Question

Consider the following reaction

The correct statement for product $B$ is It is

• A. optically active and adds one mole of bromine
• B. optically active alcohol and is neutral
• C. racemic mixture and gives a gas with saturated $NaHCO _3$ solution
• D. racemic mixture and is neutral

Hint:

A racemic mixture is a mixture containing equal amounts of enantiomers, and a carboxylic acid can release \(\text{CO}_2\) when reacted with \(\text{NaHCO}_3\).

Answer 1

The Correct Option is C

The given reaction involves the following steps:

1. \(C_2H_5CHO + HCHO \xrightarrow{\text{dil. NaOH}} \text{A}\) 
   This step represents the aldol condensation between propanal and methanal, producing aldol product 'A'.
2. Heating (∆) which leads to dehydration of the aldol product to form an α,β-unsaturated carbonyl compound.
3. Treatment with NaCN followed by hydrolysis (\(H_3O^+\)) converts the unsaturated compound to a hydroxy acid 'B' with the molecular formula \(C_5H_8O_3\).

Let's explore the properties of the product 'B' based on the above reaction sequence.

• 'B' is a β-hydroxy acid, which implies the presence of both a hydroxyl group (-OH) and a carboxylic group (-COOH).
• The compound is expected to be optically inactive due to the formation of a racemic mixture.
• Reactive with saturated \(NaHCO_3\) solution, 'B' would release \(CO_2\) gas, characteristic of acids reacting with bicarbonates.

The correct statement is that 'B' is a racemic mixture and gives a gas with saturated \(NaHCO_3\) solution, due to the above discussed chemistry.