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

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So, the correct option is (C): racemic mixture and gives a gas with saturated NaHCO_3​ solution

Answer 2

The reaction proceeds as follows:

\[ \text{CH}_3\text{CH}_2\text{CHO} + \text{HCHO} \xrightarrow{\text{(i) NaOH, $\Delta$}} \text{CH}_3\text{CH}_2\text{CH(OH)CH}_2\text{CN} \]

In the given reaction, propanal reacts with formaldehyde in the presence of NaOH and heat to form an aldol product. The aldol product undergoes nucleophilic substitution with NaCN, forming a nitrile group.

Upon hydrolysis, the nitrile group is converted into a carboxylic acid, which then undergoes decarboxylation, resulting in a racemic mixture. The carboxylic acid will also release carbon dioxide when reacted with a saturated solution of NaHCO₃.

Thus, the correct statement for product B is that it is a racemic mixture and gives a gas with saturated NaHCO₃ solution.