Skip to content
Question 57 of 69

Q.Nitrobenzene on reduction with Zn/NaOH gives :

(a) C6H5−N=N−C6H5C_6H_5-N=N-C_6H_5
(b) C6H5−NH−NH−C6H5C_6H_5-NH-NH-C_6H_5
(c) C6H5NHOHC_6H_5NHOH
(d) C6H5−N=N−C6H5C_6H_5-N=N-C_6H_5 with an oxygen attached to the second N (azoxybenzene, C6H5−N=N(→O)−C6H5C_6H_5-N=N(\to O)-C_6H_5)
Puducherry TnboardTamil Nadu HSC (DGE) Board 2019MCQ· 1mImportance★★★★★
83% · 57/69 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Zn/NaOH is a strongly reducing alkaline system that reduces nitrobenzene completely through nitroso- and azoxy-/azo- intermediates to hydrazobenzene, C6H5−NH−NH−C6H5C_6H_5-NH-NH-C_6H_5.

The reduction of nitrobenzene proceeds through a well-defined sequence of intermediates, C6H5NO2→C6H5NO→C6H5NHOH→C_6H_5NO_2 \rightarrow C_6H_5NO \rightarrow C_6H_5NHOH \rightarrow (coupling) →C6H5−N=N(→O)−C6H5\rightarrow C_6H_5-N=N(\to O)-C_6H_5 (azoxybenzene) →C6H5−N=N−C6H5\rightarrow C_6H_5-N=N-C_6H_5 (azobenzene) →C6H5−NH−NH−C6H5\rightarrow C_6H_5-NH-NH-C_6H_5 (hydrazobenzene), and which product is isolated depends entirely on the strength and pH of the reducing system used: Zn/NH4ClNH_4Cl (neutral, mild) stops at phenylhydroxylamine; Sn/HCl or Fe/HCl (acidic, strong) goes all the way to aniline; Zn dust in strongly alkaline NaOH solution is the classical reagent that carries the reduction through to the fully reduced coupled produc …

Start your 14-day free trial to unlock the full solution →

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests
Start 14-day free trial →See Plans & Pricing →

Full access for 14 days. No credit card required.

←Previous questionQ13 — (a) Distinguish between primary, secondary and tertiary amines. (b) Wr…→Next questionQ15 — Write any three characteristics of dyes.