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Q.
The number of meso form of the given compound $\left(\right.A\left.\right)$ is
NTA AbhyasNTA Abhyas 2022
Solution:
Number of meso form $2^{\left(\right. \frac{\text{n} - 1}{2} \left.\right)}$ when compound is divisible into two identical halves and has odd number of chiral carbons.
$\text{m} = 2^{\left(\right. \frac{3 - 1}{2} \left.\right)}$
$=2^{1}=2$ .
