Textbook: Brown 6th Ed. (2011)

Chapter 16: Aldehydes and Ketones

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Individual Problems

Individual Problems

Problem # 669

In your own words, what is the major difference in the addition of a Grignard reagent to an oxidation state III carbonyl (ester/acid chloride) versus an oxidation state II carbonyl? (aldehyde/ketone)

Solution

Oxidation state III carbonyls (esters, acid chlorides) contain a built-in leaving group (such as ^-OR or ^-Cl) and so undergo nucleophilic acyl substitution reactions. The product is another carbonyl.

Oxidation state II carbonyls (aldehydes and ketones) do not contain a built-in leaving group and so undergo nucleophilc acyl addition reactions (instead of substitution). The product is an alcohol.

Because Grignards react with all carbonyls- esters and aldehydes/ketones- esters and acid chlorides will react twice with Grignards: once in a Nuc Acyl Sub mechanism to form a ketone, which will then react with another equivalent of Grignard in a Nuc Acyl Add mechanism to form an alcohol.

Problem Details

MendelSet practice problem # 669 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition

Problem # 668

Show how each alcohol can be prepared from a combination of a carbonyl and a Grignard reagent.

Solution

The trick to these retrosynthesis problems is to determine where the connections or "cuts" were made.

The carbonyl carbon becomes an alcohol after a Grignard reaction, so that's where the "cut" must be.

Note that there can often be more than one correct answer to these types of problems.

For a), adding propyl Grignard to acetone or methyl Grignard to 2-pentanone will result in the product.

We can also use two equivalents of methyl Grignard with 4-carbon ester, such as ethyl butanoate. Esters contains a build in leaving group (^-OR) and so react twice with Grignards.

For b), adding phenyl Grignard to cyclopentanone will do the job.

Problem Details

MendelSet practice problem # 668 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

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

Keyword(s): Grignard, retrosynthesis

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 # 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.

Problem Details

MendelSet practice problem # 710 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

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

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

Keyword(s): EDG and EWG, carbonyl trends

Problem # 712

Show what combination of amine and carbonyl would result in each imine or enamine.

Solution

For both imines and enamines, the sp² carbon comes from the carbonyl, and the nitrogen comes from the amine.

Problem Details

MendelSet practice problem # 712 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

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

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

Keyword(s): imines, enamines

Problem # 670

Draw out the mechanism for the addition of excess phenyl Grignard to the carbonyl compound below.

Solution

This carbonyl has two leaving groups attached to it- each of those oxygens can take part in a nucleophilic acyl substitution reaction and form a new carbonyl product.

First the Grignard attacks the oxidation state IV carbonyl carbon (4 oxygen bonds, so oxidation state 4). The carbonyl itself will act as a leaving group and form a tetrahedral intermediate. But tetrahedral intermediates don't last if there are any leaving groups attached to the carbon, so the ^-O will "come back down again", kick off an oxygen leaving group, and reform the carbonyl.

Then a second equivalent of Grignard will attack that carbonyl (an ester), and we will do another nucleophilic acyl substitution reaction to form yet another carbonyl.

Finally, the third carbonyl doesn't have any leaving groups built in (it's a ketone), so when the third equivalent of Grignard attacks it, it will do a nucleophilic acyl addition reaction, and the product will be an alcohol.

Notice that as the reaction progresses the oxidation state of the carbonyl carbon (number of oxygen bonds attached to it) goes down form 4 to 3 to 2 and then to 1.

Problem Details

MendelSet practice problem # 670 submitted by Matt on July 18, 2011.

Subject: Organic Chemistry

Subject Topic(s): Grignard and Masked Carbanions (alcohols from carbonyls), Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals), Nucleophilic Acyl Substitution (Carboxylic Acid Derivatives)

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

Keyword(s): nucleophilic acyl substitution, nucleophilic acyl addition

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 # 711

Show two ways of preparing the alkene below via the Wittig reaction starting from triphenyl phosphine (PPh₃).

Is one route better than the other? Why?

Solution

Alkenes can be prepared from a combination of ylide and aldehyde or ketone. This is the Wittig reaction.

One carbon on the alkene comes from the carbon bonded to the PPh₃ on the ylide, and the other carbon comes from the carbonyl.

There are usually two different ways to make an alkene via the Wittig reaction. So is one way better than the other?

Yes. Ylides come from the S_N2 reaction of PPh₃ with an alkyl halide. Because it's an S_N2 reaction, you want to use the least substituted alkyl halide! (1º is better than 2º). So of the two reactions below that would result in the desired alkene, the top method is better because it involves the less bulky alkyl halide.

This principle also holds true for the Williamson ether synthesis (problem 703). Less substituted alkyl halides are better for S_N2 reactions.

Problem Details

MendelSet practice problem # 711 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

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

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

Keyword(s): Wittig reaction

Problem # 714

Complete each synthesis below. All carbon sources must come from alkenes.

Each synthesis will involve protecting groups.

Solution

The protecting group most commonly used for aldehydes and ketones (in undergraduate orgo) is ethylene glycol.

It is put on using HOCH₂CH₂OH/H⁺ and removed with H₂O/H⁺.

For a), the reaction calls for the use of acetylide with an alkyl halide. But acetylides also react with carbonyls.

So before the alkyne is deprotonated using a strong base (such as NaNH₂), the carbonyl must be protected with ethylene glycol.

For b), the reaction calls for the addition of two equivalents of phenyl Grignard to the ester. The problem is that esters aren't as reactive as ketones (or aldehydes), so the Grignard would react with the ketone before it ever touched the ester! To prevent this, the ketone must be protected before Grignard is added.

Problem Details

MendelSet practice problem # 714 submitted by Matt on July 22, 2011.

Subject: Organic Chemistry

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

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

Keyword(s): protecting groups, acetals and ketals

Problem # 715

α-D-glucose is shown below.

a) Is α-D-glucose an acetal, hemiacetal, ketal, or hemiketal?

b) Draw the carbonyl form of α-D-glucose.

Solution

a) You can spot the carbonyl carbon because it's the one with two oxygens bonded to it (oxidation state II).

The carbonyl carbon has an -OH (alcohol) bonded to it, and not an -OR (ether), so it's a hemiacetal or a kemiketal.

The carbonyl carbon also has a hydrogen bonded to it, so it must have come form an aldehyde (instead of ketone). So it's a hemiacetal. (remember: aldehyde -> acetal, ketone -> ketal).

b) The hemiacetal carbon's -OH (or -OR in the case of a ketal or acetal) is the oxygen that used to be the carbonyl C=O.

Problem Details

MendelSet practice problem # 715 submitted by Matt on July 23, 2011.

Subject: Organic Chemistry

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

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

Keyword(s): acetals and ketals

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 # 713

When a carbonyl is treated with semicarbazide under acidic conditions an "imine" is produced called a semicarbazone.

Which of the two products below is the correct structure for a semicarbazone? Explain.

Solution

What this problem is really asking is "Which nitrogen on semicarbazide is more nucleophilic?"

The nitrogens bonded to the carbonyl on semicarbazide all have resonance forms, so their lone pairs aren't as available for nucleophilic attack.

The other nitrogen (on the left) is the most nucleophilic because it isn't involved in any resonance forms, so its lone pair isn't shared with another atoms.