The magnitude of the rotational barrier in propane arising on account of torsional strain is about

Options

A. `12.54KJmol^(-1)`
B. `13.79KJ mol^(-1)`
C. `14.86KJ mol^(-1)`
D. `17.86KJ mol^(-1)`

Answer

B

Explanation

In propane `(CH_(3)-CH_(2)-CH_(3))`, rotation can occur about two `C-C` single bonds, and is essentially free. Although the methyl group `(Me)` is considered larger than `H` atom, the rotational barrier `(3.3 kcal mol^(-1))` is only a little higher than for ethane. This implies that threre is still not very much crowding in the eclisped conformation and the rotational barrier is mainly due to small repulsive interactions between the electrons (in eclipsed bonding) on adjacent `C's`. These repulsive forces between bonds on adjacent atoms which restrict rotation are known as torsional strain.