MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

MP Board Class 12th Chemistry Solutions Chapter 10 Haloalkanes and Haloarenes

Haloalkanes and Haloarenes NCERT Intext Exercises

Question 1.
Write structures of the following compounds :
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert. ButyI-3-iodoheptane
(iv) 1,4-Dibromobut-2-ene
(v) 1-Bromo-4-sec. buty1-2-methylbenzene.
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
H₂SO₄ cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:

Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields:
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides.
Answer:

All the hydrogen are the same i.e. 1°. Therefore, the replacement of only one of them will give the same product.

Replacement of a, b and c hydrogen atoms give three isomeric monochlorides.

Replacement of a, b, c and d hydrogen atoms give four isomeric monochlorides.

Question 5.
Draw the structures of major monohalo products in each of the following reactions:

Answer:

Question 6.
Arrange each set of compounds in order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
(i) Chloromethane, Bromomethane, Dibromomethane, Bromoform (Boiling point increases with increase in molecular mass).
(ii) Isopropylchloride, 1-Chloropropane, 1-Chlorobutane. (Isopropylchloride being branched has lower b.p. than 1-Chloropropane).

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an SN₂ mechanism? Explain your answer.

Answer:
If the leaving group is the same in different isomers of a particular formula, the reactivity of the isomers towards the S_N2 mechanism decreases with the steric hindrance.

which is 2° alkyl halide having some steric hindrance.

which is 3° alkyl halide having much more steric hindrance (than 2°).

are 2° alkyl halides. But in (II) the CH₃ group is at the C₂ atom which is closer to Br (exerts more steric hindrance) to the attacking nucleophile at the C₁ atom as compared to (I).

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster S_N1 reaction:

Answer:
The reactivity of alkyl halide towards S_N1 reaction depends upon the stability of the intermediate carbocation formed as 3° > 2° > 1°

Question 9.
Identify A, B, C, D, E, R and R’ in the following :


Answer:

Since, D of D₂O gets attached to same C-atom on which – MgBr or Br was present so that

tert-Alkyl halides do not undergo Wurtz reaction. Therefore, the question is not correct. They undergo dehydrohalogenation to give alkenes. Hence,

Haloalkanes and Haloarenes NCERT Textbook Exercises

Question 1.
Name of the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl, or aryl halides:

1. (CH₃)₂CHCH(Cl)CH₃
2. CH₃CH₂CH(CH₃)CH(C₂H₅)Cl
3. CH₃CH₂C(CH₃)₂CH₂I
4. (CH₃)₃CCH₂CH(Br)C₆H₅
5. CH₃CH(CH₃)CH(Br)CH₃
6. CH₃C(C₂H₅)₂CH₂Br
7. CH₃C(Cl)(C₂H₅)CH₂CH₃
8. CH₃CH=C(Cl)CH₂CH(CH₃)₂
9. CH₃CH=CHC(Br)(CH₃)₂
10. p-ClC₆H₄CH₂CH(CH₃)₂
11. m-ClCH₂C₆H₄CH₂C(CH₃)₃
12. o-Br-C₆H₄CH(CH₃)CH₂CH₃.

Answer:

1. 2-Chloro-3-methylbutane (2° alkyl)
2. 3-Chloro-4-methylhexane (2° alkyl)
3. l-Iodo-2,2-dimethylbutane (1° alkyl)
4. l-Bromo-3,3-dimethyl-l-phenylbutane (2° benzylic)
5. 2-Bromo-3-methylbutane (2° alkyl)
6. 3-Bromomethyl-3-methylpentane (1° alkyl)
7. 3-Chloro-3-methylpentane (3° alkyl)
8. 3-Chloro-5-methylhex-2-ene (vinyl)
9. 4-Bromo-4-methylpent-2-ene (allylic)
10. 1-Chloro-4-(2′-methylpropyl) benzene (aryl) or p-Chloro isobutyl benzene
11. 1-Chloromethyl-3-(2’2′-diethylpropyl) benzene (benzylic) or /n-Neopentyl ben-zyl chloride
12. 1-Bromo-2-(l’-methylpropyl) benzene (aryl).

Question 2.
Give the IUPAC names of the following compounds :

1. CH₃CH(Cl)CH(Br)CH₃
2. CHF₂CBrCIF
3. ClCH₂C = CCH₂Br
4. (CCl₃)₃CCl
5. CH₃C(p-CIC₆H₄)2CH(Br)CH₃
6. (CH₃)₃CCH=ClC₆H₄I-p.

Answer:

1. 2-Bromo-3-chlorobutane
2. 1 -Bromo-1 -chloro-1,2,2-trifluoroethane
3. l-Bromo-4-chlorobut-2-yne
4. 1,1,1,2,3,3,3-heptachloro-2-(trichloromethyl) propane
5. 3-Bromo-2,2,-bis (4′-chlorophenyl) butane
6. 1-Chloro-l-(4′-iodophenyl)-3,3-dimethyl-but-1-ene.

Question 3.
Write the structures of the following organic halogen compounds :
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) Perfluorobenzene
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene.
Answer:

Question 4.
Which one of the following has the highest dipole moment:
(i) CH₂Cl₂
(ii) CHCl₃
(iii) CCl₄.
Answer:

CCl₄ (iii) is a symmetrical and has a resultant zero dipole moment. In CHCl₃ (ii), the resultant of two C – Cl dipoles is opposed by the resultant of C – H and C – Cl bonds. Since the latter result is expected to be smaller than the former, therefore, CHCl₃ has a finite dipole (1.03 D) moment.
In CH₂Cl₂ (i), the resultant of two C-Cl dipole moments is reinforced by the resultant of two C-H dipoles, therefore, CH₂Cl₂ (1.62 D) has a dipole moment higher than that of CHCl₃.
Thus, CH₂Cl₂ has the highest dipole moment.

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight Identify the hydrocarbon.
Answer:

1. The hydrocarbon with molecular formula C₅H₁₀ can be either a cycloalkane or alkene.
2. Since the hydrocarbon does not react with Cl₂ in the dark, it cannot be an alkene but it must be cyclohexane.
3. Since the compound can give a single monochloro derivative, the cycloalkane is cyclopentane.

Question 6.
Write the isomers of the compound having the formula C₄H₉Br.
Answer:
The isomers of C₄H₉Br along with their common names are given below :

Question 7.
Write the equations for the preparation of 1-iodobutane from :
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
Answer:

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
The nucleophiles with two nucleophilic centres are called ambident nucleophiles. Depending on the reagent and the reaction conditions, the reaction may take place predominantly at one of these centres.
Example: CN^– has two nucleophilic centres, one each at carbon and nitrogen and forms cyanides and isocyanides respectively.

Question 9.
Which compound in each of the following pairs will react faster in S_N2 reaction with OH-:
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl.
Answer:
(i) CH₃I reacts faster than CH₃Br in S_N2 reaction with OH^– because I^– ion is a better leaving group than Br^– ion because of its large size.
(ii) CH₃Cl reacts faster than (CH₃)₃CCl because of steric hindrance in (CH₃)₃CCl.

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene :
(i) 1-Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2,2,3-Trimethyl-3-bromopentane.
Answer:
(i) In 1-Bromo-l-methylcyclohexane, the β hydrogens on either side of the Br atom are equivalent, therefore, only 1-alkene is formed.


Since, the alkene (A) is more substituted according to SaytzefTs rule, it is more stable and will be the major product.

Question 11.
How will you bring about the following conversions :
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1-nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propyne
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-l-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
Answer:

Question 12.
Explain why:
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride ?
(ii) Alkyl halides, though polar, are immiscible with water ?
(iii) Grignard reagents should be prepared under anhydrous conditions ?
Answer:

Due to sp² hybridization of C-atom in chlorobenzene, C-atom is more electronegative (greater s-character) whereas, cyclohexyl chloride, C-atom is sp3 hybridized i.e., less electronegative (lesser 5-character). So, polarity of C—Cl bond in chlorobenzene is less than C—Cl bond in cyclohexyl chloride. Further due to delocalisation of lone pair of electrons of Cl-atom over the benzene ring, C—Cl bond in chlorobenzene acquires partial double bond character while C—Cl bond in cyclohexyl chloride is a pure single bond. Thus, C—Cl bond in chlorobenzene is shorter than in cyclohexyl chloride. As dipole moment is a product of charge and distance, therefore, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

(ii) Water molecules have enough strong intermolecular H-bonding which is difficult to be broken by alkyl halides though they are polar in nature as well. Therefore, alkyl halide do not dissolve in water and form separate layers.

(iii) Grignard reagents (R-Mg-X) are readily decomposed by water to produce alkanes. That is why, they should be prepared under anhydrous conditions. Instead, ether is used as a solvent during the preparation of the Grignard reagent.

Question 13.
Give the uses of freon-12, D.D.T., Carbon tetrachloride and Iodoform.
Answer:
Uses of freon – 12:
Freon -12 (dichlorodifluoromethane, CF₂Cl₂) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays etc. However, it damages the ozone layer. Hence its manufacture is banned.

Uses of DDT:
DDT (p, p – dichlorodiphenyltrichloroethane) is one of the best-known pesticides. It is very effective against mosquitoes and lice. However, due to its harmful effects, its manufacture is banned;
Uses of carbon tetrachlorides (CCl₄):

• It is used for manufacturing refrigerants and propellants for aerosol cans.
• It is used as feedstock in the synthesis of CFCs and other chemicals.
• It is used as a solvent in the manufacture of pharmaceutical products.
• CCl₄ was once widely used as a cleaning fluid, a degreasing agent in industries, a spot remover homes, and a fire extinguisher.

Uses of iodoform (CHI₃):
Iodoform was used as an antiseptic, but now it has been replaced by other formulations containing iodine-due to its objectionable smell. The antiseptic property of CHI₃ is only due to the liberation of free iodine when it comes in contact with the skin.

Question 14.
Write the structure of the major organic product in each of the following reactions :


Answer:

Question 15.
Write the mechanism of the following reaction :

Answer:
KCN gives CN- ion as a nucleophile in an aqueous medium which is resonance hybrid of the following :

Thus, cyanide ions is an ambident nucleophile. Therefore, it can attack the C-atom of C—Br bond in «-BuBr either through C-atom or through N-atom. Thus, two possible products are cyanides and isocyanides respectively.

But C—C bond is more stable than C—N bond, so attack occurs through C-atom and hence, cyanide is predominantly formed.

Question 16.
Arrange the compounds of each set in order of reactivity towards SN2 displacement:
(i) 2-Bromo-2-methylbutane, 1 Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methyIbutane, 2-Bromo-2-methylbutane, 3-Bromo-2- methylbutane
(iii) 1-Bromobutane, l-Bromo-2,2-dimethyl-propane, l-Bromo-2-methylbutane, l-Bromo-3-methylbutane.
Answer:
The reactivity of S_N2 reaction depends upon steric hindrance. More the steric hindrance lesser will be the reactivity. Therefore, the reactivity of different alkyl halides towards S_N2 reaction is 1° > 2° > 3°.

Question 17.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH ?
Answer:

However, under S_N2 mechanism, the reactivity depends on steric hindrance, therefore, C₆H₅CH₂Cl gets hydrolysed more easily than C₆H₅CHClC₆H₅ under S_N2 conditions.

Question 18.
p-Dichlorobenzene has higher m.p. and solubility than those of o- and m- isomers. Discuss.
Answer:
p-Dichlorobenzene is more symmetrical than o- and m- isomers. For this season, it fits more closely than o- and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has higher mp and lower solubility than o- and m- isomers.

Question 19.
How the following conversions can be carried out:
(i) Propene to propan-1-ol
(ii) Ethanol to but-l-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3,4-dimethyIhexane
(x) 2-Methyl-1-propene to 2-chloro-2-methyl-propane
(xi) Ethyl chloride to propanoic acid
(xii) But-1-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) ttert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenyl isocyanide.
Answer:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
In the presence of water, KOH dissociates completely into OH^– ions which being a strong nucleophile brings about substitution on alkyl halides and produce alcohols from alkyl halide. Further in an aqueous solution, OH^– ions are highly solvated (hydrated). This solvation reduces the basic character of OH^– ions which, therefore, fails to abstract a proton from β-carbon of alkyl chloride to form alkenes. In an alcoholic medium, (less polar than H₂O) OH- is less highly hydrated, therefore, acts as the strong base and abstracts the proton from β-carbon giving alkene as a major product (dehydrohalogenation). Moreover, the alcoholic solution contains C₂H₅O- ethoxide ion in addition to OH^– ions. Being a stronger base than OH^–, the abstract proton gives alkene.

Question 21.
Primary alkyl halide C₄H₉Br (a) is reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d). C₈H₁₈ is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.
Answer:
Two possible isomers of given 1° alkyl halide C₄H₉Br are :

According to the question, compound (a) on reaction with sodium does not give the same product produced by n-butyl bromide. So (a) cannot be (I).

Question 22.
What happens when:
(i) n-butyl chloride is treated with alcoholic KOH.
(ii) bromobenzene is treated with Mg in the presence of dry ether.
(iii) chlorobenzene is subjected to hydrolysis.
(iv) ethyl chloride is treated with aqueous KOH.
(v) methyl bromide is treated with sodium in the presence of dry ether.
(vi) methyl chloride is treated with KCN?
Answer:
(i) CH₃CH₂CH₂CH₂Cl + KOH (alc.)

Haloalkanes and Haloarenes Other Important Questions and Answers

Haloalkanes and Haloarenes Objective Type Questions

Choose the correct answer:

Question 1.
Which of the following compound gives a yellow precipitate with AgNO3 solution:
(a) KIO₃
(b) CHI₃
(c) KI
(d) CH₂I₂.
Answer:
(c) KI

Question 2.
What is formed by the reaction of Ethyl bromide with lead sodium alloy :
(a) Tetraethyl lead
(b) Tetraethyl bromide
(c) Both (a) and (b)
(d) None of these.
Answer:
(a) Tetraethyl lead

Question 3.
Reaction CH₃Br + OH → CH₃ – OH + Br- is:
(a) Electrophilic substitution
(b) Electrophilic addition
(c) Nucleophilic addition
(d) Nucleophilic substitution.
Answer:
(d) Nucleophilic substitution.

Question 4.
When acetylene added to HC1, the product formed is :
(a) CH₂ = CHCl
(b) CH₃ – CH – Cl₂
(c) Cl – CH = CHCl
(d) None of these.
Answer:
(b) CH₃ – CH – Cl₂

Question 5.
In Aryl halide, the carbon linked to the halogen atom is :
(a) sp hybridized
(b) sp hybridized
(c) sp hybridized
(d) sp³d hybridized.
Answer:
(b) sp hybridized

Question 6.
In S_N1 reaction in the first step is formed :
(a) Free radical
(b) Carbanion
(c) Carbocation
(d) Final product.
Answer:
(c) Carbocation

Question 7.
Chlorobenzene reacts with chloral and concentrated H₂SO₄ to form :
(a) PVC
(b) TNT
(c) B.H.C.
(d) DDT.
Answer:
(d) DDT.

Question 8.
Reaction CH₃OH + OH^– → CH₃OH + Br^– is :
(a) S_N1
(b) S_N2
(C) S_E-1
(d)S_E-2.
Answer:
(b) S_N2

Question 9.
The following compound reacts with silver powder to form acetylene :
(a) CH₂I₂
(b) CH₃I
(c) CHI₃
(d) Cl₄.
Answer:
(c) CHI₃

Question 10.
Pyrene is used with any one of the following for extinguishing fire :
(a) CO₂
(b) CH₂Cl₂
(c) CCl₄
(d) CH₂ = CHCl.
Answer:
(c) CCl₄

Question 11.
Which of the following compound is known by the name freon :
(a) CHCl₃
(b) CCl₄
(c) CCl₂F₂
(d) CF₄.
Answer:
(c) CCl₂F₂

Question 12.
On heating ethyl iodide with alcoholic KOH the product formed is :
(a) Ethanol
(b) Ethane
(c) Acetylene
(d) Ethylene.
Answer:
(d) Ethylene.

Question 13.
On heating C₂H₅OH with iodine and a base the product formed is :
(a) CH₃I
(b) CHI₃
(c) CH₃CHO
(d) CHCI₃.
Answer:
(b) CHI₃

Question 14.
Which of the chemical formula is of chloropicrin :
(a) CCl₃ – CHO
(b) C(NO₂)Cl₃
(c) CH₃ – C(NO₂)Cl₂
(d) CCl₃ – NH₂.
Answer:
(b) C(NO₂)Cl₃

Question 15.
Order of polarity of CH₃I, CH₃Br and CH₃Cl molecules is :
(a) CH₃Br > CH₃Cl > CH₃I
(b) CH₃I > CH₃Br > CH₃Cl
(c) CH₃CI > CH₃Br > CH₃I
(d) CH₃CI > CH₃I > CH₃Br.
Answer:
(c) CH₃CI > CH₃Br > CH₃I

Question 16.
Correct order of reactivity of Alkyl halides is :
(a) Iodide > Bromide > Chloride
(b) Iodide < Bromide < Chloride
(c) Bromide > Iodide > Chloride
(d) Bromide < Chloride < Iodide.
Answer:
(a) Iodide > Bromide > Chloride

Question 17.
What is formed on heating iodoform with silver powder :
(a) Alkane
(b) Ethylene
(c) Acetylene
(d) Isocyanide.
Answer:
(c) Acetylene

Question 18.
Raschig method is used for the manufacture of which of the following :
(a) Chlorobenzene
(b) Benzene
(c) Toluene
(d) Nitro-benzene.
Answer:
(a) Chlorobenzene

Question 2.
Fill in the blanks :

1. The formula of the harmful product formed on keeping chloroform open is ………………
2. General formula of alkyl halide is ………………
3. On heating aromatic primary amine with chloroform and alcoholic caustic potash a bad smelling gas ……………… is formed.
4. B.H.C. is an insecticide whose commercial name is ………………
5. Chloretone is a high grade ………………
6. SN1 reaction occur in ……………… steps.
7. In haloarene substitution reactions are mainly ………………
8. Formula of refrigerant Freon is ………………

Answer:

1. 1. COCl₂

1. C_nH_2n+1X
2. Phenyl isocyanide
3. Gammaxene or lindane
4. Hypnotic medicine
5. Two
6. Electrophilic
7. CCl₂F₂.

Question 3.
Match the following :

Answer:

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

Question 4.
Answer in one word/sentence :

1. On reacting Benzene with methyl chloride in presence of AlCl₃, Toluene is formed, what is the name of the reaction ?
2. Alkyl halide is polar in nature still it is insoluble in water.
3. On heating alkyl halide with sodium metal, product formed is.
4. On heating Benzene diazonium salt with cuprous halide and its corresponding acid haloarene is formed. Write name of the reaction.
5. On heating iodobenzene with copper powder at 200°C, product obtained is.
6. What is formed on treating benzene with Cl₂ in the presence of sunlight ?
7. Write laboratory method of preparation of chlorobenzene.
8. Write the name of nucleophilic substitution reaction in primary alkyl halide.

Answer:

1. Friedel-Crafts reaction
2. Due to inability to form hydrogen bond
3. Alkane
4. Sandmeyer reaction
5. Diphenyl
6. B.H.C.
7. Raschig method
8. Bimolecular nucleophilic substitution reaction.

Haloalkanes and Haloarenes Short Answer Type Questions

Question 1.
(i) Write Iodoform reaction.
(ii) Iodoform gives yellow ppt with AgNO₃ solution but chloroform doesn’t Why ?
(iii) What happens when ethyl bromide is heated with alcoholic KOH ?
Answer:
(i) Iodoform reaction : When ethyl alcohol or acetone is heated with iodine and NaOH, yellow crystals of iodoform are formed.
C₂H₅OH +4I₂ + 6NaOH → 5NaI + HCOONa + 5 H₂O + CHI₃

(ii) When iodoform is heated with AgNO₃ solution a yellow ppt. (Agl) is obtained but chloroform doesn’t give this reaction because in chloroform C-Cl bond is more stable than C-I bond in iodoform.

(iii) On boiling Ethyl bromide with alcoholic KOH ethylene is formed.

Question 2.
Explain the Sandmeyer reaction with example.
Answer:
Decomposition of diazonium salts (Sandmeyer reaction): When a diazonium salt solution is added to a solution of cuprous halide dissolved in the corresponding halogen acid, the diazo group is replaced by a halogen atom.

Question 3.
Give chemical reaction between chlorobenzene and chloral in presence of cone. H₂SO₄.
Or,
How is D.D.T. formed ? Write its one application.
Answer:
DDT (Dichlorodiphenyl trichloroethane) is formed by the condensation of one molecule of chloral with two molecules of chlorobenzene in presence of cone. H₂SO₄.

Application : It is an important pesticide.

Question 4.
What are Gem-dihalide and Vicinal-dihalide ?
Answer:
When both halogen atoms are linked to one carbon atom of hydrocarbon then it is known as Gem-dihalide. Gem means geminal i.e., same position.

When both halogen atoms are connected to two different neighbouring carbon atoms, then it is known as vicinal dihalide. Vic means vicinal which means adjacent position.

Question 5.
What is Lucas reagent? Give its application.
Answer:
Solution of ZnCl₂ in conc. HCl is known as Lucas reagent.
Application: It is used to differentiate primary, secondary and tertiary alcohols. On adding the alcohol to Lucas reagent, a tertiary alcohol reacts immediately forming a ppt. of alkyl chloride. If the ppt. appears after few minutes, then the alcohol is secondary. If no ppt. is obtained in cold the alcohol is primary.

Question 6.
Explain, Carbylamine reaction and give one application of this reaction.
Answer:
Carbylamine reaction: On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenyl isocyanide or carbylamine is formed which has a very bad smell and is poisonous.

Application: Chloroform and primary amine can be tested by this reaction.

Question 7.
What is 666 (lindane)? Explain its preparation and use in agriculture.
Answer:
It is 1, 2, 3, 4, 5, 6-Hexachlorocyclohexane. It is obtained by heating benzene with chlorine in presence of sunlight.
Preparation: It is prepared by the chlorination of benzene in the presence of ultra-violet light.

1, 2, 3, 4, 5, 6-Hexachlorocyclohexane (B.H.C.)
Uses : Benzene hexachloride is an addition compound and its 7 -isomer is called gammexane. It is an important pesticide used in agriculture. It is also called lindane or 666.

Question 8.
Explain the following reaction of chlorobenzene :
(i) Reaction with chlorine in the presence of FeCI₃ in dark
(ii) Fittig reaction.
Answer:
(i) When Chlorobenzene reacts with Cl₂ in the presence of FeCl₃ in dark o – dichlorobenzene and P – dichlorobenzene is obtained.

(ii) Fittig reaction : When Chlorobenzene is heated at 200°C with Cu powder in a sealed tube Diphenyl is formed.

When two molecules of aryl halide reacts with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as Fittig reaction.

Question 9.
How can you obtained following compounds from chloroform ? Give equations :
(a) Methane
(b) Acetylene
(c) Carbon tetrachloride.
Answer:
(a) Chloroform reduces into methane by Zn and H₂O.

(b) When CHCl₃ heated with Ag powder it gives C₂H₂.

(c) By the chlorination of CHCl₃ in presence of sunlight CCl4 is formed.

Question 10.
Write short notes on :
(a) Hunsdiecker method and
(b) Raschig process.
Ans.
(a) Hunsdiecker method : When silver salt of a carboxylic acid is heated with bromine, in the presence of an inert solvent like CCl₄, aryl bromide is formed.

This method is called Hunsdiecker method.

(b) Raschig process : When benzene vapours mixed with air and HCl gas is passed over CuCl₂(catalyst) at 230°C, chlorobenzene is formed.

Question 11.
Write method of preparation, properties and uses of Freon.
Answer:
Freon : Dichloro, Difluoro methane.
It is formed by the action of SbF₃ with CCl₄ in presence of SbCl₅.

It has very low boiling point due to which by increasing the pressure at room temperature it can be liquefied.
It is a non-poisonous, non-combustible and inactive substance which is used as a cooling agent in the refrigerator. It is used in aerosol and foam.

Question 12.
(i) b.p. of ethyl iodide is higher than b.p. of ethyl bromide. Give reason
(ii) Explain why the m.p. of para dichlorobenzene is higher than its ortho and meta derivatives.
Answer:
(i) In alkyl halides containing same alkyl group boiling point increases with increases in atomic weights of halogen atoms. Molecular weight of ethyl iodide is more than ethyl bromide and therefore boiling point of ethyl iodide is also high.
(ii) Para derivatives of dichlorobenzene is more symmetrical than its ortho and meta derivatives therefore its m.p. is higher.

Question 13.
Give main nucleophilic substitution reaction of alkyl halides.
Answer:
Nucleophilic substitution Reactions :

1. Substitution by —OH group (Hydrolysis): Alkyl halide on reacting with water or aqueous KOH hydrolyse to form alcohol.
C₂H₅Br + KOH → C₂H₅ — OH + KBr

2. Substitution by —OR group : Alkyl halide reacts with sodium alkoxide or Ag₂O to form ether by substitution of halogen atom by —OR group.
C₂H₅Br + NaOC₂H₅ → C₂H₅ — OC₂H₅ + NaBr
2C₂ H₅I + Ag₂O → (C₂H₅)₂ O + 2AgI

3. Substitution by —CN group: Alkyl halide reacts with aqueous or alcoholic KCN to form alkyl cyanide.
C₂H₅Cl + KCN → C₂H₅CN + KCl

4. Substitution by ammonia (Hofmann method) : On heating alkyl halide with aqueous or alcoholic solution of NH₃ in a sealed tube at 100°C, a mixture of various amine is obtained.

Question 14.
Give the laboratory method for the preparation of chlorobenzene and explain its nitration and sulphonation reactions.
Answer:
Laboratory method: Chlorobenzene is prepared by direct halogenation of arene. Aryl chloride may be prepared from arenes by the action of chlorine or presence of halogen carrier like FeCl₃, FeBr₃ and AlCl₃. Iodine and iron filings can also be used as halogen carrier.

Nitration : Haloarenes react with nitrating mixture to form o-nitro and p-nitro sub-stituted haloarenes.

Sulphonation : On heating with cone. H₂SO₄, 2-Chlorobenzene sulphonic acid and 4-Chlorobenzene sulphonic acid are formed.

Question 15.
Give the chemical reactions when ethyl halide reacts with the following :
(i) Alloy of Pb-Na
(ii) Mg metal
(iii) AgNO₂
(iv) Na metal.
Answer:
(i) Reaction with lead-sodium alloy: Alkyl halides, form alkyl lead when treated with lead-sodium alloy.

Tetraethyl lead (TEL) is an antiknock compound which is added in petrol.

(ii) Reaction with magnesium : Alkyl halide forms alkyl magnesium halide i.e., Grignard reagent with magnesium in presence of dry ether as a solvent.

(iii) Reaction with AgNO₂: Mainly nitroethane is formed.
C₂H₅I + AgNO₂ → C₂H₅ NO₂ + Agl

(iv) Reaction with sodium (Wurtz reaction) : When alkyl halide is heated with sodium, in the presence of dry ether, alkane is formed.

Question 16.
Give the preparation method, properties and applications of dichloroethane. Ans. Preparation of dichloroethane: It can be prepared by replacing two hydrogen of ethane by two chlorine atoms.
(i) By the passage of Cl₂ into ethene :

(ii) By the heating of mix of ethane diol and HCl in presence of anhydrous ZnCl₂.

Properties : (a) Reaction with aqueous KOH :

(b) Reaction with alcoholic KOH :

Vinyl chloride reacts with ale. KOH and form vinyl ethyl other.
CH₂ = CHCl + HOC₂H₅ + KOH → CH₂ = CH – OC₂H₅ + KCl + H₂O
(c) Reaction with KCN :

(d) Reaction with Zn powder and methanol:

Use : (i) As a solvent, (ii) As an antiknocking fuel, (iii) Removing paints.

Question 17.
Write Frankland reaction.
Answer:
Reaction with zinc (Frankland’s reaction) : This reaction is similar to the Wurtz reaction but zinc is used in place of sodium.

Question 18.
Explain Friedel-Craft’s reaction with chemical equation.
Answer:
Friedel-Craft’s reaction (alkylation) : Alkyl halides react with benzene in presence of anhydrous aluminium chloride to give alkyl benzene.

Acetylation : Acetyl chloride reacts with benzene in presence of anhydrous aluminium chloride to give acetophenone.

Question 19.
Identify ‘A’, ‘B’ ‘C’ and ‘D’

Answer:

Question 20.
An alcohol ‘A’ on reaction with cone. H₂SO₄ gives an alkene ‘B’. ‘B’ after bromination with sodamide gives dehydrogenated compound ‘C’. ‘C’ on reaction of H₂SO₄ in presence of HgSO₄ gives ‘D’ Identify ‘A’, ‘B% ‘C% and ‘D’.
Answer:

Haloalkanes and Haloarenes Long Answer Type Questions

Question 1.
Write short notes on :
(a) Hunsdiecker method
(b) Raschig process
(c) Wurtz’s reaction
(d) Westron
(e) Frankland reaction
(f) Carbylamine reaction
(g) Iodoform reaction
(h) Fittig reaction.
Answer:
(a) Hunsdiecker method : When silver salt of a carboxylic acid is heated with bromine, in the presence of an inert solvent like CCl₄, aryl bromide is formed.

This method is called Hunsdiecker method.
(b) Raschig process : When benzene vapours mixed with air and HCl gas is passed over CuCl₂(catalyst) at 230°C, chlorobenzene is formed.

(c) Wurtz’s reaction : When alkyl halide is heated with sodium, in the presence of dry ether, alkane is formed.

(d) Westron : Symmetrical tetrachloromethane or acetylene tetrachloride are known as Westron. It can be prepared by the chlorination of acetylene.
Westron, it can be prepared by the chlorination of acetylene.

It is poisonous, non- flammable liquid, which gives westrol when it is boiled with lime.
(e) Frankland reaction : When alkyl halide is heated with zinc dust, alkane is formed.

(f) Carbylamine reaction : On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has bad smell and is poisonous.

(g) Iodoform reaction : When ethyl alcohol or acetone is heated with iodine and NaOH, yellow crystals of iodoform are formed.
C₂H₅OH +4I₂ + 6NaOH → 5NaI + HCOONa + 5 H₂O + CHI₃

(h) Fittig reaction : When Chlorobenzene is heated at 200°C with Cu powder in a sealed tube Diphenyl is formed.

When two molecules of aryl halide reacts with sodium metal in presence of dry ether, then diphenyl is formed. This reaction is known as Fittig reaction.

Question 2.
Give the laboratory method for the preparation of chloroform. Describe the formation of chloroform by ethanol with labelled diagram, equation and principle.
Answer:
Laboratory method : Chloroform is prepared in the laboratory by the action of water and bleaching powder on ethyl alcohol or acetone.

Method : About 100 gm of bleaching powder made into a paste by adding about 200 ml of water and taken in a flask fitted with a condenser. Now, 25 ml of alcohol or acetone is added and the mixture is distilled, chloroform collects as a heavy liquid under water.

It is washed with dilute NaOH solution then with water, dried over fused calcium chloride and redistilled.

The available chlorine of bleaching powder acts as oxidising as well as chlorinating agent during the preparation of chloroform from alcohol and acetone.

CaOCl₂ + H₂O →(OH)₂ + Cl₂

The chemistry involved in the conversion of alcohol and acetone into chloroform is as shown below:

(A) From alcohol: The steps involved are :
(i) Ethyl alcohol is oxidized by chlorine to acetaldehyde.

(ii) Acetaldehyde reacts with chlorine to give chloral, i.e. trichloro acetaldehyde.

(iii) Two moles of chloral react with one mole of calcium hydroxide to produce chloroform.

Question 3.
What product is formed by the reduction of chloroform ? Give the chemical equation when it reacts to nitric acid and acetone.
Answer:
Reduction: (i) On heating with Zn and HCl, it reduces to form methylene dichloride.

(ii) On heating with zinc dust and water, methane is formed.

Nitration : On treating chloroform with concentrated nitric acid, the hydrogen atom of chloroform is replaced by nitro group and nitro chloroform (or chloropicrin) is formed. It is a liquid (b.p. 112°C) which is used in war as a poisonous gas.

Condensation : Chloroform condenses with acetone in presence of sodium hydroxide to form chloretone which is a hypnotic (sleep inducing drug) of high grade.

Question 4.
Give the Chemical reaction when chloroform reacts with following :
(a) Oxidation, (b) Carbylamine reaction, (c) Ag powder, (d) Nitration, (e) Reimer- Tiemann reaction.
Or,
How will you obtain the following from chloroform : (i) Carbonyl chloride, (ii) Acetylene, (iii) Chloropicrin, (iv) Phenyl isocyanide, (v) Chloretone, (vi) Salicylal- dehyde.
Or,
How trichloro methane reacts with : (a) Atmospheric air, (b) Aniline and ale. KOH, (c) Ag powder, (d) Cone. HNO₃, (e) Phenol.
Answer:
(a) Action of air and light (Oxidation): Chloroform oxidizes in presence of sunlight and air and forms a poisonous gas, phosgene (carbonyl chloride).

Ordinary chloroform contains phosgene gas and is used as a solvent. Pure chloroform which is used as an anaesthetic does not contain even traces of phosgene. While preserving chloroform which is to be used as an anaesthetic, the following precautions are taken :

(i) The chloroform is filled in blue or brown coloured bottle up to the neck. After putting a stopper, the bottle is kept in dark. As there is no empty space in the bottle, it is also free from air.
(ii) One percent ethyl alcohol is added in the bottle. If phosgene gas is formed, alcohol reacts with it to form diethyl carbonate, a non-toxic substance, i.e. (C₂H₅)₂CO₃,

(b) Carbylamine reaction : On heating chloroform with primary amine (e.g., aniline) and alcoholic KOH solution, phenylisocyanide or carbylamine is formed which has a very bad smell and is poisonous.

(c) Reaction with Ag powder or Dehalogenation : On heating chloroform with sil-ver powder, pure acetylene gas is formed.
CHCl₃ + 6 Ag + Cl₃CH → HC ≡ CH + 6 AgCl

(d) Nitration : On treating chloroform with concentrated nitric acid, the hydrogen atom of chloroform is replaced by nitro group and nitro chloroform (or chloropicrin) is formed. It is a liquid (b.p. 112°C) which is used in war as a poisonous gas.

(e) Reimer-Tiemann reaction : On heating chloroform with concentrated alkali and phenol at 60-70°C, o-hydroxy benzaldehyde (salicylaldehyde) is formed. Traces of p-hydro- xybenzaldehyde are also formed.

Question 5.
Explain nucleophilic substitution reaction of chlorobenzene (Give only equations).
Answer:
Nucleophilic substitution reaction of Chlorobenzene : Halogen atom in haloarenes is very strongly linked directly to benzene ring due to which it cannot be easily substituted by nucleophilic reagents like -OH,-OR,-NH₂,-CN etc. but at high pressure, temperature and in presence of suitable catalyst halogen can be substituted by these groups.

(i) Substitution by -OH group : On heating chlorobenzene with NaOH at 200 at-mospheric pressure and 300°C temperature phenol is formed.

(ii) Substitution by alkoxy (-OR) group :

With sodium alkoxide mixed ether is formed.

(iii) Substitution by Amino group : On heating with aqueous ammonia in presence of Cu₂O at 60°C atmospheric pressure and 200°C temperature aromatic amine is formed.

(iv) Substitution by Cyano group:

Question 6.
Explain the nucleophilic substitution reaction in alkyl halide by S_N1 and S_N2 mechanism.
Answer:
Nucleophilic substitution reaction : In the carbon halogen bond of haloalkane, halogen atom is more electronegative as compared to the carbon atom hence, the shared pair of electrons between carbon and halogen is more attracted by the halogen atom. As a result a small negative charge and an equivalent positive charge develops on halogen atom and carbon atom respectively.

Nucleophile attacks the electron deficient carbon due to the presence of partial posi-tive charge on it and replaces the weaker nucleophilic ion i.e. the halide ion. Thus, the reaction is known as nucleophilic substitution reaction.

The order of reactivity of different alkyl halide towards nucleophilic substitution reaction is:
RI > RBr > RCl > RF
Mechanism of Nucleophilic substitution reactions :
Nucleophilic substitution reaction occurs through two different mechanism :

(1) S_N1 Mechanism (Unimolecular nucleophilic substitution) : In this mechanism following steps are involved :
(a) Formation of carbocation by dissociation of substrate i.e., reactant molecule.

(b) Attack of nucleophile on carbocation forming the product.

(2) S_N2 mechanism (Bimolecular nucleophilic substitution): Reactions of this type occur in one step i.e. they are concerted reactions. These reaction nucleophilic attack results in a transition state in which both the reactant molecules are partially bonded to each other and then the halide ion escapes out forming the product.

Rate of reaction = K[(RX)OH^–]
Order of reactivity of alkyl halide is : Primaiy > Secondary > Tertiary.

Question 7.
Draw labelled diagram of laboratory method for preparation at iodoform from alcohol. Write related chemical equation.
Answer:

Chemical reaction :

Question 8.
Explain the following reactions of chlorobenzene :
(a) Reaction with chlorine in the absence of FeCl₃ in dark.
(b) Ullmann reaction.
Answer:
(a) Chlorine reacts with chlorine in dark, in the presence of FeCl3 to form ortho -dichlorobenzene and p-dichlorobenzene.

(b) On heating bromo or iodobenzene at 200°C temperature with Cu in a sealed tube, diphenyl is formed. This reaction is known as Ullmann reaction.

Question 9.
Write the equation of following reactions of chlorobenzene :
(i) Halogenation
(ii) Nitration
(iii) Sulphonation
(iv) Alkylation.
Answer:
(i) Halogenation : Haloarene reacts with halogen in presence of halogen carrier like FeCl₃ to form ortho and para substituted dihaloarene.

(ii) Nitration : Haloarenes react with nitrating mixture to form o-nitro and p-nitro substituted haloarenes.

(iii) Sulphonation : On heating with cone. H₂SO₄, 2-Chlorobenzene sulphonic acid and 4-Chlorobenzene sulphonic acid are formed.

(iv) Alkylation: Alkylation takes place with alkyl halide in presence of anhydrous AlCl₃.

Question 10.
Haloalkanes are more reactive than haloarenes. Give reason.
Or,
Why, aryl halides are less reactive than alkyl halides ?
Answer:
In aryl halides, halogen atom is attached more strongly to the nucleus therefore the nucleophilic substitution takes slowly than alkyl halides. There is two reasons for the less reactivity of aryl halides.

(i) In aryl halides sp3 hybridization takes place whereas in alkyl halides sp2 hybridization is present due to sp2 hybridization in haloarenes the halogen are attached to nucleus more strongly.

(ii) Due to the presence of resonance in aryl halides there is some double bond character in C—Cl bond. Thus, the bond length of C—Cl bond is lesser than C—Cl bond in haloalkanes. Therefore, it is difficult to replace the halogen of haloarenes.

MP Board Class 12th Chemistry Solutions