Question

When compound $X$ is oxidised by acidified potassium dichromate, compound $Y$ is formed. Compound $Y$ on reduction with $LiAlH _{4}$ gives $X . X$ and $Y$ respectively are :

• A. $C _{2} H _{5} OH , CH _{3} COOH$
• B. $CH _{3} COCH _{3}, CH _{3} COOH$
• C. $C _{2} H _{5} OH , CH _{3} COCH _{3}$
• D. $CH _{3} CHO , CH _{3} COCH _{3}$

Answer 1

Answer: (A)

When ethyl alcohol is oxidised by acidified potassium dichromate, $CH _{3} COOH (Y)$ is obtained as :$\underset{X}{3 C _{2} H _{5} OH }+2 K _{2} Cr _{2} O _{7}+8 H _{2} SO _{4} \longrightarrow $
$\underset{Y}{3 CH _{3} COOH }+2 Cr _{2}\left( SO _{4}\right)_{3}+2 K _{2} SO _{4}+11 H _{2} O$
Carboxylic acid undergo reduction with $LiAlH _{4}$ to give primary alcohol as
$\underset{(Y)}{CH _{3}- \overset{\underset{||}{O}}{C} - OH} \xrightarrow[\text{(Ether)}]{LiAlH_4} \underset{(X)}{CH _{3} CH _{2} OH }$
$So , X$ is $CH _{3} CH _{2} OH$ and $Y$ is $CH _{3} COOH$