Hydrocarbons Question 250
Question: Action of hydrogen chloride on $ \underset{CH_3}{\mathop{\underset{|}{\mathop{CH_3-C=CH_2}}}} $ and on $ CH\equiv CH $ will predominantly give the compounds, respectively
Options:
A) $ \underset{CH_3}{\mathop{\underset{|}{\mathop{CH_3-CH=CH_2Cl}}}} $ and $ CH_2Cl-CH_2Cl $
B) $ \underset{CH_3}{\mathop{\underset{|}{\mathop{CH_3-CCl=CH_3}}}} $ and $ CH_3-CHCl_2 $
C) $ \underset{CH_3}{\mathop{\underset{|}{\mathop{CH_3-CH=CH_2Cl}}}} $ and $ CH_3-CHCl_2 $
D) $ \underset{CH_3}{\mathop{\underset{|}{\mathop{CH_3-CH=CH_3}}}} $ and $ CH_2Cl-CH_2Cl $
Show Answer
Answer:
Correct Answer: B
Solution:
$ CH_3-\underset{CH_3}{\mathop{\underset{|}{\mathop{C}}}}=CH_2+HCl\to CH_3-\underset{\underset{CH_3}{\mathop{|}}}{\overset{\overset{Cl}{\mathop{|}}}{\mathop{C-}}}CH_3 $
$ CH\equiv CH+HCl\to CH_2=CH-Cl\xrightarrow{HCl}CH_3-CHCl_2 $
BETA
