Hydrocarbons Question 398
Question: In the following reaction $ CH_3-CH_2-CH_2-CH_3\underset{475K}{\mathop{\xrightarrow{H_2SO_4}}} $ [AIIMS 1983]
Options:
A) $ CH_3CH=CHCH_3 $ predominates
B) $ CH_2=CHCH_2CH_3 $ predominates
C) Both are formed in equal amounts
D) The amount of production depends on the nature of catalyst
Show Answer
Answer:
Correct Answer: A
Solution:
$ CH_3-CH_2-CH_2-CH_3\underset{475K}{\mathop{\xrightarrow{H_2SO_4}}} $
$ \underset{\text{More symmetrical}(major\text{product)}}{\mathop{CH_3-CH=CH-CH_3}} $
$ CH_3-CH_2-CH_2-CH_3\underset{475K}{\mathop{\xrightarrow{H_2SO_4}}} $
$ \underset{\begin{smallmatrix} \text{Less symmetrical or } \\ unsymmetrical \\ \text{(minor product)} \end{smallmatrix}}{\mathop{CH_2=CH-CH_2-CH_3}} $ It is based on Saytzeff’s rule.
According to this more symmetrical or more alkylated alkene predominates.
BETA
