How will you convert ethanoic acid into benzene?
(N/A) $CH_3COOH$ $\xrightarrow{NaOH_{(aq)}} CH_3COONa$ $\xrightarrow[\Delta]{Soda\ lime} CH_4$ $\xrightarrow[hv]{Cl_2} CH_3Cl$ $\xrightarrow[Wurtz\ reaction]{Na/dry\ ether} C_2H_6$ $\xrightarrow{Cl_2} C_2H_5Cl$ $\xrightarrow{alc. KOH} CH_2=CH_2$ $\xrightarrow{Br_2} CH_2Br-CH_2Br$ $\xrightarrow{alc. KOH} CH_2=CHBr$ $\xrightarrow{NaNH_2} CH \equiv CH$ $\xrightarrow[873\ K]{Red\ hot\ iron\ tube} C_6H_6$ (Benzene)
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Similar Questions
Assertion : Nitrating mixture used for carrying out nitration of benzene consists of conc. $HNO_3 +$ conc. $H_2SO_4$.
Reason : In presence of $H_2SO_4$,$HNO_3$ acts as a base and produces $NO_2^+$ ions.
Reason : In presence of $H_2SO_4$,$HNO_3$ acts as a base and produces $NO_2^+$ ions.
MediumAIIMS 2015
View SolutionExplain why the following systems are not aromatic?
Difficult
View SolutionIdentify the correct statements for a ring system to exhibit aromaticity.
$(i)$ It must not be planar.
$(ii)$ It must possess $(4n+2) \pi$-electrons.
$(iii)$ It must be planar.
$(iv)$ It must possess $4n \pi$-electrons.
The correct answer is
$(i)$ It must not be planar.
$(ii)$ It must possess $(4n+2) \pi$-electrons.
$(iii)$ It must be planar.
$(iv)$ It must possess $4n \pi$-electrons.
The correct answer is
In the given reaction,$C_6H_6 + HNO_3 \xrightarrow{H_2SO_4} C_6H_5NO_2 + H_2O$,$HNO_3$ acts as:
Medium
View SolutionSodium benzoate on heating with sodalime gives:
Medium
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