Textbook: Carey and Giuliano 8th Ed. (2010)

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"

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 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.

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.

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.