Question

In the following sequence of reactions, what is the end product (D)?
C$_2$H$_5$Br $\xrightarrow{\text{KCN}}$ A $\xrightarrow{\text{H$_3$O$^+$}}$ B $\xrightarrow{\text{LiAlH$_4$}}$ C $\xrightarrow{\text{Cu, 573 K}}$ D

• A. Acetaldehyde
• B. Acetone
• C. Propionaldehyde
• D. Propanol-1

Hint:

Reduction of acids gives alcohols; mild oxidation of 1° alcohols gives aldehydes.

Answer 1

The Correct Option is C

Step 1: C$_2$H$_5$Br + KCN $\rightarrow$ C$_2$H$_5$–CN (nitrile A)
Step 2: Hydrolysis of nitrile (A) with H$_3$O$^+$ gives propanoic acid (B)
Step 3: Reduction with LiAlH$_4$ gives 1-propanol (C)
Step 4: Oxidation with Cu at 573 K converts 1-propanol to propionaldehyde (D)