carboxylic acids

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.

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.

This is a good mechanism to know (nucleophilic acyl substitution). If you don't understand it, see problems 706 and 708.

Vinyl benzene is a common starting material and intermediate in second semester organic chemistry synthesis problems, so you should be familiar how to make vinyl benzene from benzene (see problem 722).

The trick to many synthesis problem is to realize how many carbons need to be added, as that will determine which route you use.

a) Adding zero carbons: alcohol then oxidation

All you need is an alcohol added to the last carbon, which you can oxidize to a carboxylic acid. So from the alkene do an anti-Markovnikov addition of H₂O via hydroboration, followed by oxidation using Jones reagent (aqueous chromium, Cr⁶⁺ in acid).

b) Adding one carbon: nitrile (cyanide, CN)

When you add one carbon that screams "cyanide!" The most common way to add CN is using NaCN with an alkyl halide in an S_N2 reaction.

Because you want to add the extra carbon (CN) to the end of the chain, you want to do an anti-Markovnikov addition using HBr and perioxides. Then add NaCN to form the nitrile.

To convert a nitrile to a carboxylic acid, hydrolyze under acidic conditions (add water with acid, H₃O⁺).

c) Adding two carbons: epoxide

When you need to add two carbons, ethylene oxide is the way to go. But ethylene oxide is an electrophile, so first you must transform vinyl benzene into a nucleophile, specifically, a Grignard reagent.

We want the Grignard part (MgBr) on the end, so do an anti-Markovnikov addition with HBr/peroxides followed by Mg⁰/ether. Then add the epoxide which adds two carbons with an alcohol on the end. Finally, oxidize using Jones reagent.

Each carboxylic acid can be converted to its methyl ester via Fischer esterification: add methanol under acidic conditions (CH₃OH/H⁺).

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)