The dehydrohalogenation of neopentyl bromide with alcoholic KOH mainly gives

Options

A. 2-methyl-1-butene
B. 2-methyl-2-butene
C. 2, 2-dimethyl-1-butene
D. 2-butene

Answer

B

Explanation

`CH_(3)-overset(CH_3)overset(|)underset(CH_3)underset(|)C-CH_(2)-Br + underset((alc))(KOH) rarr CH_(3)-overset(CH_3)overset(|)C=CH-CH_(3) + KBr + H_(2)O`
In this reaction `1^@` carbonium ion is formed which rearranges to form `3^@` carbonium ion from which base obstruct proton. Hence 2-methyl-2-butene is formed as a main product.
`underset(1^@ "carbonium less stable")(CH_(3) - overset(CH_3)overset(|)underset(CH_3)underset(|)C-overset(+)(CH_2)) overset("Methyl shift")rarr CH_(3) - overset(CH_3)overset(|)underset(+)C-CH_(2)-CH_(3)`
`overset("Elimination of proton from " beta)overset("carbon which is less hydrogenated")rarr underset("2-Methyl-2- Butene")(CH_3 - overset(CH_3)overset(|)C=CH-CH_(3))`