Alcohols and Ethers

An alcohol is R−OH. An ether is R−O−R′. Both have an oxygen with two lone pairs, but only the alcohol has a hydrogen on that oxygen, and that single bond is responsible for almost every difference between the two.

One bond, four consequences. The O−H is why alcohols react and ethers do not.

There are two supply routes, and they come from opposite directions. From alkenes, you are adding water and the only question is regiochemistry. From carbonyls, you are adding hydride and the only question is what carbonyl you started with.

From the carbonyl side, hydride reduction is the move. An aldehyde reduces to a primary alcohol and a ketone reduces to a secondary alcohol, and NaBH₄ handles both. Carboxylic acids and esters are less electrophilic, so they need the stronger reagent: LiAlH₄ takes them down to primary alcohols. That difference in strength is itself a common exam question, since it lets you reduce a ketone in a molecule while leaving an ester untouched.

NaBH₄ handles aldehydes and ketones. LiAlH₄ is needed for acids and esters.

Oxidizing an alcohol means removing a hydrogen from the carbinol carbon, the one bearing the OH. Count those hydrogens first and the whole rule falls out on its own.

1° → aldehyde or acid. 2° → ketone. 3° → no normal oxidation.

The reason PCC stops is worth knowing rather than memorizing. Strong aqueous oxidants convert the aldehyde into a hydrate, which has a fresh C−H on a carbon bearing OH, so oxidation continues. PCC runs anhydrous, so no hydrate forms and the aldehyde survives.

Heat an alcohol with H₂SO₄ and it loses water to give an alkene. The mechanism explains everything you need to predict: the acid protonates the OH, turning it into water, which is an excellent leaving group. Water leaves, giving a carbocation, and a base removes a neighboring proton to form the π bond.

Because 2° and 3° alcohols run through that free carbocation (E1), three consequences follow at once. The major product is the Zaitsev alkene, the more substituted and more stable one. Carbocation rearrangements are possible, so a 2° cation next to a quaternary carbon will shift before eliminating. And primary alcohols are the exception: they resist E1 and need harsher conditions, often proceeding with more E2 character.

Protonate, lose water, eliminate. E1 means Zaitsev plus possible rearrangements.

Worked example: 2-butanol with H₂SO₄ and heat gives 2-butene as the major product rather than 1-butene, because the internal alkene is more substituted. Contrast this with the E2 dehydrohalogenation from Lesson 9: same Zaitsev preference, but there you had a real leaving group already and no carbocation to rearrange.

This is the section that unlocks the rest of the course. OH⁻ is a strong base, and strong bases are poor leaving groups, so an alcohol cannot do SN2 or E2 as written. You have two ways to fix it: swap the OH for a halogen, or convert it into a sulfonate ester.

Halide routes invert the carbon. Tosylation keeps it, because the C−O bond survives.

Deprotonate an alcohol with NaH to make an alkoxide, then let that alkoxide attack an alkyl halide. The alkoxide is the nucleophile, the halide is the electrophile, and the mechanism is plain SN2.

Alkoxide plus a methyl or 1° halide. SN2 means the halide must be unhindered.

Because it is SN2, the halide has to be methyl or primary. A tertiary halide will not do backside attack, and since the alkoxide is also a strong base, you get E2 elimination and an alkene instead of your ether. The fix for a hindered target is always the same: reverse the assignment. To make tert-butyl methyl ether, use tert-butoxide as the nucleophile and methyl iodide as the electrophile, not the other way around.

Ethers are inert to base, to oxidants, and to most nucleophiles. What does break them is a strong acid with a good nucleophilic counterion: HI or HBr. The oxygen is protonated first, which turns the alkoxide half into a neutral alcohol, a perfectly good leaving group. Then iodide attacks a carbon and the C−O bond breaks.

Which carbon gets attacked depends on the substitution, exactly as you would expect from Lesson 8. If one side is methyl or primary, iodide does an SN2 there. If one side is tertiary, the protonated ether ionizes to a stable carbocation and the reaction goes SN1 at that carbon instead.

Protonate, then attack the less hindered carbon. Excess HI converts the alcohol too.

Worked example: CH₃−O−CH₂CH₃ plus HI gives CH₃I and CH₃CH₂OH, because iodide attacks the methyl side. With excess HI and heat, that ethanol is itself converted into CH₃CH₂I, so you end up with two alkyl iodides. Read the equivalents before you decide where to stop.

An epoxide is technically just a cyclic ether, but squeezing that C−O−C into a three-membered ring introduces serious ring strain. Opening the ring releases that strain, so epoxides react with nucleophiles that ordinary ethers would completely ignore.

The regiochemistry flips depending on conditions, and this is the single most tested idea in the ether half of the chapter.

Base goes for the roomy carbon, acid goes for the substituted one. Both are backside attack, so both are anti.