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  4. In the Hunsdiecker reaction,

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Q. In the Hunsdiecker reaction,

Haloalkanes and Haloarenes

Solution:

In the Hunsdiecker reaction $\to$
$CH_{3}CH_{2}COOAg + Br_{2} \xrightarrow{CCl_{4}} CH_{3} CH_{2}Br + CO_{2} + AgBr$
Silver Propionate$\,\,\,\,\,\,\,$ Ethyl Bromide
(No. of C = 3)$\,\,\,\,\,\,\,$ (No. of C = 2)
No. of carbon Atoms decrease Therefore, option ‘a’ is correct.