JEE Main Halogen Derivative Previous Year Questions with Solutions

If one or more hydrogen atom is replaced in hydrocarbon by an equivalent number of halogen, the compounds formed are called halogen derivatives of hydrocarbons. Halogen derivatives are classified as aliphatic halogen compounds and aromatic halogen compounds. Aliphatic halogen compounds are formed by replacing one or more hydrogen of aliphatic hydrocarbons (alkanes, alkenes, and alkynes). Aliphatic halogen compounds are again divided as haloalkanes, haloalkenes and haloalkynes. Aromatic hydrocarbons are known as arenes. Halogen derivatives of arenes are termed as aromatic halogen compounds.

As far as the JEE exam is concerned, halogen derivative is an important topic. The important concepts include haloalkanes, haloarenes, SN1 and SN2 substitution reaction, etc. The questions given here give you an idea about what type of questions to expect from this chapter in the JEE exam. Halogen Derivative previous year questions and solutions are given in this article. This article helps students to revise the previous year questions and score higher rank in the JEE exam. BYJU’S provides accurate solutions created by our subject experts. Students can easily download these solutions in PDF format for free.

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JEE Main Previous Year Solved Questions on Halogen Derivative

1. The synthesis of alkyl fluorides is best accomplished by

(a) Finkelstein reaction

(b) Swart’s reaction

(c) free radical fluorination

(d) Sandmeyer’s reaction.

Solution:

Alkyl fluorides are more conveniently prepared indirectly by heating suitable chloro or bromoalkanes with inorganic fluorides.

CH3Br + AgF → CH3F + AgBr

This reaction is called Swart’s reaction.

Hence option (b) is the answer.

2. The optically inactive compound from the following is

(a) 2-chloropropanal

(b) 2-chloropentane

(c) 2-chlorobutane

(d) 2-chloro-2-methylbutane.

Solution:

Since there is no chiral center, 2-chloro-2-methylbutane is optically inactive.

Hence option (d) is the answer.

3. Which one of the following is likely to give a precipitate with AgNO3 solution?

(a) (CH3)3CCl

(b) CHCl3

(c) CH2=CH-Cl

(d) CCl4

Solution:

Tert-butyl chloride forms most stable 3° carbocation. So will give white precipitate of AgCl with AgNO3 solution immediately.

Hence option (a) is the answer.

4. A solution of (–)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5, due to the formation of

(a) free radical

(b) carbanion

(c) carbene

(d) carbocation

Solution:

During racemisation, carbocation intermediate is formed.

Hence option (d) is the answer.

5. What is DDT among the following?

(a) A fertilizer

(b) Biodegradable pollutant

(c) Non-biodegradable pollutant

(d) Greenhouse gas

Solution:

DDT (Dichloro diphenyl trichloroethane) is a non-biodegradable pollutant. The non-biodegradable pollutants cannot be broken down into simpler, harmless substances in nature.

Hence option (c) is the answer.

6. The compound formed on heating chlorobenzene with chloral in the presence of concentrated sulphuric acid is

(a) gammexene

(b) DDT

(c) freon

(d) hexachloroethane.

Solution:

The compound formed on heating chlorobenzene with chloral in the presence of concentrated sulphuric acid is DDT.

Hence option (b) is the answer.

7. Iodoform can be prepared from all except

(a) isopropyl alcohol

(b) 3-methyl-2-butanone

(c) isobutyl alcohol

(d) ethyl methyl ketone.

Solution:

All the compounds except isobutyl alcohol will form iodoform.

Hence option (c) is the answer.

8. Fluorobenzene (C6H5F) can be synthesised in the laboratory

(a) by heating phenol with HF and KF

(b) from aniline by diazotization followed by heating the diazonium salt with HBF4

(c) by direct fluorination of benzene with F2 gas

(d) by reacting bromobenzene with NaF solution.

Solution:

Fluorobenzene (C6H5F) can be synthesised in the laboratory from aniline by diazotization followed by heating the diazonium salt with HBF4.

Hence option (b) is the answer.

9. Alkyl halides react with dialkyl copper reagents to give

(a) alkenes

(b) alkyl copper halides

(c) alkanes

(d) alkenyl halides.

Solution:

R2CuLi + R’X → R – R’ + RCu + LiX

Alkyl halide reacts with dialkyl copper reagents to give alkanes. It is called Corey House synthesis of alkane.

Hence option (c) is the answer.

10. The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is

(a) 2-butene

(b) acetylene

(c) ethene

(d) 2-butyne

Solution:

2- butyne is formed by the reaction of 1,1,1-trichloroethane with silver powder.

Hence option (d) is the answer.

11. Elimination of bromine from 2-bromobutane results in the formation of

(a) equimolar mixture of 1 and 2-butene

(b) predominantly 2-butene

(c) predominantly 1-butene

(d) predominantly 2-butyne.

Solution:

2-bromobutaneCH3CH = CHCH3 + CH3CH2CH=CH2

The major product follows Saytzeff rule.

Hence option (b) is the answer.

12. Which of the following will have a meso-isomer also?

(a) 2-chlorobutane

(b) 2,3-dichlorobutane

(c) 2,3-dichloropentane

(d) 2-hydroxypropanoic acid

Solution:

Because of the presence of plane of symmetry, 2,3-dichlorobutane have meso isomer.

Hence option (b) is the answer.

13. The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction, is

(a) CH3Cl

(b) (C2H5)2CHCl

(c) (CH3)3CCl

(d) (CH3)2CHCl

Solution:

In SN2 reactions, the nucleophile attacks from the backside resulting in the inversion of molecule. As we move from 1° alkyl halide to 3° alkyl halide, the crowding increases and +I effect increases. This causes the carbon-bearing halogen less positively polarised. So it is less readily attacked by the nucleophile.

Hence option (a) is the answer.

14. Which of the following on heating with aqueous KOH produces acetaldehyde?

(a) CH3COCl

(b) CH3CH2Cl

(c) CH2ClCH2Cl

(d) CH3CHCl2

Solution:

CH3CHCl2 on heating with aqueous KOH produces acetaldehyde.

Hence option (d) is the answer.

15. In SN2 reactions, the correct order of reactivity for the following compounds: CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is

(a) (CH3)2CHCl > CH3CH2Cl > CH3Cl > (CH3)3CCl

(b) CH3Cl > (CH3)2CHCl > CH3CH2Cl > (CH3)3CCl

(c) CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl

(d) CH3CH2Cl > CH3Cl > (CH3)2CHCl > (CH3)3CCl

Solution:

Reactivity in SN2 is inversely proportional to steric hindrance.

Hence the correct order of reactivity is CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl

Hence option (c) is the answer.

Also Read:- Haloalkanes and Haloarenes