Hydrocarbons Question 401
Question: The dehydrohalogenation of neopentyl bromide with alcoholic KOH mainly gives [IIT-JEE (Screening) 1990; MP PET 1993]
Options:
A) 2-methyl-1-butene
B) 2-methyl-2-butene
C) 2, 2-dimethyl-1-butene
D) 2-butene
Show Answer
Answer:
Correct Answer: B
Solution:
$ \begin{matrix} \ \ CH_3 \\ CH_3-\underset{|}{\overset{|}{\mathop{C}}}\ -CH_2-Br+\underset{\text{(alc)}}{\mathop{KOH}}\xrightarrow{{}} \\ \ \ \ \ \ \ \ \ CH_3 \\ \end{matrix} $
$ \begin{matrix} CH_3 \\ CH_3-\overset{|}{\mathop{C}}=\ CH-CH_3+KBr+H_2O \\ \end{matrix} $
In this reaction $ 1^{o} $ carbonium ion is formed which rearranges to form $ 3^{o} $ carbonium ion from which base obstruct proton.
Hence 2-methyl-2-butene is formed as a main product.
$ \underset{{1^{\text{o }}}\text{carbonium less stable}}{\mathop{\begin{matrix} \ \ CH_3 \\ CH_3-\underset{|}{\overset{|}{\mathop{C}}}\ -\ \overset{+}{\mathop{C}}H_2 \\ \ \ \ \ \ \ \ \ CH_3 \\ \end{matrix} }}\xrightarrow{\text{Methyl shift}}CH_3-\underset{{}}{\overset{CH_3}{\mathop{\underset{+}{\overset{|}{\mathop{C-}}}}}}CH_2-CH_3 $
$ \underset{\text{2-Methyl-2-Butene}}{\mathop{CH_3-\overset{CH_3}{\mathop{\overset{|}{\mathop{C=}}}}CH-CH_3}} $
BETA
