Describe how naphthalene can be prepared from the given starting material.

The presence of activating substituents enhances the reactivity of the benzene towards electrophilic substitution reactions. The presence of deactivating substituents reduces the reactivity of the benzene ring.

Some activating substituents are $-{\mathrm{NH}}_2$and $-\mathrm{OR}$, and some deactivating substituents are $-\mathrm{COOH}$,$-\mathrm{CHO}$, etc.

The given compound, 4-phenylbutanoyl chloride, reacts with ${\mathrm{AlCl}}_3$and ${\mathrm{H}}_2\mathrm{O}$to form 3,4-dihydronaphthalen-1(2H)-one. This reacts with Pd/C and to $H_2$ form 1,2,3,4-tetrahydronaphthalene.

This reacts with $B{r}_2$and light to form 1-bromo-1,2,3,4-tetrahydronaphthalene. This further reacts with tert-ButoO-form 1,2-dihydronaphthalene. This reacts with NBS, peroxide, and heat to form 2-bromo-1,2-dihydronaphthalene. This reacts with tert-Bu O-to form naphthalene.

Preparation of naphthalene from the given starting material