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Chapter 18: Q 81 P (page 919)

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

Short Answer

Expert verified

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 -NH2 and -OR, and some deactivating substituents are -COOH,-CHO, etc.

Step by step solution

01

Electrophilic aromatic substitution

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 -NH2 and -OR, and some deactivating substituents are -COOH,-CHO, etc.

02

Preparation of naphthalene from the given starting material

The given compound, 4-phenylbutanoyl chloride, reacts with AICH3 and H2O to form 3,4-dihydronaphthalen-1(2H)-one. This reacts with Pd/C and H2 to form 1,2,3,4-tetrahydronaphthalene.

This reacts with Br2 and light to form 1-bromo-1,2,3,4-tetrahydronaphthalene. This further reacts with tert-BuO- to form 1,2-dihydronaphthalene. This reacts with NBS, peroxide, and heat to form 2-bromo-1,2-dihydronaphthalene. This reacts with tert-BuO- to form naphthalene

Preparation of naphthalene from the given starting material

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Most popular questions from this chapter

Rank the following compounds from largest Keqto smallest Keq for hydrate formation:

Question: Why does hydration inactivates FeBr3 ?

a. Rank the following esters from most reactive to least reactive in the first slow step of a nucleophilic acyl substitution reaction (formation of the tetrahedral intermediate):

b. Rank the same esters from most reactive to least reactive in the second slow step of a nucleophilic acyl substitution reaction (collapse of the tetrahedral intermediate).

A mixture of 0.10 mol benzene and 0.10 mol p-xylene was allowed to react with 0.10 mol nitronium ion until all the nitronium ion was gone. Two products were obtained: 0.002 mol of one and 0.098 mol of the other.

  1. What was the major product?
  2. Why was more of one product obtained than of the other?

Question: Which is a more stable intermediate in each pair?

a.

b.
See all solutions

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