concentration of H₂SO₄ decreases as sulphate ions are consumed to form PbSO₄ during the working of the cell. As a result of this the density of solution also decreases.

Recharging the cell / battery: Lead storage battery can be recharged by connecting it to an external source of direct current. This reverses the flow of electron with the deposition of Pb on the anode and PbO₂ on the cathode. That is, during recharging operation the cell behaves as electrolytic cell. Following reaction occurs during recharging.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer:
Methanol (CH₃OH), propane (C₃H₈)

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Answer:
Formation of carbonic acid takes place on the surface of iron

In presence of H⁺ ion, oxidation of iron takes place

The electrons are used at other spot where reduction takes place.

Overall reaction,

Electrochemistry NCERT Textbook Exercises

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Answer:
A metal with lesser standard potential (more reactive) can displace the other metal from solution of its salts.

Question 2.
Given the standard electrode potentials, K⁺/K = -2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = -0.74V. Arrange these metals in their increasing order of reducing power.
Answer:
The lower the reduction potential, the higher is the reducing power. Hence, the reducing power of the given metals increases in the following order.
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq)+ 2Ag_(s) takes place. Further show :
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Answer:

(i) Anode i.e., Zn electrode is negatively charged.
(ii) Electrons are current carriers.

Question 4.
Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

Calculate the ∆_rG°, and equilibrium constant of the reactions.
Solution:

Question 5.
Write the Nernst equation and emf of the following cells at 298 K :

Solution:

Question 6.
In the button cells widely used in watches and other devices the following reaction takes place:

Solution:

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:
Conductivity: The conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section.

Molar conductivity: Molar conductivity of a solution at a dilution (V) is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm³ of the solution when the electrodes are one cm apart and the area of cross-section of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by Λm.

Variation with concentration: The conductivity of a solution (Both for strong and weak electrolytes) decreases with a decrease in the concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increase in the decrease in the concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increases with decrease in the concentration of the electrolyte. This is because both the number of ions, as well as mobility of ions, increases with dilution. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

Question 8.
The conductivity of 0.20 M solution of KCI at 298 K is 0.0248 Scm^-1. Calculate its molar conductivity.
Solution:

Question 9.
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω What is the cell constant if the conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1.
Solution:

Question 10.
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below :
Concentration/M 0.001 0.010 0.020 0.050 0.100 10² × kSm^-1 1.237 11.85 23.15 55.53 106.74 for all concentrations and draw a plot between Λ_m and C^1/2 Find the value calculate Λ_m and C^1/2 Find the value calculate Λ_m of Λ⁰_m.
Answer:

Λ⁰_m = intercept on Λ_m axis = 124-0 S cm² mol^-1 (on extrapolation to zero concentration)

Question 11.
The conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity and if Λ⁰_m for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant?
Solution:

Question 12.
How much charge is required for the following reductions :
(i) 1 mol of Al³⁺ to Al.
(ii) 1 mol of Cu²⁺ to Cu.
(iii) 1 mol of MnO^–₄ to Mn²⁺.
Solution:
(i) The electrode reaction is :

(ii) The electrode reaction is :

(iii) The electrode reaction is :

Question 13.
How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl₂.
(ii) 40.0 g of Al from molten Al₂O₃.
Solution:

Question 14.
How much electricity is required in coulomb for the oxidation of
1. 1 mole of H₂O to O₂
2. 1 mole of FeO to Fe₂O₃.
Answer:
1. 2H₂O → 4H⁺ + O₂ + 4e
2F of electricity is required for the oxidation of 1 mol of H₂O

2. Fe²⁺ → Fe³⁺ + e
1F of electricity is required for the oxidation of 1 mole of FeO

Question 15.
A solution of Ni(NO₃)₂ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Answer:
According to Faraday’s first law:
W = ZIr [z = \(\frac { M }{ nF } \)]

Question 16.
Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO_3, and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Solution:

Question 17.
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible :
(i) Fe³⁺_(aq) and I^–_(aq)
(ii) Ag⁺_(aq) and Cu_(s)
(iii) Fe³⁺_(aq) and Br^–_(aq)
(iv) Ag_(s) and Fe³⁺_(aq)
(v) Br_2(aq) and Fe²⁺_(aq)
Answer:
A reaction is feasible if EMF of the cell is +ve.
Cathode : At which reduction occurs.
Anode : At which oxidation occurs.

Question 18.
Predict the products of electrolysis in each of the following:
(a) An aqueous solution of AgNO₃ with silver electrodes.
(b) An aqueous solution of AgNO₃ with platinum electrodes.
(c) A dilute solution of H₂SO₄ with platinum electrodes.
(d) An aqueous solution of CuCl₂ with platinum electrodes.
Answer:
(i) At cathode:
The following reduction reactions compete to take place at the cathode
Ag⁺(aq) + e^– → Ag(s); E^θ = 0.80 V
H⁺ (aq) + e^– → \(\frac { 1 }{ 2 }\) H₂ (g); E^θ = 0.00V
The reaction with a higher value of E_θ takes place of the cathode. Therefore, deposition of silver will take place at the cathode.

At anode:
The Ag anode is attacked by NO₃^– ions. Therefore, the silver electrode at the anode dissolves in the solution to from Ag⁺.

(ii) At cathode: Same as above
At anode: Anode is not attackable and hence OH^– ions have lower discharge potential than NO₃^– ions and OH^– ions react to give O₂
OH^– → OH + e^–
4OH → 2H₂O + O₂ (g)
(iii) H₂SO₄ → 2H⁺ + SO^2-₄
HO₂ ⇌ H⁺ + OH^–

At cathode:
2H⁺+ 2e^– → H₂
At anode: 4OH^– → 2H₂O + O₂ + 4e^–
i. e., H₂ will be liberated at cathode and O₂ at the anode.

(iv) CuCl₂ → Cu²⁺+2Cl^–

At Cathode: Cu²⁺ ions will be reduced in preference to H⁺ ions
Cu²⁺ + 2e → Cu
At anode: Cl’ ions will be oxidised in preference to OH^– ions.
2Cl^– → Cl₂ + 2e^–
i.e., Cu will be deposited on the cathode and Cl₂ will be liberated at the anode.

Electrochemistry Other Important Questions and Answers

Electrochemistry Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
If specific conductance = K, R = Resistance, l = distance between the electrodes A = Cross-sectional area of a conductor C_m = mol L^-1, C_eq = g.eq L^-1 then specific resistance of electrolyte will be equal to :

Question 2.
K is equal to :

Question 3.
Molar conductivity Λ_m is equal to :

Question 4.
Cell constant is :

Question 5.
For electrolyte (v₊ = v_– = 1) like NaCI:

Question 6.
Which among the following is not a conductor of electricity :
(a) NaCl_(aq)
(b) NaCl_(s)
(c) NaCl_(mol)
(d) Ag.

Question 7.
By which of the following the resistance of the solution is multiplied to obtain cell constant:
(a) Specific conductance (K^–)
(b) Molar conductance (Λ_m)
(c) Equivalent conductance (Λ_eq)
(d) None of these.

Question 8.
Increase in equivalent conductance of the solution of an electrolyte by dilution is due to:
(a) Increase in ionic attraction
(b) Increase in molecular attraction
(c) Increase in association of electrolyte
(d) Increase in ionization of electrolyte.

Question 9.
If the specific conductance and observed conductance of an electrolyte is same then its cell constant will be :
(a) 1
(b) 0
(c) 10
(d) 1000

Question 10.
Unit of cell constant is :
(a) ohm^-1 cm^-1
(b) cm
(c) ohm cm
(d) cm^-1.

Question 11.
Unit of specific conductance is :
(a) ohm^-1
(b) ohm^-1 cm^-1
(c) ohm^-2 cm^-1 equivalent^-1
(d) ohm^-1 cm^-2.

Question 12.
If the concentration of any solution is C gram equivalent/ litre and specific resistance is A, its equivalent conductance will be :

Question 13.
The coating of layer of zinc on iron to prevent it from corrosion is called :
(a) Galvanization
(b) Cathodic protection
(c) Electrolysis
(d) Photoelectrolysis.

Question 14.
A saturated solution of KNO₃ is used for salt bridge because :
(a) Speed of K⁺ is more than NO^–₃
(b) Speed of NO^–₃ is more than K⁺
(c) Speed of both is nearly same
(d) Solubility of KNO₃ is high in water.

Question 15.
Acts as dipolar in dry cells :
(a) NH₄Cl
(b) Na₂CO₃
(c) pbSO₄
(d) MnO₂

Question 16.
Reduction is known as :
(a) Electronation
(b) De-electronation
(c) Protonation
(d) De-protonation.

Question 17.
What is the path of electric current in a Daniel cell when Zn and Cu electrodes are connected:
(a) From Cu to Zn inside the cell
(b) From Cu to Zn outside the cell
(c) From Zn to Cu inside the cell
(d) From Zn to Cu outside the cell.

Question 18.
In a cell containing Zn electrode and normal hydrogen electrode (NHE), Zn acts like:
(a) Anode
(b) Cathode
(c) Neither cathode nor anode
(d) Both anode and cathode.

Question 19.
If salt bridge is removed from half cells then voltage :
(a) Reduces and becomes zero
(b) Increases
(c) Immediately increases
(d) Does not change.

Question 20.
When lead storage battery is discharged :
(a) SO₂ is released
(b) Pb is manufactured
(c) PbSO₄ is used
(d) H₂SO₄ is used.

Question 21.
Process of Rusting of iron is :
(a) Oxidation
(b) Reduction
(c) Corrosion
(d) Polymerisation.

Question 22.
Value of standard potential of hydrogen electrode is :
(a) Positive
(b) Negative
(c) Zero
(d) No definite value.

Answers:
1. (b), 2. (c), 3. (d), 4. (b), 5. (a), 6 (b), 7. (a), 8. (d), 9. (a), 10. (d), 11. (b), 12. (a), 13. (a), 14. (c), 15. (d), 16. (a), 17. (d), 18. (a), 19. (a), 20. (d), 21. (c), 22. (c).

Question 2.
Fill in the blanks :

1. Acetic acid is a ………………… electrolyte.
2. Conductance of electrolyte ………………… with the increase in temperature.
3. On increasing dilution the value of specific conductance of a solution …………………
4. ………………… decreases with increase in size of ion.
5. Unit of specific resistance is …………………
6. Primary cells cannot be ………………… again.
7. Device which converts chemical energy into electrical energy is known as ………………… cell.
8. Amount of electric current which produces one gram equivalent of a substance is known as …………………
9. In metallic conduction ………………… property remains unchanged.
10. Reciprocal of resistance is known as …………………
11. Conductance of 1 cm cube of a conductor is called …………………
12. 1 Faraday is equal to ………………… coulomb.
13. Rusting of iron is an example of …………………
14. Potential of standard hydrogen is assumed to be …………………

Answers:

1. Weak
2. Increases
3. Decreases
4. Conductance
5. ohm cm
6. Charged
7. Electrochemical
8. Faraday
9. Chemical property
10. Conductance
11. Specific conductance
12. 96500 coulomb
13. Corrosion process
14. 0.0 volt
15. Electrochemical cell.

Question 3.
Match the following :

I.

Answers:

1. (e)
2. (c)
3. (d)
4. (b)
5. (a).

II.


Answer:

1. (c)
2. (d)
3. (b)
4. (a)
5. (e)
6. (f).

Question 4.
Answer in one word / sentence :

1. Give two examples of strong electrolyte.
2. Give two examples of weak electrolytes.
3. What is the effect of temperature on electrolytic conductivity ?
4. Write the formula of Kohlrausch’s law.
5. Write the formula of equivalent conductance.
6. Write the formula of molar conductance.
7. Device which converts electrical energy into chemical energy.
8. What is the potential of both the electrodes of the cell due to which electric current flows in the cell ?
9. What is the potential produced due to redox reaction between a metal electrode and its ions called ?
10. What is the unit of potential difference ?
11. Write the formula of cell constant.
12. Cell which can be recharged are known as.
13. State the unit of Equivalent conductance.
14. What is the chemical composition of rust ?
15. Write the relation between Electromotive force and Equilibrium constant of a cell.
16. What is the name of the reaction in which oxidation and reduction occur simultaneously ?
Answers:
1. Strong electrolyte : HCl, NaOH, NaCl
2. Weak electrolyte : CH₃COOH, H₂CO₃,
3. Conductivity increases
4.

5. Equivalent conductance

6.

7. Electrolytic cell
8. Electromotive force
9. Electrode potential
10. Volt
11. Cell constant = \(\frac { l }{ A } \)
12. Secondary cell
13. Ohm cm² gm eq^-1
14. Fe₂O₃ . xH₂O
15.

16. Redox reaction.

Electrochemistry Very Short Answer Type Questions

Question 1.
Write the definition of Electrochemical cell.
Answer:
System in which chemical energy is converted to electrical energy by oxidation reduction is known as electrochemical cell or voltaic cell.

Question 2.
What is an Electrolytic cell ?
Ans.
The container or system in which electrical energy is passed by which chemical reaction takes place thus, electrical energy is converted to chemical energy is known as electrolytic cell.

Question 3.
What is Electrode potential ?
Answer:
The potential difference developed between the electrodes and electrolyte of an Electrolytic cell is known as Electrode potential.

Question 4.
What is a strong electrolyte ? Write two examples.
Answer:
Electrolyte which completely dissociate in aqueous solution are known as strong electrolyte.
Example : NaCl, KCl, NH₄Cl etc.

Question 5.
What is a weak Electrolyte ? Write two examples.
Answer:
Electrolyte which dissociate partially in aqueous solution are known as weak electrolyte.
Example : NH₄OH, CH₃COOH, HCN etc.

Question 6.
What is meant by standard electrode potential ?
Answer:
Standard electrode potential (E°) of a half cell is the potential difference when one electrode is dipped in molar solution of its ion at 298 K. If electrode is gaseous the pressure of gas must be one atmosphere.

In IUPAC system, reduction potential are known as standard electrode potential.

Question 7.
Write Ohm’s law.
Answer:
According to Ohm’s law “It states that potential difference across the conductor is directly proportional to the current (I) flowing through it” i.e.,
Mathematically, it can be written as :
I α V
V = IR (R = Resistance, unit = ohm, Ω)

Question 8.
What is cell constant ?
Answer:
For a conductivity cell, the ratio of distance between two electrodes (l) and area of cross-section of electrode (A) is called as cell constant.
∴ Cell constant = \(\frac { l }{ A } \) or x = \(\frac { l }{ A } \)
Unit of cell constant = cm^-1.

Question 9.
What is galvanization ? Explain.
Answer:
Iron is coated with the layer of zinc to protect it from rusting. This process is known as galvanization. The galvanized iron articles keep their lustre due to the coating of invisible protective layer of basic zinc carbonate, ZnCO₃.Zn(OH)₂.

Question 10.
What is Electrochemical Equivalent ?
Answer:
Electrochemical equivalent of a substance is that mass of substance released or deposited on an electrode when a current of one ampere is passed for one second.

Electrochemistry Short Answer Type Questions

Question 1.
What is salt bridge ? Write its two functions.
Answer:
‘U’ shaped tube filled with KCl or KNO₃ in Agar-Agar solution or gelatin, is known as salt bridge. It connects the two half cell.
Functions : (i) It allows the flow of current by completing the circuit.
(ii) It maintains the electrical neutrality.

Question 2.
Derive relation between standard electromotive force and equilibrium constant
Answer:
Relation between standard electromotive force and equilibrium constant can be derived using van’t Hoff isochore. For any given reaction equilibrium constant IQ is equal to the ratio of rate constant of forward reaction and rate constant of backward reaction.

Value of equilibrium constant Kc can be calculated using standard free energy change (∆G°) because

Question 3.
What do you understand by oxidation-reduction reactions ?
Answer:
Oxidation-Reduction reactions: Chemical reactions in which valency of elements changes are known as oxidation-reduction reactions. In this process both oxidation and reduction reactions occur simultaneously, in which one of the substance is oxidized and the other substance is reduced. Like

In this reaction FeCl₃ is reduced to FeCl₂ and SnCl₂ is oxidized to SnCl₄. In the reaction valency of Fe decreases and valency of Sn increases.

Question 4.
Write difference between Metallic conduction and Electrolytic conduction.
Answer:
Differences between Metallic conduction and Electrolytic conduction :

Metallic conduction:

1. Metallic conduction takes place by movement of electrons.
2. There is no chemical change.
3. There is no transfer of matter.
4. In metallic conduction conductivity decreases with increase in temperature.

Electrolytic conduction

1. Electrolytic conduction takes place by movement of ions.
2. Due to chemical change decomposition of electrolyte takes place.
3. Transfer of matter takes place as ions.
4. In electrolytic conduction conductivity increases with increase in temperature.

Question 5.
What are the difference between emf (Cell potential) and potential difference
Answer:
Difference between EMF and Potential difference :

EMF / Cell potential:

1. It is the potential difference between the two terminals of the cell when no current is flowing in the circuit, i.e., in an open circuit.
2. It is the maximum voltage which can be obtained from a cell.
3. It can be measured by potentiaometrie method.
4. Work performed by electromotive force is the maximum work done by a cell.
5. It is responsible for continuous flow of current in electric circuit.

Potential difference:

1. It is the difference of the electrodes potentials of the two electrodes when the cell is sending current through the circuit.
2. It is the less than the maximum voltage as it is the difference of electrode potential.
3. It can be measured by simple voltmeter also.
4. Work performed by potential difference is less than the maximum work done by a cell.
5. It is not responsible for the continuous flow of current in circuit.

Question 6.
What is specific conductance ? Give its unit.
Answer:
Specific conductivity: The reciprocal of resistivity is called specific conductiv¬ity. It is defined as the conductance between the opposite faces of one centimeter cube of a conductor. It is denoted by K (kappa).


The specific conductivity of a solution at a given dilution is the conductance of one cm cube of the solution. It is represented by K (kappa).
Note : The specific conductivity of a solu¬tion of electrolyte depends upon the dilution or molar concentration of the solution.

Question 7.
What is resistivity of any solution ?
Answer:
Resistivity : When current flow in the solution through two electrodes the resistance is proportional to length and inversely proportional to cross-sectional area A.

The constant p (rho) is called resistivity or specific resistance.
Unit: If l is expressed in cm, A in cm2 and R in ohm, the unit of resistivity will be

or Resistivity of any solution is the resistance of 1 cm cube.

Question 8.
Differentiate between Electrochemical cell (Galvanic cell) and Electrolytic cell.
Answer:
Differences between Electrochemical and Electrolytic cells :

Electrochemical cell or Galvanic cell:

1. It is a device to convert chemical energy into electrical energy.
2. It consists of two electrodes in different compartments joined by a salt bridge.
3. Redox reactions occurring in the cell are spontaneous.
4. Free energy decreases with operation of cell, i.e., ∆G < 0.
5. Useful work is obtained from the cell.
6. Anode works as negative and cathode as positive electrodes.
7. Electrons released by oxidation process at anode go into external circuit and pass to cathode.
8. To set-up this cell, a salt bridge/porous pot is used.

Electrolytic cell:

1. It is a device to convert electrical energy into chemical energy.
2. Both the electrodes are in same solution.
3. Redox reactions occurring in the cell are non-spontaneous.
4. Free energy increases with operation of cell, i.e., ∆G > 0.
5. Work is done on the system.
6. Anode is positive and cathode is negative.
7. Electrons enter into cathode electrode from external source and leave the cell at anode.
8. No salt bridge is used in this cell.

Question 9.
What is equivalent conductance ?
Answer:
Equivalent conductance:
“Conductance of total ion produced by one gram equivalent of electrolyte in the solution is called equivalent conductance.” It is denoted by Λ_eq.

Question 10.
What is molar conductance ?
Answer:
Molar Conductivity : The molar conductivity of a solution at definite concentration (or dilution) and temperature is the conductivity of that volume which contains one mole of the solute and is placed between two parallel electrodes 1 cm apart and having sufficient area to hold whole of the solution. It is denoted by Λ_m.
Mathematically,
Λ_m = K × V …(1)
Where V is the volume in ml in which one gram mole of substance is dissolved.
If M is molarity or m moles are dissolved in 1000 ml.

Question 11.
Define cell constant Develop a relation between specific conductance and cell constant.
Answer:
Cell constant: In any conductive cell, the distance between two electrodes and surface area of electrode A are constant. The ratio of l and a is called cell constant i.e.

Unit of cell constant is cm^–1 and it is generally expressed by x.
Relation between specific conductance and cell constant : For a conductor, the resistance R is directly proportional to length R and inversely proportional to area of cross – section of electrolyte.

Question 12.
What are the factors which influence the electrical conductance of electrolytes ?
Answer:
Factor which influence electrical conductivity of electrolytes :

The main factor which influence the electrical conductivity are following:

1. Temperature: It influence following interactions.
(a) Interionic attractions: It depends upon the solute-solute interactions. Which is found between the ions of solute.
(b) Solvation of ions: It depends upon solute-solvent interactions. It is relation between ions of solute and solvent molecules.
(c) Viscosity of solvent: It depends upon solvent-solvent interactions. Solvent molecules are related with each other.
With increase in temperature all these three effects decrease and average kinetic energy of ions increases. Thus, with increase of temperature, resistance of solution decreases and hence conductance increases.

2. Nature of electrolyte : The conductance of solution depends upon the nature of electrolyte. On the basis of conductance measurement electrolytes are classified as strong electrolyte and weak electrolyte. Strong electrolytes have high value of conductance even at higher concentration also.

3. Dilution or concentration : It is main factor which influence electrical conductance. Effect of dilution or concentration can be studied indivisually in equivalent conductance, specific conductance and molar conductance. But for a general concept of electrical conductance of solution as the concentration is lowered or dilution increases, electrical conductance of whole solution increases.

Question 13.
On what factors does the various conductivities of an electrolytic solution depend ?
Answer:
Conductivities of electrolytic solution depend on the following factors :

1. Dilution : On increasing dilution, value of specific conductance of a solution decreases, value of equivalent conductance and molar conductance increases.
2. Nature of solvent: A solvent with high dielectric constant has high conductivity and with low dielectric constant has low conductivity.
3. Number of ions present in solution: Conductivity of strong electrolytes is higher than the conductivity of weak electrolytes.
4. Size of ion : In aqueous solution, small ions are heavily hydrated due to which their conductivity decreases.
5. Effect of Temperature: With the increase in temperature conductivity increases.

Question 14.
With the increase in dilution how do specific conductance, Equivalent conductance and molar conductance change ?
Answer:
With the increase in dilution, specific conductance decreases. This is because by the increase in dilution number of ions present in 1 cm cube of solution decreases.
But, Equivalent conductance Λ_eq = K × V
and Molar conductance Λ_m = K × V
Equivalent conductance and molar conductance are the product of specific conductance and dilution. By the increase in dilution (or decrease in concentration) magnitude of K decreases but that of V increases.
Increase in magnitude of V is comparatively much more than the decrease in magnitude of K. Thus, by the combined effect of both, by the increase in dilution Λ_eq and Λ_m increases.

Question 15.
What is an Electrolytic cell and how does it work ?
Answer:
Electrolytic cells : In these cells electric current is supplied through an external source, as a result of which chemical reactions take place which is called electrolysis like : Electrolysis of water, NaCl, Al₂O₃ etc. For example in Solvay trough cell electrode is immersed in sodium chloride solution and electric current is passed due to which NaCl electrolyses.

At mercury cathode sodium is released and at anode chlorine is released. Sodium forms amalgam with mercury and is taken out of the cell.

Question 16.
What is meant by electromotive force of an electrochemical cell ?
Answer:
The difference in electrode potentials of the two electrodes of an electro- chemical cell is known as electromotive force or cell potential. It is expressed in volt.

Due to difference in potential electric current flows from an electrode of lower potential to an electrode of higher potential. EMF of the cell can be expressed in terms of reduction potential as :

Cell potential = Standard electrode potential – Standard electrode potential
of R.H.S. electrode of L.H.S electrode

EMF of a cell is measured by connecting the voltmeter between the two electrodes of a cell. EMF of a cell depend on the concentration of solutions of both half cells and nature of the two electrodes. For example, In Daniel cell, concentration of CuSO₄ and ZnSO₄ solutions in the two half cells is 1M and at 298 K EMF of the cell is 1.10 volt.

Electrochemistry Long Answer Type Questions

Question 1.
What is standard hydrogen electrode ? How is it prepared ?
Answer:
Standard hydrogen electrode : This consists of gas at 1 atmospheric pressure bubbling over a platinum electrode immersed in 1 M HC1 at 25°C (298 K) as shown in figure. The platinum electrode is coated with platinum black to increase its surface. The hydrogen electrode thus con¬structed forms a half cell which on coupling with any other half cell begins to work on the principle of oxidation or reduction. Electrode depending upon the circumstances works both as anode or cathode.

Standard hydrogen electrode (SHE) is arbitrarily assigned a potential of zero.

Question 2.
Derive Nernst Equation for single electrode potential.
Answer:
Value of standard electrode potential given in electrochemical series is applicable only when the concentration of electrolyte is 1M and temperature is 298 K. But in electrochemical cells the concentration of electrolyte is not definite and electrode potential depends on concentration and temperature. In such condition single electrode potential can be expressed by Nernst equation.
For a reduction half reaction, Nernst equation can be expressed as follows :

Where E = Reduction electrode potential
E° = Standard electrode potential (Mⁿ⁺ concentration 1M and at 298 K)
R = Gas constant = 8.31 JK^-1 mol^-1 = Temperature (in kelvin) = 298 K
n = Valency of metal ion, F = 1 Faraday (96,500 coulomb)

Equation (2) is Nernst equation for single electrode potential.

Question 3.
Write the Faraday’s laws of electrolysis.
Answer:
Faraday’s first law of electrolysis : The law states that, “The mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed.”
Thus, if W gm of the substance is deposited on passing Q coulomb of electricity, then
W α Q or W = ZQ
Where, Z is a constant of proportionality and is called electrochemical equivalent of the substance deposited. If a current of I ampere is passed for t second, then Q = I × t. So that,
W = Z × Q = Z × I × t
Thus, if Q = 1 coulomb, I = 1 ampere and t = 1 second, then W = Z. Hence, electrochemical equivalent of a substance may be defined as, “The mass of the substance deposited when a current of one ampere is passed for one second.”

As one faraday (96500 C) deposits one gram equivalent of the substance, hence electrochemical equivalent can be calculated from the equivalent mass.
Faraday’s second law of electrolysis : It states that, “When the same quantity of electricity is passed through solutions of different electrolytes connected in series, the weight of the substances produced at the electrodes are directly proportional to their equivalent mass.”

For example, for CuSO₄ solution and AgNO₃ solution connected in series, if the same quantity of electricity is passed, then

Question 4.
What is rusting of iron ? Describe Electrochemical theory of rusting.
Answer:
Corrosion: Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric condition are called corrosion. Rusting of iron is an example of corrosion, chemically it is Fe₂O₃xH₂O.

Electrochemical theory of rusting :
Anode reaction : On one spot of iron sheet, oxidation takes place and this spot behaves as an anode.

The electrons which are released at this spot travel through the metal and reach another spot on the metal which acts as cathode. These electrons cause the reduction of oxygen in the presence of hydrogen ions (H⁺). H⁺ ions are formed due to decomposition of carbonic acid formed by dissolution of CO₂ in H₂O.

Question 5.
What is corrosion ? Write three factors affecting it and any three methods to prevent it.
Answer:
Corrosion : Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric conditions are called corrosion.

Factors affecting corrosion :

1. Nature of metal: Reactive metal corrodes readily.
2. Impurities in metal: Impure metal corrodes quickly to a greater extent.
3. Environment: If environment around metal contains oxygens, carbon dioxide, moisture, salts and acidic gases like CO₂, SO₂, SO₃ etc. then corrosion occurs quickly.

Methods to prevent corrosion :

1. Barrier protection : In this method, the surface of the metal is coated with paint, oil or grease. Due to this the surface of the metal remain unexposed to atmospheric conditions and hence corrosion is prevented. Surface of the metal can also be protected by :
(i) Coating metal surface with non-corroding metals like nickel and chromium is called electroplating.
(ii) Dipping iron article in molten metal like zinc. This process is called galvanization.

2. Sacrificial protection : The rusting of iron can be prevented by covering the iron with more electropositive metals like zinc. Zinc metal has more tendency to get oxidized as compared to iron. So, iron articles will not be harmed till the layer of zinc present on its surface hence zinc metal is called the sacrificial metal.

3. Antirust solution : The alkaline antirust solution are employed to prevent rusting, alkaline solution prevent the availability of H⁺ ions.

In this method, iron articles are dipped in alkaline sodium phosphate or chromate solution. Due to this an insoluble sticking film of iron phosphate is formed on the surface which prevents rusting.

Question 6.
Describe dry cell with labelled diagram.
Answer:
Dry cell: It is a primary cell based on Leclanche cell invented by G. Leclanche in 1868. In a primary cell, the electrode reactions cannot be reversed by an external source of electrical energy. In this cell, the cell reaction takes place only once i.e., this cell is not rechargeable.

It is generally used in torches, transistors, radios, calculators, tape recorders, etc. It consists of a hollow zinc cylinder which is filled with a paste of NH₄Cl and a little ZnCl₂. This paste is made with the help of water. The zinc cylinder acts as anode while cathode is a graphite rod (Carbon). The carbon rod is surrounded by a black paste of MnO₂ and carbon powder. The zinc cylinder has an outer insulation of cardboard case.

Dry cells are sealed with wax or other material to protect the moisture from evaporation. When the electrodes are connected, the cell operates.

The electrode reactions are complex. Metallic Zn is oxidized to Zn²⁺ and the electrons liberated are left on the container. The reactions which take place at electrodes can be represented as :

This reaction prevents polarization due to formation of ammonia. It also prevents the substantiaL increase of concentration of Zn²⁺ ions which would decrease the cell potential. This potential of dry cell is approximately 1.5 V.

Defect: Due to acidic nature of NH₄Cl zinc container corrodes due to which holes develop through which the chemicals come out.

Nowadays, the cells are made leakage resistant. In it KOH is used in place of NH₄Cl by which zinc does not corrode.

Question 7.
What is Kohlrausch law ? Give its two applications.
Answer:
Kohlrausch in 1875 gave a generalisation known as Kohlrausch’s law, “At infinite dilution when the dissociation of the electrolyte is complete, each ion makes a definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated.”
Or
“The value of molar conductance at infinite dilution is given by the sum of the contributions of ions (cation and anion).”

Where, λ^∞₊ and λ^∞_– are ionic contributions or ionic conductances of cation and anion while v₊ and v_– are the number of cations and anions in the formula unit of electrolyte.

Applications of Kohlrausch’s law :

(i) Calculation of molar conductance at infinite dilution for weak electrolytes :

Molar conductance or equivalent conductance of weak electrolytes cannot be obtained graphically by extrapolation method, since these are feebly ionized. Kohlrausch’s law enables indirect evaluation in such cases. For example, molar conductances of acetic acid can be obtained from the knowledge of molar conductances at infinite dilution of HCl, CH₃COONa and NaCl which are strong electrolytes.

From Kohlrausch’s law, it is clear that

(ii) Determination of degree of dissociation :

Question 8.
Draw a labelled diagram of Daniel cell and explain cell reaction.
Or,
Draw a labelled diagram of electrochemical cell and write cell reaction.
Answer:
Electrochemical cell: In the redox reactions, the transfer of electrons between oxidizing and reducing agents occurs through wire and thus chemical energy changes into electrical energy. The device on which chemical energy changes into electrical energy is called electro chemical cell. These are also known as galvanic or voltaic cells. Working of these cells can be understood with the example of Daniel cell.

Daniel cell : In this cell, Zn rod is dipped in ZnSO4 solution and Cu rod in copper sulphate solution. Both solutions are connected through KC1 salt bridge. When Zn and Cu electrodes are connected by wire and galvanometer, flow of electrons from Zn to Cu occurs. Zinc atoms change into Zn2+ and electrons reach at Cu electrode, where Cu2+ changes into Cu metal and this copper deposits on electrode.

Electrochemistry Numerical Questions

Question 1.
On passing 5 ampere electric current for 30 minute through a container filled with AgNO₃ 10.07 gram silver is deposited, then determine the chemical equivalent of silver. If electrochemical equivalent of hydrogen is 0-00001036 then calculate the equivalent mass of Silver.
Solution:
Given : W = 10-07 gm, i = 5 ampere, t = 30 × 60 second
According to Faraday’s first law W = Zit
∴ 10.7 = Z × 30 × 60

Question 2.
What are weak electrolytes ? Give one example. Find out molar conductivity of LiBr aqueous solution infinite dilution when joint conductance of Li⁺ ion and Br^– ion are 38.7 Scm² mol^-1 and 78.40 Scm² mol^-1 respectively.
Solution:
Weak electrolytes : These are the substances which dissociate only to a small extent.
Examples: CH₃COOH,NH₄OH

Question 3.
What are strong electrolytes ? Find out the molar conductivity of aqueous solution of BaCl₂ at infinite dilution when ionic conductance of Ba⁺² ion and Cl^– ion are 127.30 Scm² mol^-1 and 76.34 Scm² mol^-1 respectively.
Solution:
Strong electrolytes : These are substances which dissociate almost completely into ions under all dilutions.
Examples: NaCl, HCl,CH₃COONa

Question 4.
Calculate the molar conductance of Al₂(SO₄)₃ if


Solution:
By the ionisation of Al₂(SO₄)₃ solution

Question 5.
Molar conductivity of \(\frac { M }{ 30 } \) CH₃COOH is 9.625 mho and molar conductance of CH₃COOH at infinite dilution (Λ^∞_m) is 385 mho. Calculate the percentage of dissociation of \(\frac { M }{ 30 } \) CH₃COOH.
Solution:
Degree of dissociation

Question 6.
Calculate molar conductance of acetic acid at infinite dilution from following values:

Solution:
From Kohlrausch’s law,

MP Board Class 12th Chemistry Solutions

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 9 Differential Equations Ex 9.5 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 9 Differential Equations Ex 9.5

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 Integrals Ex 7.8 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.8

In this article, we share MP Board Class 12th Maths Book Solutions Chapter 7 Integrals Ex 7.10 Pdf, These solutions are solved by subject experts from the latest MP Board books.

MP Board Class 12th Maths Solutions Chapter 7 Integrals Ex 7.10

MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers

Alcohols, Phenols and Ethers NCERT Intext Exercises

Question 1.
Classify the following as primary, secondary and tertiary alcohols :

Answer:
(i) 1°, (ii) 1°, (iii) 1°, (iv) 2°, (v) 2°, (vi) 3°.

Question 2.
Identify allylic alcohols in the above examples.
Answer:
Allylic alcohols are (ii) and (vi).

Question 3.
Name the following compounds according to IUPAC system :

Answer:
(i) 3-chloromethyl-2-isopropyl pentan-1-ol.
(ii) 2,5-Dimethyl hexane-1,3-diol.
(iii) 3-Bromocyclohexan-l-ol.
(iv) Hex-l-en-3-ol.
(v) 2-Bromo-3 -methyl but-2-en- l-ol.

Question 4.
Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?

Answer:

Question 5.
Write structures of the products of the following reactions:

Answer:

(ii) NaBH₄ is weak reducing agent, it reduces aldehyde/ketones and not the esters.
Thus,

Question 6.
Give structures of the products you would expect when each of the following alcohol reacts with (a) HCl -ZnCl₂, (b) HBr and (c) SOCl₂ : (i) Butan-l-ol , (ii) 2-Methylbutan-2-ol.
Answer:
(a) With ZnCl₂-HCl (Lucas reagent) : Butan-l-ol (1° alcohol) does not react with Lucas reagent at room temperature. However, turbidity appears only after heating but- 2-methyl butan-2-ol (3° alcohol) reacts with Lucas reagent at room temperature immediately gives turbidity

(b) With HBr : Both the alcohols react with HBr to give corresponding alkyl bromides.

(c) With SOCl₂ : Both the alcohols react with SOCl₂ to give corresponding alkyl chlorides.

Question 7.
Predict the major product of acid catalysed dehydration of:
(i) 1-methylcyclohexanol and (ii) butan-l-ol.
Answer:

According to Saytzeff rule, the highly substituted product is the major product.

Question 8.
ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer:
(i) Phenoxide ion:


Resonating structures of p-nitro phenoxide ion. In substituted phenols, the presence of electron withdrawing group (-R effect) such as -NO₂ group, increases the acidic strength of phenol, ortho and para nitrophenoxide ions are more stable (because of additional resonance structures show in boxes) than phenoxide ion due to effective delocalisation of negative charge in phenoxide ion. As a result o-and p- nitrophenols are more acidic than phenols.

Question 9.
Write the equations involved in the following reactions :
(i) Reimer – Tiemann reaction
(ii) Kolbe’s reaction.
Answer:
(i) Reimer-Tiemann reaction : When phenol is treated with chloroform in presence of aqueous sodium hydroxide at 60°C, o-Hydroxy benzaldehyde (Salicylaldehyde) and p-Hydroxy benzaldehyde are formed. The ortho-isomer is the major product. This reaction is called Reimer-Tiemann reaction.

If carbon tetrachloride is used in place of chloroform, salicylic acid is obtained as the main product.

(ii) Kolbe- Schmidt reaction : When sodium salt of a phenol is heated with CO₂ at 130°C. (403K) and 4-7 atm pressure, sodium salicylate is formed. This on acidification gives salicylic acid.

Salicylic acid is the starting material for the manufacture of aspirin which is an important analgesic.

Question 10.
Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-oI.
Answer:
Williamson’s synthesis is reaction of alkyl halide (1°) with sodium alkoxide to give ether by S_N2 mechanism. Thus, alkyl halide should be derivde from ethanol and alkoxide ion from 3-methyl pentan-2-ol. The complete reaction is as follows :

Alkyne-reactions-cheat-sheet-summary-for-organic-chemistry-reactions.

Question 11.
Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4-nitrobenzene and why ?

Answer:
Chemically both sets are equally probable. In set (A) the Br group is activated by the electron withdrawing effect of —NO₂ group. Therefore, nucleophilic attack of CH₃ONa followed by elimination of NaBr gives the desired ether.

In set (B) nucleophilic attack of 4-nitrophenoxide ion on methylbromide gives the desired product.

Illustration about Set line Chemical formula Calculator, Electrical panel and Light bulb with concept of idea.

Question 12.
Predict the products of the following reactions:

Answer:

When one of the groups in unsymmetrical ether is tertiary, then the halide formed is tertiary halide.

Alcohols, Phenols and Ethers NCERT Textbook Exercises

Question 1.
Write IUPAC names of the following compounds :


Answer:
(i) 2,2,4-Trimethyl pentan-3-ol
(ii) 5-Ethylheptane-2,4-diol
(iii) Butan-2,3-diol
(iv) Propane-1,2,3-triol
(v) 2-Methylphenol
(vi) 4-Methylphenol
(vii) 2,5-Dimethylphenol
(viii) 2,6-Dimethylphenol
(ix) l-Methoxy-2-methylpropane
(x) Ethoxybenzene
(xi) 1-Phenoxyheptane
(xii) 2-Ethoxybutane.

Question 2.
Write structures of the compounds whose IUPAC names are as follows :
(i) 2-Methylbutan-2-ol
(ii) l-Phenylpropan-2-ol
(iii) 3,5-Dimethylhexane -1, 3, 5-triol
(iv) 2,3 – Diethylphenol
(v) 1 – Ethoxypropane
(vi) 2-Ethoxy-3-methylpentane
(vii) Cyclohexylmethanol
(viii) 3-CycIohexylpentan-3-ol
(ix) Cyclopent-3-en-l-
(x) 4-Chloro-3-ethylbutan-l-ol.
Answer:

Question 3.
(i) Draw the structures of all isomeric alcohols of molecular formula C5H12O and give their IUPAC names.
(ii) Classify the isomers of alcohols in question (i) as primary, secondary and tertiary alcohols.
Answer:

Isomers (ii), (vii) and (viii) contain chiral centers that can exhibit enantiomerism.

Question 4.
Explain, why propanol has a higher boiling point than hydrocarbon, butane?
Answer:
The molecules of butane are held together by weak van der Waals’ force of attraction while those of propanol are held together by stronger intermolecular hydrogen bonding. Therefore, the b.p. of propanol is much higher than that of butane.

Question 5.
Alcohols are comparatively more soluble in water than hydrocarbons of comparable molecular masses. Explain this fact.
Answer:
Hydrogen bonding between alcohol and water molecules is the reason.

Question 6.
What is meant by the hydroboration-oxidation reaction? Illustrate it with an example.
Answer:
The addition of borane followed by oxidation is known as the hydroboration – oxidation reaction. For example, propan- l-ol is produced by the hydroboration – oxidation reaction of propene. In this reaction, propene reacts with deborane (BH₃)₂ to form trialkyl borane as an additional product. This additional product is oxidized to alcohol by hydrogen peroxide in the presence of aqueous sodium hydroxide.

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C₇H₈O.
Answer:
Three isomers are :

Question 8.
While separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which will be steam volatile. Give reason.
Answer:
Intramolecular H-bonding is present in o-hitro phenol and p-nitrophenol. In p- nitrophenol, the molecules are strongly associated due to the presence of intramolecular bonding. Hence, o-nitrophenol is steam volatile.

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Direct addition of water to ethene in presence of acid does not occur. Indirectly, ethene is first passed through cone. H₂SO₄ at room temperature to form ethyl hydrogen-sulfate, which is decomposed by water on heating to form alcohol.

Mechanism : H₂SO₄ → H⁺ + O^Θ SO₂OH

Step-I: Protonation of alkene to form carbo-cation by the electrophilic attack of hydronium ion (H₃O⁺).

Step-II: Nucleophilic attack by water on carbo-cation to yield protonated alcohol.

Step-III: Deprotonation (loss of proton) to form an alcohol:

Question 12.
You are given benzene, cone. H₂SO₄ and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:

Question 13.
Show how will synthesis:
(i) 1-Phenylethanol from a suitable alkene?
(ii) Cyclohexylmethanol using an alkyl halide by an S_N2 reaction?
(iii) Pentan-1-ol using a suitable alkyl halide?
Answer:

Question 14.
Give two reactions that show the acidic nature of phenol. Compare acidity of phenol with that of ethanol.
Answer:
The reactions showing acidic character of phenols are:
(i) Reaction with sodium : Phenol reacts with Na to give H₂ gas.

(ii) Reaction with NaOH : Phenol dissolves in NaOH to give sodium phenoxide and OH

Phenol is more acidic than ethanol. This is due to the reason that phenoxide ion left after the loss of a proton from phenol is stabilized by resonance (for structure refer to text acidic nature of phenol) while ethoxide ion (left after the loss of a proton from ethanol) is not.

Question 15.
Explain, why ortho nitrophenol is more acidic than ortho methoxy phenol?
Answer:
Due to the strong -R and -I effect of -NO₂ group, the electron density in O-H bond decreases, and hence the loss of proton becomes easy.

-R effect results in +ve charge on O-atom and hence facilitates the release of the proton. Moreover, o-nitrophenoxide formed after the loss of a proton is stabilized by resonance.

o-nitrophenoxide ion is stabilized by resonance and hence o-nitrophenol is a stronger acid. On the other hand, due to the +R effect of the —OCH₃ group the electron density in the O—H bond increases and this makes the loss of proton difficult.

Furthermore, after the loss of proton o-methoxy- phenoxide ion left is destabilized by resonance.

The two -ve charge repel each other and therefore destabilize the o-methoxyphenoxide ion. Thus, it is less acidic than o-nitrophenol.

Question 16.
Explain, how does the -OH group attached to a carbon of benzene ring activate it towards electrophilic substitution ?
Answer:
The – OH group is an electron-donating group. Thus, it increases the electron density in the benzene ring as shown in the given resonance structure of phenol. As a result, the benzene ring is activated towards electrophilic substitution.

Question 17.
Give equations of the following reactions:
(i) Oxidation of propan-l-ol with alkaline KMnO₄ solution.
(ii) Bromine in CS₂ with phenol.
(iii) Dilute HNO₃ with phenol.
(iv) Treating phenol with chloroform in presence of aqueous NaOH.
Answer:

Question 18.
Explain the following with an example:
(i) Reimer-Tiemann reaction
(ii) Kolbe’s reaction
(iii) Williamson-ether synthesis
(iv) Unsymmetrical ether.
Answer:
(i) Reimer-Tiemann reaction: When phenol is treated with chloroform in presence of aqueous sodium hydroxide at 60°C, o-Hydroxy benzaldehyde (Salicylaldehyde) and p-Hydroxy benzaldehyde are formed. The ortho-isomer is the major product. This reaction is called the Reimer-Tiemann reaction.

If carbon tetrachloride is used in place of chloroform, salicylic acid is obtained as the main product.

(ii) Kolbe- Schmidt reaction: When sodium salt of phenol is heated with CO₂ at 130°C. (403K) and 4-7 atm pressure, sodium salicylate is formed. This on acidification gives salicylic acid.

Salicylic acid is the starting material for the manufacture of aspirin which is an important analgesic.

(iii) Williamson-1 ether synthesis: This is the best method for the preparation of ethers because both symmetrical and unsymmetrical (Aliphatic as well as aromatic) ethers can be prepared. When haloalkane is treated with sodium alkoxide then ether is formed. It is an example of a nucleophilic substitution reaction in which halide ion (X-^–) of haloalkane is replaced by alkoxy or aroxy group.

Example : Reaction of ethyl bromide with sodium ethoxide gives diethyl ether
C₂H₅Br + C₂H₅ONa → C₂H₅OC₂H₅ + NaBr
(iv) Unsymmetrical ether: An unsymmetrical ether is an ether where two groups on the two sides of an oxygen atom differ (i.e., have an unequal number of carbon atoms.
For example; ethyl methyl ether (CH₃—O—CH₂CH₃)

Question 19.
Write the mechanism of acid dehydration of ethanol to yield ethene.
Answer:
The mechanism of acid dehydration of ethanol to yeild ethene involes the following three steps:

Step 1: Protonation of ethanol to form ethyl oxonium ion :

Step 2: Formation of carbocation (rate determinning step):

Step 3: Elimination of a proton to form ethene:

The acid consumed in step I is released in Step 3. After the formation of ethene, it is removed to shift the equilibrium in a forward direction.

Question 20.
How are the following conversions carried out:
(i) Propene → Propan-2-ol.
(ii) Benzyl chloride → Benzyl alcohol.
(iii) Ethyl magnesium chloride → Propan-1-ol.
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol.
Answer:

Question 21.
Name the reagents used in the following reactions:

1. Oxidation of primary alcohol to a carboxylic acid.
2. Oxidation of a primary alcohol to aldehyde.
3. Bromination of phenol to 2,4,6-tribromo-phenol.
4. Benzyl alcohol to benzoic acid.
5. Dehydration of propan-2-ol to propene.
6. Butan-2-one to butan-2-ol.

Answer:

1. Alkaline KMnO₄
2. PCC (Pyridinium chlorochromate)
3. Aqueous Br₂
4. Na₂Cr₂O₇/H₂SO₄
5. 85% H₃PO₄
6. NaBH₄ or LiAlH₄

Question 22.
Give a reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethanol undergoes intermolecular H-boding due to the presence of the -OH group, resulting in the association of molecules. Extra energy is required to break these hydrogen bonds. On the other hand, methoxymethane does not undergo H-bonding. Hence, the bp of ethanol is higher than that of methoxymethane.

Question 23.
Give IUPAC names of the following ethers:

Answer:
(i) 1-Ethoxy-2-methylpropane
(ii) 2-Chloro-1-methoxyethane
(iii) 4-Nitro anisole
(iv) 1-Methoxypropane
(v) 1 -Ethoxy-4,4-dimethylcyclohexane
(vi) Ethoxybenzene.

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
Answer:

Question 25.
Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.
Answer:
Limitations of Williamson synthesis : (i) Better results are obtained, if the alkyl halide is primary. In case of secondary and tertiary alkyl halides, elimination completes over substitution. If a tertiary alkyl halide is used, an alkene is the only reaction product and no ether is formed. For example, the reaction of CH₃ONa with (CH₃)₃C—Br gives exclusively 2-methylpropene.

It is because alkoxides are not only nucleophile but strong bases as well. They react with alkyl halide leading to elimination reaction. Thus, in order to prepare methyl tertiary butyl ether, we must use methyl halide (primary) and sodium tertiary butoxide.

(ii) Aryl halides and vinyl halides cannot be used as substrate for the preparation of aromatic aliphatic ether because aryl halide and vinyl halides are less reactive towards nucleophilic substitution reaction.

Question 26.
How is 1-propoxypropane synthesized from propan-l-ol ? Write mechanism of this reaction.
Answer:
(a) It can be prepared by Williamson’s synthesis.

(b) It can also be prepared by dehydration of propan-l-ol

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohol is not a suitable method. Give reason.
Answer:
The 1° alcohol gets protonated. Now it is attacked by another alcohol molecule. The mechanism is S_N2 (Nucleophilic attack)

The 2° and 3° alcohols also get protonated. But another molecule of alcohol cannot attack it due to steric hindrance. The protonated alcohol loses a molecule of H₂O to form a stable 2° or 3° carbo-cation. The carbocation prefers to lose a proton to form alkene.

Similarly, the 3° alcohol (CH₃)₃COH forms isobutane.

Question 28.
Write the equation of the reaction with hydrogen iodide :
(i) 1-Propoxypropane
(ii) Methoxybenzene and
(iii) Benzyiethylether.
Answer:

Question 29.
Explain the fact that in aryl alkyl ethers:
(i) The aikoxy group activates the benzene ring towards electrophilic substitution and
(ii) It directs the incoming substituents to ortho and para positions in benzene ring.
Answer:

In aryl alkyl ethers, due to the +R effect of the aikoxy group, the electron density in the benzene ring increases as shown in the following resonance structure.

Thus, benzene is activated towards electrophilic substitution by the aikoxy group.

(ii) It can also be observed from the resonance structure that the electron density increases more at the ortho para positions than at the meta position. As a result, the incoming substituents are directed to the ortho and para positions in the benzene ring.

Question 30.
Write the mechanism of the reaction of HI with methoxymethane.
Answer:

Protonated ether undergoes S_N2 attack by 1- ion and gives a mixture of methyl iodide and methyl alcohol. However, if HI is taken in excess, the methyl’alcohol formed in (ii) is also converted into methyliodide by the following mechanism.

Question 31.
Write equations of the following reactions:
(i) Friedel-Craft’s reaction – alkylation of anisole.
(ii) Nitration of anisole.
(iii) Bromination of anisole in ethanoic acid medium.
(iv) Friedel-Craft’s acetylation of anisole.
Answer:
(i) Friedel-Crafts reaction – alkylation of anisole : Anisole undergoes Friedel- Crafts reaction i.e., the alkyl and aryl group is introduced at ortho and para-positions by reaction with alkyl halide and aryl halide in the presence of anhydrous aluminium chloride (Lewis acid) as catalyst.

(ii) Nitration of anisole : Anisole reacts a mixture of concentrated sulphuric and nitric acids to yield a mixture of ortho and para nitroanisole.

(iii) Bromination of anisole in ethanoic acid medium : Phenetole or anisole undergoes bromination with bromine in ethanoic acid even in absence of iron (III) bromide catalyst. It is due to the activation of benzene ring by the ethoxy group, para-isomer is obtained in 90% yield.

(iv) Friedel-Craft’s acetylation of anisole :

Question 32.
Show how would you synthesize the following alcohols from appropriate alkenes ?

Answer:
(i) Alcohols on dehydration give alkene. Dehydrate the given alcohol. The alkene formed will give the desired alcohol on adding a molecule of H₂O. The addition of H₂O molecule takes place according to MarkownikofFs rule. In case dehydration of the given alcohol gives two alkenes, then see which alkene will give the desired alcohol.

Both the alkene give the desired alcohol on adding a molecule of H₂O.

(ii) Two alkenes are formed. They will give the desired alcohol on adding a H₂O molecule.

Addition of H₂O molecule to pent-1-ene gives the desired alcohol. Remember that -OH group comes to that carbon atom of the double bond which contains less number of H-atoms. In case of pent-2-one, both the carbon atoms of double bond have one H-atom. Therefore, -OH group may come to either of the carbon atom. This alkene will give pentan-2-ol as well as pentan-3-ol.

Addition of H₂O molecule to 2-methylcy-clohexyl but-2-ene will give the desired alcohol -OH group comes to that carbon atom of double bond which contains less number of H- atom.

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

Give a mechanism for this reaction.
(Hint : The secondary carbocation formed in step-II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)
Answer:
Alcohol gets protonated and then a H2O molecule is given out to give a carbocation.

Now, the 2° carbocation rearrange to form more stable 3° carbocation, one H—atom migrates from the adjacent carbon atom to C+. It is called 1,2-shift. Then addition of Br^– takes place.

Alcohols, Phenols and Ethers Other Important Questions and Answers

Alcohols, Phenols and Ethers Objective Type Questions

Choose the correct answer:

Question 1.
Sodium dissolves easily in alcohol because :
(a) Alcohol has higher density than water
(b) Alcohol is lighter than water
(c) Alcohol is neutral
(d) Alcohol is amphoteric.
Answer:
(d) Alcohol is amphoteric.

Question 2.
Most acidic among the four compound is :
(a) Phenol
(b) o-nitrophenol
(c) m – nitrophenol
(d) p – nitrophenol.
Answer:
(c) m – nitrophenol

Question 3.
The reaction for the formation of salicylaldehyde from phenol is:
(a) Rosenmund reaction
(b) Friedel Crafts reaction
(c) Reimer-Tiemann reaction
(d) Wurtz’s reaction.
Answer:
(c) Reimer-Tiemann reaction

Question 4.
Most effective reagent which converts propanol-2 to propanone:
(a) LiAlH₄
(b) Cu/300°C
(c) CO₂
(d) K₂Cr₂O₇.
Answer:
(b) Cu/300°C

Question 5.
Carbolic acid is :
(a) Phenol
(b) Phenyl benzoate
(c) Phenylacetate
(d) Methyl salicylate.
Answer:
(a) Phenol

Question 6.
Which compound is known as oil of wintergreen:
(a) Phenyl benzoate
(b) Phenyl salicylate
(c) Phenylacetate
(d) Salol.
Answer:
(d) Salol.

Question 7.
The reaction of Lucas reagent is fastest with:
(a) (CH₃)₃ C-OH
(b) (CH₃)₂ CHOH
(c) CH₃-(CH₂)₂ OH
(d) CH₃CH₂OH.
Answer:
(a) (CH₃)₃ C-OH

Question 8.
At low temperature phenol reacts with Br₂ in CS₂ to give

Answer:

Question 9.
Which compound is obtained on passing vapours of ethanol on hot Al₂O₃:
(a) Ethyl ether
(b) Acetone
(c) Acetaldehyde
(d) Ethane.
Answer:
(a) Ethyl ether

Question 10.
Which of the compound is Aspirin :
(a) Acetylsalicylic acid
(b) Salicylic acid
(c) Acetamide
(d) Salicylamide.
Answer:
(a) Acetylsalicylic acid

Question 11.
The following compound reacts with phthalic acid to give an acid-base indicator:
(a) Chlorobenzene
(b) Phenol
(c) Alcohol
(d) Ether.
Answer:
(b) Phenol

Question 12.
Bakelite is formed when phenol is condensed with:
(a) HCHO
(b) CH₃CHO
(c) C₆H₅CHO
(d) CH₃COCH₃.
Answer:
(a) HCHO

Question 13.
Used as an anaesthetic:
(a) CH₃OH
(b) C₂H₅OH
(c) CH₃—CHO
(d) (C₂H₅)₂ O.
Answer:
(d) (C₂H₅)₂ O.

Question 14.
Lucas reagent is:
(a) Conc. HCl
(b) Conc. H₂SO₄
(c) Anhydrous ZnCl₂
(d) Conc. HCl and anhydrous ZnCl₂.
Answer:
(d) Conc. HCl and anhydrous ZnCl₂.

Question 15.
Ether and alcohol can be distinguished by the following:
(a) Reaction with Na
(b) Reaction with PCl₅
(c) Reaction with 2, 4 dinitrophenyl hydrazine
(d) None of these.
Answer:
(a) Reaction with Na

Question 16.
Which is used for poisoning the alcohol:
(a) Methyl alcohol
(b) Ethyl alcohol
(c) Glycerine
(d) All the above.
Answer:
(a) Methyl alcohol

Question 17.
Gives Libermann’s nitroso test:
(a) C₆H₅OH
(b) CH₃-OH
(c) C₂H₅—OH
(d) CH₃-O-CH₃
Answer:
(a) C₆H₅OH

Question 18.
Which is identified by Lucas reagent:
(a) Phenol
(b) Ether
(c) Aldehyde
(d) Alcohol.
Answer:
(d) Alcohol.

Question 19.
Alcohols are soluble in water. Its main reason is:
(a) O—H bond
(b) Hydrogen bond
(c) Covalent bond
(d) Electrovalent bond.
Answer:
(b) Hydrogen bond

Question 20.
Which is formed on heating ethyl alcohol with bleaching powder:
(a) Diethyl ether
(b) Phenol
(c) Chlorobenzene
(d) Chloroform.
Answer:
(d) Chloroform.

Question 2.
Fill in the blanks :

1. The general formula of ether is ……………….
2. The main product of Kolbe-Schmidt reaction of phenol is ……………….
3. By the hydrogenation of phenol ………………. is formed.
4. Alcohol reacts with I₂ and a base to give a yellow precipitate of ……………….
5. Phenol on being heated with Zn powder forms ……………….
6. On heating formaldehyde with ………………. bakelite is formed.
7. Diethyl ether is used as an ……………….
8. On heating RX with NaOR, ROR is formed. Name of this reaction is ……………….
9. Alcohol is ………………. whereas phenol is of ………………. nature.
10. On heating alcohol with cone. H₂SO₄ for 160 – 170°C ………………. is formed.
11. By the dehydration of ethyl alcohol ………………. and ………………. are obtained.
12. Rectified spirit is a mixture of ………………. % alcohol and ………………. water.

Answer:

1. C_nH_2n+1OC_nH_2n+1
2. Salicylic acid
3. Cyclohexanol
4. Iodoform (CHI₃),
5. Benzene
6. Phenol
7. Anesthetic
8. Williamson synthesis
9. Neutral, acidic
10. Alkene
11. Ethylene, diethyl ether
12. 95-5%, 4-5%.

Question 3.
Match the following :

Answer:

1. (f)
2. (d)
3. (e)
4. (g)
5. (b)
6. (a)
7. (j)
8. (c)
9. (h)
10. (i).

Question 4.
Answer in one word/sentence:

1. Diethyl ether does not reacts with Na. Why?
2. Fire caused due to ether cannot be extinguished by water. Why?
3. Which is formed on burning ether?
4. Write name of the enzyme which converts maltose to glucose.
5. Sulphuric ether is known as.
6. Reaction of ether with HI is used for the detection of what?
7. Phenol reacts with Br₂ in presence of CS₂ to form.
8. In Victor Meyer method, 1° alcohol gives which colour with base.
9. Phenol reacts with phthalic anhydride in presence of H₂SO₄ to form.
10. Which gas is obtained during fermentation?
11. Name the primary alcohol which gives the iodoform test.
12. Name the reaction in which phenol reacts with chloroform and sodium hydroxide to form salicylaldehyde.

Answer:

1. Absence of acidic H-atom
2. Lighter than water and insoluble
3. CO₂ and H₂O
4. Maltase
5. Diethyl ether
6. Alkoxy (Ziesel)
7. o-and p-bromophenol,
8. Red
9. Phenolphthalein
10. CO₂
11. C₂H₅—OH
12. Reimer-Tiemann reaction.

Alcohols, Phenols and Ethers Short Answer Type Questions

Question 1.
Why ether is less soluble in a saturated solution of NaCl? Explained.
Answer:
Ether is a weak polar compound and a saturated solution of NaCl decreases the polarity of the ether. Na⁺ and Cl^– ions attract H₂O molecules more strongly than ether molecules, thus the solubility of ether decreases in the saturated solution of NaCl.

Question 2.
Your teacher gives you two bottles, one with ethanol and the other with methanol without labelling.
(i) Write a suitable test to distinguish them.
(ii) By using the same test, can you distinguish between propan-1-ol and ethanol? Write the chemical equation.
Answer:
(i) The liquid which gives iodoform test is ethanol. Methanol does not give iodoform test. Ethanol reacts with alkaline solution of Iodine (I₂/NaOH) to form yellow precipitate of iodoform.
CH₃CH₂OH + 3I₂ + 4NaOH → CHI₃ + HCOONa + 3H₂O + 3NaI
CH₃OH + 3I₂ + 4NaOH -» No reaction

(ii) Yes. Ethanol gives iodoform test while propan -1 – ol does not give iodoform test.
CH₃ – CH₂CH₂OH + I₂ + NaOH → No reaction

Question 3.
Give equations for the preparation of ethyl alcohol by starch and write name of enzymes.
Answer:

Enzymes : (1) Diastase, (2) Maltase, (3) Zymase.

Question 4.
What is Lucas reagent ? How are primary, secondary and tertiary alcohol identified by it ? Explain.
Answer:
Mixture of anhydrous ZnCl₂ and cone. HCl is known as Lucas reagent.

1. Tertiary Alcohol : On adding Lucas reagent in alcohol at normal temperature, immediately white oily precipitate of Alkyl chlorides is formed, then it is tertiary alcohol.
2. Secondary Alcohol: If on adding Lucas reagent in alcohol, at normal temperature, a white oily precipitate of alkyl chloride is obtained after 5 minutes, then it is secondary alcohol.
3. Primary Alcohol: Primary alcohol does not show any reaction with Lucas reagent at normal temperature.

Question 5.
Why the b.p. of alcohol are higher than ethers and alkene ?
Or,
C₂H₅OH and CH₃OCH₃ both have same molecular formula (C₂H₆O) but the b.p. of alcohol is 78.4°C and b.p. of ether is -240°C. Explain the reason.
Answer:
In case of C₂H₅OH there is strong intermolecular hydrogen bonding between the molecules of alcohol. So alcohols (C₂H₅OH) required much energy to evaporate than ether molecules. In other words, we can say that the C₂H₅OH molecules are in associated form due to H-bonding so the b.p. of C₂H₅OH is very higher than ether and alkene.

Question 6.
What do you understand by Methylated spirit or denaturing alcohol ?
Answer:
Methylated Spirit : Ordinary rectified spirit is known as industrial alcohol.

Methylated spirit is 90% ethanol to which nauseating materials like methyl alcohol, pyridine or mineral naphtha have been added. These materials are added so that the ethanol will not be used for beverage purposes. This process is called denaturing. It is of two types :

It is used for the preparation of spirit, varnish etc. By it misuse of ethanol as a beverage is controlled.

Question 7.
Explain the manufacture of CH₃OH by water gas.
Answer:
From water gas : Steam is passed over red hot coke when water gas is formed.

Water gas is mixed with half its volume of hydrogen, compressed to about 200 atm and passed over a catalyst which is a mixture of oxides of copper, zinc and chromium at 300°C.

Question 8.
Give the chemical equation of the following conversion:
(i) Diethyl ether from ethanol
(ii) Ethanol from diethyl ether
(iii) Ethyl acetate from ethanol
(iv) Ethanol from glucose.
Answer:
(i) Diethyl ether from Ethanol:

(ii) Ethanol from Diethyl ether :

(iii) Ethyl acetate from Ethanol :

(iv) Ethanol from Glucose :

Question 9.
Differentiate between Phenol and Alcohol and write Libermann’s reaction related to phenol.
Answer:
Differences between Phenol and Alcohol:

Phenol:

1. Physical properties: Characteristic phenolic odour, sparingly soluble in water.
2. It is acidic and dissolves in bases to form salt.
3. On oxidation, hybrid coloured product is formed.
4. Produce characteristic colour with Ferric chloride.
5. It does not react with halogen acid.
6. With PCl₅, mainly form triaryl phosphate.

Alcohol:

1. Pleasant odour, fairly soluble in water.
2. It is neutral and do not reacts with bases.
3. It can easily oxidize to Aldehydes and ketones.
4. It does not reacts with ferric chloride.
5. Forms Alkyl halide.
6. Alkyl chloride are formed.

Libermann’s Reaction: On adding few drops of concentrated sulphuric acid and little sodium nitrite in phenol first dark blue colour is produced on adding water colour becomes red and on adding an alkali red colour again changes to blue colour.

Question 10.
Pure phenol is a colourless solid but why it is converted into pink after some time?
Or,
What change in colour is observed in phenol in presence of oxygen? Explain with reaction.
Answer:
In the presence of air pure phenol oxidises into quinone.

This quinone again combines with two molecules of phenol by H-bond and gives pink phylloquinone.

Question 11.
Name the enzymes present in yeast which can convert sucrose into glucose and fructose and then to ethyl alcohol.
Answer:
Invertase can convert sucrose to glucose and fructose Zymase can convert glucose and fructose to ethanol and carbon dioxide.

Question 12.
Boiling point of higher than the corresponding alkane. Why ?
Answer:
Boiling point of alcohols is much higher than hydrocarbons of nearly similar molecular mass due to intermolecular hydrogen bond. Alcohols molecules associate kilo calories mole^-1. Thus, extra energy is required for the separation of these molecules, which lead to an increase in boiling point. Hydrocarbons do not form hydrogen bond, thus their boiling point is comparatively less.

Question 13.
Ethyl alcohol and phenol both contain — OH group. What is the reason that phenol is acidic and alcohol has alkaline effect?
Or,
Ethyl alcohol and phenol both contain — OH group. What is the reason that phenol is acidic and alcohol is neutral in nature?
Answer:
Explanation of acidic nature of phenol: One possible explanation why phenols are stronger acids as compared to alcohols is that phenols exist as a resonance hybrid.

Due to resonance, the oxygen atom gets a positive charge and attracts the electron pair of the O—H bond and thus facilitates the release of a proton. The phenoxide ion formed after the release of a proton is also stabilized by resonance.

In alcohols, no resonance is possible hence the hydrogen atom is more firmly linked to the oxygen.

Question 14.
Write Victor Meyer method to distinguish primary, secondary and tertiary alcohol.
Answer:
Victor Meyer’s method:

• The given alcohol is converted into an iodide by concentrated HI or red phosphorus and iodine.
• The iodide is treated with silver nitrite to form nitroalkane.
• Nitroalkane is finally treated with nitrous acid (NaNO₂ + H₂SO₄) and made alkaline with KOH.

If a blood red colour is obtained, the original alcohol is primary.
If a blue colour is obtained, the alcohol is secondary.
If no colour is produced, the alcohol is tertiary.

Question 15.
What is Williamson’s continuous etherification process? Is it a continuous process? Explain. Give labeled diagram.
Or,
Describe the laboratory method of preparation of diethyl ether. How ether thus obtained is purified?
Answer:
Laboratory Method for the Preparation of Diethyl Ether (Sulphuric Ether):
Diethyl ether is prepared in the laboratory and industry by the Williamson continuous etherification process, i.e., by heating ethanol (in excess) with concentrated sulphuric acid.

Sulphuric acid is regenerated in the reaction hence, it appears as if only a small amount of acid may convert excess alcohol into the ether. So, this method is called Williamson’s continuous etherification process but actually, we cannot get ether continuously.

This is due to the following two reasons:

1. Water formed in the reaction dilutes the acid and its reactivity decreases.
2. A part of sulphuric acid is reduced by alcohol into sulphur dioxide.

Method: Ethanol and H₂SO₄ (2:1) are taken in a flask and heated on sand bath at ‘ 140°C. Ethanol is added at the same rate at which ether distilled over and is collected in a receiver cooled in ice-cold water.

Purification: Ether contain ethanol, water and sulphuric acid as impurities. It is washed with NaOH to remove sulphuric acid and then agitated with 50% solution of calcium chloride to remove the alcohol. It is then washed with water, dried over anhydrous calcium chloride and redistilled.

Question 16.
Write a note on fermentation.
Answer:
Fermentation: In this method, molasses or starch are used as raw materials. This is the old method and is used at present also. Fermentation (Latin-fermentare means to boil) proceeds with fast evolution of CO₂ producing much foam giving the appearance as if the solution is boiling. “It is a process of decomposition of complex large molecules into simple and small molecules slowly and by the enzyme.” Enzymes are of many kinds and are present in yeast which is a living and complex substance containing several types of bacteria which are called enzymes. It is a good ferment. The enzymes present in yeast are zymase, maltase, invertase, etc.

When yeast is mixed in glucose solution and kept at proper conditions, ethyl alcohol is formed as a result of fermentation.

Favourable conditions for Fermentation:

1. Favourable temperature: It is between 25-35°C.
2. Other substances: Some inorganic salts like ammonium sulphate or ammonium nitrate function as food for fermentation.
3. Concentration: Solution should be dilute (Concentration 8-10%).
4. Air: The process occur in presence of air.

Question 17.
(i) How can we obtain phenol from benzene diazonium chloride?
(ii) What is the reaction of diethyl ether with HI acid?
Answer:
(i) Phenols are prepared by hydrolysis of diazonium salts by water, dil. acids etc.

(ii) The reaction of diethyl ether with cone. HI acid, on heating gives one molecule of ethyl iodide and one molecule of ethyl alcohol.

Question 18.
Give two reactions that show the acidic nature of phenol.
Answer:
(a) The reactions showing acidic character of phenol are:
(i) Reaction with sodium: Phenol reacts with sodium to give H₂ gas.

(ii) Reaction with NaOH: Phenol dissolve in NaOH to give sodium phenoxide.

(b) Comparison of Acidity of Phenol with alcohol.
C₂H₅OH + NaOH → No reaction
But phenol reacts with NaOH and exhibits its strong acidic nature.
Ionization of Ethanol and Phenol is as follows :
On the other hand ethoxide ion and ethanol do not represent resonance thus negative charge is on oxygen of ethoxide ion whereas charge displacement occur in phenoxide ion.
pK_a value of ethanol is 15.9 where of phenol is 10.0. Thus, phenol is many more times’ more acidic than ethanol.

Alcohols, Phenols and Ethers Long Answer Type Questions

Question 1.
Explain the mechanism of dehydration of alcohol.
Answer:
Dehydration of alcohol:
(i) When ethyl alcohol is heated in excess of cone. H₂SO₄ molecule of water is eliminated and alkene is formed.

Mechanism : (i) Protonation of alcohol by H₂SO₄

(ii) Removal of water

(iii) Elimination of β-hydrogen in the form of proton by base (bisulphate ion)

Stability of the carbocation (I) determines the case of dehydration and order of stability of carbocation is :
CH₃ < C₂H₅ < Isopropyl < Tertiary butyl

Question 2.
How is ethyl alcohol obtained by molasses? Explain in brief.
Or
What are molasses? How is alcohol obtained by fermentation? Explain. Tell favourable conditions of fermentation. Draw labelled diagram of coffee still.
Answer:
From molasses: Molasses is the syrupy solution of sugar left after the separa¬tion of cane sugar or beet sugar crystals from the concentrated juice.

The different steps of the manufacture processes are:

(a) Dilution: The molasses is diluted with water so that a concentration of 8-10 percent sugar is obtained in the solution. This is acidified with dilute sulphuric acid to retard other bacterial growth. A solution of ammonium salts is also added which acts as food for the ferment.

(b) Alcoholic fermentation: The dilute solution obtained above [From step (a)] is taken in big fermentation tanks and some yeast is added. The mixture is kept for a few days and the temperature is maintained at about 30°C. The fermentation reaction starts and the enzyme Invertase (From Yeast) converts sucrose into glucose and fructose which are then converted into ethanol by Zymase (From Yeast).

The fermentation is completed in about 3 days. The carbon dioxide is collected as a by-product.

(c) Distillation: The fermented liquor is technically called wash or wort which contains about 9-10 percent ethanol. It is then distilled in a continuous still called Coffey’s still. It consists of two tall fractionating columns which are called analyser and the rectifier. It works on the counter-current principle and the steam and washes travel in opposite directions through the still.

The steam goes upwards in the analyser and takes away the alcohol vapours from the down coming dilute alcohol. The mixture leaves the analyser from the top and enters the rectifier at the base. Here it heats the wash flowing through the pipes on its way to the analyser. Most of the steam condenses and the alcohol vapours condense in the condenser. The distillate contains 90% alcohol.

(d) Rectification: Wash is rectified by fractional distillation.

Question 3.
How can you change the following:
(i) Methanol to ethanol
(ii) Ethanol to methanol
Answer:
(i) Methanol to ethanol:

(ii) Ethanol to methanol:

Question 4.
Differentiate primary, secondary and tertiary alcohol by oxidation and dehydrogenations method.
Answer:
1. Oxidation: The oxidizing agents generally used for oxidation of alcohols are acid dichromate, acid or alkaline KMnO_4, and dilute HNO₃.

(i) A primary alcohol is easily oxidized to an aldehyde and then to an acid both containing the same number of carbon atoms as the original alcohol.

(ii) A secondary alcohol on oxidation gives a ketone with the same number of carbon atoms as the original alcohol, ketones are oxidized with difficulty but prolonged action of oxidizing agents produce carboxylic acids containing fewer carbon atoms than the original alcohol.

(iii) A tertiary alcohol is resistant to oxidation in neutral or alkaline solutions but is readily oxidized by an acid oxidizing agent giving a mixture of ketone and acid each having lesser number of carbon atoms than the original alcohol.

2. Dehydrogenation (Action of hot reduced copper at 300°C): Different types of alcohols give different products when their vapours are passed over Cu gauze at 300°C.
Primary alcohols lose hydrogen and yield an aldehyde.

Secondary alcohols lose hydrogen and yield a ketone.

Tertiary alcohols are not dehydrogenated but lose a water molecule to give alkenes.

Question 5.
How can you obtain the following compounds from phenol :
(i) 2, 4, 6-Tribromophenol
(ii) Picric acid
(iii) Aniline
(iv) Benzene
(v) Phenolphthalein
(vi) p-cresol, o-cresol.
Answer:
(i) Phenol to Tribromophenol: Phenols readily react with halogens to give polyhalogen substituted compounds. Phenol gives white precipitate of 2, 4, 6-tribromo- phenol with bromine water.

(ii) Phenol to Picric acid : Nitration : On nitration, phenols give a variety of products depending upon the conditions.

Nitration of phenol with conc. HNO₃ in presence of conc. H₂SO₄ gives, 2,4,6-Trinitro-phenol (Picric acid).

(iii) Phenol to Aniline:

(iv) Phenol to Benzene:

(v) Phenol to Phenolphthalene: Phenol condenses with phthalic anhydride in presence of conc. H2SO4 give phenolphthalein which is an indicator for acid-base titrations and is used as a laxative in medicine.

(vi) Phenol to ortho and para cresol:

Question 6.
Give the chemical equation of the following conversion:
(i) Diethyl ether from ethanol
(ii) Ethanol from diethyl ether
(iii) Ethyl acetate from ethanol
(iv) Ethanol from glucose.
Answer:
(i) Diethyl ether from Ethanol:

(ii) Ethanol from Diethyl ether:

(iii) Ethyl acetate from Ethanol:

(iv) Ethanol from Glucose :

Question 7.
Describe the manufacture method of methanol by the destructive distillation of wood.
Answer:
Manufacture of methanol: By the destructive distillation of wood: Dried wood is heated (350°C) in closed retorts for about 3 hours. Different products obtained are given ahead:

Recovery of methyl alcohol from pyroligneous acid : Pyroligneous acid contains methyl alcohol (2-4%), acetone (0-5 to 1%), acetic acid (about 10%) and water.

The pyroligneous acid is taken in a copper vessel and distilled. The vapours are passed through hot milk of lime which retains acetic acid as non-volatile calcium acetate. Methyl alcohol and acetone vapours pass over and are condensed. This mixture is subjected to fractional distillation to separate methyl alcohol (b.p. 65°C) and acetone (b.p. 56°C). Methyl alcohol obtained by fractional distillation is about 95% pure. To purify further methyl alcohol is treated with anhydrous calcium chloride when it forms solid derivative CaCl₂.4CH₃OH. This solid derivative is separated and methyl alcohol is recovered by distillation.

Question 8.
Give equations for three methods of preparation of phenol.
Answer:
Methods of preparation of phenol:
(i) By the hydrolysis of Benzene diazonium salts: Benzene diazonium salt is formed by aromatic primary amine (aniline) with nitrous acid at 0-5°C. On boiling aqueous solution of this salt phenol is formed.

(ii) By alkaline fusion of sodium benzene sulphonate : On fusing sodium benzene sulphonate with NaOH, sodium phenoxide is formed which on acidification forms phenol.

(iii) Rasching method : On heating benzene with mixture of HCl and air to 230°C in the presence of Cu catalyst chlorobenzene is formed which on hydrolysis form phenol.

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Chemistry Solutions Chapter 14 Biomolecules

Biomolecules NCERT Intext Exercises

Question 1.
Glucose or sucrose are soluble in water but cyclohexane or benzene (simple six membered ring compounds) are insoluble in water. Explain.
Answer:
Glucose and sucrose have – OH groups while cyclohexane and benzene do not. Thus glucose and sucrose can form hydrogen bonds with water and are soluble in water. While cyclohexane and benzene do not form hydrogen bonds with water and thus insoluble in water.

Question 2.
What are the expected products of hydrolysis of lactose?
Answer:
Galactose and glucose are the products of hydrolysis of lactose.

Question 3.
How do you explain the absence of aldehyde group in the penta acetate of D-glucose?
Answer:
The pentacetate of glucose does not react with hydroxylamine indicating the absence of the free – CHO group.

Question 4.
The melting points and solubility in water of amino acids are generally higher than that of the corresponding haloacids. Explain.
Answer:
The amino acids exist as Zwitter ionsDue to this dipolar salt like character they have strong dipole – dipole attractions or electrostatic attractions. There fore, their melting points are higher than haloacids which do not have salt like character. Further, due to salt like character, they interact strongly with H₂O. As a result, solubility in water of amino acids is higher than that of the corresponding haloacids which do not have salt like character.

Question 5.
Where does the water present in the egg go after boiling the egg ?
Answer:
About three-fourth of an egg is plain water in which the albumin protein and fats are suspended. At room temperature, the protein strands are tightly folded in a complex three-dimensional shape. Because the individual proteins are able to move freely, both the egg white and yolk remain liquid. But when egg is boiled, the protein strands straighten out.

The protein ends, normally protected within the folds, become exposed and the open ends join together in bridge-like bonds forming a giant structure. These new bonds prevent the molecules to move thereby making the egg material solid and the water seems to be disappeared.

Question 6.
Why cannot vitamin C be stored in our body ?
Answer:
Vitamin C is soluble in water. It cannot be stored in our body because it is easily excreted in the urine.

Question 7.
What products would be formed when a nucleotide from DNA containing thiamine is hydrolysed?
Answer:
D – 2 – deoxyribose (sugar), thymine (nitrogenous base) and phosphoric acid are formed when a nucleotide from DNA containing thymine is hydrolysed.

Question 8.
When RNA is hydrolysed, there is no relationship among the quantities of different bases obtained. What does this fact suggest about the structure of RNA?
Answer:
When RNA is hydrolysed, there is no relationship between the quantities of four bases, i.e., cytosine (C), guanine (G), adenine (A) and uracil (U) are obtained, therefore, the base-pairing principle, i.e., (A) pairs with (T) and (C) pairs with (G) is not followed (as in DNA). This suggest that unlike DNA (a double-stranded structure), the RNA has a single-stranded molecule.

Biomolecules Ncert Textbook Exercises

Question 1.
What are monosaccharides?
Answer:
Monosaccharides are carbohydrates which cannot be hydrolysed further to give simpler units of polyhydroxy aldehydes or ketones.

Question 2.
What are reducing sugars?
Answer:
Carbohydrates which reduce Fehling’s solution to red p.pt of Cu₂O or Tollen’s reagent to shining metallic Ag are called reducing sugars. All monosaccharides (both aldoses and ketoses) and disaccharides except sucrose are reducing sugars.

Question 3.
Write two main functions of carbohydrates in plants.
Answer:

• Cellulose is a predominant constituent of cell wall of plant cells.
• Starch is the main food storage polysaccharide of plants.

Starch is stored in seeds and acts as the reserve food material for the small plant unless it is capable of making its own food by photosynthesis.

Question 4.
Classify the following into monosaccharides and disaccharides: Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
Answer:
Monosaccharide: Ribose, 2-deoxyribose, galactose and fructose.
Disaccharide: Maltose and lactose.

Question 5.
What do you understand by the term glycosidic linkage?
Answer:
The oxygen (ethereal) linkage through which two monosaccharide units are joined together by the loss of a water molecule to form a molecule of a disaccharide is called glycosidic linkage. For example, sucrose (a disaccharide) is composed of C₁ of α – glucose and C₂ of β – fructose through the glycosidic linkage.

Question 6.
What is glycogen? How is it different from starch?
Answer:
Glycogen-like starch is a condensation polymer of α – D glucose. Just as glucose is stored in plants in the form of starch it is stored in the form of glycogen in human beings. It is present in liver, muscles and brain. When body needs glucose, enzymes breakdown glycogen to provide glucose. It is also present in yeast and fungi.

The main difference between glycogen and starch is that starch is made up of amylose and amylopectin whereas glycogen consists of a structure almost same as amylopectin with large number of branching, (amylose has a linear structure and amylopectin has a branched structure with branching after every 25-30 molecules). Glycogen has branching after every 10-15 molecules.

Question 7.
What are the hydrolysis products of:
1. Sucrose
2. Lactose
Answer:
1. Glucose and fructose:

2. D-Glucose and D-Galactose:

Question 8.
What is the basic structural difference between starch and cellulose ?
Answer:
Starch consists of two components :
amylose and amylopectin. Amyiose is a long linear polymer of 200-1000 α-D-(-) glucose units held by C₁ – C₄ glycosidic linkage. It is soluble in water. Amylopectin is a branched chain polymer of α-D-(+) glucose linkage whereas branching occurs by C₁– C₆ glycosidic linkage. It is insoluble in water. On the other hand cellulose is a straight chain polysaccharide composed only of β – D- (+) glucose units which are formed by glycosidic linkage between C₁ of one glucose unit and C₄ of next glucose unit.

Question 9.
What happens when D – glucose is treated with the following reagents:
(i) HI
(ii) Bromine water
(iii) HNO₃?
Answer:
(i) When glucose is treated with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in straight chain.

(ii) On heating glucose with bromine water, it gets oxidized to six carbon carboxylic acid, gluconic acid.

(iii) Glucose on treatment with nitric acid gives a dicarboxylic acid, saccharic acid.

Question 10.
Enumerate the reactions of D-glucose which cannot be explained by its open chain structure.
Answer:
Following reactions could not be explained by open chain structure of D – glucose:

1. The penta-acetate of glucose does not react with hydroxylamine indicating the absence of free – CHO group.
2. Despite having the aldehyde group
   • Glucose does not form the hydrogen sulphide addition product with NaHSO₃.
   • Glucose does not give Schifts test.

Question 11.
What are essential and non-essential amino acids ? Give two examples of each type.
Answer:
(a) Essential amino acids:
The amino acids which our body cannot make and must be obtained through diet.
Examples:
Valine, Isoleucine, Arginine, Lysine, Threonine etc.

(b) Non-essential amino acids:
These are the amino acids which our body can make.
Examples:
Glycine, Alanine, Glutamic acid, Aspartic acid, Glutamine, Serine etc.

Question 12.
Define the following as related to proteins :
(i) Peptide linkage
(ii) Primary structure
(iii) Denaturation.
Answer:
(i) Peptide linkage:
A peptide bond is an amide linkage formed between – COOH group of one a-amino acid and NH₂ group of the other a-amino acid by loss of a molecule of water. It units two amino acids unit in a peptide bond (molecule).

(ii) Proteins are the polymers of a-amino acids. These polymers (also known as polypeptide) consist of amino acids linked with each other in a specific sequence. This sequence of amino acids is known as the primary structure of proteins. Any change in this sequence of amino acids (i.e., primary structure) creates a different proteins.

(iii) A process that changes the physical and biological properties of proteins without affecting the chemical composition of proteins is called denaturation. The denaturation is caused by certain physical and chemical treatment such as changes in pH, temperature, presence of some salts or certain chemical agents.

Question 13.
What are the common types of secondary structure of proteins ?
Answer:
The conformation of polypeptide chain assumed as a result of hydrogen bonding is called secondary structure of proteins. The two types of secondary structures are a-helix and β – pleated sheet structure. (For detail refer your NCERT text-book.)

Question 14.
What type of bonding helps in stabilizing the a-helix structure of proteins ?
Answer:

(Refer your NCERT text-book for details and structure.)

Question 15.
Differentiate between globular and fibrous proteins.
Answer:
Difference between globular and fibrous protein:
Globular protein:

• They have foilded ball like structure.
• They have three-dimensional structure
• They are soluble in water and aq. solution of salt and base.
• These proteins are inactive towards temperature and pH value.

Fibrous protein:

• These molecule have long threads like structure.
• They have sheet like structure.
• These are insoluble in water.
• Fibrous protein are active towards temperature and pH value.

Question 16.
How do you explain the amphoteric behaviour of amino acids ?
Answer:
Due to dipolar or Zwitter’s ion structure, amino acids are amphoteric in nature. The acidic character of the amino acids is due to the – NH₃ group and the basic character is due to the – COOH group as shown below:
Acidic character:

Basic character:

Question 17.
What are enzymes ?
Answer:
The enzymes are globular proteins which act as bio-catalyst. They are very specific and efficient in their action. Almost all the reactions that occur in the body are catalyzed by enzymes.

Question 18.
What is the effect of denaturation on the structure of proteins ?
Answer:
During denaturation, 2° and 3° structures of proteins are destroyed but 1° structure remains intact. As a result of denaturation, the globular proteins (soluble in H₂O) are converted into fibrous proteins (insoluble in H₂O)) and their biological activity is lost. The coagulation of egg white on boiling is a common example of denaturation.

Question 19.
How are vitamins classified ? Name the vitamin responsible for the coagulation of blood.
Answer:
On the basis of solubility in water or fat, the vitamins are generally classified into two types:

• Water soluble vitamins:
  These include vitamin B-complex. (B₁ B₂, B₃, B₄, B₆, B₁₂ and nicotinic acid etc.) and vitamin C etc.
• Fat soluble vitamins:
  These include vitamins A, D, E and K. Liver cells are rich in fat soluble vitamins. Vitamin K is responsible for coagulation of blood.

Question 20.
Why are vitamin-A and vitamin-C essential to us ? Give their important sources.
Answer:
Vitamin – A:
It is essential for us because its deficiency causes Night blindness and Xerophthalmia (Hardening of cornea of eye).

Sources:
Carrots, milk, butter, fish liver oil, egg yolk, green and yellow vegetables.

Vitamin – C:
It is essential for us because its deficiency causes Scurvy (bleeding gums), Tooth decay’s and Pyorrhoea etc.

Sources:
Lemon, orange (citrous fruits), amla, tomatoes, potatoes and green leafy vegetables.

Question 21.
What are nucleic acids ? Mention their two important functions.
Answer:
Nucleic acids are long chain polymers of nucleotides. They are also called poly-nucleotides. Nucleic acids are mainly of two types, the deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).

Functions:
(i) DNA is responsible for transmission of hereditary effects from one generation to another. This is due to unique property of replication, during cell division and two identical DNA strands are transferred to the daughter cells.

(ii) DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body. Actually, the proteins are synthesized by various RNA molecules (m-RNA and t-RNA etc.) in the cell but the message for the synthesis of a particular protein is present in DNA.

Question 22.
What is the difference between a nucleoside and a nucleotide ?
Answer:
A nucleoside contains only basic component of nucleic acids namely a pentose sugar and a nitrogenous base. A nucleotide contains all the three basic components of nucleic acids namely a phosphoric acid group, a pentose sugar and a nitrogenous base.

Question 23.
The two strands in DNA are not identical but are complementary. Explain.
Answer:
The two strands in DNA molecule are held together by hydrogen bonds beetween purine base of one strand and pyrimidine base of the other and vice-versa. Because of different sizes and geometries of the bases, the only possible pairing in DNA are G (guanine) and C (cytosine) through three H-bonds i.e., (C = G) and between A (adenine) and T (thiamine) through two H-bonds i.e., (A=T) (for figure refer your NCERT text book). Due to this base pairing principle, the sequence of bases in one strand automatically fixes the sequence of bases in the other strand. Thus, the two strands are complimentary and are not-identical.

Question 24.
Write the important structural and functional differences between DNA and RNA.
Answer:
Differences between DNA and RNA:
DNA:

• Occurs mainly in the nucleus of the cell.
• It contains the sugar deoxyribose.
•  Does not contain nitrogenous base, uracil.
• It has a double strand helix
• It is responsible for the transmission of heredity character.
• Alkaline hydrolysis is quite slow.
• Ratio A/T = 1 and G/C = 1.

RNA:

• Occurs in the cytoplasm of the cell.
• It contains the sugar ribose.
• Does not contain nitrogenous base thymine.
• It has double as well as single strand helix.
• It helps in protein biosynthesis.
• Alkaline hydrolysis takes place readily.
• Such ratio is not present.

Question 25.
What are the different types of RNA found in the cell ?
Answer:
There are three types of RNAs:

• Ribosomal RNA (r-RNA)
• Messenger RNA (m-RNA)
• Transfer RNA (t-RNA).

Biomolecules Other Important Exercises

Biomolecules Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Which protein transports oxygen in blood flow:
(a) Haemoglobin
(b) Insulin
(c) Albumin
(d) Myoglobin.
Answer:
(d) Myoglobin.

Question 2.
Beri-beri diseases is caused due to deficiency of which vitamin:
(a) Vitamin – A
(b) Vitamin – C
(c) Vitamin – B₂
(d) Vitamin – D
Answer:
(c) Vitamin – B₂

Question 3.
Enzyme which enhances the conversion of glucose to ethanol is:
(a) Zymase
(b) Invertase
(c) Maltase
(d) Diastase.
Answer:
(a) Zymase

Question 4.
In human body carbohydrate is stored:
(a) In the form of glucose
(b) In the form of glycogen
(c) In the form of starch
(d) In the form of fructose.
Answer:
(b) In the form of glycogen

Question 5.
Change in optical rotation of a freshly prepared solution of sugar after some time is called:
(a) Optical activity
(b) Inversion
(c) Specific rotation
(d) Mutation.
Answer:
(b) Inversion

Question 6.
Formula of most familiar disachharide is:
(a) C₁₀H₁₈O₉
(b) C₁₀H₂₀O₁₀
(c)C₁₈H₂₂O₁₁
(d) C₁₂H₂₂O₁₁.
Answer:
(d) C₁₂H₂₂O₁₁.

Question 7.
The following statement is false in relation to Ribose:
(a) It is a polyhydroxy compound
(b) It is a aldehydic sugar
(c) It contain 6 carbon atoms
(d) It has optical rotation.
Answer:
(c) It contain 6 carbon atoms

Question 8.
In the formation of carbohydrate are required:
(a) 2 carbon
(b) 3 carbon
(c) 4 carbon
(d) 6 carbon.
Answer:
(d) 6 carbon.

Question 9.
Haemoglobin is:
(a) Enzyme
(b) Globular protein
(c) Vitamin
(d) Carbohydrate.
Answer:
(d) Carbohydrate.

Question 10.
Which carbohydrate is an essential constituent of plant cells:
(a) Cellulose
(b) Starch
(c) Sucrose
(d) Vitamin.
Answer:
(b) Starch

Question 11.
How many subunits are present in haemoglobin:
(a) 2
(b) 3
(c) 4
(d) 5.
Answer:
(b) 3

Question 12.
Starch is a polymer of:
(a) Glucose
(b) Sucrose
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Glucose

Question 13.
Which sugar is present in maximum amount in human blood:
(a) d-fructose
(b) d-glucose
(c) Sucrose
(d) Lactose.
Answer:
(b) d-glucose

Question 14.
Element present in Vitamin B₁₂ is:
(a) Pb
(b) Zn
(c) Fe
(d) CO.
Answer:
(d) CO.

Question 15.
Amount of glucose in blood is determined by:
(a) Tollen’s reagent
(b) Benedict solution
(c) Alkaline iodine solution
(d) Bromine water.
Answer:
(b) Benedict solution

Question 16.
Deficiency of which Vitamin causes rickets:
(a) Vitamin C
(b) Vitamin B
(c) Vitamin A
(d) Vitamin D.
Answer:
(d) Vitamin D.

Question 17.
In metabolic processes which of the following provide maximum energy :
(a) Protein
(b) Vitamin
(c) Thiamine
(d) Pyrimidine.
Answer:
(d) Pyrimidine.

Question 18.
Vitamin B₁ is:
(a) Riboflavin
(b) Cobaltamine
(c) Thiamine
(d) Pyrimidine.
Answer:
(a) Riboflavin

Question 19.
Deficiency of Vitamin C leads to:
(a) Scurvy
(b) Rickets
(c) Pyorrhoea
(d) Anaemia.
Answer:
(a) Scurvy

Question 20.
Most effective energy reservoir in a living cells is:
(a) A.M.P.
(b) A.T.P.
(c) A.D.P.
(d) U.D.P.
Answer:
(b) A.T.P.

Question 21.
Disaccharide present in milk is:
(a) Sucrose
(b) Lactose
(c) Maltose
(d) Cellulose.
Answer:
(b) Lactose

Question 22.
Which is not glyceroid:
(a) Fat
(b) Oil
(c) Phospholipid
(d) Soap.
Answer:
(d) Soap.

Question 23.
Which is not found in R.N.A.:
(a) Thymine
(b) Uracil
(c) Adenine
(d) Guanine.
Answer:
(a) Thymine

Question 24.
Enzymes are:
(a) Nitrogen containing complex compound
(b) Carbohydrate
(c) Co-ordination compounds
(d) Metallic compound.
Answer:
(a) Nitrogen containing complex compound

Question 25.
Chemical name of vitamin C is:
(a) Cyano cobaltamine
(b) Ascorbic acid
(c) Tocopherol
(d) Biotin.
Answer:
(b) Ascorbic acid

Question 2.
Fill in the blanks:

1. By the oxidation of glucose …………… molecules of ATP are produced.
2. The breaking of complex molecules in organisms is known as ……………
3. In hyperglycaemia the amount of …………… in blood increases.
4. Deficiency of …………… leads to eye disease.
5. Deficiency of iodine leads to …………… disease.
6. Blood …………… the temperature of the entire body.
7. …………… hormone balances the amount of sugar in blood.
8. …………… is responsible for the clotting of blood.
9. Denaturation does not affect …………… structure of protein.
10. Protein is a polymer of ……………
11. …………… is the basic unit of protein.
12. …………… is not present in D.N.A.
13. Haemoglobin is a …………… compound of iron.
14. Oil and fats obtained from organisms are known as ……………

Answers:

1. 38
2. Catabolism
3. Sugar
4. Vitamin A
5. Goitre
6. Balance
7. Insulin
8. Vitamin K
9. Primary
10. Amino acids
11. Amino acid
12. Uracil
13. Complex
14. Lipid.

Question 3.
Match the following:

Answer:

1. (f) Vitamin A
2. (c) Vitamin D
3. (b) Vitamin E
4. (e) Vitamin B
5. (d) Vitamin B₁₂
6. (a) Biotin
7. (g) Lactose.

Question 4.
Answer in one word / sentence:

1. Write the chemical name of Vitamin C.
2. Tell the source of Vitamin K.
3. Is responsible for clotting of blood ?
4. Which bond links amino acids together ?
5. How many amino acids are synthesized by human body ?
6. Cellulose is a linear polymer of which glucose ?
7. In RNA molecule which pyrimidine is present in place of Thymine ?
8. Lactose on hydrolysis gives.
9. Glucose contains Pyranose ring whereas Fructose contain.
10. In polysaccharides, monosaccharide units are linked to each other by which bond ?
11. Which protein helpful for clotting of blood known as ?
12. Write one example of Monosaccharide Carbohydrate.
13. What is the name of Disaccharides sugar present in milk ?

Answers:

1. Ascorbic acid
2. Green leafy vegetables
3. Vitamin K (Phylloquinone)
4. Peptide bond
5. Ten
6. 2-glucose
7. Uracil
8. Glucose and Lactose
9. Furanose ring
10. Glycosidic
11. Fibrinogen
12. Glucose or Fructose
13. Lactose.

Biomolecules Short Answer Type Questions

Question 1.
What is the effect of deficiency of proteins on our body ?
Answer:
Deficiency of protein causes following diseases:

• Anaemia:
  The deficiency of haemoglobin protein in the body creates lack of enough blood which causes paleness in body.
• Kwashiorkar:
  This disease is mainly found in children, in this body of patient swells up and deforms.

Question 2.
Define carbohydrates.
Answer:
Substances which are polyhydroxy aldehydes or polyhydroxy ketones or which give these compounds on hydrolysis are called carbohydrates.

Question 3.
Sucrose on enzymatic hydrolysis gives one mole of glucose and one mole of fructose.

1. Which enzyme catalyses this hydrolysis?
2. Write down the monosaccharides present in maltose and lactose.
3. Which is the sweetest sugar, glucose, or fructose?

Answer:

1. Invertase
2. Maltose contains two units of glucose while lactose is glucose + galactose.
3. Fructose

Question 4.
Write the diseases caused by the following Vitamins:
(a) Vitamin-A, (b) Vitamin-B, (c) Vitamin-D, (d) Vitamin-E.
Or,
Write the functions of the following vitamins:
(a) Vitamin-A, (b) Vitamin-D, (c) Vitamin-E, (d) Vitamin-K.
Answer:
Diseases caused by the above Vitamins:
(a) Vitamin-A: Night blindness
(b) Vitamin-B: Beri-beri, loss of appetite and vigour
(c) Vitamin-D: Rickets, osteomalacia
(d) Vitamin-E: Loss of reproductive ability.

Functions caused by the above Vitamins:
(a) Vitamin-A: For vision and growth develops resistance against diseases.
(b) Vitamin-D: For bones, control of metabolism of calcium and phosphorus.
(c) Vitamin-E: Virility in man and reproduction.
(d) Vitamin-K: Coagulation of blood.

Question 5.
Write a note on the structure of DNA.
Answer:
The three-dimensional structure of DNA was elucidated by James Watson and Francis Crick. They proposed that the DNA chain consists of two strands of polynucleotides coiled around each other in the form of the double helix.

The nucleic acid backbone consists of alternating sugar-phosphate residues. One of the four bases is attached to each of the sugar residues in the backbone. The bases are held together by hydrogen bonding between them. The hydrogen bonding is very specific as the structure of bases is such that they permit only one mode of pairing with other bases.

The pairing is A – T, C – G. i.e., adenine pairs only with thymine, cytosine pairs only with guanine. Two strands of DNA are complementary to each other in the sense that the sequence of bases in one strand automatically decides that of the other.

Question 6.
What are enzymes? Give applications.
Answer:
Enzymes are complicated organic molecules which are produced by living cells and are also called biological catalysts. They are complex nitrogenous molecules, unlike inorganic catalysts, they take part in the reaction and get destroyed.
Example:
Enzyme invertase catalyse the hydrolysis of sucrose.

Factors affecting the activity of catalyst:

• Temperature
• pH
• Concentration of enzyme
• Concentration of substance
• Concentration of the product formed.

Functions of enzymes:
Enzymes play a very important role in many biological reactions. The names of some important enzymes and the biochemical reactions catalysed by them are given below:

1. Role of enzymes in the process of digestion:
Enzymes play an important role in digestion resulting in the breakdown of macromolecules into simple molecules.

2. Industrial applications:
Enzymes play an important role in many industrial processes such as the manufacture of ethanol from sugar or starch, the manufacture of acetic acid, the manufacture of cheese, leather, tanning, etc.

Question 7.
What are monosaccharides? Explain with examples.
Answer:
These are the simplest carbohydrates which cannot be hydrolysed into small molecules. Their general formula is C_nH_2nO_n (there are some exceptions also). Here n = 2 to 10. These are sweet in taste and soluble in water. These contain either one – CHO group or one > C = O group.

Aldopentose – Ribose, Zylose etc. (C₅H₁₀O₅)
Aldohexose – Glucose, Galactose, Mannose etc. (C₆H₁₂O₆)
Ketohexose – Fructose, Sorbose etc. (C₆H₁₂O₆).

Protein Molecular Weight Calculator makes it easy for you to determine the theoretical yield value of the chemical reaction in fraction of seconds.

Question 8.
What are proteins?
Answer:
The word protein is derived from the greek word protious (Protious = to take the first) i.e. first or very important. Proteins are high molecular mass nitrogen containing complex organic compounds found in the protoplasm of all animal and plants.
Chemically protein is the condensation polymer of a-amino acid.

Question 9.
What are carbohydrates? Which unit of carbohydrate provide energy to human body?
Answer:
Carbohydrates are compound of carbon, hydrogen and oxygen. General formula of carbohydrate is (C_xH₂O)_y where x and y are integers. These compounds have ratio of hydrogen and oxygen 2: 1 like water (H₂O). Therefore the name of these compounds is given carbohydrates. Examples of carbohydrates are glucose (C₆H₁₂O₆), fructose (C₆H₁₂O₆), sucrose (C₁₂H₂₂O₁₁) etc.

Glucose is the unit of carbohydrates which is responsible to provide energy. Glucose decomposes slowly with the help of oxydase enzyme present in the body into CO₂ and water. Energy is released in this process.

Question 10.
Fibrous protein present in hair is elastic while in silk it is not elastic. Explain.
Answer:
Protein present in hair has α – helical structure. If it is stretched, the hydrogen bonder broken, thereby increasing the length of the helix. On releasing the tension, the hydrogen bonus are formed again, giving back the original shape. So proteins having α – helix structure are elastic. Proteins in silk have β – pleated structure and they are not elastic because stretching leads to pulling of the peptide (covalent) bonds.

Question 11.
Explain the formation of amide and peptide bonds in the structure of proteins.
Answer:
(1) Amide bond:
Protein is a complicated organic compound which is formed by the combination of different amino acids. When the – CONH – link resulting from the reaction between a compound containing a carboxylic group (- COOH) and that with an amine group (- NH₂) it forms an amide bond.

(2) Peptide linkage:
When the – CONH – like resulting from the reaction between two molecules of a-amino acids, is called peptide bond. Such a bond involves only the alpha-amino and a-carboxylic group. Thus, two molecules of a-amino acid on reaction yield a peptide bond.

Question 12.
Define polysaccharides. Give examples also.
Answer:
Polysaccharides are natural isomers which have a molecular weight of up to many lacks, General formula of polysaccharides is (C₆H₁₀O₅)_n  where the value of n is from 12 to thousands. These are complex material which is formed by condensations of monosaccharides. These compounds contain glycosidic bonds.
Example: Starch, cellulose, etc.

Question 13.
What is invert sugar?
Answer:
Cane – sugar is dextrorotatory [D or +] which gives an equimolar mixture of monosaccharides. This mixture is laevorotatory [D or-]. Hence the mixture of glucose and fructose obtained as a result of hydrolysis of cane-sugar is called invert sugar.

Question 14.
What are disaccharides? Write the general formula of disaccharides.
Answer:
Disaccharides are sugar which is formed by the combination of two monosaccharides by the removal of one molecule of water. Both monosaccharides are of hexose type in which one unit is glucose. Thus, disaccharides are of aldose-aldose or aldose-ketose type. The general formula of disaccharides is C₁₂H₂₂O₁₁
Example: Sucrose, maltose, lactose, etc.

Question 15.
Give the pyranose structure of sucrose and maltose.
Answer:
In sucrose two monosaccharides (glucose and fructose) are combined through a glycosidic bond which is present between C₁ of α-glucose and C₂ of β-fructose.

Whereas maltose molecule is obtained by the condensation of two glucose molecules.

Question 16.
Clarify the structure of protein.
Answer:
Molecule of protein is made up of amino acids. Basically protein molecules are linear polymers of amino acids. Their entire structure can be expressed in four steps:

1. Primary structure:
   It provides knowledge of mutual combination of various amino acids in a polypeptide chain.
2. Secondary structure:
   It provides knowledge of conformations of peptide chains of protein.
3. Tertiary structure:
   It explains how protein molecule coils to obtain a specific shape.
4. Quarternary structure:
   By this, arrangement of two polypeptide chains with respect to each other is known.

Question 17.
What do you understand by nucleoside and nucleotide ?
Answer:
Nucleoside: When a purine or pyrimidine base joins with a pentose sugar molecule, then it is known as nucleoside.

Question 18.
What are carbohydrates ? Explain mono, di and polysaccharides with example.
Answer:
Carbohydrates are polyhydric aldehydes or polyhydric ketones or substances which on hydrolysis produce such compounds. The carbohydrates are classified into three major classes on the basis of hydrolysis.

1. Monosaccharides:
These are the simples carbohydrates. They cannot be hydrolysed further to give more simple carbohydrates. Monosaccharides have the general formula (CH₂O)_n These are crystalline solids, soluble in water and are sweet in taste. Example: Glucose, Fructose.

2. Disaccharides:
Disaccharides are sugar which are formed by combination of two monosaccharides by removal of one molecule of water. Both monosaccharides are of hexose type – in which one unit is glucose. General formula of disaccharides is C₁₂H₂₂O₁₁.
Example: Sucrose, Maltose etc.

3. Polysaccharides:
General formula of polysaccharides is (C₆H₁₀O₅)_n where value of n is from 12 to thousands. These are complex material which are formed by condensation of monosaccharides. Example: Starch, Cellulose etc.

Question 19.
What happens when protein is denatured ?
Answer:
Denaturation of Protein:
Disruption of tertiary structure of protein is called denaturation. These reactions take place by heating in presence of acids or highly concentrated salts or heavy metals. Denaturation does not affect the primary structure of protein. Denaturation takes place in the rearrangement of secondary and tertiary structures. As a result of this, protein losses its biological actvity.

During denaturation the protein molecule uncoils from an ordered and specific conformation into a more disordered conformation. Denaturation takes place when proteins are heated or treated with acids, bases, alcohols, KI, urea, acetone etc. or when exposed to UV or X-ray radiations.

Question 20.
Write functions of vitamin B and name two diseases caused due to its deficiency.
Answer:
Vitamin B include a group of vitamins which are as follows:
‘Vitamin-B₁ (Thiamine), Vitamin – B₂ (Riboflavin), Vitamin B₃ (Pentothenic acid), Vitamin-B₆ (Pyridoxine), Vitamin B₁₂ (Cyano cobalamin)

Vitamin B₁ (Thiamine)- Source: husk of rice, green vegetables, egg.
Function: Maintenance of activity of nervous system.

Deficiency disease:
(i) Beri-Beri (swelling in hands and legs).
(ii) Gastric (disorder in digestion).

Vitamin B₂ (Riboflavin)- Source: egg yolk, milk etc.
Function: Growth of body.
Deficiency disease: Dermatitis, weakness in eye sight.

Question 21.
What are vitamins ? Name the vitamins whose deficiency lead to following diseases:

• Haemorrhage (No clotting of blood)
• Night blindness
• Anaemia
• Ricket
• Pyrohea
• Loss of reproductive ability
• Convulsions.

Answer:
Vitamins are complex organic molecules which act as essential nutrients for body. Though these vitamins are not formed in our body but deficiency of these vitamins causes various diseases in our body. The vitamins are classified according to solubility.

• Fat soluble vitamins: Vitamin A, D, E and K.
• Water soluble vitamins: Vitamin C and vitamin B complex.

Name the vitamins deficiency lead to following diseases :

• Haemorrhage (No clotting of blood) : Vitamin K (Phylloquinone)
• Night blindness: Vitamin A (Retinol)
• Anaemia: Vitamin B₁₂ (Cyanocobalamin)
• Ricket: Vitamin D (Calciferol)
• Pyorrhoea: Vitamin C (Ascorbic acid)
• Loss of reproductive ability: Vitamin E (a -Tocopherol)
• Convulsions: Vitamin B6 (Pyridoxine).

Question 22.
Write about the following:

1. Testosterone
2. Thyroxine
3. Insulin
4. Cortisone.

Answer:

1. Testosterone:
   Testosterone hormone is secreted by the gland testes and its function is to control the reproductive organs in males.
2. Thyroxine:
   This hormone is secreted by the thyroid gland. Its function is to control metabolic activity and growth.
3. Insulin:
   Insulin hormone is secreted by pancreas and its function is to control the amount of glucose in the blood.
4. Cortisone:
   Cortisone is secreted by adrenal cortex and its function is to control the metabolism of protein and water.

Question 23.
Write the name of two-two diseases caused due to deficiency of retinol, thiamine, ascorbic acid and riboflavin.
Answer:
Retinol: Night blindness, Xerophthalmia
Thiamine : Beri-Beri, growth of body stops
Ascorbic acid: Scurvy, Tooth loss
Riboflavin: Weak eye sight, dermititis.

Question 24.
Write two-two source of vitamin A and C. Write one-one disease caused due to its deficiency.
Answer:

• Vitamin Source Disease
• Vitamin-A Milk, butter, cheese Night blindness
• Vitamin-C Lemon, Tomato Scurvy.

Question 25.
Write the functions of vitamin A, D, E and K.
Answer:

• Vitamin-A : It participate in the formation of visible pigment named Rhodopsin and Iodopsin.
• Vitamin-D : It is useful in the formation of bones.
• Vitamin-E : It prevents the breaking of RBCs.
• Vitamin-K : It is helpful in the clotting of blood.

Question 26.
Write functions and sources of the following bio-molecules/elements :

1. Protein
2. Carbohydrates
3. Fat
4. Calcium.

Answer:

1. Protein:
   Formation of new tissues, and their repairing with the body
   Source: Milk, egg, meat, cheese, fish.
2. Carbohydrates: Carbohydrates provides energy to the body.
   Source: Rice, potato, fruits, cane sugar etc.
3. Fats: They provides energy to the body.
   Source: Ghee, oils, milk, egg. etc.
4. Calcium: Increase the bones and teeth.
   Source: Green leafy vegetables, milk.

Question 27.
Write two Differences between a – Amino acid and Protein.
Answer:
Differences between a – Amino acid and Protein:
α – Amino acid:

• They are simple compounds having amino and Carboxylic acid group.
• On combining amino acid gives dipeptide,polypeptide and then protein e.g. glucose, lysine etc.

Protein:

• Proteins are complex nitrogenous compounds.
• Protein on hydrolysis gives amino acid. e.g. Haemoglobin, casein etc.

Question 28.
Explain in brief:
(a) Name of two enzymes and their function.
(b) Name of two water soluble enzymes and disease caused due to their deficiency.
Answer:
(a) Name of enzyme and function:
1. Amylase (ptyalin): It converts starch into glucose:

2. Pepsin: It converts protein into amino acid:

(b) Water soluble vitamin:
1. Vitamin B₁: Thiamine
Deficiency disease: Beri-Beri

2. Vitamin C: Ascorbic acid
Deficiency disease: Scurvy Other examples are vitamin B₂, B₆, B₁₂ and K.

Biomolecules Long Answer Type Questions

Question 1.
Write a note on Nucleic acid.
Answer:
Nucleic acid:
It is found in nucleus of the cell. It has large amount of phosphorus. Nucleic acid are poly nucleotides which is formed by the combination of various nucle otide units. Each nucleotide is formed by three chemical components:

1. Phosphate group
2. Pentose ribose sugar or De – oxyribose
3. Heterocyclic base: Like derivative of pyrimidine (Thiamine, uracil, cytosine) and derivatives of purine (Adinine and Guanine).

Nucleic acid are of two types:
(A) DNA: De-oxy ribonucleic acid.
(B) RNA: Ribonucleic acid.

Components of DNA:

• De-oxyribose sugar molecule
• Phosphoric acid molecule
• Nitrogenous base:

It is of two types:

1. Pyrimidine base: It includes cytosine (C) and Thymine (T)
2. Purine base: It includes Adenine (A) and Guanine (G)

Components of RNA:
RNA contains Ribose and nitrogen base like Adenine (A), Guanine (G), Uracil (U) and Cytosine (C).

Question 2.
What are carbohydrates ? Write its classification and four main functions.
Answer:
Definition:
Optically active polyhydroxy aldehydes or ketones or substances which yield these on hydrolysis are known as carbohydrates.
Example: Glucose, starch, cellulose, sucrose etc.

Classification of Carbohydrate:

Functions of carbohydrates:

• It is the main structural component of cell.
• It acts as a bio-fuel and provides energy to organisms for doing work.

• Carbohydrate is stored in the liver as glycogen, which hydrolysis to provide the energy required.
• Cellulose is found in grass and plants which provide energy to animals grazing grass because these animals possess specific enzymes which hydrolyses cellulose to glucose.

Question 3.
Give the diseases caused by ascorbic acid, thiamine retinol and nicotinic acid.
Or
Give the source and diseases caused by Vitamin A, B, C and D.
Answer:
Name of Vitamins, its functions and deficiency diseases are given in the ahead chart: Chemical names of Vitamins, their sources, functions and diseases due to deficiency

Question 4.
Give differences between monosaccharide, disaccharide and polysaccharide.
Answer:
Difference between monosaccharide, disaccharide and polysaccharide:

MP Board Class 12th Chemistry Solutions