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Question 67 of 89

Q.Complete and rewrite the balanced chemical equations —

(a) Chlorobenzene →473K, pressureNaCN+CuCN\xrightarrow[473K,\ pressure]{NaCN + CuCN} ?
(b) Isobutyraldehyde →50% KOH\xrightarrow{50\%\ KOH} ?
(c) Butanone + 2,4 dinitro-phenyl hydrazine →H+\xrightarrow{H^+} ?
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 3mImportance★★★★★
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Chlorobenzene is cyanated to benzonitrile; isobutyraldehyde undergoes aldol addition (not condensation, since its alpha carbon becomes fully substituted); butanone forms its 2,4-DNP hydrazone.

a. Chlorobenzene + NaCN/CuCN, 473 K, pressure: this is a copper-catalysed cyanation (Rosenmund–von Braun-type) of the aryl chloride, replacing −Cl-Cl with −CN-CN:

C6H5Cl+NaCN→pressureCuCN, 473KC6H5CN (benzonitrile)+NaClC_6H_5Cl + NaCN \xrightarrow[\text{pressure}]{CuCN,\ 473K} C_6H_5CN\ (\text{benzonitrile}) + NaCl

b. Isobutyraldehyde + 50% KOH: isobutyraldehyde, (CH3)2CH−CHO(CH_3)_2CH-CHO, has exactly ONE α\alpha-hydrogen (on the carbon bonded to −CHO-CHO), so unlike formaldehyde/benzaldehyde (no α\alpha-H, which undergo Cannizzaro's reaction), it undergoes aldol addition under concentrated base: base removes the single α\alpha-H, the resulting carbanion/enolate attacks the carbonyl carbon of a second molecule, giving a β\beta-hydroxy aldehyde:

2(CH3)2CH−CHO→50% KOH(CH3)2C(CHO)−CH(OH)−CH(CH3)22(CH_3)_2CH-CHO \xrightarrow{50\%\ KOH} (CH_3)_2C(CHO)-CH(OH)-CH(CH_3)_2

(3-hydroxy-2,2,4-trimethylpentanal). Notably, in this product the α\alpha-carbon that reacted is now bonded to two methyl groups, the −CHO-CHO, and the new C–C bond — a fully-substituted (quaternary) carbon with no α\alpha-hydrogen left, so the usual base-catalysed dehydration to an α,β\alpha,\beta-unsaturated aldehyde cannot proceed further; the reaction stops at the aldol (addition) stage rather than going on to full 'condensation'.

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