All Practice Problems

Problem # 699

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

All carbon sources must contain three carbons or less.

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

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 # 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 # 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 # 708

The overall mechanism for Fischer esterification is shown below. This isn't a real mechanism, just an outline.

Methanol (the nucleophile) attacks the carbonyl carbon, forming a tetrahedral intermediate, which then loses a water to reform the carbonyl. This mechanism is called nucleophilic acyl substitution.

Use curved arrows to draw a full mechanism for this reaction. I've included structures for you to use as a guide.

This reaction takes place under acidic conditions, so the mechanism you draw will be similar to those in problem 706.

Solution

Acidic mechanisms only appear complicated because they contain several proton transfer steps.

Nucleophilic acyl substitution mechanisms have only three real steps- the "up, down, and kick."

First, the nucleophile attacks the carbonyl carbon, forming a tetrahedral intermediate (the "up").

Then the carbonyl reforms (the "down") and a leaving group leaves (the "kick").

Problem Details

MendelSet practice problem # 708 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

Subject Topic(s): Carboxylic Acids (Inductive Effects, Esterification), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution, Fischer esterification, up-down-kick carbonyl mechanism

Problem # 710

Rank the carbonyls A-D below in order of decreasing electrophilicity (reactivity with nucleophiles).

(1 = Most reactive). Explain your reasoning.

Solution

The carbonyl carbon is electrophilic because it has a partial positive charge.

Are there any groups that make a carbon more positive? Yes, electron withdrawing groups (EWG). They pull away electron density, which increases electrophility. So C is the most reactive.

Conversely, electron donating groups (EDG) add electron density, and so make the carbonyl carbon less positive, and less electrophilic. Alkyl groups (carbon chains) are mildly EDG so ketone B will be less reactive than C.

Hydrogen is neither a EDG or EWG, so the aldehyde A will be in between the B and C. Also, the hydrogen is very small, so the carbonyl carbon is easily to get to (less steric bulk to block an attack).

Ester D is the least reactive because it has a resonance (the lone pair on the oxygen gets involved).

So overall, the order form most reactive to least reactive is C > A > B > D.