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MS 918 - Epoxide and Ether Practice

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 # 697

Rank the following compounds in order of decreasing boiling point.

Also, make a guess about their relative solubilities in water. Explain your reasoning.
Solution
Water is very polar, so the more polar a molecule is, the more soluble it will be in water.

Alcohols can hydrogen bond, and so will be the most polar, have the highest boiling point, and be the most soluble in water.

Aldehydes (carbonyls) can't hydrogen bond but the C=O double bond is very polar, so aldehydes are somewhat soluble in water.

The ether is the least polar of the three functional groups, and so will have the lowest boiling point, and be the least soluble in water.

So the overal trend for polarity, water solubility, and boiling point is:

alcohol > aldehyde > ether

Problem Details
MendelSet practice problem # 697 submitted by Matt on July 20, 2011.
Subject: Organic Chemistry
Subject Topic(s): Ethers and Epoxides, Intermolecular Forces and Functional Groups
Question Type: Concept (explain why this is so..)
Keyword(s): boiling point, polarity, solubility
Problem # 698

When propyl bromine is treated with KF in benzene no reaction takes place. But when the crown ether 18-Crown-6 is added to the reaction mixture the desired propyl fluoride product is produced. Explain.
Solution
KF is insoluble in propyl bromide or benzene, so the two compounds never "touch" each other, and no reaction takes place. (F^- and propyl bromide are in different phases and so don't come into contact).

But when the crown ether is added, it solvates the potassium ion, allowing the K⁺ and F^- ions to dissolve in the solvent, so the fluoride ion and propyl bromide are able to "talk" to each other, and the rection can take place.

Problem Details
MendelSet practice problem # 698 submitted by Matt on July 21, 2011.
Subject: Organic Chemistry
Subject Topic(s): Ethers and Epoxides
Question Type: Concept (explain why this is so..)
Keyword(s): solubility, crown ethers
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.

Problem Details
MendelSet practice problem # 699 submitted by Matt on July 21, 2011.
Subject: Organic Chemistry
Subject Topic(s): Ethers and Epoxides
Question Type: Synthesis (show how to prepare X from Y..)
Keyword(s): epoxides, Williamson ether synthesis
Problem # 700

Write out a mechanism for the reaction below using curved arrows. Be sure to include formal charges.
Solution
Ethers react with 2 equivalents of H-X to form water and two equivalents of alkyl halide.

In this case, the ether was cyclic, so the ring had to open up.

The reaction can go through either an S_N1 or S_N2 mechanism. Since this was a primary ether, it will go through an S_N2 mechanism (the carbocation is too unstable for the reaction to go S_N1).

Problem Details
MendelSet practice problem # 700 submitted by Matt on July 21, 2011.
Subject: Organic Chemistry
Subject Topic(s): Ethers and Epoxides
Question Type: Mechanism (show a mechanism using curved arrows..)
Keyword(s): SN2 mechanism, ether hydrolysis
Problem # 701

The acid-catalyzed condensation of alcohols to form ethers is reversable; ethers can be hydrolyzed back to alcohols. How can the direction of this equilibrium be controlled to preferentially form ethers?
Solution
To push an equilibrium to one side, add starting material and remove product. This is Le Chatelier's principle from general chemistry.

So to push this reaction to the right and form ether, add alcohol and remove ether and water as they form.

To push this reaction to the left and form alcohol, add water to ether and remove alcohol as it forms.

How do you "remove something as it forms?" Alcohols and ethers have (relatively) low boiling points, and can be removed by hooking up a vacuum line and condenser to your reaction. The ether (or alcohol) boils off under the reduced pressure, and then recondenses in a separate piece of glassware. (Sort of like in distillation.)

Water has a relatively high boiling point and so is difficult to remove under reduced pressure. To remove water, molecular sieves are used. They're like tiny sponges that only absorb water (and not other solvents), removing it from the reaction.

Problem Details
MendelSet practice problem # 701 submitted by Matt on July 21, 2011.
Subject: Organic Chemistry
Subject Topic(s): Ethers and Epoxides, Alcohol Reactions (PBr3, SOCl2, HX, Sulfonate ester formation)
Question Type: Concept (explain why this is so..)
Keyword(s): alcohol condensation, ether hydrolysis, equilibrium and Le Chatelier's principle
Problem # 702

Show how to prepare each compound starting from propylene oxide.

(Propylene oxide image below courtesy of Wikipedia.)
Solution
Epoxides can open up in two different ways.

To add a nucleophile to the less substituted side of an epoxide, use basic conditions. This is done in #2 below.

To add a nucleophile to the more substituted side of an epoxide, use acidic conditions. This is done is #1 below.

Why do the conditions matter? Epoxides have two electrophilic carbons. Normally nucleophiles will preferentially attack the less substituted carbon, as they do in S_N2 reactions. Recall that S_N2 reactions usually happen with strong nucleophiles- that is, negative charges (basic conditions).

When an epoxide reacts under acidic conditions, the transition state has carbocation character, and so it's sort of like an S_N1 reaction. That is, instead of less substituted carbons being favored due to less steric bulk, more substituted carbons are favored do to a more stable carbocation. So acidic conditions cause an epoxide to open up on the more substituted side.

Problem Details
MendelSet practice problem # 702 submitted by Matt on July 21, 2011.
Subject: Organic Chemistry
Subject Topic(s): Ethers and Epoxides
Question Type: Synthesis (show how to prepare X from Y..)
Keyword(s): epoxides
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

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