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Chapter 16: Q16-69E (page 524)

p-Bromotoluene reacts with potassium amide to give a mixture of m and p-methylaniline. Explain.

Short Answer

Expert verified

The reaction followed as:

Step by step solution

01

Step 1: Benzyne formation

When the elimination reaction occur in the substituted aromatic rings the unreactive and the unstable intermediate is formed called benzyne which is attacked by nucleophiles and followed by protonation to form the desired product.

02

Reaction of p-Bromotoluene with potassium amide

When p- bromo toluene which is a substituted benzene react with potassium amide in which hydrogen is protonated by amide followed by Br ion removal benzyne formation takes place as an intermediate and then again ammonia is added to the triple bond of benzyne and the formation of two methylaniline observed.

03

Mechanism

Mechanism followed as:

  1. Abstraction of the proton with the elimination of Br ion
  2. Formation of benzyne intermediate
  3. Addition of ammonia to benzyne
  4. Formation of two methylaniline occur

Formation of methylaniline

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Most popular questions from this chapter

Refer to Table 6-3 on page 170 for a quantitative idea of the stability of a benzyl radical. How much more stable (in kJ/mol) is the benzyl radical than a primary alkyl radical? How does a benzyl radical compare in stability to an allylradical?

4- chloropyridine undergoes reaction with dimethylamine to yield 4-dimethylaminopyridine. Propose a mechanism for the reaction.

Monobromination of toluene gives a mixture of three bromotoluene products. Draw and name them.

How many products might be formed on chlorination of o-xylene (o-dimethylbenzene), m-xylene, and p-xylene?

You know the mechanism of HBr addition to alkenes, and you know the effects of various substituent groups on aromatic substitution. Use this knowledge to predict which of the following two alkenes reacts faster with HBr. Explain your answer by drawing resonance structures of the carbocation intermediates.
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