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NEET CHEMISTRYMedium

The product 'D' in the following reaction is: CH3CNH3O+ANH3,ΔBNaOBrCNaNO2/HClDCH_3CN \xrightarrow{H_3O^+} A \xrightarrow{NH_3, \Delta} B \xrightarrow{NaOBr} C \xrightarrow{NaNO_2/HCl} D

A

HCHO

B

CH₃COOH

C

CH₃NH₂

D

CH₃OH

Step-by-Step Solution

The reaction sequence proceeds as follows:

  1. Hydrolysis (A): Methyl cyanide (CH3CNCH_3CN) undergoes acid hydrolysis (H3O+H_3O^+) to form Ethanoic acid (CH3COOHCH_3COOH) [A]. CH3CN+2H2O+H+CH3COOH+NH4+CH_3CN + 2H_2O + H^+ \rightarrow CH_3COOH + NH_4^+
  2. Amide Formation (B): Heating ethanoic acid with ammonia (NH3,ΔNH_3, \Delta) removes water to form Ethanamide (CH3CONH2CH_3CONH_2) [B]. CH3COOH+NH3CH3COONH4ΔCH3CONH2+H2OCH_3COOH + NH_3 \rightarrow CH_3COONH_4 \xrightarrow{\Delta} CH_3CONH_2 + H_2O
  3. Hoffmann Bromamide Degradation (C): Reaction with sodium hypobromite (NaOBrNaOBr or Br2/NaOHBr_2/NaOH) converts the amide to a primary amine with one less carbon atom. Thus, Methylamine (CH3NH2CH_3NH_2) [C] is formed. CH3CONH2+Br2+4NaOHCH3NH2+Na2CO3+2NaBr+2H2OCH_3CONH_2 + Br_2 + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O
  4. Reaction with Nitrous Acid (D): Primary aliphatic amines react with nitrous acid (NaNO2+HClNaNO_2 + HCl) to form unstable diazonium salts which liberate nitrogen gas and form alcohols. Thus, Methanol (CH3OHCH_3OH) [D] is the final product. CH3NH2+HNO2[CH3N2+Cl]H2OCH3OH+N2+HClCH_3NH_2 + HNO_2 \rightarrow [CH_3N_2^+Cl^-] \xrightarrow{H_2O} CH_3OH + N_2 + HCl
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