Class-XII

Chemistry

Chapter-10

Haloalkanes and Haloarenes

NCERT INTEXT QUESTION

10.1 Write structures of the following compounds:

(i) 2-Chloro-3-methylpentane

(ii) 1-Chloro-4-ethylcyclohexane

(iii) 4-tert. Butyl-3-iodoheptane

(iv) 1,4-Dibromobut-2-ene

(v) 1-Bromo-4-sec. butyl-2-methylbenzene.

Answer:

10.2 Why is sulphuric acid not used during the reaction of alcohols with KI?

Answer:

H2SO4 is an oxidising agent. It oxidises HI produced during the reaction to I2 and

thus prevents the reaction between an alcohol and HI to form alkyl iodide. To

prevent this, a nonoxidizing acid like H3PO3 is used.

10.3 Write structures of different Di halogen derivatives of propane.

Answer:

Four isomers are possible. These are:

10.4 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:

10.5 Draw the structures of major monohalo products in each of the following

reactions:

Answer:

10.6. Arrange each set of compounds in order of increasing boiling points :

(i) Bromomethane, bromoform, chloromethane, Di bromomethane

(ii) 1- Chloropropane, isopropyl chloride, 1- Chlorobutane.

Answer:

(i)Chloromethane < Bromomethane < Dibromo methane < Bromoform

The reason is:

(a)for same alkyl group, Boiling Point increases with size of halogen atom.

(b)Boiling Point increases as number of halogen atoms increase.

(ii) Isopropyl chloride < 1 – Chloropropane < 1 – Chlorobutane

Reason:

(a)For same halogen, Boiling Point increases as size of alkyl group increases.

(b)Boiling Point decreases as branching increases

10.7. Which alkyl halide from the following pairs would you expect to react

more rapidly by an SN2 mechanism? Explain your answer.

Answer:

10.8. In the following pairs of halogen compounds, which compound undergoes

faster SN1 reaction?

Answer:

10.9. Identify A, B, C, D, E, R and R1 in the following:

Answer:

NCERT EXERCISES

10.1 Name the following halides according to IUPAC system and classify them as

alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(i) (CH3)2CHCH(Cl)CH3

(ii) CH3CH2CH(CH3) CH(C2H5) Cl

(iii) CH3CH2C(CH3)2CH2I

(iv) (CH3)3CCH2CH(Br)C6H5

(v) CH3CH(CH3) CH(Br)CH3

(vi) CH3C(C2H5)2CH2Br

(vii) CH3C(Cl)(C2H5) CH2CH3

(viii) CH3CH=C(Cl)CH2CH(CH3)2

(ix) CH3CH=CHC(Br)(CH3)2

(x) p-ClC6H4CH2CH(CH3)2

(xi) m-ClCH2C6H4CH2C(CH3)3

(xii) o-Br-C6H4CH(CH3) CH2CH3

Answer:

(i)

2-Chloro-3methylbutane, (2° alkyl halide)

(ii)

3-Chloro-4-methyl hexane, (2° alkyl halide)

(iii)

1 -Iodo-2,2-dimethylbutane, (1° alkyl halide)

(iv)

l-Bromo-3, 3-dimethyl -1-phenylbutane, (2° benzylic halide)

(v)

2-Bromo-3-methylbutane, (2° alkyl halide)

(vi)

1-Bromo-2-ethyI-2-methylbutane, (1° alkyl halide)

(vii)

3-Chloro-3-methylpentane, (3° alkyl halide)

(viii)

3-Chloro-5-methylhex-2-ene, (vinylic halide)

(ix)

4-Bromo-4-methylpent-2-ene, (allylic halide)

(x)

1-Chloro-4-(2-methylpropyl) benzene, (aryl halide)

(xi)

1-Chloromethyl-3- (2,2-dimethylpropyl) benzene, (1° benzylic halide).

(xii)

1-Bromo-2- (l-methyl propyl) benzene, (aryl halide).

10.2 Give the IUPAC names of the following compounds:

(i) CH3 CH(Cl) CH(Br) CH3

(ii) CH F2 CBr ClF

(iii) Cl CH2 C ≡ C CH2 Br

(iv) ( CCl3)3 CCl

(v) CH3C(p – ClC6H4 )2 CH(Br) CH3

(vi) (CH3 )3 C CH = C Cl C6 H4 I – p

Answer:

(i)

2−Bromo−3−chlorobutane

(ii)

1−Bromo−1−chloro−1, 2, 2−trifluoroethane

(iii)

1 – Bromo – 4 – chloro but – 2 − yne

(iv)

2−(Tri chloromethyl) −1,1,1,2,3,3,3−heptachloropropane

(v)

2−Bromo−3, 3−bis (4 − chlorophenyl) butane

(vi)

1−chloro−1−(4−iodophenyl) −3, 3−dimethylbut−1−ene

10.3 Write the structures of the following organic halogen compounds.

(i) 2 -Chloro-3 -methyl pentane

(ii) p -Bromo chlorobenzene

(iii) 1 -Chloro-4-ethylcyclohexane

(iv) 2 – (2 -Chlorophenyl) -1 -iodooctane

(v) 2 – Bromobutane

(vi) 4 – tert – Butyl -3 -iodoheptane

(vii) 1 – Bromo – 4 – sec – butyl – 2 – methylbenzene

(viii) 1 ,4 – Dibromo but – 2 – ene

Answer:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

10.5 A hydrocarbon C5H10 does not react with chlorine in dark but gives a single

monochloro compound C5H9CI in bright sunlight. Identify the hydrocarbon.

Answer:

The hydrocarbon with molecular formula C5H9 0 can either a cycloalkane or an

alkene.

Since the compound does not react with Cl2 in the dark, therefore it cannot be an

alkene but must be a cycloalkane. Since the cycloalkane reacts with Cl2 in the

presence of bright sunlight to give a single monochloro compound, C5H9Cl,

therefore, all the ten hydrogen atoms of the cycloalkanes must be equivalent.

Thus, the cycloalkane is cyclopentane.

10.6 Write the isomers of the compound having formula C4H9Br.

Solution:

The possible isomers of C4H9Br are

10.7 Write the equations for the preparation of 1 -iodobutane from

(i) 1-butanol

(ii) 1-chlorobutane

(iii) but-1-ene

Answer:

10.8 What are ambident nucleophiles? Explain with an example.

Answer:

Ambident nucleophiles are nucleophiles that are capable of attacking the

substrate (alkyl halide) through two different atoms.

It so ‘happens due to the presence of two nucleophilic centres which arise from

the contributing (resonance) structures that are possible for the ion.

e.g., In NO–2 ion, there is a lone pair of electrons on N and therefore makes it

nucleophilic while oxygen by virtue of the negative charge acts as a nucleophile.

Thus, NO–2 can attack via O or N atom thereby making it ambidentate.

10.9 Which compound in each of the following pairs will react faster in SN2

reaction with OH–?

(i) CH3Br or CH3l

(ii) (CH3)3CCI or CH3Cl

Answer:

(i) Between CH3Br and CH3I, CH3I will react faster via the SN2 mechanism. In

SN2 mechanism, C – X bond breaks and the faster it breaks faster is the reaction.

I- is a better leaving group. Owing to its large size, the C – I bond breaks faster

than the C – Br bond and reaction proceeds further at a greater rate.

(ii) The order of reactivity in an SN2 reaction depends on minimum steric

hindrance around the carbon involved in the C – X bond. Lesser the steric

hindrance felt by the incoming nucleophile, more reactive will be the alkyl halide

towards SN2 reaction.

Based on this, CH3Cl will react faster than (CH3)3CCl.

10.10 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:

10.11 How will you bring about the following conversions?

(i) Ethanol to but-1 -yne

(ii) Ethane to bromoethane

(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-1 -ene to but-2-ene

(ix) 1-Chlorobutane ton-octane

(x) Benzene to biphenyl

Answer:

10.12 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?

Reason:

(i)(a) In order to understand the lower dipole moment of chlorobenzene we need

to look into the contributing structures of the molecules.

(b) From the above structures we find that the C – Cl bond in chlorobenzene has a

partial double bond character (structure II, III and IV). As a result, the C – Cl bond

length here is shorter than the C – Cl single bond but longer than the C – Cl double

bond.

(c) Also evident is the positive charge on Cl atom which reduces the partial

negative (δ-) charge which it is expected to carry by the virtue of its

electronegativity.

(d) Consequently, the dipole moment, which is a product of bond length and

partial negative charge on Cl atom, reduces. However, in cyclohexyl chloride this

does not happen. It is an alkyl halide and carbon is purely sp3 hybridised and C –

Cl bond has the bond length of a single bond and 8“ appearing on Cl is also higher,

thus, the greater dipole moment.

(ii) Only those compounds which can form hydrogen bonds with water are

miscible with it. Alkyl halides, though polar due to the presence of electronegative

halogen atom, are immiscible since they cannot form hydrogen bonds.

(iii) Grignard reagents R – Mg – X is a class of highly reactive compounds which

can extract a proton even from water molecule. They thus, turn into the

corresponding alkanes and render any other desired reaction ineffective.

This is why Grignard reagents are prepared in the absolute absence of water

(anhydrous conditions), (e.g.,)

10.14 Write the structure of the major organic product in each of the following

reactions:

Answer:

10.15 Write the mechanism of the following reaction

Solution:

10.16 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-methylbutane, 2-Bromo-2- methyl butane, 3-Bromo-2-

methylbutane

(iii)1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methyl butane,

1-Bromo-3- methyl butane.

Answer:

SN2 reaction proceeds via the formation of transition state where the carbon

atom is surrounded by 5 other atoms (groups). Thus, for such a transition state to

form, the steric interactions have to be minimum. Therefore, the most favourable

substrates for SN2 reactions are 1° alkyl halides followed by 2° and 3° alkyl halide.

Order of reactivity towards SN2: 1° > 2° > 3° > aryl halide.

In (iii), although all the given alkyl halides are 1° but the steric hindrance around

the carbon bearing the -Br atom decides the order of reactivity. More the number

of bulky groups around this carbon lower will be its reactivity towards SN2.

10.17 Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by

aqueous KOH?

Answer:

C6H5CHClC6 H5 is hydrolysed faster.

(a) Hydrolysis of an alkyl halide is an example of nucleophilic substitution

reaction. In case of aryl halides this follows the SN1 pathway i.e., via the

formation of carbocation.

(b) C6H5CH2Cl or benzyl chloride gives

(c) Out of I & II, carbocation II is more stable. The reason is the presence of two

phenyl rings attached to the carbon carrying the positive charge.

(d) As a result, the delocalisation of the +ve charge is greater and the carbocation

is more stable. Due to this, (II) is formed faster and the corresponding halide is

hydrolysed with greater ease as compared to benzyl chloride.

10.18 p-Dichlorobenzene has higher m.p. and solubility than those of o- and m-

isomers. Discuss.

Answer:

The para-isomers have high melting points and solubility as compared to their

ortho and meta isomers due to symmetry of para-isomers that fits into crystal

lattice better than ortho and para isomers.

10.19 How can the following conversions be carried out?

(i) Propene to propan-1 -ol

(ii) Ethanol to but-1 -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 propane nitrile

(viii) Aniline to chlorobenzene

(ix) 2-Chlorobutane to 3,4-dimethylhexane

(x) 2-Methyl-1 -propene to 2-chloro-2- methylpropane

(xi) Ethyl chloride to propanoic acid

(xii) But-1-ene to n-butyl iodide

(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) tert-Butyl bromide to isobutyl bromide

(xx) Aniline to phenyl isocyanide

Solution:

10.20 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: Formation of alcohols from the reaction between alkyl chlorides and aqueous KOH is an example of simple nucleophilic substitution.

But when aqueous KOH is replaced by alcoholic KOH, alkenes are formed instead of alcohols due to elimination of HCl from an alkyl halide.

10.21 Primary alkyl halide C4H9Br (A) 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 which 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. Solution:

10.22 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:

Chlorobenzene is highly unreactive towards nucleophilic substitution. However, it can be hydrolysed to phenol by heating in aqueous sodium hydroxide solution at a temperature of 623 K and 300 atm pressure. The presence of an electron withdrawing group increases the reactivity of haloarenes.