Textbook: Smith 3rd Ed. (2010)

Chapter 9: Alcohols, Ethers, and Epoxides

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Individual Problems

Individual Problems

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 # 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 # 347

For the reaction below, draw the structures of the carbocation intermediate and the final product.

Solution

This is a substitution reaction. Because the reaction takes place in acid and the leaving group is on a 2º carbon, it will probably form a carbocation (S_N1 mechanism).

^-OH is a poor leaving group so the alcohol will protonate first, so it can leave as H₂O, which is neutral.

Problem Details

MendelSet practice problem # 347 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1), Alcohol Reactions (PBr3, SOCl2, HX, Sulfonate ester formation)

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): SN1 mechanism

Problem # 333

Let's go over how a carbocation can form from an alcohol.

Write in the curved arrows to show the formation of the protonated alcohol, and water acting as a leaving group to form a carbocation.

Solution

Problem Details

MendelSet practice problem # 333 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbocation Fundamentals (Stability, Rearrangement, SN1), Acid-Catalyzed Dehydration of Alcohols

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): carbocation, carbocation formation

Problem # 348

For the reaction below, draw the structures of the carbocation intermediate and the final product.

Solution

The delta (Δ) in the reaciton arrows means that heat is being added to this reaction, which tends to favor elimination over substitution. Also, the reaction is using a non-nucleophilic acid (H₂SO₄), which tends to favor elimination reactions (H₃PO₄ is another common reagent for E1 reactions, while HCl or HBr tend to go S_N1).

Because this reaction is taking place in acid, a carbocation is likely to form, so this is an E1 reaction. Since water is lost over the course of the reaction, this is a dehydration, which is a type of elimination reaction.

Problem Details

MendelSet practice problem # 348 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, SN1/SN2/E1/E2 Trends and Competition Reactions, Acid-Catalyzed Dehydration of Alcohols

Question Type: Reactions (predict the product, show the starting material..)

Keyword(s): E1 mechanism, dehydration

Problem # 519

Let's work through an elimination reaction. Draw the structures for each of the species in the three boxes below (protonated thiol, carbocation, and alkene). Also draw curved arrows to show electron movement.

Solution

The first step in this reaciton, like many reactions, is an acid-base reaction- when you see an H⁺ (acid), the first step is usually something getting protonated. This starting material is a thiol (sulfur), but the same thing would happen with an alcohol or ether (oxygen).

When you have a protonated thiol (or alcohol, ether, etc.), you know the reaction isn't finished yet- products are rarely charged. There are two legal moves to get that positive charge off of the protonated thiol. Either the thiol deprotonates, or it takes off as a neutral leaving group. It can't deprontate because that would be going backwards, so your only "legal move" is for the HSCH₂CH₃ to act as a leaving group (See problem 518).

But then you're left with a carbocation, which is definitely not the final product. How can we get rid of it? (See problem 335 for general ways carbocations react). If the HSCH₂CH₃ attacked in an S_N1 type reaction that would be OK, but it would be going backwards! So the only way to get rid of the carbocation is to do a beta-elminiation (E1).

Problem Details

MendelSet practice problem # 519 submitted by Matt on June 30, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Acid-Catalyzed Dehydration of Alcohols

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): E1 mechanism, alcohols

Problem # 677

Show a mechanism for the acid-catalyzed cyclization (condensation) of 1,4-butanediol.

Solution

This is just an S_N2 reaction.

The first step is to turn one of the -OH groups into a better leaving group by having it pick up a proton. (I choose the hydroxyl on the right side).

Then the other -OH group can attack carbon 1, and the protonated hydroxyl group leaves as water.

Finally, the proton on the -OH group that acted as a nucleophile will "fall off" (deprotonate).

Problem Details

MendelSet practice problem # 677 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Ethers and Epoxides, Alcohol Reactions (PBr3, SOCl2, HX, Sulfonate ester formation)

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): SN2 mechanism, alcohol condensation

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 # 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 # 341

Predict the product(s) of the reaction below, and used curved arrows to show a mechanism.

Solution

You know this is an elimination reaction because no nucleophile is present; H₃PO₄ and H₂SO₄ are non-nucleophilic acids. IF the reagent were HCl or HBr on the other hand, this would be a substitution reaction.

Since acid is present, a carbocation will probably form, so we know that this is an E1 mechanism.

The 2º carbocation that is initially formed will undergo a 1,2-hydride shift to become a 3° carbocation, but that doesn't affect the final product of the reaction; with either the 2° or 3° carbocation, the most stable alkene product is 2-methyl-2-butene.

Problem Details

MendelSet practice problem # 341 submitted by Matt on June 7, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Acid-Catalyzed Dehydration of Alcohols

Question Type: Mechanism (show a mechanism using curved arrows..)

Keyword(s): Zaitsev's rule, E1 mechanism, dehydration

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.