JEE Main 2022 · 29 Jul · Shift 1 · Chemistry Q78

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ChemistrySingle correct+4 / -1Question 78 of 90

Consider the following reaction sequence : 4-aminobenzonitrile is treated with (i) AlH(i-Bu) $_2$ , (ii) H $_2$ O to give 'A', which with CH $_3$ CHO / dil NaOH, Δ gives B (Major Product).

The product 'B' is :

Scheme: 4-aminobenzonitrile → ((i) AlH(i-Bu)2, (ii) H2O) 'A' → (CH3CHO, dil NaOH, Δ) B (major product).

\mathrm{OHC-C_6H_4-N{=}CH-CH_3}

(para)

Option A: 4-(ethylideneamino)benzaldehyde.

\mathrm{H_2N-C_6H_4-CH{=}CH-CHO}

(para)

\mathrm{H_2N-C_6H_4-CH_2-N{=}CH-CH_3}

(para)

Option C: N-ethylidene-(4-aminophenyl)methanamine.

\mathrm{H_2N-C_6H_4-CO-N{=}CH-CHO}

(para)

Option D: 4-amino-N-(2-oxoethylidene)benzamide.

Answer and worked solution

Official answer: B

DIBAL-H reduces the nitrile to the aldehyde: 4-aminobenzaldehyde. Cross aldol with acetaldehyde (the aryl aldehyde has no α-H) then dehydrates: 4-aminocinnamaldehyde, $\mathrm{H_2N-C_6H_4-CH{=}CH-CHO}$ .

(B)

Topics:Aldol and related condensations Oxidation and reduction

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JEE Main 2022 · 29 Jul · Shift 1, Chemistry question 78