Ethers and Epoxides

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

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).