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Chapter 9: Q16. (page 350)

Question: Explain why two substitution products are formed in the following reaction.

Short Answer

Expert verified

Answer

The mechanism of the given reaction is shown below

Step by step solution

01

Step-by-Step Solution Step 1: Carbocation formation

Whenever a more electronegative atom attached to the carbon atom attracts the charge density to it, it results in the formation of the carbocation.

02

Formation of the stable carbocation in the reaction

In the first step of the reaction, heterolytic cleavage takes place between C and Cl as Cl is more electronegative than carbon.

This leads to the formation of a secondary carbocation. The shifting of hydride takes place to form the tertiary carbocation (stable). The mechanism is shown below.

Formation of stable carbocation

03

Formation of the major products

There would be two carbocations formed. The attack of the nucleophile on both carbocations takes place, followed by deprotonation.

This results in the formation of two major products, as shown below.

Formation of the major compounds

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Most popular questions from this chapter

Question: a) What is the major alkene formed when A is dehydrated with H2SO4 ? (b) What is the major alkene formed when A is treated with POCl3and pyridine? Explain why the major product is different in these reactions.

Question: Name each epoxide.

a.

b.

c.

Question: Draw a stepwise, detailed mechanism for the following intramolecular reaction that forms a cyclic ether.

Question: If the reaction of an alcohol with SOCl2and pyridine follows an SN2 mechanism, what is the stereochemistry of the alkyl chloride formed from (R)-butan-2-ol?

Question:Rearrangements can occur during the dehydration of alcohols even though no carbocation is formed-that is, a 1,2-shift occurs as the C-O+H2 bond is broken, forming a more stable 2掳or3掳carbocation, as shown. Using this information, draw a stepwise mechanism that shows how CH3CH2CH2CH2OH is dehydrated with H2SO4to form a mixture of CH3CH2CH=CH2and the cis and trans isomers of CH3CH=CHCH3. We will see another example of this type of rearrangement in Section 18.5C.
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