`CH_3CH=CH_2underset(C Cl_4)overset(Br_2)tounderset(excess)overset(NaNH_2)tooverset(H_2O)to`
Identify the end product of the above sequence of reactions.

Options

A. `CH_2=C=CH_2`
B. `CH_3C-=CH`
C. `CH_3-=overlineCNa^(+)`
D. `CH_2=CH-CH_2Cl`

Answer

B

Explanation

`CH_3CH=CH_2+Br_2tounderset("A vic-dihalide")(CH_3CHBrCH_2Br)overset(3NaNH_2)underset(-2HBr)toCH_3C-=overline(C)Na^(+)overset(H_2O)toCH_3C-=`
Vie-dihalides undergo a double dehydrohalogenation, requiring two moles of `NaNH_2` dehydrohalogenation of vicinal dihalides is particularly useful because the dihalides are themselves readily obtained from the corresponding alkenes by the addition of halogen. Thus alkynes can be synthesized from alkenes via vic-dihalides in this method an alkene is first treated with `Br_2` to form a vic-dibromo compound:
`RCH-=CHR+Br_(2)tounderset("Vic-dibromide")(Runderset(Br)underset(|)(C)H-underset(Br)underset(|)(C)HR)`
Then vic-dibromide is dehydrohalogenated through its reaction with a strong base such as `NaNH_2`. The dehydrohalogenation occurs in two steps. The first step yields a bromoalkene.
`R-CHBrCHBr-Roverset(-HBr)toR-underset("A vinyl halide")(CBr=CH)-Roverset(-HBr)toR-C-=C-R`
Since it is more difficult to dehydrohalogenate the intermediate vinyl bromide than an alkyl bromide, a much stronger base `NaNH_2` is used in place of alc. KOH. This is due to double bond character on account of resonance in vinyl bromide:
Than an alkyl bromide, a much stronger base `NaNH_2` is used in place of alc. KOH. This is due to double bond character on account of resonance in vinyl bromide the terminal alkyne `(CH_3C-=CH)` is converted to the alkynyl anion as soon as it forms, thus, three moles of the base are used in the overall reaction. The strong basic amide ion brings about an E2 elimination reaction.