Question

$PhCHO+(C{H}_{3}CO{)}_{2}O\stackrel{C{H}_{3}COONa}{\to }A\stackrel{HBr}{\to }B$
The product $B$ is :

• A. $PhCH=CHC{H}_{2}Br$
• B. $PhCH\left(Br\right)C{H}_{2}COOH$
• C. $PhC{H}_{2}CH\left(Br\right)COOH$
• D. $PhCH=CH-COBr$

Answer 1

The correct option is

C

$PhCH\left(Br\right)C{H}_{2}COOH$

Benzaldehyde and acetic anhydride condense in the presence of sodium acetate to form compound $A$, which is cinnamic acid.

This is a perkin condensation reaction.
It is followed by Markownikov's addition of $HBr$ to form $3-bromo-3-phenylpropanoic$ acid.
$PhCHO+(C{H}_{3}CO{)}_{2}O\stackrel{C{H}_{3}COONa}{\to }\underset{A}{PhCH=CH-COOH}\stackrel{HBr}{\to }\underset{B}{PhCHBr-C{H}_2-COOH}$

The reactions is as follows:

Conversion of Cinnamic acid to 3-bromo-3-phenylpropanoic acid

Hence, the correct option is C.