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.

To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism.

In this mechanism, the nucleophile (methanol) becomes the -OCH₃ group in the ester.

At least 80% of second semester organic chemistry is two mechnanisms: nucleophilic acyl addition and nucleophilic acyl substitution. It's worth your time to become familiar with these mechanisms. See problems 705 (basic conditions), 706 (acidic conditions), 707, and 708.

Esters are prepared from a carboyxlic acid and an alcohol (Fischer esterification).

So the problem becomes, "how can you prepare this ester from an alochol and a carboxylic acid, each containing three carbons? (propene)"

The "pair" of carboxylic acid and alcohol to make this ester is propanoic acid and propanol.

From propene, to make propanol, just do an anti-Markovnikov H₂O addition using borane (BH₃ or B₂H₆) then hydrogen peroxide (H₂O₂).

To make propanoic acid, oxidize propanol to propanoic acid using Jones reagent (aqueous chromium).

You can also convert propanoic acid to its acid chloride using SOCl₂, and then add propanol.

To answer this problem, you must be familiar with the nucleophilic acyl substitution mechanism. (see problem 725)

Water reacts with an ester to form a carboxylic acid. This is what happens here. 

If there is only trace amounts of water, it's possible that the water reacts with the ester to form a carboxylic acid, and then goes back to reform the ester. But, in this process, leaves an isotopically labeled oxygen in the carbonyl position.

The Hoffman elimination first turns an amine into a better leaving group by methylating it with excess methly iodide, and then eliminates the methylated amine using base (such as ^-OH fomed from Ag₂O in H₂O). The products are alkenes.

This reaction is similar to an E2 elimination in that a beta proton is ripped off, and the alkene forms between alpha and beta. One notable difference from the E2 is that the less substituted product is favored, so Hoffman products are usually anti-Zaitsev.

a) There are three different types of beta protons, so three possible products can form. Product A is the least substituted alkene, so it's the major product.

b) There is only one possible product. The Hoffman elimination can't be done on aryl hydrogens (benzene rings).

When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).

Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.

sp² carbons are more electronegative (electron withdrawing) than sp³ carbons due to increased s-character. Electron withdrawing groups (EWG) make acids more acidic and bases less basic.

For this reason, piperidine (compound A) is the more basic than pyridine (compound B):

(more basic) A > B (less basic)

Nitro (-NO₂) is an EWG, so 4-nitropyridine (compound C) will be less basic than compound A.

Dimethylamine is an electron donating group (EDG), so it adds electron density to pyridine and increases basicity. So dimethylamino pyridine (DMAP, compound D) will be more basic than compound A:

(more basic) D > B > C (less basic)

So how does compound A compare with compound D? There is no way of predicting this based on EWG/EDG rules alone. After all, what should be more basic: piperidine, which has sp³ carbons (increases base strength), or DMAP, which has sp² carbons (decreases base strength) but also an EDG (increases base strength) ?

There's no way to tell. But it turns out that Piperidine is a little more basic than DMAP. So the overal order is:

(more basic) A > D > B > C (least basic)

a) The species on the left are much more stable than the species on the right because they have complete octets.

Nitrogen is less electronegative than oxygen, so a positive nitrogen (with a full octect) is more stable than a positive oxygen; oxygen is electronegative and so doesn't "like" being positive (have low electron density). 

The carbocation and the "nitrocation" are both very unstable because they only have 6 valence electrons. Because nitrogen is more electronegative than carbon, the nitrocation is even less stable than the carbocation, because its worse at handling a positive charge.

b) The cation intermediates that result from substitution at C-3 only contain carbocations, while the intermdiates from C-2 or C-4 substitution contain a "nitrocation," which is less stable than a regular carbocation, and so are not favored. So pyridine undergoes EAS reactions at C-3.

One more note: For substitution at C-2 and C-4, it's tempting to draw a resonance form where the positive charge is on the nitrogen, which is flanked by two double bonds. But this cyclic compound is too strained to exist, and so doesn't contribute to this analysis.

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.

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^-).

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.