Organic Chemistry

Epoxides can open up in two different ways.

To add a nucleophile to the less substituted side of an epoxide, use basic conditions. This is done in #2 below.

To add a nucleophile to the more substituted side of an epoxide, use acidic conditions. This is done is #1 below.

Why do the conditions matter? Epoxides have two electrophilic carbons. Normally nucleophiles will preferentially attack the less substituted carbon, as they do in S_N2 reactions. Recall that S_N2 reactions usually happen with strong nucleophiles- that is, negative charges (basic conditions).

When an epoxide reacts under acidic conditions, the transition state has carbocation character, and so it's sort of like an S_N1 reaction. That is, instead of less substituted carbons being favored due to less steric bulk, more substituted carbons are favored do to a more stable carbocation. So acidic conditions cause an epoxide to open up on the more substituted side.

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.

Ethers react with 2 equivalents of H-X to form water and two equivalents of alkyl halide.

In this case, the ether was cyclic, so the ring had to open up.

The reaction can go through either an S_N1 or S_N2 mechanism. Since this was a primary ether, it will go through an S_N2 mechanism (the carbocation is too unstable for the reaction to go S_N1).

a) When you see 2 carbons and 1 oxygen, that the tell-tale sign that you're adding ethylene oxide (the simplest epoxide).

But that would only leave you with an alcohol. How do you get to the ether? Using the Williamson ether synthesis.

b) As I mentioned in problem 673, when you see an alcohol you are also looking at a carbonyl, because you can interconvert the two (alcohol to aldehyde/ketone using PCC, aldehyde/ketone to alcohol using NaBH₄).

To add the methyl group, convert the alcohol to a ketone (which is an electrophile), and then add methyl Grignard (a nucleophile). But once again, you are only left with an alcohol. How to convert it to an ether? The Williamson ether synthesis.

When you see an ether in a synthesis problem, remember the Williamson ether synthesis. It will come in handy.

KF is insoluble in propyl bromide or benzene, so the two compounds never "touch" each other, and no reaction takes place. (F^- and propyl bromide are in different phases and so don't come into contact).

But when the crown ether is added, it solvates the potassium ion, allowing the K⁺ and F^- ions to dissolve in the solvent, so the fluoride ion and propyl bromide are able to "talk" to each other, and the rection can take place.

Water is very polar, so the more polar a molecule is, the more soluble it will be in water.

Alcohols can hydrogen bond, and so will be the most polar, have the highest boiling point, and be the most soluble in water.

Aldehydes (carbonyls) can't hydrogen bond but the C=O double bond is very polar, so aldehydes are somewhat soluble in water.

The ether is the least polar of the three functional groups, and so will have the lowest boiling point, and be the least soluble in water.

So the overal trend for polarity, water solubility, and boiling point is:

alcohol > aldehyde > ether

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!

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).

Hydride reagents such as LiAlH₄ and NaBH₄ behave like hydride nucleophiles (H^-), so that's what I used as shorthand. The real mechanism is very similar but involves aluminum coordinating to the oxygen.

Notice that the the ester will reform the carbonyl after the first hydride attacks. This is because esters have a built in leaving group, and so undergo nucleophilic acyl substitution reactions. The aldehyde that forms then undergoes a nucleophilic acyl addition reaction with the second equivalent of hydride.

Also note that you can't stop the reaction halfway at the aldehyde- LiAlH₄ will take an ester all the way down to an alcohol.