Alcohol Oxidation Reactions (PCC, Jones Reagent)

Let's solve this NMR structure elucidation problem using steps similar to those used in problem 662.

1. Are there any hints?

Compound A has one oxygen and after treatment with aqueous chromium becomes compound B, which has two oxygens. This means A is probably an alcohol, B is probably a carboxylic acid.

Compound B is then treated with methanol under acidic conditions to form compound C. These are conditions for a Fischer esterification, so C is probably the methyl ester.

2. How many IHD are there?

Compound A: C₅H₁₂O = C₅H₁₂ should be C₅H₁₂ (C_nH_2n+2) so 0 IHD.

Compound B: C₅H₁₀O₂ = C₅H₁₀ should be C₅H₁₂. Missing 2H, so 1 IHD.

Compound C: C₆H₁₂O₂ = C₆H₁₂ should be C₆H₁₄. Missing 2H, so 1 IHD.

These IHD counts fit our assumptions from part 1).

3. Draw some structures and eliminate, learn, repeat.

Some clues from the NMR:

• The isopropyl splitting pattern is present: d(6) (signal c at ~0.9 ppm) and multiplet(1) (signal b at ~2.4 ppm).
• The s(3) at ~3.7 ppm is probably the methyl group from the methyl ester.
• We know from before we have one IHD, and it's probably an ester.

So start drawing structures and eliminate those that don't fit the data!

The purpose of this problem is to illustrate that there are often several routes to prepare a single molecule.

For the left route:

1. Br₂/FeBr₃ adds a bromine to benzene via EAS.
2. Mg/ether converts bromobenzene to phenyl Grignard (a nucleophile).
3. Phenyl Grignard attacks the aldehyde.
4. A protic workup gets rid of all negative charges (and quenches the Grignard reagent).

For the right route:

1. Benzene undergoes Friedel-Crafts acylation (an EAS reaction) to form the ketone.
2. The ketone is then reduced to an alcohol using NaBH₄.

Note that both methods add a carbon chain to benzene, a step that is quite common in synthesis problems!