Consider the following reaction Propanal+Methanal= $\xrightarrow[\substack{\text { (ii) } \Delta \mathrm{NaCN} \\ \text { (iii) } \mathrm{NaCN} \\ \text { (iv) } \mathrm{H}_{3} \mathrm{O}^{+}}]{\text {(i)dil. } \mathrm{NaOH}} \underset{\left(\mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}_{3}\right)}{\text { Product B }}$ The correct statement for product B is. It is

(C) racemic mixture and gives a gas with saturated $\mathrm{NaHCO}_{3}$ solution

(A) optically active and adds one mole of bromine

(B) racemic mixture and is neutral

(D) optically active alcohol and is neutrall

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

Consider the following reaction

Propanal+Methanal=

(i)dil. NaOH (ii) Δ NaCN (iii) NaCN (iv) H_{3} O^{+} (C_{5} H_{8} O_{3}) Product B

The correct statement for product B is. It is

(A)optically active and adds one mole of bromine
(B)racemic mixture and is neutral
(C)racemic mixture and gives a gas with saturated $NaHCO_{3}$ solution✓ Correct
(D)optically active alcohol and is neutrall

Answer: (C)

Solution

Propanal, which has

α

-hydrogens, undergoes a crossed aldol reaction with methanal in dilute alkali, and on warming the aldol loses water to give 2-methylprop-2-enal. Addition of sodium cyanide to this aldehyde forms the cyanohydrin, and acid hydrolysis of the nitrile gives the carboxylic acid

CH X_{2} = C (CH X_{3}) - CH (OH) - COOH

(

C X_{5} H X_{8} O X_{3}

). The carbon bearing the hydroxyl, carboxyl, hydrogen and vinyl groups is a stereocentre formed without any chiral influence, so the product is obtained as a racemic mixture, and being a carboxylic acid it liberates carbon dioxide with saturated sodium hydrogencarbonate solution. Product B is a racemic acid that gives a gas with saturated sodium hydrogencarbonate solution.

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Aldehydes Ketones and Carboxylic Acids

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