Organic Chemistry 202-SNU-RE • CEGEP, Montreal

Revision sheet: alcohols, phenols and ethers (202-SNU-RE)

The alcohol is the pivotal functional group of organic chemistry: we arrive at it from alkenes, carbonyl compounds and Grignard reagents, and we leave it toward alkenes, halides, aldehydes and acids. Almost all the syntheses of the course go through it.

This sheet leaves the lecture material in your course notes and gathers the decisions that pay off: the class above all, the choice of oxidant, the transformation of the $-OH$ into a good leaving group, and the medium that decides how an epoxide opens.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next, tells you which chapter to tackle next and lets you ask for the method or exercise you are missing. Create your space, an email is enough.

The thread of the chapter

Three questions decide everything: which CLASS of alcohol, which leaving group we build from the $-OH$ , and in which medium we open an epoxide.

This chapter is part of 202-SNU-RE, Organic Chemistry

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Classes of alcohols
Hydrogen bonding
Acidity of alcohols
Acidity of phenols
Markovnikov hydration
Hydroboration-oxidation
Acid dehydration
Zaitsev's rule
Controlled oxidation
Jones reagent and PCC
Tosylate
Epoxide ring opening

The essentials

Classes, and who associates with whom

• An alcohol is PRIMARY, SECONDARY or TERTIARY according to the number of carbons bonded to the carbon bearing the $-OH$ .
• A PHENOL carries the $-OH$ directly on an aromatic ring and behaves quite differently; an ETHER is $R-O-R'$ .
• An alcohol is both a DONOR and an ACCEPTOR of hydrogen bonds; an ether is an acceptor ONLY.
• Hence boiling points of ethers close to those of alkanes of the same mass, and far below those of alcohols.

The alcohol has a hydrogen to donate and a lone pair to offer: two molecules attach. The ether has nothing to donate, and two molecules ignore each other.

The class is read on the carbon bearing the OH group, not on the whole molecule. It is what decides oxidation, the substitution mechanism and the ease of dehydration.

Acidity, from the most to the least acidic

• $pK_{a}$ values to know: ethanoic acid $4.76$ ; phenol $10.0$ ; water $15.7$ ; ethanol $16.0$ ; $2$ -methylpropan- $2$ -ol $18.0$ .
• Phenol is SIX orders of magnitude more acidic than an alcohol, because its conjugate base is delocalized over the whole ring.
• Among alcohols, acidity DECREASES from methanol to the tertiary alcohol: donor alkyl groups destabilize the alkoxide.
• A withdrawing group on the ring strengthens the acidity of phenol further: $7.15$ for $4$ -nitrophenol, $0.38$ for trinitrophenol.

On the

pK_{a}

scale, phenol sits between the carboxylic acid and water: it is far more acidic than an alcohol, without being a carboxylic acid.

Acidity is never read on the acid but on the STABILITY of its conjugate base. It is the same rule as for carboxylic acids, and it explains every gap in the table.

Four routes to an alcohol

• Acid hydration of an alkene: MARKOVNIKOV, goes through a carbocation, so rearrangements are possible.
• Oxymercuration-demercuration: Markovnikov WITHOUT rearrangement.
• Hydroboration-oxidation: ANTI-Markovnikov and SYN addition.
• Reduction of a carbonyl compound by $NaBH_{4}$ or $LiAlH_{4}$ , or addition of a Grignard reagent.

The same alkene, two reagents: acid hydration puts the

-OH

on the more substituted carbon, hydroboration-oxidation on the terminal carbon.

The first two routes give the same product, but the second avoids rearrangement: that is exactly why it exists, and the question chooses it when a rearrangement threatens.

What we do next: dehydrate, oxidize, make it leave

• DEHYDRATION by concentrated $H_{2}SO_{4}$ at $180$ °C: $E1$ mechanism, reactivity tertiary > secondary > primary, ZAITSEV's rule.
• At $140$ °C on a primary alcohol, the symmetrical ETHER is obtained rather than the alkene.
• OXIDATION: primary to aldehyde with PCC, to acid with dichromate; secondary to ketone; tertiary: NOTHING.
• $HO^{-}$ is a poor leaving group: we PROTONATE, we TOSYLATE, or we go through $SOCl_{2}$ and $PBr_{3}$ .

The three classes facing the oxidant: only carbons still carrying a hydrogen are oxidized, which totally excludes the tertiary alcohol.

Tosylation does NOT break the $C-O$ bond: the configuration of the carbon is therefore retained in this step, and inversion takes place only at the time of the substitution.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Making ether molecules associate with each other

1.5 marks, and a reversed ranking of boiling points

What not to write

“The ether carries an oxygen, so it forms hydrogen bonds and boils high.”

What to write

“The ether is an ACCEPTOR only: it has no hydrogen on its oxygen, hence no association between its own molecules.”

Two ethers do not associate, for lack of a hydrogen to donate; with water, the oxygen of the ether ACCEPTS the hydrogen bond, hence its solubility.

Why: A hydrogen bond needs a donor AND an acceptor. The ether can accept the bond of an alcohol or of water, which explains its solubility, but it cannot associate with itself.

2. Believing the tertiary alcohol is the most acidic

1.5 marks, and the whole order of the table

What not to write

“The more alkyl groups, the more the alkoxide is stabilized, so the tertiary alcohol is the most acidic.”

What to write

“Alkyl groups are DONORS: they destabilize the alkoxide. The order is methanol $15.5$ , ethanol $16.0$ , tertiary $18.0$ .”

Why: An anion is stabilized by WITHDRAWING groups, which disperse its charge. A donor group does the opposite and concentrates the charge on the oxygen.

3. Oxidizing a tertiary alcohol

2 marks, and an invented product in a synthesis chain

What not to write

“Dichromate oxidizes all alcohols, so the tertiary alcohol gives a ketone.”

What to write

“The carbon bearing the OH group has no hydrogen left: there is nothing to remove, and the tertiary alcohol does not oxidize.”

The carbon bearing the OH group of the secondary alcohol keeps a circled hydrogen: oxidation is possible. That of the tertiary alcohol has none: nothing to remove, no oxidation.

Why: Controlled oxidation removes a hydrogen from the carbon bearing the OH group and one from the oxygen. With no hydrogen on the carbon, the operation is impossible, which gives a test for recognizing the classes.

4. Over-oxidizing a primary alcohol

2 marks, and the product announced is not the one obtained

What not to write

“We want the aldehyde, so we use dichromate in acid medium.”

What to write

“Dichromate goes all the way to the carboxylic ACID: to stop at the aldehyde, PCC in anhydrous medium is needed.”

The same primary alcohol: anhydrous PCC stops at the aldehyde, dichromate in aqueous medium goes all the way to the acid.

Why: The aldehyde formed hydrates in aqueous medium and is immediately re-oxidized. Anhydrous PCC prevents this hydration and so stops the reaction halfway.

5. Making a hydroxide ion leave as it is

2 marks, and a reaction that does not take place

What not to write

“The alcohol reacts with bromide: the $HO^{-}$ leaves and the bromine replaces it.”

What to write

“ $HO^{-}$ is a strong base, hence a poor leaving group: it must first be protonated, tosylated, or converted with $SOCl_{2}$ .”

Why: A good leaving group is a weak base. Converting the $-OH$ into $-OH_{2}^{+}$ or into a tosylate turns it precisely into a weak base, and that is the whole purpose of these reagents.

6. Believing that tosylation inverts the configuration

2 marks on the stereochemistry question

What not to write

“The tosylate forms by substitution at the carbon, so with inversion.”

What to write

“The $C-O$ bond is NOT broken: tosylation retains the configuration, and inversion happens only at the next step.”

The wedge stays hashed after tosylation: the

C-O

bond is untouched. Only the

S_{N}2

that follows turns the carbon over.

Why: It is the $O-H$ bond that is attacked, not the carbon-oxygen bond. The asymmetric carbon is therefore never touched in this step.

7. Applying Markovnikov to hydroboration

2 marks, and the regioisomer announced is the other one

What not to write

“Hydroboration adds water, so the $-OH$ goes on the more substituted carbon.”

What to write

“Hydroboration is ANTI-Markovnikov: the $-OH$ goes on the LESS substituted carbon, and the addition is syn.”

Why: Boron, bulky, attaches to the least hindered carbon, and oxidation then replaces the boron by $-OH$ while keeping the position. No carbocation is involved, so no rearrangement either.

8. Forgetting the rearrangement in a dehydration

2 marks, and it is the question that distinguishes

E1

from

E2

What not to write

“The secondary alcohol dehydrates and gives the alkene right next to it.”

What to write

“The mechanism is $E1$ : the carbocation formed can rearrange before losing its proton, and the major alkene comes from the rearranged cation.”

The secondary cation rearranges before losing its proton: a hydride migrates, the charge passes to the tertiary carbon, and the major alkene follows from it.

Why: A carbon skeleton that changes during the reaction is the signature of a carbocation intermediate. This is even how the mechanism is proved experimentally.

9. Opening an epoxide on the wrong side

2 marks, and the regioisomer is wrong half the time

What not to write

“The nucleophile always attacks the less crowded carbon.”

What to write

“In BASIC medium yes, it is a classic $S_{N}2$ ; in ACID medium, it attacks the MORE substituted carbon.”

The same epoxide, two media: the attack aims at the less substituted carbon in base, the more substituted in acid, and stays ANTI in both cases.

Why: In acid medium, the protonated oxygen polarizes the bond so strongly that the more substituted carbon carries a partial charge close to that of a carbocation. It is then the one that is attacked.

Which method to choose

Which oxidant, for which class

We read the class of the alcohol, then the product wanted.

If primary alcohol, and we want the ALDEHYDE → PCC in anhydrous medium, which stops there

Example: dichromate would go all the way to the acid
If primary alcohol, and we want the carboxylic ACID → dichromate and sulfuric acid, or permanganate

Example: the aldehyde is only an intermediate
If secondary alcohol → ketone, with any of these oxidants

Example: the ketone is not over-oxidized
If tertiary alcohol → no oxidation: the carbon has no hydrogen left

Example: it is a test for recognizing the class
If phenol → oxidation to a quinone, which is not an ordinary controlled oxidation

Example: a special case, not to be confused with alcohols

Controlled oxidation requires a hydrogen on the carbon bearing the OH group. This single criterion explains the first four branches without anything to learn by heart.

How to make the hydroxyl group leave

We look at what we want to obtain and whether the stereochemistry matters.

If we want a halide and the stereochemistry does not matter → protonate with concentrated $HX$

Example: on a tertiary alcohol, the reaction is immediate
If we want to control the stereochemistry → tosylate first: the configuration is retained, then an $S_{N}2$ inverts it once

Example: $TsCl$ in pyridine
If we want a chloride without rearrangement → $SOCl_{2}$

Example: the by-products are gases, hence easy to remove
If we want a bromide without rearrangement → $PBr_{3}$

Example: same logic as thionyl chloride
If we want an alkene → concentrated sulfuric acid when hot, accepting Zaitsev and a possible rearrangement

Example: $180$ °C

On a primary alcohol, $HX$ works poorly and concentrated acid gives an ether rather than an alkene below $150$ °C. The sulfur and phosphorus reagents exist precisely for these cases.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Choosing an alcohol preparation and justifying it

When to use it: The question asks you to prepare a specific alcohol from an alkene, and expects a choice of reagent.

1 Locate which carbon must carry the $-OH$ in the desired product.
2 Compare with the position Markovnikov would give: same carbon or the other.
3 Choose the route: acid hydration or oxymercuration for Markovnikov, hydroboration for the opposite.
4 Check whether a rearrangement threatens; if so, rule out acid hydration.

Concluding sentence

“The $-OH$ must occupy the less substituted carbon: we use hydroboration followed by oxidation, which is anti-Markovnikov and syn.”

The trap: Choosing acid hydration on a substrate where a secondary carbocation could become tertiary: the product changes skeleton.

Marking: 1 mark the targeted position, 1 mark the reagent chosen, 1 mark the justification of the rearrangement ruled out.

Justifying an acidity by the conjugate base

When to use it: The question asks you to rank acidities, or to explain the gap between a phenol and an alcohol.

1 Write the two conjugate bases, marking the charge.
2 Look for the resonance forms of each that delocalize this charge.
3 Add the inductive effect of the substituents, withdrawing or donating.
4 Conclude: the most stabilized base corresponds to the strongest acid.

Concluding sentence

“The phenoxide delocalizes its charge over three ring carbons, whereas the ethoxide keeps it entirely on the oxygen: phenol is therefore six orders of magnitude more acidic.”

The trap: Comparing the neutral molecules with each other: it is the stability of the ANION that decides, never that of the acid.

Marking: 1 mark the conjugate bases, 1 mark the delocalization, 1 mark the numerical conclusion.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Does the class allow the reaction?

Before writing an oxidation product, check that a hydrogen remains on the carbon bearing the $-OH$ .

A ketone obtained from a tertiary alcohol is impossible.
Has the carbon skeleton changed?

Compare the chains of the reactant and the product. A change of skeleton is possible only if a carbocation was involved.

A hydroboration that gave a rearranged skeleton is wrong: it goes through no cation.
Is the leaving group a weak base?

Look at what leaves: a tosylate ion, a halide or water are fine; a hydroxide or alkoxide ion is not.

A substitution written with direct departure of $HO^{-}$ does not take place.
Does the regioisomer match the medium?

For an epoxide opening, reread whether the medium is acidic or basic before placing the nucleophile.

In basic medium, attack at the more substituted carbon is wrong.

The typical problem, taken apart

Two alcohols from the same alkene

We have $2$ -methylbut- $1$ -ene. We want to prepare the tertiary alcohol on the one hand, the primary alcohol on the other.

Choose the reagents, justify each orientation, and say which of the two routes risks a rearrangement.

The 2-methylbut-1-ene of the question: the double bond joins a terminal

CH_{2}

to a carbon that already carries two alkyl groups.

Step 1

Tertiary alcohol: hydration in acid medium, or oxymercuration-demercuration.

Why

Both follow Markovnikov and put the $-OH$ on the more substituted carbon, the one that already carried two alkyl groups.

Step 2

The carbocation formed is already TERTIARY: no rearrangement can improve it.

Why

A rearrangement takes place only if it gains a step of stability. Here, the top of the ladder is reached from the first step, so acid hydration is safe.

Step 3

Primary alcohol: hydroboration with $BH_{3}$ , then oxidation with $H_{2}O_{2}$ in basic medium.

Why

Hydroboration is anti-Markovnikov: the bulky boron attaches to the terminal carbon, and oxidation replaces it by $-OH$ without moving it.

Step 4

The addition is SYN: the hydrogen and the hydroxyl group arrive on the same side of the plane.

Why

This is the stereochemical difference with acid hydration, which goes through a planar carbocation and loses all information about the face.

Step 5

Check: both products have the same molecular formula, $C_{5}H_{12}O$ , with a degree of unsaturation of zero.

Why

Both routes add water across the same double bond. Only the POSITION of the $-OH$ changes, so the molecular formula must be identical.

Step 6

Later oxidation: the primary alcohol will give an aldehyde then an acid, the tertiary one will give nothing.

Why

This is the test that distinguishes the two products in the laboratory, and it is often the next question.

The conclusion, written out

“Acid hydration gives the tertiary alcohol, with no risk of rearrangement since the carbocation is already tertiary; hydroboration followed by oxidation gives the primary alcohol, with syn addition.”

The classic mistake on this problem: Using hydroboration in the belief of obtaining the Markovnikov alcohol: it gives exactly the other regioisomer.

Learn by heart

• The alcohol is a donor AND an acceptor of hydrogen bonds; the ether is an acceptor ONLY, so it boils low.
• $pK_{a}$ : ethanoic acid $4.76$ , phenol $10.0$ , water $15.7$ , ethanol $16.0$ , tertiary alcohol $18.0$ .
• Among alcohols, acidity DECREASES as the number of alkyl groups increases.
• Markovnikov: acid hydration or oxymercuration. Anti-Markovnikov and syn: hydroboration-oxidation.
• Oxidation: primary to aldehyde with PCC or to acid with dichromate; secondary to ketone; tertiary NOTHING.
• $HO^{-}$ does not leave on its own: protonate, tosylate, or use $SOCl_{2}$ and $PBr_{3}$ . Tosylation RETAINS the configuration.
• Epoxide: in basic medium attack the LESS substituted carbon, in acid medium the MORE substituted one, and always from the back.

Frequently asked questions

Why does an ether boil much lower than an alcohol?

Because it cannot form hydrogen bonds with its own molecules. A hydrogen bond needs a donor, that is a hydrogen carried by an oxygen, and the ether has none. It remains an acceptor, which explains its solubility in water, but its molecules do not attach to each other.

Why is phenol more acidic than an alcohol?

Because its conjugate base, the phenoxide, delocalizes its negative charge over three carbons of the aromatic ring, which stabilizes it greatly. The alkoxide, for its part, keeps all the charge on its oxygen. The gap reaches six pKa units, a factor of one million on the acidity constant.

Why does a tertiary alcohol not oxidize?

Because controlled oxidation removes a hydrogen from the carbon bearing the hydroxyl group, and this carbon has none left: its four bonds go to three carbons and the oxygen. This is what allows a dichromate solution to be used as a test for recognizing the classes of alcohols.

How do you obtain an aldehyde without going all the way to the acid?

By using pyridinium chlorochromate, called PCC, in an anhydrous solvent. In aqueous medium, the aldehyde formed hydrates and the hydrated product is immediately re-oxidized to a carboxylic acid. The absence of water blocks this second step and allows the aldehyde to be isolated.

On which side does an epoxide open?

It depends on the medium. In basic medium, the nucleophile attacks the less crowded carbon, as in a one-step substitution. In acid medium, the protonated oxygen strongly polarizes the bond and it is the more substituted carbon that is attacked. In both cases, the attack is from the back, hence anti.

Practise it

Corrected exercises: Alcohols, phenols and ethers, 202-SNU-RE

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

15 corrected exercises
150 points
225 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

Previous sheet Aromatic compounds Next sheet Aldehydes and ketones

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-snu-alcohols-ethers. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

Looking for a 202-SNU-RE tutor in Montreal?

Get in touch for a first session. The whole chapter comes down to one question asked at the right moment, “can this leaving group really leave?”, and that is the method we set up here.

Book a session WhatsApp

Site by Studio Squalli

Revision sheet: alcohols, phenols and ethers (202-SNU-RE)

The essentials

Classes, and who associates with whom

Acidity, from the most to the least acidic

Four routes to an alcohol

What we do next: dehydrate, oxidize, make it leave

The mistakes that cost marks

1. Making ether molecules associate with each other

2. Believing the tertiary alcohol is the most acidic

3. Oxidizing a tertiary alcohol

4. Over-oxidizing a primary alcohol

5. Making a hydroxide ion leave as it is

6. Believing that tosylation inverts the configuration

7. Applying Markovnikov to hydroboration

8. Forgetting the rearrangement in a dehydration

9. Opening an epoxide on the wrong side

Which method to choose

Which oxidant, for which class

How to make the hydroxyl group leave

How the answer is expected to be written

Choosing an alcohol preparation and justifying it

Justifying an acidity by the conjugate base

Check before you hand in

The typical problem, taken apart

Two alcohols from the same alkene

Learn by heart

Frequently asked questions

Why does an ether boil much lower than an alcohol?

Why is phenol more acidic than an alcohol?

Why does a tertiary alcohol not oxidize?

How do you obtain an aldehyde without going all the way to the acid?

On which side does an epoxide open?

Corrected exercises: Alcohols, phenols and ethers, 202-SNU-RE

See also

Looking for a 202-SNU-RE tutor in Montreal?