Question

$Ph-C{H}_2-\underset{OH}{\underset{|}{C}H}-C{H}_3\stackrel{K}{\to }\stackrel{C_2{H}_{5}Br}{\to }\left(A\right)$

Product $\left(A\right)$ in above reaction is :

• A. $Ph-C{H}_2-\underset{OEt}{\underset{|}{C}H}-C{H}_3$, (inversion)
• B. $Ph-C{H}_2-\underset{OEt}{\underset{|}{C}H}-C{H}_3$, (retention)
• C. $Ph-C{H}_2-\underset{OEt}{\underset{|}{C}H}-C{H}_3$, (racemic)
• D. $Ph-CH=CH-C{H}_3$

Answer 1

Answer: (B)

$Ph-C{H}_2-\underset{OEt}{\underset{|}{C}H}-C{H}_3$, (retention)

Potassium removes the most acidic hydrogen present in the compound. Here the hydrogen attached to oxygen is highly acidic. So,

(Refer to Image)

Here, the configuration of carbon does not change so, retention.