Compound A formed in the following reaction reacts with B gives the product C. Find out A and B. CH 3 - C equiv CH + Na rightarrow A xrightarrow B CH 3 - C equiv C - CH 2 - CH 2 - CH ( CH 3) - CH 2 + NaBr

A = CH 3 - C equiv overline C Na , B = CH 3 - CH 2 - CH 2 - Br

A = CH 3 - CH = CH 2 , B = CH 3 - CH 2 - CH 2 - Br

A = CH 3 - CH 2 - CH 3 , B = CH 3 - C equiv CH

A=C H 3 -Cequiv bar C Noverset + mathop a \,,B=C H 3 -C H 2 -C H 3

Chemistry · Hydrocarbons

JEE Main 2024 — 30 January, Shift 1 — Question 66

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

\text{CH}_3 - \text{C}\equiv\text{CH} + \text{Na} \rightarrow \text{A} \xrightarrow{\text{B}} \text{CH}_3 - \text{C}\equiv\text{C} - \text{CH}_2 - \text{CH}_2 - \text{CH}(\text{CH}_3) - \text{CH}_2 + \text{NaBr}

Option A:
$\mathrm{A}=\mathrm{CH}_{3}-\mathrm{C} \equiv \overline{\mathrm{C}} \mathrm{Na}, \mathrm{B}=\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}$
Correct
Option B:
$\mathrm{A}=\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2}, \mathrm{~B}=\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{Br}$
Option C:
$\mathrm{A}=\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{3}, \mathrm{~B}=\mathrm{CH}_{3}-\mathrm{C} \equiv \mathrm{CH}$
Option D:
$A=C{{H}_{3}}-C\equiv \bar{C}N\overset{+}{\mathop{a}}\,,B=C{{H}_{3}}-C{{H}_{2}}-C{{H}_{3}}$

\text{CH}_3 - \text{C}\equiv\text{CH} \xrightarrow{\text{Na}} \text{CH}_3 - \text{C}\equiv\text{C}^-\text{Na}^+ \xrightarrow{\text{CH}_3\text{CH}_2\text{CH}_2 - \text{Br}} \text{NaBr} + \text{CH}_3 - \text{C}\equiv\text{C} - \text{CH}_2\text{CH}_2\text{CH}_3

Answer key and solution verified before publishing.

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