OxfordAQA A-Level Chemistry: Alcohols (9620)

Section 3.3.5 Alcohols is the subject of this guide, written for the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. It takes in both parts of the section, 3.3.5.1 Oxidation of alcohols and 3.3.5.2 Elimination, together with Required practical 4 (distil a product from a reaction). Everything here is International AS content. The Unit 2 exam (Organic 1 and Physical 1) tests sections 3.3.1 to 3.3.6, and the practical and synoptic Unit 5 exam may test any content.

Condensed recall for the last weeks is in the 9620 alcohols notes, and marked questions are in the 9620 alcohols question set. Every 9620 topic is gathered on the course page for OxfordAQA chemistry; mark off what you have learned on the printable checklist. A short diagnostic test (free, about 10 minutes) points to weak spots across the course.

Specification points in this topic

Point	Skills required	Stage
3.3.5.1 Oxidation of alcohols	Classify alcohols as primary, secondary or tertiary; know that acidified potassium dichromate(VI) is a suitable oxidising agent; write oxidation equations (using [O] is acceptable); explain how the method decides between an aldehyde and a carboxylic acid; distinguish aldehydes from ketones with Fehling’s solution and Tollens’ reagent	International AS
3.3.5.2 Elimination	Form alkenes from alcohols by acid-catalysed elimination; outline the mechanism; know that these alkenes can make addition polymers without monomers derived from crude oil	International AS
Required practical 4	Distil a product from a reaction	International AS

Naming and the general formula of alcohols: see the 9620 organic naming page. Hydrogen bonding, which explains several boiling-point points below, is in the 9620 intermolecular forces guide.

3.3.5.1 Classifying alcohols

Find the carbon atom that carries the –OH group. Count how many other carbon atoms are bonded to it.

Class	Carbons bonded to the C–OH carbon	H atoms on the C–OH carbon	Example
Primary	one (methanol, CH₃OH, has none but is treated as primary)	two	pentan-1-ol, CH₃CH₂CH₂CH₂CH₂OH
Secondary	two	one	cyclopentanol, C₅H₉OH (–OH on a ring carbon)
Tertiary	three	none	2-methylpentan-2-ol, (CH₃)₂C(OH)CH₂CH₂CH₃

The number of hydrogen atoms on the C–OH carbon is what controls oxidation, so it is worth writing it in.

3.3.5.1 Oxidation with acidified potassium dichromate(VI)

The specification names acidified potassium dichromate(VI) as a suitable oxidising agent: K₂Cr₂O₇ solution made acidic with dilute H₂SO₄. When the alcohol is oxidised, the orange dichromate(VI) ions are reduced to green chromium(III) ions. So a colour change from orange to green shows that oxidation has happened.

You may write the oxygen supplied by the oxidising agent as [O]. The specification states that equations showing [O] as the oxidant are acceptable.

The three outcomes:

• Primary alcohol → aldehyde → carboxylic acid. Each step uses one [O]. The first step also makes water.
• Secondary alcohol → ketone. One [O]; water also forms. Ketones are not oxidised further by this reagent.
• Tertiary alcohol: not easily oxidised. There is no hydrogen atom on the C–OH carbon to remove, so the mixture stays orange.

Worked example 1: products and equations

Give the organic product, an equation using [O] and the observation for each alcohol warmed with acidified potassium dichromate(VI).

Pentan-1-ol (primary). The –CH₂OH group becomes –CHO, then –COOH. The chain stays five carbons long.

CH₃(CH₂)₃CH₂OH + [O]  → CH₃(CH₂)₃CHO + H₂O       pentanal
CH₃(CH₂)₃CHO + [O]    → CH₃(CH₂)₃COOH           pentanoic acid
CH₃(CH₂)₃CH₂OH + 2[O] → CH₃(CH₂)₃COOH + H₂O     overall

Observation: orange to green.

Cyclopentanol (secondary). The –CH(OH)– ring carbon becomes C=O.

C₅H₉OH + [O] → C₅H₈O + H₂O       cyclopentanone

Observation: orange to green.

2-methylpentan-2-ol (tertiary). No reaction under these conditions. Observation: the mixture stays orange.

Check every [O] equation by counting atoms on both sides.

3.3.5.1 How the method decides aldehyde or acid

A primary alcohol gives two possible products, so you control the outcome with the apparatus and the amounts.

Product wanted	What you do	Why it works
Aldehyde	Use the alcohol in excess (a limited amount of oxidising agent), heat gently, and distil off the aldehyde as soon as it forms	The aldehyde boils at a lower temperature than the alcohol, because aldehyde molecules cannot hydrogen bond to each other. It leaves the flask before it can be oxidised again
Carboxylic acid	Use excess oxidising agent and heat under reflux, then distil the acid out at the end	In reflux the vapour condenses and runs back into the flask, so any aldehyde made is trapped with the excess dichromate and is converted to the acid

Worked example 2: choosing the set-up

A technician has pentan-1-ol and wants (a) a sample of pentanal and (b) a sample of pentanoic acid. Describe the method for each.

• (a) Put acidified potassium dichromate(VI) in a flask with the pentan-1-ol in excess. Warm gently with the flask connected to a condenser set for distillation (sloping down to a collecting vessel). Pentanal is the most volatile organic substance in the flask, so it distils over as it forms and escapes further oxidation.
• (b) Mix pentan-1-ol with excess acidified potassium dichromate(VI). Heat under reflux (condenser vertical, open at the top) until the reaction is complete, then rearrange the apparatus and distil off the pentanoic acid.

Pair each method with its reason: distil immediately for the aldehyde, reflux with excess oxidising agent for the acid.

3.3.5.1 Telling aldehydes from ketones

An aldehyde still has an H atom on its carbonyl carbon, so it is easy to oxidise; a ketone has none and resists oxidation. Fehling’s solution and Tollens’ reagent are both mild oxidising agents that react only with the aldehyde, and the specification names both.

Reagent	Method	Result: aldehyde	Result: ketone
Fehling’s solution (blue, contains Cu²⁺ complexes)	Warm in a water bath	Blue solution forms a red precipitate of copper(I) oxide, Cu₂O	Stays blue
Tollens’ reagent (contains [Ag(NH₃)₂]⁺)	Warm in a water bath	A silver mirror coats the glass	No change

Each positive result means the aldehyde has become a carboxylic acid (as its salt, since both reagents are alkaline), while the metal ion is reduced: Cu²⁺ to copper(I), Ag⁺ to silver. Use a water bath, not a naked flame, because many aldehydes and ketones are flammable. The silver complex in Tollens’ reagent appears again in the 9620 transition metal chemistry guide.

Worked example 3: identifying two liquids

Two colourless liquids are pentanal and cyclopentanone. Explain how Fehling’s solution distinguishes them.

1. Warm a few drops of each with Fehling’s solution in a water bath.
2. Pentanal: the blue solution gives a red precipitate, because the aldehyde is oxidised.
3. Cyclopentanone: the solution stays blue, because ketones are not oxidised by Fehling’s solution.
4. Equation for the positive result: CH₃(CH₂)₃CHO + [O] → CH₃(CH₂)₃COOH.

Always give an observation for both compounds. “No reaction” is not an observation; “stays blue” is.

3.3.5.2 Elimination

Alkenes can be formed from alcohols by acid-catalysed elimination. Heat the alcohol with concentrated phosphoric acid (concentrated sulfuric acid also works) as the catalyst, and a molecule of water is removed. This is also called dehydration.

The mechanism

1. The –OH oxygen uses one of its lone pairs to bond to an H⁺ ion supplied by the acid. This gives a protonated alcohol with a positive oxygen, R–OH₂⁺.
2. The C–O bond breaks heterolytically. Both electrons go to the oxygen, and a water molecule leaves. A carbocation forms on the carbon that held the –OH.
3. A carbon adjacent to C⁺ gives up a proton: the pair of electrons from its C–H bond becomes the second bond of C=C. Because H⁺ comes back out, the acid is not used up; it is a catalyst.

Draw each curly arrow from a lone pair or a bond, never from a positive charge or from H⁺ itself.

Worked example 4: how many alkenes?

Which alkenes form from 2-methylpentan-2-ol with hot concentrated phosphoric acid?

• Step 1: C2 becomes the positive carbon. It carries two methyl groups and a propyl group.
• Step 2: look at each neighbouring carbon that has a hydrogen atom. The two methyl carbons are equivalent; C3 is different.
• Step 3: losing H⁺ from a methyl carbon gives 2-methylpent-1-ene, CH₂=C(CH₃)CH₂CH₂CH₃.
• Step 4: losing H⁺ from C3 gives 2-methylpent-2-ene, (CH₃)₂C=CHCH₂CH₃.
• Step 5: check for E/Z isomers. C1 of the first alkene carries two hydrogens and C2 of the second carries two methyl groups, so neither shows E/Z isomerism.

Answer: two alkenes, 2-methylpent-1-ene and 2-methylpent-2-ene.

Polymers without crude oil

The specification notes that alkenes made this way can be used to produce addition polymers without using monomers derived from crude oil. If the alcohol comes from a non-petroleum source (for example, from plant material), the alkene and the polymer made from it do not depend on crude oil.

Required practical 4: distil a product from a reaction

Distillation separates a volatile product from a reaction mixture. In this topic it is how you collect an aldehyde before it is oxidised further, and how you collect an alkene from an elimination.

Points of technique to explain:

• Anti-bumping granules in the flask give smooth boiling.
• Thermometer bulb level with the side arm that leads to the condenser, so it reads the temperature of the vapour that is distilling.
• Condenser water in at the bottom and out at the top, so the condenser stays full of cold water.
• Heat with an electric heater or a water bath, not a naked flame, because organic vapours are flammable.
• Never heat a sealed system: the collecting vessel must be open to the air (or vented).
• Collect the fraction that distils close to the boiling point of the product you want.

Worked example 5: percentage yield of a distilled product

6.16 g of pentan-1-ol (Mr 88.0) is oxidised and the pentanal (Mr 86.0) is distilled off as it forms. The student collects 2.71 g of pentanal. Calculate the percentage yield.

n(pentan-1-ol)       = 6.16 ÷ 88.0 = 0.0700 mol
theoretical n(pentanal) = 0.0700 mol            (1 : 1 ratio)
actual n(pentanal)   = 2.71 ÷ 86.0 = 0.03151 mol
percentage yield     = 0.03151 ÷ 0.0700 × 100 = 45.0 %

Answer: 45.0 %. Losses include incomplete reaction, further oxidation and liquid left in the apparatus. For more mole practice, see the 9620 moles and equations guide.

Common errors

• Counting all the carbons in the molecule to classify an alcohol, instead of only those bonded to the C–OH carbon.
• Saying “reflux” for an aldehyde, or “distil” without saying the aldehyde is removed as it forms.
• Explaining the low boiling point of an aldehyde as “weaker bonds”. It is the absence of hydrogen bonding between aldehyde molecules.
• Writing that a tertiary alcohol “turns green”. It stays orange.
• Giving “no reaction” for the ketone in Fehling’s or Tollens’ tests instead of an observation.
• Starting the first elimination arrow at H⁺. It starts at a lone pair on oxygen.
• Forgetting a second alkene, or E/Z isomers, when the C–OH carbon has two different neighbours with H atoms.

Where this links

Next, try the alcohols question set. Redox language for the dichromate colour change is in the 9620 redox guide.

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.5 Alcohols (International AS), 3.3.5.1 Oxidation of alcohols and 3.3.5.2 Elimination, with Required practical 4.