Question

Consider the following reaction sequence,
${CH}_{3}C\equiv CH\underset{HCl}{\overset{2mole}{\to }}A\underset{aq.KOH}{\overset{Heat}{\to }}B$
The products (A) and (B) are, respectively :

• A. (A): ${CH}_3{CH}_2{CHCl}_2$ and (B): ${CH}_3{CH}_{2}CHO$
• B. (A): ${CH}_3{C\left(Cl\right)}_2{CH}_3$ and (B): ${CH}_3{COCH}_3$
• C. (A): ${CH}_3{C\left(Cl\right)}_2{CH}_3$ and (B): ${CH}_{3}C\equiv CH$
• D. (A): ${CH}_3{CH\left(Cl\right)CH}_{2}Cl$ and (B): ${CH}_3{CH\left(OH\right)CH}_{2}Cl$

Answer 1

The correct option is B (A): ${CH}_3{C\left(Cl\right)}_2{CH}_3$ and (B): ${CH}_3{COCH}_3$

${CH}_{3}C\equiv CH\underset{HCl}{\overset{2mole}{\to }}\underset{A}{{CH}_3{C\left(Cl\right)}_2{CH}_3}\underset{aq.KOH}{\overset{Heat}{\to }}\underset{B}{{CH}_3{COCH}_3}$

In the first step, two moles of $HCl$ are added to carbon carbon triple bond of propyene to form a geminal dichloride.

Chlorine atoms are added to the more substituted internal $C$ atom. In the next step, the hydrolysis of two $C-Cl$ bonds gives a geminal diol which then loses a molecule of water to form acetone.

For the given reaction sequence, the products (A) and (B) are ${CH}_3{C\left(Cl\right)}_2{CH}_3$ and ${CH}_3{COCH}_3$ respectively.

Option B is correct.