Textbook: Loudon 5th Ed. (2009)

Chapter 10: The Chemistry of Alcohols And Thiols

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Individual Problems

Individual Problems

Problem # 678

Draw the structure of the major organic product from each reaction sequence.

Solution

The purpose of this problem is to illustrate that there are often several routes to prepare a single molecule.

For the left route:

Br₂/FeBr₃ adds a bromine to benzene via EAS.
Mg/ether converts bromobenzene to phenyl Grignard (a nucleophile).
Phenyl Grignard attacks the aldehyde.
A protic workup gets rid of all negative charges (and quenches the Grignard reagent).

For the right route:

Benzene undergoes Friedel-Crafts acylation (an EAS reaction) to form the ketone.
The ketone is then reduced to an alcohol using NaBH₄.

Note that both methods add a carbon chain to benzene, a step that is quite common in synthesis problems!

Problem Details

MendelSet practice problem # 678 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Electrophilic Aromatic Substitution, Alcohol Oxidation Reactions (PCC, Jones Reagent)

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

Keyword(s): hydride reduction

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 # 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 # 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 # 673

Show how each compound can be prepared from an alkene containing 3 carbons (or less).

Each answer will involve the reaction of a Grignard with either a carbonyl or epoxide.

Note: epoxides are prepared from alkenes using a peroxy acid (epoxidation) such as mCPBA.

Solution

The trick to synthesis problems in second semester organic chemistry to recognize that alcohols ARE ketones ARE carboxylic acids. What do I mean? Alcohols, ketones/aldehydes, and carboxylic acids can all be easily converted using PCC or Jones Reagent (NaCr₂O₇/H₂SO₄).

For example, for a), the product is a ketone, but it may as well be an alcohol, because alcohols can be converted to ketones with PCC.

b) is similar, except the position of the alcohol (one away from the "bond cut", instead of directly connected to the cut as in a) ) indicates the starting material was an epoxide and not a carbonyl.

c) is just like b), except instead of PCC, use Jones Reagent to oxidize the alcohol all the way to a carboxylic acid.

Problem Details

MendelSet practice problem # 673 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alcohol Oxidation Reactions (PCC, Jones Reagent)

Question Type: Synthesis (show how to prepare X from Y..)

Keyword(s): epoxides, oxidation

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 # 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 # 679

Compound A (C₅H₁₂O) is oxidized using aqueous chromium (Jones reagent) to compound B (C₅H₁₀O₂), which is then treated with methanol under acidic conditions to yield compound C (C₆H₁₂O₂) and water.

The ¹H NMR of compound C is shown below. Determine the structures of compounds A, B, and C.

Solution

Let's solve this NMR structure elucidation problem using steps similar to those used in problem 662.

1. Are there any hints?

Compound A has one oxygen and after treatment with aqueous chromium becomes compound B, which has two oxygens. This means A is probably an alcohol, B is probably a carboxylic acid.

Compound B is then treated with methanol under acidic conditions to form compound C. These are conditions for a Fischer esterification, so C is probably the methyl ester.

2. How many IHD are there?

Compound A: C₅H₁₂O = C₅H₁₂ should be C₅H₁₂ (C_nH_2n+2) so 0 IHD.

Compound B: C₅H₁₀O₂ = C₅H₁₀ should be C₅H₁₂. Missing 2H, so 1 IHD.

Compound C: C₆H₁₂O₂ = C₆H₁₂ should be C₆H₁₄. Missing 2H, so 1 IHD.

These IHD counts fit our assumptions from part 1).

3. Draw some structures and eliminate, learn, repeat.

Some clues from the NMR:

The isopropyl splitting pattern is present: d(6) (signal c at ~0.9 ppm) and multiplet(1) (signal b at ~2.4 ppm).
The s(3) at ~3.7 ppm is probably the methyl group from the methyl ester.
We know from before we have one IHD, and it's probably an ester.

So start drawing structures and eliminate those that don't fit the data!