Question

In the following sequence of reactions a compound $A$, (molecular formula $C _{6} H _{12} O _{2}$ ) with a straight chain structure gives a $C _{4}$ carboxylic acid. $A$ is:
$A \xrightarrow[H_3O^+]{LiAlH_4} B \xrightarrow{\text{Oxidation}} C_4$ -carboxylic acid

• A. $CH _{3}- CH _{2}- COO - CH _{2}- CH _{2}- CH _{3}$
• B. $CH_3 - CH_2- \overset{\underset{|}{O}H}{CH} -CH_2 - O - CH = CH_2$
• C. $CH _{3}- CH _{2}- CH _{2}- COO - CH _{2}- CH _{3}$
• D. $CH _{3}- CH _{2}- CH _{2}- O - CH = CH - CH _{2}- OH$

Answer 1

Answer: (C)

Correct answer is (c) $CH _{3}- CH _{2}- CH _{2}- COO - CH _{2}- CH _{3}$