\(CH_3-\underset{CH_3}{\underset{|}CH}-C ≡ CH\)\(\overset{excess\ HBr}\longrightarrow\)
The product of the above reaction is:
(a) \(CH_3-\overset{CH_3}{\overset{|}CH}-\overset{Br}{\overset{|}C}=\overset{Br}{\overset{|}CH_2}\)
(b) \(CH_3-\overset{CH_3}{\overset{|}CH}-\overset{Br}{\overset{|}C}=CH_2\)
(c) \(CH_3-\overset{CH_3}{\overset{|}CH}-\overset{Br}{\overset{|}{\underset{Br}{\underset{|}{C}}}}-CH_3\)
(d) \(CH_3-\overset{CH_3}{\overset{|}CH}-CH_2-\overset{Br}{\overset{|}{\underset{Br}{\underset{|}{CH}}}}\)
Correct answer: (c) \(CH_3-\overset{CH_3}{\overset{|}CH}-\overset{Br}{\overset{|}{\underset{Br}{\underset{|}{C}}}}-CH_3\)
Explanation:
\(CH_3-\overset{CH_3}{\overset{|}CH}-C ≡ CH \) \(\overset{H^+Br^-}\longrightarrow\)
\(CH_3-\overset{CH_3}{\overset{|}CH}-\underset{Br}{\underset{|}CH} = CH_2\)\(\overset{H^+Br^-}\longrightarrow\) \(CH_3-\overset{CH_3}{\overset{|}CH}-\overset{Br}{\overset{|}{\underset{Br}{\underset{|}{C}}}}-CH_3\)