Williamson ether synthesis

MS 918 - Epoxide and Ether Practice Submitted by Matt on August 6, 2011.

Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 16

Keywords: epoxides, Williamson ether synthesis

Description: Covers the major topics dealing with ethers and epoxides:

Boiling points and water solubilities of alcohols, ethers, and carbonyls

Crown ethers

Synthesis problems involving the Williamson ether synthesis and epoxides

Acidic and basic ring opening of epoxides

Acid catalized hydrolysis of ethers

Total Problems: 7

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Problem # 703

Show two ways to prepare the ether below from a combination of an alcohol and an alkyl halide via the Williamson ether synthesis.

Is one way better than the other? Why?

Solution

The Williamson ether synthesis takes place in two steps. First an alcohol is deprotonated to form a strong nucleophile (RO^-, this step isn't shown in the image below). Then the alkoxide (negative alcohol) attacks an alkyl halide in an S_N2 reaction.

So this problem is really asking, which step of conditions is most favorable for an S_N2 reaction?

Recall that S_N2 reactions compete with E2 reactions. If the nucleophile is too basic, or if there is too much bulk, it will go E2 instead of S_N2. (See problem 560 for a full explanation of these competition reactions)

Below, the top combination uses the less substituted (1º) alkyl halide, and so is the best for an S_N2 reaction.

The bottom reaction uses a bulkier (2º) alkyl halide, and will probably give a higher percentage of E2 side reaction.

Problem Details

MendelSet practice problem # 703 submitted by Matt on July 21, 2011.

Subject: Organic Chemistry

Subject Topic(s): SN1/SN2/E1/E2 Trends and Competition Reactions, Ethers and Epoxides

Question Type: Concept (explain why this is so..)

Keyword(s): SN1/SN2/E1/E2, Williamson ether synthesis

Problem # 699

Show how each compound can be prepared from the indicated starting material.

All carbon sources must contain three carbons or less.

Solution

a) When you see 2 carbons and 1 oxygen, that the tell-tale sign that you're adding ethylene oxide (the simplest epoxide).

But that would only leave you with an alcohol. How do you get to the ether? Using the Williamson ether synthesis.

b) As I mentioned in problem 673, when you see an alcohol you are also looking at a carbonyl, because you can interconvert the two (alcohol to aldehyde/ketone using PCC, aldehyde/ketone to alcohol using NaBH₄).

To add the methyl group, convert the alcohol to a ketone (which is an electrophile), and then add methyl Grignard (a nucleophile). But once again, you are only left with an alcohol. How to convert it to an ether? The Williamson ether synthesis.

When you see an ether in a synthesis problem, remember the Williamson ether synthesis. It will come in handy.