My Problems

Submitted by Matt

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List of user's published problems.

Problem # 707

The overall mechanism for imine formation is shown below. (This isn't a real mechanism, just an outline)

Use curved arrows to draw the full mechanism for imine formation under acidic conditions. (I've added outlines of the intermediate structures for you to use as a guide). This mechanism is similar to that in problem 706 (carbonyl hydrate equilibria).

Solution

This reaction takes place under acidic conditions, so oxygen should never be negative (only neutral or positively charged).

Acidic mechanisms tend to be a bit of pain, because you have to include many proton transfer steps (protonation and deprotonation).

Problem Details

MendelSet practice problem # 707 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

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

Keyword(s): nucleophilic acyl addition, up-down-kick carbonyl mechanism, imines

Problem # 706

Carbonyls are in equilibrium with their hydrate forms. This equilibrium happens in both acid and base.

Let's go through this equilibrium under acidic conditions. Draw a mechanism using curved arrows for each reaction below.

Remember that under acidic conditions, most species are either neutral or positively charged, and rarely negatively charged. So your structures will contain either ROH or ROH₂⁺, but not RO^-.

a) Carbonyl to Hydrate (acidic)

b) Hydrate to Carbonyl (acidic)

Solution

The interconversion between a carbonyl (sp² carbon) and a tetrahedral intermediate (sp³ carbon) is the most common mechanism you will encounter in second semester organic chemistry.

You should be familiar drawing it under both acidic (this problem) and basic (problem 705) conditions.

In a), the carbonyl "goes up" to form a tetrahedral intermediate.

In b), an oxygen "comes back down" to reform the carbonyl and kick off a leaving group.

You will see this "up, down, kick" pattern in most mechanisms that involve attack at the carbonyl carbon, which is most of the reactions in second semester orgo!

a) Carbonyl to Hydrate (acidic) "UP"

Because this reaction takes place under acidic conditions, the carbonyl must protonate before the nucleophile attacks, to prevent oxygen from ever being negative (ROH instead of RO^-).

b) Hydrate to Carbonyl (acidic) "DOWN"

The leaving group must protonate before it leaves, so it doesn't leave as a negative molecule (H₂O instead of HO^-).

Problem Details

MendelSet practice problem # 706 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

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

Keyword(s): nucleophilic acyl addition, up-down-kick carbonyl mechanism, carbonyl hydrates

Problem # 705

Carbonyls are in equilibrium with their hydrate forms. This equilibrium happens in both acid and base.

Let's go through this equilibrium under basic conditions. Draw a mechanism using curved arrows for each reaction below.

Remember that under basic conditions, most species are either neutral or negatively charged, and rarely positively charged. So your structures will contain either ROH or RO^-, but not ROH₂⁺.

a) Carbonyl to Hydrate

Notice that no oxygen is ever positive during these basic mechanisms (always negative or neutral).

b) Hydrate to Carbonyl

Solution

The mechanisms in this problem and problem 706 are the most common mechanisms you will draw during second semester organic chemistry, and so it's a good idea to draw them out a few times.

In a), the nucleophile attacks the carbonyl carbon, and the double bond goes "up" to form a tetrahedral (sp³) carbon.

In b), one of the oxygen atoms acts as leaving group, and a lone pair on the other oxygen comes "down" to reform the double bond (and the sp² carbon).

You will see this "up, down, kick" pattern in most mechanisms that involve attack at the carbonyl carbon, which is most of the reactions in second semester orgo!

a) Carbonyl to Hydrate (basic conditions) "UP"

b) Hydrate to Carbonyl (basic conditions) "DOWN"

Problem Details

MendelSet practice problem # 705 submitted by Matt on July 21, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

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

Keyword(s): nucleophilic acyl addition, up-down-kick carbonyl mechanism, carbonyl hydrates

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 # 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 # 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 # 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 # 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 # 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 # 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 # 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!

Problem Details

MendelSet practice problem # 679 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Nuclear Magnetic Resonance Spectroscopy (NMR), Alcohol Oxidation Reactions (PCC, Jones Reagent)

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

Keyword(s): proton NMR, oxidation, Fischer esterification

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 # 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 # 674

Show a mechanism for the reduction of butyrolactone using LiAlH₄.

Solution

Hydride reagents such as LiAlH₄ and NaBH₄ behave like hydride nucleophiles (H^-), so that's what I used as shorthand. The real mechanism is very similar but involves aluminum coordinating to the oxygen.

Notice that the the ester will reform the carbonyl after the first hydride attacks. This is because esters have a built in leaving group, and so undergo nucleophilic acyl substitution reactions. The aldehyde that forms then undergoes a nucleophilic acyl addition reaction with the second equivalent of hydride.

Also note that you can't stop the reaction halfway at the aldehyde- LiAlH₄ will take an ester all the way down to an alcohol.

Problem Details

MendelSet practice problem # 674 submitted by Matt on July 19, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carbonyl Hydride Reductions (NaBH4, LiAlH4)

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

Keyword(s): hydride reduction

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.