Explanation
Correct Answer: Option B
Explanation:
The direct iodination reaction of benzene is,
$${{C}_{6}}{{H}_{6}}+{{I}_{2}}\rightleftharpoons {{C}_{6}}{{H}_{5}}I+HI$$
This reaction is a reversible reaction. The product $$HI$$ reacts with iodobenzene to get back benzene.
Hence, direct iodination of benzene is not possible because the product $${{C}_{6}}{{H}_{5}}I$$ is reduced by $$HI$$.
$$R C-C=R \xrightarrow[Lindlar catalsyst]{H_2}$$?
$$C_{10}H_{12}$$
$$\downarrow$$
$$C_nH_{2n + 2}$$
$$n = 10$$
$$ n – 1 = 9 \rightarrow C_9H_{20}$$
$$ n + 1 = 11 \rightarrow C_{11}H_{24}$$
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