Q.CH2=CH2 + Br2 --> A --(alc. KOH)--> B --(NaNH2)--> C. Find A, B and C.
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Start your 14-day free trial to unlock the full solution →CH2=CH2 + Br2 gives 1,2-dibromoethane (A); alcoholic KOH removes one HBr to give vinyl bromide (B); NaNH2 removes the second HBr to give acetylene (C).
Step 1 — Formation of A: Ethylene (CH2=CH2) reacts with bromine (Br2) in an electrophilic addition reaction — the pi bond of the alkene attacks Br2, and both bromine atoms add across the double bond:
CH2=CH2 + Br2 --> CH2Br-CH2Br (A = 1,2-dibromoethane, ethylene dibromide)
Step 2 — Formation of B: Treating the vicinal dibromide A with alcoholic KOH (a base, used specifically to favour ELIMINATION over substitution) removes one molecule of HBr (dehydrohalogenation), forming a carbon-carbon double bond again — this time with one bromine remaining:
CH2Br-CH2Br --(alc. KOH)--> CH2=CHBr (B = vinyl bromide, bromoethene)
Step 3 — Formation of C: Vinyl bromide B, when treated with sodamide (NaNH2, a very strong base), undergoes a second dehydrohalogenation — this time eliminating HBr from the vinylic system to introduce a second degree of unsaturation, converting the C=C double bond into a C-tripleBond-C triple bond:
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