JEE MainChemistry

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+4 marks−1 if incorrectSingle correctPrevious-year question

Cross-Aldol Condensation Followed by Cyanohydrin Formation and Hydrolysis

Consider the following reaction Propanal + Methanal xrightarrow[beginsubarrayl (ii) Δ; (iii) NaCN; (iv) H_3O^(+) endsubarray](i) dil.NaOH underset(C_5H_8O_3)Product B The correct statement for product B is. It is

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Aoptically active alcohol and is neutralBracemic mixture and gives a gas with saturated NaHCO_3 solutionCoptically active and adds one mole of bromineDracemic mixture and is neutral

Step-by-step solutionView

Correct answer

Product B is 2-hydroxy-2-methylbut-3-enoic acid (or 2-ethylacrylic acid derivative depending on aldol orientation; specifically, methanal plus propanal gives 2-methylpropenal via cross-aldol condensation, which with NaCN/H3O+ gives 2-hydroxy-2-methylbut-3-enoic acid, a racemic carboxylic acid that liberates CO2 with NaHCO3).

Option analysis

Why each option works or fails

A · optically active alcohol and is neutral

Believing that possessing a chiral center makes the synthesized product optically active without an asymmetric induction source, and confusing the carboxylic acid group for a neutral alcohol. Recognize that creating a stereocenter from achiral reactants yields an equimolar racemic mixture (optically inactive), and acidic $- COOH$ groups react with bases rather than being neutral.

B · racemic mixture and gives a gas with saturated NaHCO_3 solution

This is the correct statement. Cross-aldol condensation of methanal and propanal followed by dehydration yields 2-methylpropenal (methacrolein). Cyanohydrin formation and hydrolysis yields 2-hydroxy-2-methylbut-3-enoic acid (formula $C_{5} H_{8} O_{3}$ ), which contains a carboxylic acid group (effervesces ${CO}_{2}$ with saturated ${NaHCO}_{3}$ ) and forms as a racemic mixture.

C · optically active and adds one mole of bromine

Assuming that a chiral center automatically means the sample shows macroscopic optical activity. Remember that reactions of achiral starting materials without chiral reagents or catalysts always produce racemic modifications which show net zero optical rotation.

D · racemic mixture and is neutral

Overlooking that cyanohydrin hydrolysis converts the nitrile into a carboxylic acid, assuming instead that the product remains neutral. Account for acidic hydrolysis of nitriles ( $- CN \overset{H_{3} O^{+}}{\to} - COOH$ ), which introduces an acidic functional group that reacts with ${NaHCO}_{3}$ .

Reviewed route

Solution

StepWorking

01Identify

Propanal ( $e x t {C H}_{3} e x t {C H}_{2} e x t C H O$ ) has $α$ -hydrogens, whereas methanal ( $e x t H C H O$ ) has none. Under dilute $e x t N a O H$ and heating ( $Δ$ ), a crossed Aldol condensation occurs between the enolate of propanal and methanal to give an $\alpha,eta$ -unsaturated aldehyde, followed by reaction with $e x t N a C N$ and acidic hydrolysis ( $e x t H_{3} e x t O^{+}$ ).

02Mechanism

Step 1 & 2: $e x t {C H}_{3} e x t {C H}_{2} e x t C H O + e x t H C H O \overset{e x t d i l . N a O H, Δ}{\to} e x t {C H}_{2} = e x t C (e x t {C H}_{3}) e x t C H O$ (2-methylpropenal, methacrolein). Step 3 & 4: Methacrolein undergoes 1,4-conjugate addition (Michael addition) or 1,2-addition followed by rearrangement/hydrolysis. Given the molecular formula $e x t C_{5} e x t H_{8} e x t O_{3}$ , let us evaluate the addition: $e x t {C N}^{-}$ adds as a nucleophile. If $e x t {C N}^{-}$ adds to the carbonyl (1,2-addition), we get cyanohydrin $e x t {C H}_{2} = e x t C (e x t {C H}_{3}) e x t C H (e x t O H) e x t C N$ . Hydrolysis of $- e x t C N$ gives carboxylic acid: $e x t {C H}_{2} = e x t C (e x t {C H}_{3}) e x t C H (e x t O H) e x t C O O H$ with molecular formula $e x t C_{5} e x t H_{8} e x t O_{3}$ .

03Product

Product B is 2-hydroxy-3-methylbut-3-enoic acid, $e x t {C H}_{2} = e x t C (e x t {C H}_{3}) - \overset{*}{extC} e x t H (e x t O H) - e x t C O O H$ . It contains a chiral center at C-2 ( $e x t C^{*}$ ). Since the starting materials are achiral and no chiral catalyst/reagent is used, nucleophilic addition of cyanide to the planar aldehyde carbonyl occurs equally from both faces (re/si), generating an equimolar pair of enantiomers (a racemic mixture, optically inactive). Furthermore, the presence of the $- e x t C O O H$ group means it is acidic and releases $e x t {C O}_{2}$ gas when treated with saturated aqueous $e x t {N a H C O}_{3}$ .

✓Verify

Formula check: $e x t C_{5} e x t H_{8} e x t O_{3}$ has $5 i m e s 12 + 8 i m e s 1 + 3 i m e s 16 = 60 + 8 + 48 = 116 e x t g / m o l$ . The formula matches $e x t {C H}_{2} = e x t C (e x t {C H}_{3}) e x t C H (e x t O H) e x t C O O H$ (5 carbons, 8 hydrogens, 3 oxygens, degree of unsaturation = $5 - 8 / 2 + 1 = 2$ , corresponding to 1 double bond and 1 carbonyl). Carboxylic acid group effervesces with $e x t {N a H C O}_{3}$ , and generation of a single chiral center from an achiral precursor yields a racemic mixture.

Hints that build this answer step by step

What intermediate forms when propanal reacts with methanal in the presence of dilute $NaOH$ followed by heating (steps i and ii)?
2-Methylpropenal (methacrolein), ${CH}_{2} = C ({CH}_{3}) CHO$
What product B ( $C_{5} H_{8} O_{3}$ ) is obtained after nucleophilic addition of $NaCN$ to 2-methylpropenal followed by acidic hydrolysis (steps iii and iv)?
2-Hydroxy-2-methylbut-3-enoic acid, ${CH}_{2} = C ({CH}_{3}) CH (OH) COOH$ (or its 1,2-addition isomer)
Given that product B has a stereocenter and a carboxylic acid group, what are its optical activity and reaction with saturated ${NaHCO}_{3}$ ?
It is a racemic mixture and releases ${CO}_{2}$ gas with saturated ${NaHCO}_{3}$ .

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is the product a racemic mixture and not an optically active single enantiomer?

All starting materials and reagents are achiral. Attack of cyanide ion on the planar $e x t {s p}^{2}$ carbonyl group of methacrolein has equal probability from either face, forming an equimolar mixture of $(R)$ and $(S)$ enantiomers, which is optically inactive by external compensation.

Answer

Product B is 2-hydroxy-2-methylbut-3-enoic acid (or 2-ethylacrylic acid derivative depending on aldol orientation; specifically, methanal plus propanal gives 2-methylpropenal via cross-aldol condensation, which with NaCN/H3O+ gives 2-hydroxy-2-methylbut-3-enoic acid, a racemic carboxylic acid that liberates CO2 with NaHCO3).

Why each option works or fails

A: optically active alcohol and is neutral - Believing that possessing a chiral center makes the synthesized product optically active without an asymmetric induction source, and confusing the carboxylic acid group for a neutral alcohol. Recognize that creating a stereocenter from achiral reactants yields an equimolar racemic mixture (optically inactive), and acidic $- COOH$ groups react with bases rather than being neutral.
B · correct: racemic mixture and gives a gas with saturated NaHCO_3 solution - This is the correct statement. Cross-aldol condensation of methanal and propanal followed by dehydration yields 2-methylpropenal (methacrolein). Cyanohydrin formation and hydrolysis yields 2-hydroxy-2-methylbut-3-enoic acid (formula $C_{5} H_{8} O_{3}$ ), which contains a carboxylic acid group (effervesces ${CO}_{2}$ with saturated ${NaHCO}_{3}$ ) and forms as a racemic mixture.
C: optically active and adds one mole of bromine - Assuming that a chiral center automatically means the sample shows macroscopic optical activity. Remember that reactions of achiral starting materials without chiral reagents or catalysts always produce racemic modifications which show net zero optical rotation.
D: racemic mixture and is neutral - Overlooking that cyanohydrin hydrolysis converts the nitrile into a carboxylic acid, assuming instead that the product remains neutral. Account for acidic hydrolysis of nitriles ( $- CN \overset{H_{3} O^{+}}{\to} - COOH$ ), which introduces an acidic functional group that reacts with ${NaHCO}_{3}$ .

Step-by-step solution

Identify: Propanal ( $e x t {C H}_{3} e x t {C H}_{2} e x t C H O$ ) has $α$ -hydrogens, whereas methanal ( $e x t H C H O$ ) has none. Under dilute $e x t N a O H$ and heating ( $Δ$ ), a crossed Aldol condensation occurs between the enolate of propanal and methanal to give an $\alpha,eta$ -unsaturated aldehyde, followed by reaction with $e x t N a C N$ and acidic hydrolysis ( $e x t H_{3} e x t O^{+}$ ).
Mechanism: Step 1 & 2: $e x t {C H}_{3} e x t {C H}_{2} e x t C H O + e x t H C H O \overset{e x t d i l . N a O H, Δ}{\to} e x t {C H}_{2} = e x t C (e x t {C H}_{3}) e x t C H O$ (2-methylpropenal, methacrolein). Step 3 & 4: Methacrolein undergoes 1,4-conjugate addition (Michael addition) or 1,2-addition followed by rearrangement/hydrolysis. Given the molecular formula $e x t C_{5} e x t H_{8} e x t O_{3}$ , let us evaluate the addition: $e x t {C N}^{-}$ adds as a nucleophile. If $e x t {C N}^{-}$ adds to the carbonyl (1,2-addition), we get cyanohydrin $e x t {C H}_{2} = e x t C (e x t {C H}_{3}) e x t C H (e x t O H) e x t C N$ . Hydrolysis of $- e x t C N$ gives carboxylic acid: $e x t {C H}_{2} = e x t C (e x t {C H}_{3}) e x t C H (e x t O H) e x t C O O H$ with molecular formula $e x t C_{5} e x t H_{8} e x t O_{3}$ .
Product: Product B is 2-hydroxy-3-methylbut-3-enoic acid, $e x t {C H}_{2} = e x t C (e x t {C H}_{3}) - \overset{*}{extC} e x t H (e x t O H) - e x t C O O H$ . It contains a chiral center at C-2 ( $e x t C^{*}$ ). Since the starting materials are achiral and no chiral catalyst/reagent is used, nucleophilic addition of cyanide to the planar aldehyde carbonyl occurs equally from both faces (re/si), generating an equimolar pair of enantiomers (a racemic mixture, optically inactive). Furthermore, the presence of the $- e x t C O O H$ group means it is acidic and releases $e x t {C O}_{2}$ gas when treated with saturated aqueous $e x t {N a H C O}_{3}$ .
Verify: Formula check: $e x t C_{5} e x t H_{8} e x t O_{3}$ has $5 i m e s 12 + 8 i m e s 1 + 3 i m e s 16 = 60 + 8 + 48 = 116 e x t g / m o l$ . The formula matches $e x t C^{*}$ (5 carbons, 8 hydrogens, 3 oxygens, degree of unsaturation = $5 - 8 / 2 + 1 = 2$ , corresponding to 1 double bond and 1 carbonyl). Carboxylic acid group effervesces with $e x t {N a H C O}_{3}$ , and generation of a single chiral center from an achiral precursor yields a racemic mixture.

Cross-Aldol Condensation Followed by Cyanohydrin Formation and Hydrolysis

What feels right?

Why each option works or fails

Solution

01Identify

02Mechanism

03Product

✓Verify

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