10 haloalkanes and haloarenes - abhyaasclasses.in

10 Haloalkanes and Haloarenes

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

Solution

I.U.P.A.C Name: 2-Chloro-3-methylbutane

Type: 20 – Alkyl halide

I.U.P.A.CName:3-Chloro-4-methylhexane

Type:20–Alkylhalide

iii) CH3CH2C(CH3)2CH2I iv) (CH3)3CCH2CH(Br)C6H5

Solution

iii) CH3CH2C(CH3)2CH2I

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IUPAC Name: 1-Iodo-2,2-dimethylbutane


IUPAC Name: 1-Bromo-1-phenyl -3,


v) CH3CH(CH3)CH(Br)CH3 vi) CH3C(C2H5)2CH2Br

Solution

v) CH3CH(CH3)CH(Br)CH3

IUPAC Name: 2-Bromo-3-methylbutane ,


vi) CH3C(C2H5)2CH2Br

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IUPAC Name: 3- Bromoethyl-3-methylpentane


vii) CH3C(Cl)(C2H5)CH2CH3 viii) CH3CH=C(Cl)CH2CH(CH3)2

Solution

IUPAC Name: 3- Chloro-3-methylpentane


IUPAC Name: 3- Chloro-5-methyl hex-2-ene

Type: 10 –Vinyl halide

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ix) CH3CH=CHC(Br)(CH3)2 x)p-ClC6H4CH2CH(CH3)2

Solution

IUPAC Name: 4-Bromom-4-methyl pent-2-ene

Type: Allyl halide

IUPAC Name: 1-Chloro -4-(2-methylpropyl) benzene

Type: Aryl halide

xi) m-ClCH2C6H4CH2C(CH3)3 x)o-Br-C6H4CH(CH3)CH2CH3

Solution

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IUPAC Name: 1-Chloromethyl -3(,22-dimethylpropyl) benzene

Type: Alkyl halide

IUPAC Name: 1-Bromo-2-(1-methylpropyl) benzene

Type: Aryl halide

2) Give the IUPAC names of the following compounds:

i) CH3CH(Cl)CH(Br)CH3

ii) CHF2CBrClF

Solution

I.U.P.A.C Name: 2-Bromo-3-chlorobutane

ii)

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I.U.P.A.C Name: 1-Bromo-1-chloro 1,2,2-tri fluoro ethane

iii) ClCH2C ≡ CCH2Br

iv) (CCl3)3CCl

Solution

iii)

I.U.P.A.C Name: 1-Bromo4-chlorobut-2-yne

I.U.P.A.C Name: 1,1,1,2,3,3,3- hepta chloro-2-trichloromethyl propane

v) CH3C(p-ClC6H4)2CH(Br)CH3

vi) (CH3)3CCH=CClC6H4I-p

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Solution

I.U.P.A.C Name: 3,3, -(4,41-dichlorophenyl) -2- bromobutane

or

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

I.U.P.A.C Name: 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

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

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Solution

2-Chloro-3-methylpentane p-Bromochlorobenzene

1-Chloro-4-ethylcyclohexane 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-Dibromobut-2-ene

Solution

2-Bromobutane 4-Tertbutyl-3-iodoheptane

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1- Bromo 4-secbutyl-2-methylbenzene 1,4-Dibromobut-2-ene

4) Which one of the following has the highest dipole moment?

i) CH2Cl2

ii) CHCl3

iii) CCl4

Solution

ii) CHCl3

Tetrachloromethane (CCl4) is a symmetrical molecule and has zero dipole moment.

In chloroform (CHCl3), the resultant dipole moment of two C-Cl bonds is cancelled

by resultant dipole moment of C-Cl and C-H bonds. Therefore, the resultant dipole

moment (µ= 1.03 D ) is smaller.

In dichloromethane (CH2Cl2) , the resultant dipole moments of two C-Cl and two C-

H bonds reinforce one another . Therefore, dichloromethane molecule has maximum

dipole moment (µ) of 1.62 D.

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

monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

Solution

Cyclopentane is the hydrocarbon which gives monochloro compound C5H9Cl in

bright sunlight.

6) Write the isomers of the compound having formula C4H9Br.

Solution

The isomers of the compound having formula C4H9Br are:

7) Write the equations for the preparation of 1-iodobutane from

i) 1-butanol ii) 1-chlorobutane iii) but-1-ene

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Solution

8) What are ambident nucleophiles ? Explain with an example.

Solution

Nucleophiles which possess two nucleophilic (electron pair donor) centers are called

ambident nucleophiles .

Ex:

CN- nucleophile by linking through carbon atom results in alkyl cyanides and through

nitrogen atom forms isocyanides.

9) Which compound in each of the following pairs will react faster in SN2 reaction

with OH-?

i) CH3Br or CH3I

ii) (CH3)3CCl or CH3Cl

Solution

i) CH3I

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ii) CH3Cl

10) Predict all the alkenes that would be formed by dehydrohalogenation of the

following halides with sodium ethoxide in ethanol and identify the major alkenes:

i) 1-Bromo-1-methylcylohexane.

Solution

ii) 2-Chloro-2-methylbutane

Solution

iii) 2,2,3-Trimethyl-3-bromopentane.

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Solution

2- Ethyl-3,3-dimethylbut-1-ene

( Minor Product)

In dehydrogenation reactions, the preferred product is that alkene which has the

greater number of alkyl groups attached to the doubly bonded carbon atoms.

In the above case total 3 products are possible out of which 2.2.3-trimethyl-2-pentane

is major product.

11) How will you bring about the following conversions?

i) Ethanol to but-1-yne

ii) Ethane to bromoethene

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Solution

i)

iii) Propene to 1-nitropropane

iv) Toluene to benzyl alcohol

Solution

iii)

Propene Propane 1-Nitropropane

iv)

Toulene Benzyl bromide Benzyl alcohol

v) Propene to propyne

vi) Ethanol to ethyl fluoride

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Solution

v)

Propene 1,2-Dibromopropane Propyne

vi)

Ethanol Bromoethane Ethyl fluoride

vii) Bromomethane to propanone

viii) But-1-ene to but-2-ene

Solution

vii)

Methylbromide Ethane 1,1-Dibromoethane Ethyne

Monosodium Propene Propanone

acetylide

viii)

But-1-ene 2- Bromobutane

But- 2-ene

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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?

Solution

i) In chlorobenzene C-Cl bond is less polar because of sp2 –carbon and ‘Cl’-bond

possess double bond character due to resonance, whereas in cyclohexyl chloride sp3 –

carbon and ‘Cl’ –bond is polar.

ii) Hydrogen bond is not possible in case of alkyl halides.

(Covalent character of “C-X’ bond).

iii) Grignard reagents are hydrolysed easily to give hydrocarbons(alkanes)

13) Give the uses of Freon 12, DDT, carbon tetrachloride and iodoform.

Solution

Freon12- Refrigeration gas

DDT- Fungicide and Insecticide

CCl4 – Fire extinguishes and as solvent

CHI3- as antiseptic (in preparation of Tincture of iodine)

14) Write the structure of the major organic product in each of the following

reactions:

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Solution

15) Write the mechanism of the following reaction:

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Solution

Butyl bromide Intermediate Butyl cyanide

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-methylbutane, 3-Bromo-2-methylbutane

iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-

Bromo-3-methylbutane.

Solution

i) 1-Bromopentane> 2-Bromopentane> 2-Bromo-2methylbutane

(10) (2

0) (3

0)

ii) 1-Bromo-3-methyl butane> 2-Bromo-3-methyl-butane> 2-Bromo-2

(10) (2

0) (3

0)

methylbutane.

iii) 1-Bromobutane> 1-bromo-3-methylbutane>1-Bromo-2methyl Butane> 1-Bromo-

2,2 dimethyl propane.

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NOTE: SN2 –order of reaction 1

0 > 2

0 > 3

0.

As the bulky groups hinder the approaching nucleophiles.

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

aqueous KOH?

Solution

C6H5CHClC6H5 is more easily hydrolysed because the intermediate carbo cation

formed is more resonance stabilised.

18) p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-

isomers . Discuss.

Solution

p-Dichlorobenzene has symmetric structure close packing of molecules results in

high m.pt and solubility is more because of intermolecular hydrogen bonding which

is absent in o- and m-isomers.

19) How the following conversions can be carried out?

i) Propene to propan-1-ol

ii) Ethanol to but-1-yne

iii) 1-Bromopropane to 2-bromopropane

Solution

i)

Propene 1-Bromo propane Propan-1-ol

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ii)

Ethanol Ethene 1,2-Dibromoethane Ethyne

Monosodium But-1-yne

iii)

iv) Toluene to benzyl alcohol

v) Benzene to 4-bromon nitrobenzene

Toluene Benzyl chloride Benzyl alcohol

Benzene Bromobenzene 4-Bromonitrobenzene

ix) 2-Chlorobutane to 3,4-dimethylhexane

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

xi) Ethyl chloride to propanoic acid

Solution

2 – Chlorobutane 3,4 –dimethylhexane

2-Methyl-1-propene 2-chloro-2-methylpropane

Ethyl chloride Ethyl cyanide Propanoic acid

vi) Benzyl alcohol to 2-phenylethanoic acid

vii) Ethanol to propanenitrile

viii) Aniline to chlorobenzene

Solution

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Benzyl alcohol Benzyl chloride Benzyl cyanide 2-phenylethanoic

acid

Ethanol Chloro ethane Propanenitrile

Aniline Benzene diazonium chloride Chlorobenzene

xii) But-1-ene to n-butyliodide

xiii) 2-Chloropropane to 1-propanol

xiv) Isopropyl alcohol to iodoform

Solution

But-1-ene 1-Bromobutane

n-butyliodide

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2-Chloropropane Propene

1-Bromopropane 1-Propanol

Isopropyl alcohol Propanone Iodoform

xviii) Benzene to diphenyl

xix) tert-Butyl bromide to isobutyl bromide

xx) Aniline to phenylisocyanide

Solution

Benzene Chlorobenzene Diphenyl

Tert-butyl bromide Isobutylene Isobutyl bromide

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Aniline Phenylisocyanide

xv) Chlorobenzene to p-nitrophenol

xvi) 2-Bromopropane to 1-bromopropane

xvii) Chloroethane to butane

Chlorobenzene p-nitrochlorobenzene p-nitrophenol

2-Bromopropane Propene 1-bromopropane

Chloroethane Butane

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.

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Solution

Reaction of alkyl chlorides with aqueous KOH proceeds through nucleophilic

substitution reaction mechanism; hence a stronger nuclephile OH- , replaces weaker

nucleophile X- of alkyl halide. While alcoholic KOH is a dehydrohalogenating agent

and forms an alkene by eliminating hydrogen halide. C2H5O- will act as strong base

rather than nucleophile hence elimination takes place.

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

[B] [C]

[A] [D]

A is Isobutyl bromide

B is 2-methylpropene

C is 2-Bromo-2-methylpropane

D is 2,5-Dimethylhexane

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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.

Solution

i) When n-butyl chloride is treated with alcoholic KOH , 1-butene is formed

(H2C=CH-CH2-CH3) by dehydrochlorination.

CH3-CH2-CH2-CH2-Cl + KOH(alc) → CH2=CH-CH2-CH3+ KCl + H2O

n-Butyl chloride 1 – Butene

ii) When bromobenzene is treated with Mg in the presence of dry ether , Phenyl

magnesium bromide is formed.

Bromobenzene Phenyl magnesium bromide

iii) When chlorobenzene is subjected to hydrolysis , phenol is formed

Chlorobenzene Phenol

iv) When ethyl chloride is treated with aqueous KOH, ethanol is formed.

Chloroethane Ethanol

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v) methyl bromide is treated with sodium in the presence of dry ether

vi) methyl chloride is treated with KCN

Solution

v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is

formed. This is called wurtz reaction

Methyl bromide Ethane

vi) When methyl chloride is treated with KCN, methyl cyanide is formed.

H3C-Cl + KCN → H3C-CN + KCl

Methyl chloride Methyl cyanide

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10 haloalkanes and haloarenes - abhyaasclasses.in

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