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JEE Main 202430 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
30 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Hydrocarbons

JEE Main 2024 · 30 January · Morning Shift · Chemistry · Hydrocarbons

Compound A formed in the following reaction reacts with B gives the...

Compound AA formed in the following reaction reacts with BB gives the product CC. Find out AA and B.B .

Compound A formed in the following reaction reacts with B gives the product C . Find out A and B .

Options

  1. AA=CH3−C≡C−Na+,B=CH3−CH2−CH2−BrA = {CH}_{3} - C ≡ {C}^{-}{Na}^{+}, B = {CH}_{3} - {CH}_{2} - {CH}_{2} - Br
  2. BA=CH3−CH=CH2,B=CH3−CH2−CH2−BrA = {CH}_{3} - CH = {CH}_{2}, B = {CH}_{3} - {CH}_{2} - {CH}_{2} - Br
  3. CA=CH3−CH2−CH3,B=CH3−C≡CHA = {CH}_{3} - {CH}_{2} - {CH}_{3}, B = {CH}_{3} - C ≡ CH
  4. DA=CH3−C≡C−Na+,B=CH3−CH2−CH3A = {CH}_{3} - C ≡ {C}^{-}{Na}^{+}, B = {CH}_{3} - {CH}_{2} - {CH}_{3}

Correct answer

A. A=CH3−C≡C−Na+,B=CH3−CH2−CH2−BrA = {CH}_{3} - C ≡ {C}^{-}{Na}^{+}, B = {CH}_{3} - {CH}_{2} - {CH}_{2} - Br

Step-by-step solution

When CH3C≡CH{CH}_{3}C ≡ CH reacts with sodium an intermediate CH3C≡CNa{CH}_{3}C ≡ CNa is formed which on further reaction with propyl bromide gives CH3CH2CH2C≡CCH3{CH}_{3}{CH}_{2}{CH}_{2}C ≡ {CCH}_{3} as the product. The second step of the reaction follows, SN2{S}_{N}2 mechanism.

The reaction is as follows,

Compound A formed in the following reaction reacts with B gives the product C . Find out A and B .

Hence in the above reaction,

A=CH3−C≡C−Na+,B=CH3−CH2−CH2−BrA = {CH}_{3} - C ≡ {C}^{-}{Na}^{+}, B = {CH}_{3} - {CH}_{2} - {CH}_{2} - Br

Hence option A is the answer.

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