Textbook: Bruice 6th Ed. (2010)

The general rule is that the pK₁ or a dicarboxylic acid is lower (more acidic) than the pK_a of a regular carboxylic acid, and the pK₂ is a little higher (less acidic).

Why is this? Because carboxylic acids are electron withdrawing groups (EWG). They stabilize the conjugate base of an acid, making the acid more acidic. So Both D and A should be more acidic than C. Because A has its carboxylic acid groups closer together, the EWG effect is stronger, and it will be the most acidic.

B will be the least acidic of the group. Why? Because a carboxylate group has a negative charge, which is electron donating and destablizing to the conjugate base (is a molecule is already -1, it doesn't help to become -2.)

So the overall order is:

(strongest acid) A > D > C > B (weakest acid)

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.

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.

Nitriles react with Grignard reagents to form ketones, so the carbonyl carbon on each product must have came from a nitrile.

You also have the restriction that the longest carbon chain building must be three carbons.

a) Working backwards, the left side must have come from a nitrile, so the left side came from a Grignard.

How can we make each starting from propene?

To make the Grignard, prepare propyl bromide from propene via anti-Markovnikov HBr addition (HBr/peroxides), and then add Mg⁰/ether.

To make the nitrile, add sodium cyanide (NaCN) to propyl bromide.

Then add the nitrile and Grignard together, which forms the imine intermediate, which will hydrolyze to the ketone after you add water (H₃O⁺).

b) This problem is harder because you're limited to starting materials with three carbons or less, so we have to make three "cuts" instead of two, like in a).

Working backwards again, prepare propyl Grignard from propene as described in a), and then add ethylene oxide to form 1-pentanol. Form the 6-carbon nitrile by first converting the alcohol to a bromide using PBr₃, and then adding NaCN. Finally, add methyl Grignard followed by H₃O⁺ to form the desired ketone.

Carboxylic acids are much more acidic than alcohols (because of resonance in their conjugate bases), so A and D are the least acidic of the group. The conjugate base of D has resonance, so D is more acidic than A:

(everything) > D > A

Electronegative atoms increase acid strength (by stabilizing the conjugate base), and sp² carbons are more electronegative than sp³ carbons (due to higher s-character). So B (all sp³ carbons) will be the least acidic of the carboxylic acids, and H will be more acidic than B (two sp² carbons):

(everything) > H > B > D > A

Of the remaining 4 benzoic acids, C has the most EWG (two nitros) and so is the most acidic. F has no EWG and so is the least acidic. Both E and G have only one EWG and so will be similar in acidity, with E slightly more acidic than G because the EWG affect is strongest in the ortho position (closer to the action). So the overal order is:

(strongest acid) C > E > G > F > H > B > D > A (weakest acid)

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.