Question

$Ph-CH=CH-\stackrel{O}{\stackrel{||}{C}}-C{H}_3\to Ph-CH=CH-C{O}_{2}H$
Given conversion can be achieved by:

• A. $KMn{O}_4$, $\Delta$ followed by $H^{+}$
• B. $I_2/NaOH$ followed by $H^{+}$
• C. $H_2/Pt$
• D. $LiAl{H}_4$

Answer 1

The correct option is B $I_2/NaOH$ followed by $H^{+}$

$O$

Iodoform test do not affect double bond adjacent to the ketone group and as haloform test converts $-\stackrel{||}{C}-C{H}_3$ into $CO{O}^{⊝}$. This can be used for conversion.

$ph-CH=CH-\underset{||}{C}-C{H}_3\stackrel{I_2/NaOH}{\to }ph-CH=CH-\underset{||}{C}-\stackrel{⊝}{O}N{a}^{+}+CH{I}_3$

$O$ $O$

$ph-CH=CH-\underset{||}{C}-\stackrel{⊝}{O}N{a}^{+}\stackrel{H^{+}}{\to }ph-CH=CH-\underset{||}{C}-OH$

$O$ $O$