Question

Which would undergoes sn2 reaction faster c6h5-ch2-ch2-br and c6h5-ch-ch3-br

Answer 1

Answer:

c6h5-ch2-ch2- br is react faster

rate of SN2 inversely proportional to steric hindrance

Answer 2

Answer:

$\[{C_6}{H_6} - C{H_2} - C{H_2} - Br\]$ will undergo a fast $\[{S_N}^2\]$ reaction also known as a second-order substitution reaction.

Explanation:

In the case of second-order substitution reaction($\[{S_N}^2\]$) nucleophile attacks in 1° carbon fast because of less steric hindrance.

The order of second-order substitution reaction ($\[{S_N}^2\]$) is :

1°>2°>3° carbons.

The reason behind this is, that at 3° carbon there is a lot of steric hindrance due to which reaction gets slow.

At 2° carbon steric hindrance reduces and the reaction gets fast.

At 1° carbon, there is the least steric hindrance and the reaction is fastest.

Therefore,  In $\[{C_6}{H_6} - C{H_2} - C{H_2} - Br\]$ Br is connected at 1° carbon and $\[{S_N}^2\]$ will be the fastest.