Explanation
Given,
Step 1.
($$\because N{ O }_{ 2 }$$ is $$m$$-directing group while $$C{ H }_{ 3 }$$ group is $$o-$$ and $$p-$$ directing group. Thus, $$Br$$ group attach at $$o-$$ position with respect to $$C{ H }_{ 3 }$$ and $$m$$-position with respect to $$N{ O }_{ 2 }$$ group)
Step 2. On reduction with $${ Sn }/{ HCl }, N{ O }_{ 2 }-$$ group changes to $$-N{ H }_{ 2 }$$ group.
Step 3. On treatment with $$NaN{ O }_{ 2 }+HCl$$; $$N{ H }_{ 2 }$$ changes to diazo group (i.e. $$Q$$ shows diazotisation).
Step 4. On treatment with $${ H }_{ 3 }P{ O }_{ 2 }$$, only diazo group $$\left( { N }_{ 2 }Cl \right) $$ reacts and gets removed.
Step 5. On oxidation with $${ KMn{ O }_{ 4 } }/{ O{ H }^{ - } } - C{ H }_{ 3 }$$ group changes to $$-COOH$$ group.
Correct Option: $$B$$
Hence, the above compound is hydrated maximum at $$2$$ position
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