Haloalkanes and Haloarenes — Page 5 | NEET Chemistry

Q41

Iodoform can be prepared from all except :

A Ethyl methyl ketone

B Isopropyl alcohol

C 3–Methyl – 2– butanone

D Isobutyl alcohol

Correct Answer

Option D

Solution

Iodoform test is given by methyl ketones, acetaldehyde and methyl secondary alcohols. isobutyl alcohol is a primary alcohol hence does'nt give positive iodoform test.

Q42

Given below are two statements : Statement I : C2H5OH and AgCN both can generate nucleophile. Statement II : KCN and AgCN both will generate nitrile nucleophile with all reaction conditions. Choose the most appropriate option :

A Statement I is false but statement II is true.

B Both statement I and statement II are true.

C Both statement I and statement II are false.

D Statement I is true but statement II is false.

Correct Answer

Option D

Solution

Both C2H5OH and AgCN can generate nucleophile.

KCN generates nitriles on substitution reactions with haloalkanes where AgCN generates isonitriles on substitution reactions with haloalkanes.

Because KCN is ionic and has ‘C’ nucleophilic centre whereas AgCN is covalent and has ‘N’ nucleophilic centre.

Q43

The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is ;

A Retention occurs in

S_N 1

reaction and inversion occurs in

\mathrm{S}_{\mathrm{N}} 2

reaction.

B Racemisation occurs in

\mathrm{S}_{\mathrm{N}} 1

reaction and retention occurs in

\mathrm{S}_{\mathrm{N}} 2

reaction.

C Racemisation occurs in both

\mathrm{S}_{\mathrm{N}} 1

and

\mathrm{S}_{\mathrm{N}} 2

reactions.

D Racemisation occurs in

S_N 1

reaction and inversion occurs in

\mathrm{S}_{\mathrm{N}} 2

reaction.

Correct Answer

Option D

Solution

The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is Option D.

Let's break down why this is the case by understanding both the

S_N 1

and

S_N 2

reactions and their effects on the chirality of the alkyl halide.

S_N 2

Reaction: The

S_N 2

(Substitution Nucleophilic Bimolecular) reaction is a one-step process where the bond formation (nucleophile attacking the substrate) and the bond-breaking (leaving group leaving the substrate) processes occur simultaneously.

This reaction type is associated with an inversion of configuration at the chiral center.

This means that if the reactant molecule is chiral and the nucleophile attacks from the side opposite to the leaving group, the product will have the opposite configuration to the reactant.

This phenomenon is known as Walden inversion.

Therefore, it does not lead to racemisation but rather to inversion of the stereochemistry at the chiral center.

S_N 1

Reaction: The

S_N 1

(Substitution Nucleophilic Unimolecular) reaction is a two-step process.

The first step involves the formation of a carbocation intermediate by the departure of the leaving group.

This intermediate is planar, and hence, the nucleophile can attack from either side of the plane with equal probability.

This results in the formation of products with both configurations - one retaining the original configuration and the other being its mirror image (inverted configuration).

As both enantiomers are formed, the process leads to racemisation, especially in situations where the carbocation can freely rotate and is attacked with equal probability from both sides.

Over time, if this reaction goes to completion and there's no bias in nucleophile attack direction, you'd expect a racemic mixture of products.

Thus, Option D is correct: Racemisation occurs in

S_N 1

reaction, and inversion occurs in the

S_N 2

reaction.

Q44

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

A CHCl3

B (CH3)3CCl

C CCl4

D CH2=CH–Cl

Correct Answer

Option B

Solution

The molecule which will produce stable carbocation will react with AgNO3 solution. (CH3)3CCl

\overset{{AgN{O_3}}}\longrightarrow

(CH3)3C+NO3- + AgCl( $\downarrow$ )

Q45

3-Methyl-pent-2-ene on reaction with HBr in presence of peroxide forms an addition product. The number of possible stereoisomers for the product is :

A Zero

B Two

C Four

D Six

Correct Answer

Option C

Solution

A compound having two chiral centres can exist in 4 stereoisomeric forms (2n).

Q46

The reaction :

{(CH)_3}C - Br\overset{{{H_2}O}}\longrightarrow {(CH)_3}C - OH

A elimination reaction

B substitution reaction

C free radical reaction

D displacement reaction

Correct Answer

Option B

Solution

The hydrolysis of

t

-butyl bromide is an example of

{S_N}1

reaction. The reaction consists of two steps.

Q47

How many chiral compounds are possible on monochlorination of 2–methyl butane ?

A 8

B 2

C 4

D 6

Correct Answer

Option B

Solution

The reaction involved is Out of all the four isomers formed only two compounds are optically active.

Q48

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom. Reason R : Phenols react with halogen acids violently. In the light of the above statements, choose the most appropriate from the options given below :

A A is true but R is false

B Both A and R are true but R is NOT the correct explanation of A

C A is false but R is true

D Both A and R are true and R is the correct explanation of A

Correct Answer

Option A

Solution

Assertion (A): Given statement is correct because in phenol hydroxyl group cannot be replaced by halogen atom.

Reason (R) : Given reason is false.

Hence Assertion (A) is correct but Reason (R) is false.

Q49

In SN2 reactions, the correct order of reactivity for the following compounds: CH3Cl, CH3CH 2Cl, (CH3)2CHCl and (CH3)3CCl is:

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

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

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

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

Correct Answer

Option D

Solution

Steric congestion around the carbon atom undergoing the inversion process will slow down the

{S_N}2

reaction, hence less congestion faster will the reaction. So, the order is

C{H_3}Cl > \left( {C{H_3}} \right)C{H_2} - Cl >

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {C{H_3}} \right)_2}CH - Cl > {\left( {C{H_3}} \right)_3}CCl

Q50

The synthesis of alkyl fluorides is best accomplished by:

A Sandmeyer’s reaction

B Finkelstein reaction

C Swarts reaction

D Free radical fluorination

Correct Answer

Option C

Solution

Alkyl fluorides are more conveniently prepared by heating suitable chloro $-$ or bromo-alkanes with organic fluorides such as

As{F_3},\,\,Sb{F_3},\,\,Co{F_2},\,\,AgF,\,H{g_2}{F_2}\,\,

etc. This reaction is called Swarts reaction.

C{H_3}Br + AgF\overset{\,}\longrightarrow C{H_3}F + AgBr

2C{H_3}C{H_2}Cl + H{g_2}{F_2}\overset{\,}\longrightarrow

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2C{H_3}C{H_2}F + H{g_2}Cl{}_2