1.

An alkyl halide with beta-hydrogen atoms on reaction with a base or a nucleophile has two competing paths. Substitutions (S_(N^(1)) and S_(N^(2))) and elimination (E_(1) and E_(2)). The path adopted by them depends upon the nature of the alkyl halide, strength and size of the base/nucleophile and reaction conditions. The bulkiers nucleophile prefers to act as a base and abstracts a proton rather than approaching a tetravalent carbon atom (due to steric reasons) and vise versa. A primary alkyl halide can react by any of the four mechanism (S_(N^(2)), S_(N^(1)), E_(2) and E_(1)) depending upon the stability or the intermediate carbocation or the substituted alkene formed and the reaction conditions and tertiary alkyl halide. The three possible paths (S_(N^(1)), E_(1) and E_(2)) Neopentyl bromide undergoes dehydrohalogenations to give alkene though it has no beta-hydrogen. This is due

Answer»

`E_(2)` mechanism
`E_(1)` mechanism
due to rerrangement of carbocation by `E_(1)` mechanism
HOFMANN elimination

Solution :The initially formed neopentyl carbocation being `1^(@)` undergoes REARRANGEMENT to form more stable `3^(@)` carbocation. Since the carbocation thus formed has `beta`-hydrogen, it undergoes dehydrohalogenation to give as alkene according to SAYTZEFF rule
`underset(3^(@)"carbocation")(overset(beta)(C )H_(3)-overset(Br)overset(|)underset(+)(C )-overset(beta)(C )H_(2))-CH_(3)overset(-H^(+))rarr underset("2-Methyl-2-butene")(CH_(3)-overset(CH_(3))overset("|")("C")=CHCH_(3))`


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