Question

Identify the final product ($B$) of the reaction
$C_6{H}_{5}COOH+NaH\stackrel{\ast }{C}{O}_3⟶gas\left(A\right)\underset{2.H_3{O}^{+}}{\overset{1.{CH}_{3}MgBr}{\to }}\left(B\right)(\stackrel{\ast }{C}=C^{14})$

• A. $\stackrel{\ast }{C}{H}_{3}COOH$
• B. $C_6{H}_5\stackrel{\ast }{C}OOH$
• C. ${CH}_3\stackrel{\ast }{C}OOH$
• D. $\stackrel{\ast }{H}COO{CH}_3$

Answer 1

The correct option is C ${CH}_3\stackrel{\ast }{C}OOH$

It is an acid base reaction. Benzoic acid gives $H^{+}$ to $H\stackrel{\ast }{C}{O_3}^{-}$. The $H_2\stackrel{\ast }{C}{O}_3$ formed breaks into $\stackrel{\ast }{C}{O}_2$ and $H_{2}O$.

$C_6{H}_{5}COOH+NaH\stackrel{\ast }{C}{O}_3\to$ $C_6{H}_{5}COONa+H_{2}O+\stackrel{\ast }{C}{O}_2$

The Grignard reagent attacks $\stackrel{\ast }{C}$ of $\stackrel{\ast }{C}{O}_2$ to form acetic acid.

$\stackrel{\ast }{C}{O}_2\underset{2.H_3{O}^{+}}{\overset{1.C{H}_{3}MgBr}{\to }}C{H}_3\stackrel{\ast }{C}OOH$