Hydrocarbons Question 46
Question: In a reaction $ CH_2=CH_2\underset{acid}{\mathop{\xrightarrow{Hypochlorous}}}M\xrightarrow{R}\begin{matrix} CH_2 & - & OH \\ | & {} & {} \\ CH_2 & - & OH \\ \end{matrix} $ Where M = molecule; R = reagent M and R are [CBSE PMT 1997; CPMT 2001]
Options:
A) $ CH_3CH_2Cl $ and NaOH
B) $ CH_2Cl-CH_2OH $ and $ aq\text{. }NaHCO_3 $
C) $ CH_3CH_2OH $ and HCl
D) $ CH_2=CH_2 $ and heat
Show Answer
Answer:
Correct Answer: B
Solution:
$ CH_2=CH_2\xrightarrow{HOCl}\underset{\underset{Cl}{\mathop{|}}}{\mathop{CH_2}}-\underset{\underset{OH}{\mathop{|}}}{\mathop{CH_2}}\xrightarrow{aqNaHCO_3} $
$ \underset{Glycol}{\mathop{\underset{\underset{CH_2-OH}{\mathop{|}}}{\mathop{CH_2-OH}}}} $
BETA
