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Chapter 25: PROBLEM 25.27 (page 1027)

Question: Draw the product formed by treating each compound with excess CH3I , followed by Ag2O, and then heat.

a.

b.

c.

Short Answer

Expert verified

Answer

a.

b.

c.

Step by step solution

01

Hoffmann elimination 

In Hofmann elimination, the base removes a proton from the less substituted and more accessible 伪carbon atom, because of the bulky leaving group on the nearby alpha carbon.

The bulky leaving group is obtained as quaternary ammonium salt by the attack of the amine to the methyl iodide via SN2 reaction.

02

Explanation

a. The reaction of the given amine is completed via the Hoffmann elimination process as:

Hoffmann elimination in compound a.

b.The given amine produces the product alkene via the Hoffmann elimination process as:

Hoffmann elimination in compound b.

c. The given amine produces the product alkene via the Hoffmann elimination process as:

Hoffmann elimination in compound c.

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Most popular questions from this chapter

Explain why pyrimidine is less basic than pyridine

The pKa of the conjugate acid of guanidine is 13.6, making it one of the strongest neutral organic bases. Offer an explanation.

Question:The pKa of the conjugate acid of guanidine is 13.6, making it one of the strongest neutral organic bases. Offer an explanation.

Alkyl diazonium salts decompose to form carbocations, which go on to form products of substitution, elimination, and (sometimes) rearrangement. Keeping this in mind, draw a stepwise mechanism that forms all of the following products.

What is the major Hofmann elimination product formed from each amine?

a

b

c
See all solutions

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