Textbook: McMurry 8th Ed. (2011)

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 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!

To push an equilibrium to one side, add starting material and remove product. This is Le Chatelier's principle from general chemistry.

So to push this reaction to the right and form ether, add alcohol and remove ether and water as they form.

To push this reaction to the left and form alcohol, add water to ether and remove alcohol as it forms.

How do you "remove something as it forms?" Alcohols and ethers have (relatively) low boiling points, and can be removed by hooking up a vacuum line and condenser to your reaction. The ether (or alcohol) boils off under the reduced pressure, and then recondenses in a separate piece of glassware. (Sort of like in distillation.)

Water has a relatively high boiling point and so is difficult to remove under reduced pressure. To remove water, molecular sieves are used. They're like tiny sponges that only absorb water (and not other solvents), removing it from the reaction.

The trick to synthesis problems in second semester organic chemistry to recognize that alcohols ARE ketones ARE carboxylic acids. What do I mean? Alcohols, ketones/aldehydes, and carboxylic acids can all be easily converted using PCC or Jones Reagent (NaCr₂O₇/H₂SO₄).

For example, for a), the product is a ketone, but it may as well be an alcohol, because alcohols can be converted to ketones with PCC.

b) is similar, except the position of the alcohol (one away from the "bond cut", instead of directly connected to the cut as in a) ) indicates the starting material was an epoxide and not a carbonyl.

c) is just like b), except instead of PCC, use Jones Reagent to oxidize the alcohol all the way to a carboxylic acid.

This is just an S_N2 reaction. 

The first step is to turn one of the -OH groups into a better leaving group by having it pick up a proton. (I choose the hydroxyl on the right side).

Then the other -OH group can attack carbon 1, and the protonated hydroxyl group leaves as water.

Finally, the proton on the -OH group that acted as a nucleophile will "fall off" (deprotonate).

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

1. Are there any hints?

Yes, from the IR peaks. The starting material is a carbonyl (sharp IR peak at ~1,700 cm^-1) and the product is an alcohol (broad IR peak ~3,300 cm^-1).

2. How many IHD are in each compound? (also known as degrees of unsaturation or DBE).

C₆H₁₂O is the same as C₆H₁₂ (ignore oxygen) which should be C₆H₁₄ if fully saturated (C_nH_2n+2). It's missing 2 hydrogens so C₆H₁₂O has 1 IHD. This 1 IHD must be the carbonyl.

C12H18O is the smae as C12H18 which should be C12H26 if fully saturated. It's missing 8 hydrogens so it has 4 IHD. The benzene from the Grignard reagent must account for all 4 IHD.

3. Draw some C₆H₁₂O and C₁₂H₁₈O structures and elminate those that don't fit the data, then learn and repeat.

Things we know from the problem and NMR:

• The starting material is a carbonyl and we're adding a Grignard, so we expect the product to be an alcohol. The NMR shows a peak that disappears with D₂O addition, which also confirms that the product is an alcohol.
• The product must have a benzene ring because the reagent was phenyl Grignard. That accounts for the aromatic signals (~7 ppm) and the 4 IHD in the product.
• the NMR shows a doublet with an integration of 6 and a multiplet with an integration of 1. This is the splitting pattern on an isopropyl group.
• The NMR also shows a quartet with an integration of 2 and a triplet with an integration of 3. This is the splitting pattern of an ethyl group.

Once you have a few clues from the NMR, start drawing structures! And then elminiate those that do not fit the data (too many signals, wrong multiplicity/integrations etc.).