Question

An organic compound $A\left(C_5{H}_{10}{O}_2\right)$ reacts with $B{r}_2$ in presence of phosphorus to produce $B$ which can be resolved into enantiomers. $B$ on dehydrobromination yields $C\left(C_5{H}_8{O}_2\right)$ which does not show stereoisomerism. $C$ on decarboxylation gives an alkene which on further ozonolysis gives $D$ and $E$. $D$ gives Schiff's test but $E$ does not. The correct identity of these compounds are:

• A. $A$ is $(C{H}_3{)}_{2}CHC{H}_{2}COOH$
• B. $B$ is $(C{H}_3{)}_{2}CHCHBrCOOH$
• C. $C$ is $(C{H}_3{)}_{2}C=CHCOOH$
• D. $E$ is acetone

Answer 1

The correct option is D $E$ is acetone

Bromination occurs next to the COOH group as the COOH group makes the carbocation stable.

The compound C does not show geometrical isomerism as both groups on the left side are the same of the double bond. Ozonolysis will break the bond and form secondary carbon to ketone and primary carbon to aldehyde.

Here the compound E is acetone that is $(C{H}_3{)}_{2}CO$. Here compound C has no geometrical isomer.