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

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 :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 1
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 :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 2
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 ?
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 3
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 4

Question 5.
Write structures of the products of the following reactions:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 5
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 6
(ii) NaBH4 is weak reducing agent, it reduces aldehyde/ketones and not the esters.
Thus,
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 7

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
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 8

(b) With HBr : Both the alcohols react with HBr to give corresponding alkyl bromides.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 9

(c) With SOCl2 : Both the alcohols react with SOCl2 to give corresponding alkyl chlorides.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 10
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 11

MP Board Solutions

Question 7.
Predict the major product of acid catalysed dehydration of:
(i) 1-methylcyclohexanol and (ii) butan-l-ol.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 12
According to Saytzeff rule, the highly substituted product is the major product.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 13

Question 8.
ortho and para nitrophenols are more acidic than phenol. Draw the resonance structures of the corresponding phenoxide ions.
Answer:
(i) Phenoxide ion:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 14
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 15
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 16
If carbon tetrachloride is used in place of chloroform, salicylic acid is obtained as the main product.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 17

(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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 18
Salicylic acid is the starting material for the manufacture of aspirin which is an important analgesic.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 19

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 SN2 mechanism. Thus, alkyl halide should be derivde from ethanol and alkoxide ion from 3-methyl pentan-2-ol. The complete reaction is as follows :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 20

MP Board Solutions

Question 11.
Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4-nitrobenzene and why ?
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 21
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 22
In set (B) nucleophilic attack of 4-nitrophenoxide ion on methylbromide gives the desired product.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 23

Question 12.
Predict the products of the following reactions:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 24
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 25
When one of the groups in unsymmetrical ether is tertiary, then the halide formed is tertiary halide.

MP Board Solutions

Alcohols, Phenols and Ethers NCERT Textbook Exercises

Question 1.
Write IUPAC names of the following compounds :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 26
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 27
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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 28
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 29

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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 30
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 31

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 32

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 33

Question 7.
Give the structures and IUPAC names of monohydric phenols of molecular formula, C7H8O.
Answer:
Three isomers are :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 34

MP Board Solutions

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 35

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 36

Question 10.
Write chemical reaction for the preparation of phenol from chlorobenzene.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 37

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 38
Mechanism : H2SO4 → H+ + OΘ SO2OH

Step-I: Protonation of alkene to form carbo-cation by the electrophilic attack of hydronium ion (H3O+).
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 39
Step-II: Nucleophilic attack by water on carbo-cation to yield protonated alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 40
Step-III: Deprotonation (loss of proton) to form an alcohol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 41

Question 12.
You are given benzene, cone. H2SO4 and NaOH. Write the equations for the preparation of phenol using these reagents.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 42

Question 13.
Show how will synthesis:
(i) 1-Phenylethanol from a suitable alkene?
(ii) Cyclohexylmethanol using an alkyl halide by an SN2 reaction?
(iii) Pentan-1-ol using a suitable alkyl halide?
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 43
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 44

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 45
(ii) Reaction with NaOH : Phenol dissolves in NaOH to give sodium phenoxide and OH
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 46
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 47
-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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 48

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 49
Furthermore, after the loss of proton o-methoxy- phenoxide ion left is destabilized by resonance.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 50
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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 51
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 52

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.

MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 141
If carbon tetrachloride is used in place of chloroform, salicylic acid is obtained as the main product.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 142

(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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 143
Salicylic acid is the starting material for the manufacture of aspirin which is an important analgesic.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 144

(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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 53
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 :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 54
Step 2: Formation of carbocation (rate determinning step):
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 55
Step 3: Elimination of a proton to form ethene:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 56
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.

MP Board Solutions

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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 57

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 58

Question 23.
Give IUPAC names of the following ethers:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 59
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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 60

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 61

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 62
(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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 63

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 64
(b) It can also be prepared by dehydration of propan-l-ol
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 65

MP Board Solutions

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 SN2 (Nucleophilic attack)
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 66
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 67
Similarly, the 3° alcohol (CH3)3COH forms isobutane.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 68

Question 28.
Write the equation of the reaction with hydrogen iodide :
(i) 1-Propoxypropane
(ii) Methoxybenzene and
(iii) Benzyiethylether.
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 69

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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 70
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 71
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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 72
Protonated ether undergoes SN2 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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 73

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 74

(ii) Nitration of anisole : Anisole reacts a mixture of concentrated sulphuric and nitric acids to yield a mixture of ortho and para nitroanisole.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 75

(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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 76

(iv) Friedel-Craft’s acetylation of anisole :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 77

Question 32.
Show how would you synthesize the following alcohols from appropriate alkenes ?
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 78
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 79
Both the alkene give the desired alcohol on adding a molecule of H2O.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 80
(ii) Two alkenes are formed. They will give the desired alcohol on adding a H2O molecule.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 81

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 82
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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 83
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 84
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 85

MP Board Solutions

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
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 86
Answer:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 87

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. CnH2n+1OCnH2n+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 :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 88
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.

MP Board Solutions

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 89

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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 90
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 91

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 92
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 93

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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 94
(ii) Ethanol from Diethyl ether :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 95
(iii) Ethyl acetate from Ethanol :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 96
(iv) Ethanol from Glucose :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 97

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 98
This quinone again combines with two molecules of phenol by H-bond and gives pink phylloquinone.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 99

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 100
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 101
In alcohols, no resonance is possible hence the hydrogen atom is more firmly linked to the oxygen.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 102

MP Board Solutions

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 103

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 104
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 105
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 106

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 107
(ii) The reaction of diethyl ether with cone. HI acid, on heating gives one molecule of ethyl iodide and one molecule of ethyl alcohol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 108

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 109
(ii) Reaction with NaOH: Phenol dissolve in NaOH to give sodium phenoxide.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 110

(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.
pKa value of ethanol is 15.9 where of phenol is 10.0. Thus, phenol is many more times’ more acidic than ethanol.

MP Board Solutions

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 111
Mechanism : (i) Protonation of alcohol by H2SO4
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 112
(ii) Removal of water
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 113

(iii) Elimination of β-hydrogen in the form of proton by base (bisulphate ion)
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 114
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).
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 115
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 116
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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 117
(ii) Ethanol to methanol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 118

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 KMnO4, 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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 119

(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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 120

(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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 121

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 122
Secondary alcohols lose hydrogen and yield a ketone.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 123
Tertiary alcohols are not dehydrogenated but lose a water molecule to give alkenes.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 124

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 125
(ii) Phenol to Picric acid : Nitration : On nitration, phenols give a variety of products depending upon the conditions.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 126
Nitration of phenol with conc. HNO3 in presence of conc. H2SO4 gives, 2,4,6-Trinitro-phenol (Picric acid).
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 127
(iii) Phenol to Aniline:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 128
(iv) Phenol to Benzene:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 129
(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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 130
(vi) Phenol to ortho and para cresol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 131

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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 132
(ii) Ethanol from Diethyl ether:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 133
(iii) Ethyl acetate from Ethanol:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 134
(iv) Ethanol from Glucose :
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 135

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:
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 136
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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 137

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.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 138
(ii) By alkaline fusion of sodium benzene sulphonate : On fusing sodium benzene sulphonate with NaOH, sodium phenoxide is formed which on acidification forms phenol.
MP Board Class 12th Chemistry Solutions Chapter 11 Alcohols, Phenols and Ethers 139
(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 Chapter 11 Alcohols, Phenols and Ethers 140

MP Board Class 12th Chemistry Solutions

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