acid strength

a) Bromine is an electron withdrawing group (EWG), which makes nearby protons more acidic. So the carbonyl with two bromines is the most acidic.

b) To perform alpha bromination, the enolate (or enol) must first be formed, which then attacks Br₂.

The problem with doing this under basic conditions is that each successive bromination leads to a carbonyl that is more acidic, and so forms an enolate even more easily. So once this reaction starts, it's difficult to control.

Enols are less nucleophilic than enolates, so this react can be controlled under acidic conditions. (Br₂/H3O⁺ instead of Br₂/NaOH).

Hydrogens D and E are alpha to two carbonyls and so will be the more acidic than C, which is only alpha to one carbonyl. This is because the enolates that arise from deprotonate at D and E have more resonance forms than the enolate that arrises at C.

Because ketones are more electron withdrawing than esters, D will be more acidic than E. So far we have:

D > E > C > (other)

The carbanions that would arrise from deprotonation of carbons A and B both do not have any resonance forms, so we don't expect either to be acidic. But because A is next to a fluorine atom, which is an electron withdrawing group, A will be more acidic than B. So the overall order is:

(most acidic) D > E > C > A > B (least acidic)

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)

a) The pK_a of formic acid is less than that of acetic acid, so formic acid is the stronger acid.

The pK_a difference is about ~1, and pK_a is logorithmic, so formic acid is ~10x as acidic as acetic acid.

Why? Because formate (formic acid's conjugate base) is more stable than acetate (acetic acid's conjugte base).

Why? Probably because of EDG. Alkyl groups such as methyl as weakly electron donating. Carboxylate is already negatively charged, and adding an EDG (the methyl group in acetate) makes it less stable, so acetate is a strong base, and acetic acid is the weaker acid. Formate has an hydrogen, which is neither EDG nor EWG, so doesn't have this problem.

b) The Henderson–Hasselbalch equal is pH = pK_a + log (A^-/HA). In this case, A^- is ^-OAc (acetate) and HA is HOAc (acetic acid). So plugging in the numbers:

pH = pK_a + log (^-OAc/HOAc)

4.7 = 4.7 + log (^-OAc/HOAc)

0 = log (^-OAc/HOAc),

so (^-OAc/HOAc) =1, and ^-OAc = HOAc.

So the ratio of acid to conjugate base (acetic acid to acetate) would be about 1:1.

The math above illustrates a rule you might have learned in general chemistry- when pH = pK_a, the concentration of acid equals the concentration of conjugate base.

c) The pKa of methoxy acetic acid (~3.6) is less than that of formic acid (~3.8), which means methoxy acetic acid is the strong acid, so methoxy acetate must be more stable than formate. This implies that methoxy must be somewhat electron withdrawing (an EWG). 

This might be suprising because during the aromatic reactions chapter(s) (EAS) methoxy is considered an EDG. Oxygen is electron donating in terms of resonance. But remember that oxygen is also very electronegative. So in this case, it seems that the inductive EWG effect (electronegativity) outweighs the EDG effect.

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)

The benzylic proton (middle compound) is more acidic than the allylic proton (left compound) because its conjugate base is more stable. This is because it has more resonance forms.

Cyclopentadiene (right compound) is the strongest of the three because it has the most stable conjugate base. Why is it the most stable? Because it's aromatic! To be aromatic, a compound must:

1. be cyclic and planar
2. be sp² hybridized
3. Have a Huckel number of pi electrons- 2, 6, 10, 14, etc.

The cyclopentadienyl anion meets all of these criteria, and so is aromatic, and very stable.

Compounds 1 and 2 are more acidic than compound 3 because their conjugate bases have more resonance forms than that of compound 3.

Compound 1 is more acidic than compound 2 because the resonance form of its conjugate base has two oxygen atoms and is more stable than that of compound 2, which has one oxygen and one nitrogen; a negative oxygen atom is more stable than a negative nitrogen atom.

Phenol is more acidic than cyclohexanol because the conjugate base of phenol (phenolate) has resonance while the conjugate base of cyclohexanol does not.

Thiophenol is more acidic than phenol because sulfur is larger than oxygen, and so RS^- is more stable than RO^-.