Nucleophilic Acyl Addition (Aldehydes & Ketones, Imines/Enamines, Acetals/Ketals)

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.

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

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.

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.

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).

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^-).

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"

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.

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.

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.