Alcohols

In a nutshell

Alcohols contain the hydroxyl group, -OH. That one polar O-H bond lets them hydrogen bond, so they are less volatile than alkanes and the small ones dissolve in water.

Their reactions depend on which kind of alcohol you have (primary, secondary or tertiary). You need to know five reactions: combustion, oxidation with acidified dichromate, elimination of water to make alkenes, and substitution with halide ions to make haloalkanes.

Assumed knowledge: Bonding and structure: electronegativity, polar bonds and hydrogen bonding; Basic concepts of organic chemistry: naming, displayed and skeletal formulae.

Core content

Properties of alcohols: polarity and hydrogen bonding

Alcohols are a homologous series with the general formula C_nH_2n+1OH.

Oxygen is more electronegative than carbon and hydrogen, so the C-O and O-H bonds are polar:

• The O atom carries δ-.
• The H of the O-H carries δ+.
• The O atom also has two lone pairs.

So alcohol molecules can form hydrogen bonds: the δ+ H of an O-H on one molecule is attracted to a lone pair on the O of another molecule.

Alcohols still have London forces (induced dipole-dipole interactions) like alkanes. The hydrogen bonds are extra, and they are much stronger.

Properties of alcohols: low volatility compared with alkanes

Volatility is how easily a liquid turns into a gas. A less volatile liquid has a higher boiling point.

Alcohols are less volatile than alkanes of similar size. The mark-earning chain:

1. Alkanes have only London forces between molecules.
2. Alcohols have London forces and hydrogen bonds between molecules.
3. Hydrogen bonds are stronger than London forces.
4. So more energy is needed to overcome the intermolecular forces in an alcohol.
5. So the alcohol has a higher boiling point and lower volatility.

The gap is not just because the alcohol is heavier. Compare two molecules of almost the same M_r:

Boiling only separates the molecules. No covalent bonds break when an alcohol boils.

Properties of alcohols: solubility in water

Alcohols can form hydrogen bonds with water molecules, so small alcohols dissolve in water. Alkanes cannot hydrogen bond with water, so they are insoluble.

The hydrogen bonds work both ways:

• The δ+ H of the alcohol's O-H bonds to a lone pair on a water O.
• A δ+ H of water bonds to a lone pair on the alcohol O.

Solubility decreases as the carbon chain gets longer:

Why: the hydrocarbon chain is non-polar and cannot hydrogen bond with water. As it gets longer, it makes up more of the molecule, so the single -OH group has less effect.

Still don't get it? · why hydrogen bonding changes boiling point and solubility

Think of molecules in a liquid as people in a crowded room. Alkane molecules are like people who only brush shoulders: easy to walk away from. Alcohol molecules are like people holding hands with their neighbours. To leave the room (boil), each one has to let go first, which takes more effort.

Step by step:

1. The O in an O-H bond pulls the shared electrons towards itself, because O is more electronegative. So the H end is slightly positive (δ+) and the O end slightly negative (δ-).
2. The O also has two lone pairs of electrons sitting on it.
3. The δ+ H on one molecule is attracted to a lone pair on the O of the next molecule. That attraction is a hydrogen bond. It is between molecules, not inside one.
4. To boil, molecules must be pulled apart, so the hydrogen bonds (and the London forces) between them must be overcome. More, stronger forces means more energy, so a higher boiling point.
5. Water molecules also have δ+ H atoms and lone pairs, so an alcohol can "hold hands" with water. That is why ethanol dissolves. A long alkyl chain cannot do this, so long-chain alcohols are less soluble.

Exam version: alcohols have hydrogen bonds between molecules (as well as London forces), which are stronger, so more energy is needed to overcome them: higher boiling point, lower volatility. Alcohols are soluble because they form hydrogen bonds with water.

Properties of alcohols: primary, secondary and tertiary

Classify an alcohol by looking at the carbon atom bonded to the -OH group, and counting how many other carbon atoms are attached to it.

Two quick checks:

• Branching elsewhere in the chain does not matter. 2-methylpropan-1-ol, (CH₃)₂CHCH₂OH, is primary.
• Primary and secondary alcohols have at least one H on the carbon bonded to -OH. A tertiary alcohol has none. This decides how they oxidise.

Reactions of alcohols: combustion

Alcohols burn in plenty of oxygen (complete combustion) to form carbon dioxide and water:

CH₃CH₂OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)

The general equation for complete combustion is:

$\text{C}_n\text{H}_{2n+1}\text{OH} + \tfrac{3n}{2}\,\text{O}_2 \rightarrow n\,\text{CO}_2 + (n+1)\,\text{H}_2\text{O}$

When you balance, remember the alcohol already contains one O atom, so it needs less O₂ than the alkane with the same number of carbons.

In a limited supply of oxygen, incomplete combustion gives carbon monoxide instead:

CH₃CH₂OH(l) + 2O₂(g) → 2CO(g) + 3H₂O(l)

Reactions of alcohols: oxidation with acidified dichromate

The oxidising agent is acidified potassium dichromate(VI): K₂Cr₂O₇ with dilute H₂SO₄ (written Cr₂O₇^2-/H⁺).

When an alcohol is oxidised, the colour changes from orange to green:

• Orange: Cr₂O₇^2-, chromium in the +6 oxidation state.
• Green: Cr³⁺, chromium in the +3 oxidation state.
• The dichromate is reduced, and the alcohol is oxidised.

In equations, write the oxidising agent as [O]. When an alcohol is oxidised to an aldehyde or ketone, the [O] removes two H atoms (the H of the O-H and an H from the carbon bonded to it) as H₂O, and a C=O forms.

Reactions of alcohols: oxidising primary alcohols to aldehydes or carboxylic acids

A primary alcohol is oxidised first to an aldehyde, and the aldehyde can then be oxidised further to a carboxylic acid. You control which one you get with the conditions.

To make the aldehyde: distil it off as it forms.

CH₃CH₂OH + [O] → CH₃CHO + H₂O    (ethanol → ethanal)

• Heat the alcohol with the acidified dichromate in distillation apparatus.
• The aldehyde leaves the flask as a vapour as soon as it forms, so it is not oxidised further.
• This works because the aldehyde has the lowest boiling point in the mixture: aldehyde molecules have no O-H, so there is no hydrogen bonding between them.

To make the carboxylic acid: heat under reflux with excess oxidising agent.

CH₃CH₂OH + 2[O] → CH₃COOH + H₂O    (ethanol → ethanoic acid)

• The vertical condenser turns any vapour (including the aldehyde) back into liquid, which drips back into the flask.
• So the aldehyde stays in contact with the excess oxidising agent and is oxidised further to the carboxylic acid.
• Reflux lets you heat for a long time without losing volatile reactants or products.

The second step on its own is:

CH₃CHO + [O] → CH₃COOH

Setting up and using this glassware (Quickfit, anti-bumping granules, electric heater) is covered in Organic synthesis.

Still don't get it? · distil for the aldehyde, reflux for the acid

Imagine toast under a grill. Bread (the alcohol) becomes toast (the aldehyde), and if you leave it there it becomes charcoal (the carboxylic acid). If you want toast, you pull each slice out the moment it is done. If you want charcoal, you trap it under the grill and keep the heat on.

Step by step:

1. Oxidation happens one step at a time: primary alcohol, then aldehyde, then carboxylic acid. The aldehyde is the halfway stage.
2. The aldehyde only turns into the acid if it stays in the flask with the oxidising agent.
3. The aldehyde boils at a lower temperature than everything else in the flask (it has no O-H, so no hydrogen bonds between its molecules). So when you heat, it is the first thing to boil off.
4. Distillation: the vapour goes sideways into a condenser and is collected in a separate flask. The aldehyde has escaped the oxidising agent, so it stays an aldehyde. That is "pulling the toast out".
5. Reflux: the condenser points straight up, so every vapour condenses and falls back in. The aldehyde cannot escape, and with excess oxidising agent it is oxidised to the carboxylic acid. That is "keeping it under the grill".

Exam version: aldehyde: distil (as it forms); carboxylic acid: excess oxidising agent and heat under reflux. Always name the technique, not just "heat".

Reactions of alcohols: oxidising secondary alcohols to ketones

A secondary alcohol is oxidised to a ketone. The usual conditions are heating under reflux with acidified dichromate.

CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O    (propan-2-ol → propanone)

The ketone is not oxidised any further by acidified dichromate, so there is only one product to worry about.

Reactions of alcohols: tertiary alcohols resist oxidation

Tertiary alcohols are not oxidised by acidified dichromate. The mixture stays orange.

The reason: the carbon bonded to the -OH has no hydrogen atom on it. Oxidation needs to remove an H from that carbon (with the H from the O-H) to form C=O, and there is none to remove.

PAG7: qualitative analysis of organic functional groups

This PAG uses test-tube tests to identify functional groups. For alcohols, the test is warming with acidified potassium dichromate(VI).

Method:

1. Add a few drops of the unknown to a little acidified potassium dichromate(VI) solution in a test tube.
2. Warm the tube in a hot water bath (organic liquids are flammable, so no naked flame).
3. Watch for a colour change.

This test cannot tell a primary alcohol from a secondary one, because both turn the dichromate green. Use the conditions and the product to tell them apart: a primary alcohol can give an aldehyde or a carboxylic acid; a secondary gives a ketone.

Other PAG7 work at AS: bromine water to test for a C=C (Alkenes), and aqueous silver nitrate in ethanol to compare how fast haloalkanes hydrolyse (Haloalkanes).

Reactions of alcohols: elimination of water to form alkenes

Heating an alcohol with an acid catalyst, such as concentrated H₃PO₄ or concentrated H₂SO₄, removes a molecule of water and forms an alkene.

CH₃CH₂OH → CH₂=CH₂ + H₂O    (ethanol → ethene)

Key facts:

• It is an elimination reaction (also called dehydration, because water is removed).
• The -OH and an H from a neighbouring carbon are removed, and a C=C forms between those two carbons.
• The acid is a catalyst: it is not used up.
• The mechanism is not required.

Watch out for more than one alkene. If the -OH carbon has two different neighbouring carbons with H atoms, the C=C can form on either side. Then check each alkene for E/Z isomers.

Butan-2-ol gives three alkenes:

Reactions of alcohols: substitution with halide ions to form haloalkanes

Heating an alcohol with halide ions in the presence of acid, such as NaBr and H₂SO₄, swaps the -OH for a halogen atom. The product is a haloalkane.

The acid and the sodium halide react together first to make the hydrogen halide in the flask:

NaBr + H₂SO₄ → NaHSO₄ + HBr

The hydrogen halide then substitutes the alcohol:

CH₃CH₂OH + HBr → CH₃CH₂Br + H₂O    (ethanol → bromoethane)

You can also combine them into one equation:

CH₃CH₂OH + NaBr + H₂SO₄ → CH₃CH₂Br + NaHSO₄ + H₂O

Key facts:

• It is a substitution reaction: -OH is replaced by -Br.
• The mixture is usually heated under reflux, and the haloalkane is then separated and purified (see Organic synthesis).
• The mechanism is not required.

Worked examples

1. Writing oxidation equations with [O]

Task: write equations for the oxidation of butan-1-ol, butan-2-ol and 2-methylpropan-2-ol with acidified dichromate. Give the products and conditions.

Step 1: classify each alcohol by the carbon bonded to -OH.

• Butan-1-ol, CH₃CH₂CH₂CH₂OH: that carbon has one other carbon, so primary.
• Butan-2-ol, CH₃CH(OH)CH₂CH₃: two other carbons, so secondary.
• 2-methylpropan-2-ol, (CH₃)₃COH: three other carbons, so tertiary.

Step 2: write each equation. Count the [O] and check H and O balance.

Distil: CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O    (butanal)

Reflux, excess oxidising agent: CH₃CH₂CH₂CH₂OH + 2[O] → CH₃CH₂CH₂COOH + H₂O    (butanoic acid)

Reflux: CH₃CH(OH)CH₂CH₃ + [O] → CH₃COCH₂CH₃ + H₂O    (butanone)

2-methylpropan-2-ol: no reaction, the dichromate stays orange.

Step 3: check the balance on the carboxylic acid equation, which is where most errors happen.

• Left: C₄H₁₀O plus 2 O = C 4, H 10, O 3.
• Right: C₄H₈O₂ plus H₂O = C 4, H 10, O 3. Balanced.

Where marks go: the aldehyde is written CHO (not COH), the acid needs 2[O], and each equation has H₂O as a product.

2. Balancing a combustion equation

Task: write a balanced equation for the complete combustion of propan-1-ol.

1. Formula: C₃H₇OH, which is C₃H₈O.
2. Carbon: 3 C, so 3CO₂.
3. Hydrogen: 8 H, so 4H₂O.
4. Oxygen needed on the right: (3 × 2) + 4 = 10 O atoms.
5. The alcohol already supplies 1 O, so O₂ must supply 10 - 1 = 9 O atoms, which is 4½ O₂.

C₃H₇OH(l) + 4½O₂(g) → 3CO₂(g) + 4H₂O(l)

Doubling to clear the fraction is also fine:

2C₃H₇OH(l) + 9O₂(g) → 6CO₂(g) + 8H₂O(l)

The common slip is step 5: forgetting the O inside the alcohol and writing 5O₂.

3. Predicting every alkene from an elimination

Task: pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃, is heated with concentrated H₃PO₄. Name all the organic products.

1. The -OH is on C2. Its neighbouring carbons are C1 and C3, and both carry H atoms.
2. Removing H from C1 puts the C=C between C1 and C2: pent-1-ene, CH₂=CHCH₂CH₂CH₃.
3. Removing H from C3 puts the C=C between C2 and C3: pent-2-ene, CH₃CH=CHCH₂CH₃.
4. Check E/Z for each:
   • Pent-1-ene: C1 has two H atoms, so no E/Z isomers.
   • Pent-2-ene: C2 has H and CH₃; C3 has H and CH₂CH₃. Each carbon has two different groups, so E and Z isomers exist.
5. Answer: three alkenes: pent-1-ene, E-pent-2-ene and Z-pent-2-ene (plus water).

Each one forms by the same equation:

C₅H₁₁OH → C₅H₁₀ + H₂O

4. Model answer: comparing ethanol with propane

Task: explain why ethanol (M_r 46.0) has a much higher boiling point than propane (M_r 44.0), and why ethanol is soluble in water but propane is not.

A full answer makes each of these points, in this order:

1. Ethanol has a polar O-H bond (O is more electronegative than H), and O has lone pairs.
2. So ethanol forms hydrogen bonds between its molecules, as well as London forces.
3. Propane has only London forces, which are of similar size to ethanol's because the M_r values are almost the same.
4. Hydrogen bonds are stronger, so more energy is needed to overcome the intermolecular forces in ethanol.
5. Ethanol is soluble because it forms hydrogen bonds with water molecules.
6. Propane is non-polar and cannot form hydrogen bonds with water, so it does not dissolve.

Common exam mistakes

• Saying covalent bonds break when an alcohol boils. Boiling overcomes the intermolecular forces (hydrogen bonds and London forces). The O-H and C-H bonds stay intact.
• Leaving out "more energy". "Ethanol has hydrogen bonds" is not an explanation. Say hydrogen bonds are stronger than London forces, so more energy is needed to overcome them.
• Forgetting that alcohols also have London forces. The comparison is London forces plus hydrogen bonds (alcohol) vs London forces only (alkane).
• Drawing the hydrogen bond wrongly. It goes from the δ+ H of an O-H to a lone pair on an O of another molecule. Show the δ+/δ- and the lone pair. An H on a carbon never hydrogen bonds.
• Classifying by the number in the name. Butan-2-ol is secondary because its -OH carbon has two carbons attached, not because of the "2". 2-methylpropan-1-ol is primary, despite the branch.
• Writing an aldehyde as COH. Ethanal is CH₃CHO. CH₃COH is not accepted.
• Getting the [O] count wrong. Primary to aldehyde: [O]. Primary to carboxylic acid: 2[O]. Secondary to ketone: [O]. Water is always a product.
• Giving "heat" as the only condition. Name the technique: distil for the aldehyde, reflux with excess oxidising agent for the carboxylic acid.
• Incomplete colour changes. Write orange to green, not just "turns green". For a tertiary alcohol, write stays orange (no colour change), not "no reaction" alone when asked for an observation.
• Saying tertiary alcohols "have no hydrogen". They have plenty. The point is no H on the carbon bonded to the -OH.
• Missing alkene products. Butan-2-ol and pentan-2-ol each give three alkenes once you include the E and Z isomers.
• Vague reagents for elimination. Name the acid catalyst (concentrated H₃PO₄ or H₂SO₄) and say heat. "Acid" alone is too vague.
• Using the wrong family name. OCR calls the product of the substitution a haloalkane.
• Combustion balancing slips. The alcohol contains one O atom; count it before working out the O₂.

Key definitions

• Primary alcohol: an alcohol in which the carbon atom bonded to the -OH group is attached to one other carbon atom (or none, in methanol).
• Secondary alcohol: an alcohol in which the carbon atom bonded to the -OH group is attached to two other carbon atoms.
• Tertiary alcohol: an alcohol in which the carbon atom bonded to the -OH group is attached to three other carbon atoms.
• Hydrogen bond: intermolecular bonding between a lone pair on an electronegative atom (N, O or F) of one molecule and the δ+ H atom of an -OH, -NH or HF group on another molecule.
• Volatility: the ease with which a liquid turns into a gas. Lower volatility means a higher boiling point.
• Oxidising agent: a species that oxidises another species by gaining electrons, and is itself reduced (here Cr₂O₇^2-/H⁺, written [O] in equations).
• Elimination reaction: a reaction in which a small molecule (here H₂O) is removed from a larger molecule, forming a C=C double bond.
• Substitution reaction: a reaction in which one atom or group is replaced by a different atom or group (here -OH replaced by a halogen).
• Heating under reflux: heating a mixture with a vertical condenser so that vapours condense and return to the flask, allowing prolonged heating without loss of volatile substances.
• Distillation: heating a mixture so that a volatile component boils off, is condensed and is collected separately.

Specification

• I can explain why alcohols are polar, and use hydrogen bonding to explain why they are soluble in water and less volatile than alkanes. (4.2.1 a i)
• I can classify an alcohol as primary, secondary or tertiary from its structure. (4.2.1 a ii)
• I can write balanced equations for the combustion of alcohols. (4.2.1 b)
• I can write equations, using [O], for oxidising primary alcohols to aldehydes and to carboxylic acids, and explain how distillation or reflux controls which product forms. (4.2.1 c i)
• I can write an equation, using [O], for oxidising a secondary alcohol to a ketone. (4.2.1 c ii)
• I can explain why tertiary alcohols resist oxidation, and give the colour change seen with acidified dichromate (orange to green, or stays orange). (4.2.1 c iii, PAG7)
• I can write equations for the acid-catalysed elimination of water from an alcohol, and deduce every alkene formed, including E/Z isomers. (4.2.1 d)
• I can write equations for the substitution of an alcohol with halide ions in acid (e.g. NaBr/H₂SO₄) to form a haloalkane. (4.2.1 e)

Related notes

• Bonding and structure: electronegativity, polar bonds and hydrogen bonding
• Redox: oxidation numbers, behind the dichromate colour change
• Basic concepts of organic chemistry: naming alcohols, displayed and skeletal formulae
• Alkanes: London forces and boiling points, for the alcohol vs alkane comparison
• Alkenes: E/Z isomerism, and making alcohols from alkenes with steam
• Haloalkanes: the products of the substitution reaction
• Organic synthesis: reflux, distillation, purification and two-stage routes
• Analytical techniques: using IR to show an alcohol has been oxidised
• Carbonyl compounds (A level): more on aldehydes and ketones, including telling them apart