CH_3 CHO+HCHO [ Heat ] Dil. NaOH A [H_3 O^ + ] HCN B The structure of compound B is

Correct answer: A. It undergoes aldol condensation to give (CH_2(OH) CH_2(CHO)) which dehydrates to give (CH_2 CHCHO) When this is treated with (HCN / H_3 O^ + ), it gives the carboxylic acid. aligned & CH_3 CHO+HCHO Dil. NaOH CH_2(OH) CH_2 CHO & CH_2(OH) CH_2 CHO Δ CH_2=CHCHO & CH_2=CHCHO HCN CH_2=CHCH(OH)(CN) & CH_2=CHCH(OH)(CN) H_3 O^ + CH_2=CH-CH(COOH)(OH) aligned )

JIPMER 2012

Exam: JIPMER
Year: 2012
Subject: Chemistry
Chapter: Aldehydes and Ketones

JIPMER 2012 · Chemistry · Aldehydes and Ketones

CH_3 CHO+HCHO [ Heat ] Dil. NaOH A [H_3 O^ + ] HCN B The structure...

\mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCHO} \xrightarrow[\text{ Heat }]{\text{ Dil. } \mathrm{NaOH}} A \xrightarrow[\mathrm{H}_3 \mathrm{O}^{+}]{\mathrm{HCN}} B

The structure of compound

B

Options

Correct answer

A. When a crystal of caustic soda is exposed to air, a liquid layer is deposited because :

Step-by-step solution

It undergoes aldol condensation to give (\mathrm{CH}_2(\mathrm{OH}) \mathrm{CH}_2(\mathrm{CHO})) which dehydrates to give (\mathrm{CH}_2 \mathrm{CHCHO}) When this is treated with (\mathrm{HCN} / \mathrm{H}_3 \mathrm{O}^{+}), it gives the carboxylic acid.

\begin{aligned} & \mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCHO} \xrightarrow{\text{ Dil. } \mathrm{NaOH}} \mathrm{CH}_2(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO} \\ & \mathrm{CH}_2(\mathrm{OH}) \mathrm{CH}_2 \mathrm{CHO} \xrightarrow{\Delta} \mathrm{CH}_2=\mathrm{CHCHO} \\ & \mathrm{CH}_2=\mathrm{CHCHO} \xrightarrow{\mathrm{HCN}} \mathrm{CH}_2=\mathrm{CHCH}(\mathrm{OH})(\mathrm{CN}) \\ & \mathrm{CH}_2=\mathrm{CHCH}(\mathrm{OH})(\mathrm{CN}) \xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}} \mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}(\mathrm{COOH})(\mathrm{OH}) \end{aligned}

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