Edexcel International A Level Chemistry: Alcohols (YCH11)

This guide teaches Topic 10C: Alcohols from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers learning outcomes 10.15 to 10.20, including Core Practical 7. It is International AS (IAS) content in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. The specification states that descriptions of the mechanisms of the 10.17 reactions are not required.

Course links: the Edexcel International A Level Chemistry course hub and the printable Edexcel Chemistry checklist. After reading, condense the topic with the alcohols revision notes and test yourself with the alcohols practice questions. Background: alkanes and naming, alkenes and E-Z isomerism and amount of substance. Want to find your weak spots first? Try a free diagnostic.

Topic 10C at a glance

Outcome	What you must be able to do
10.15	Name alcohols; draw structural, displayed and skeletal formulae
10.16	Tell primary, secondary and tertiary alcohols apart
10.17	Describe combustion; halogenation with PCl₅, KBr with 50% concentrated sulfuric acid, and red phosphorus with iodine; elimination with concentrated phosphoric acid
10.18	Describe oxidation by potassium dichromate(VI) in dilute sulfuric acid: primary to aldehyde (distil) or carboxylic acid (reflux), secondary to ketone; test the products
10.19	Explain reflux, solvent extraction with a separating funnel, distillation, drying with an anhydrous salt and boiling temperature determination
10.20	Core Practical 7: oxidise propan-1-ol to propanal and to propanoic acid

10.15 Naming alcohols and drawing their formulae

An alcohol contains the hydroxyl group, –OH, bonded to a saturated carbon atom. The general formula of the non-cyclic alcohols with one –OH group is CₙH₂ₙ₊₁OH.

To name an alcohol:

1. Find the longest carbon chain that includes the carbon bonded to –OH.
2. Number from whichever end puts the –OH carbon on the lowest-numbered position.
3. Replace the final “e” of the alkane name with “-ol” and put the locant before it: propan-2-ol.
4. Add side chains as prefixes, with their own locants, in alphabetical order.
5. With two –OH groups, keep the “e” and add “-diol”: propane-1,2-diol.

You need three kinds of formula.

• Structural formula shows the arrangement without drawing every bond: CH₃CH(OH)CH₃ for propan-2-ol.
• Displayed formula shows every atom and every bond, including the O–H bond:

      H   H   H
      |   |   |
  H - C - C - C - H
      |   |   |
      H   O   H
          |
          H

• Skeletal formula draws the carbon chain as a zigzag of lines. Each end and each corner is a carbon, and hydrogens on carbon are left out. The –OH group is always written in full. Propan-2-ol is a “V” of two lines with OH on the middle corner.

Worked example 1. Name CH₃CH(CH₃)CH(OH)CH₂CH₃.

The longest chain containing the C–OH carbon has five carbons, so the stem is pentan-. Numbering from the right puts –OH on C3; numbering from the left also puts it on C3. Use the side chain to decide: from the left, the methyl group is on C2; from the right it would be on C4. Choose the lower set. Answer: 2-methylpentan-3-ol.

10.16 Primary, secondary and tertiary alcohols

Focus on one atom: the carbon carrying –OH. The class depends on how many carbon atoms are joined directly to that carbon.

Class	Carbon neighbours of the –OH carbon	Group	Example
Primary	One (or none, for methanol)	–CH₂OH	2,3-dimethylbutan-1-ol
Secondary	Two	>CHOH	cyclopentanol
Tertiary	Three	C–OH with no H on that carbon	3-methylhexan-3-ol

Quick check: the C–OH carbon carries two hydrogens (primary), one (secondary) or none (tertiary).

10.17 Reactions of alcohols

Combustion

Alcohols burn in excess oxygen to give carbon dioxide and water. Balance oxygen last, remembering the oxygen atom already in the alcohol.

2CH₃CH(OH)CH₃ + 9O₂ → 6CO₂ + 8H₂O

Worked example 2. Calculate the volume of carbon dioxide formed when 4.50 g of propan-2-ol (Mr 60.0) burns completely. Use a molar gas volume of 24.0 dm³ mol⁻¹.

n(propan-2-ol) = 4.50 ÷ 60.0 = 0.0750 mol
ratio C₃H₈O : CO₂ = 1 : 3
n(CO₂) = 3 × 0.0750 = 0.225 mol
V(CO₂) = 0.225 × 24.0 = 5.40 dm³

Halogenation: swapping –OH for a halogen

Product	Reagent and conditions	Equation (R = alkyl group)
Chloroalkane	Phosphorus(V) chloride, PCl₅, solid, at room temperature	ROH + PCl₅ → RCl + POCl₃ + HCl
Bromoalkane	Potassium bromide and 50% concentrated sulfuric acid, warmed	KBr + H₂SO₄ → KHSO₄ + HBr, then ROH + HBr → RBr + H₂O
Iodoalkane	Red phosphorus and iodine, warmed	2P + 3I₂ → 2PI₃, then 3ROH + PI₃ → 3RI + H₃PO₃

PCl₅ as a test for –OH. When PCl₅ is added to a dry sample containing an –OH group, steamy fumes of hydrogen chloride are given off. The fumes turn damp blue litmus paper red and form white smoke with ammonia gas. Two cautions:

• The sample must be dry, because water also reacts with PCl₅ to give HCl.
• Carboxylic acids contain –OH too, so a positive result shows an –OH group, not necessarily an alcohol.

Why 50% concentrated sulfuric acid? The acid makes HBr from KBr in the flask. Fully concentrated sulfuric acid would oxidise much of that HBr to bromine (you met this reaction in the Group 7 work), so less HBr would be left to react with the alcohol.

Why red phosphorus and iodine? They form phosphorus(III) iodide in the flask, and this converts the alcohol into the iodoalkane.

Worked example 3. Calculate the atom economy by mass for making 2-chloropropane with PCl₅. (Mr: C₃H₇Cl 78.5, POCl₃ 153.5, HCl 36.5.)

CH₃CH(OH)CH₃ + PCl₅ → CH₃CHClCH₃ + POCl₃ + HCl
atom economy = 78.5 ÷ (78.5 + 153.5 + 36.5) × 100
             = 78.5 ÷ 268.5 × 100 = 29.2%

Elimination with concentrated phosphoric acid

Heating an alcohol with concentrated phosphoric acid removes a molecule of water: the –OH and a hydrogen from a neighbouring carbon leave, and a C=C bond forms. Phosphoric acid is used because it gives fewer side reactions than concentrated sulfuric acid, which is also a strong oxidising agent.

Worked example 4. Pentan-2-ol, CH₃CH(OH)CH₂CH₂CH₃, is warmed with concentrated phosphoric acid. Which alkenes can form?

The –OH is on C2. Its neighbours are C1 and C3, and both carry hydrogen.

• Losing H from C1 gives CH₂=CHCH₂CH₂CH₃, pent-1-ene.
• Losing H from C3 gives CH₃CH=CHCH₂CH₃, pent-2-ene. Each carbon of this C=C has two different groups, so it exists as (E)-pent-2-ene and (Z)-pent-2-ene.

That is three alkenes in total.

10.18 Oxidation with potassium dichromate(VI)

The oxidising agent is potassium dichromate(VI) in dilute sulfuric acid. In equations you may write it as [O]. When oxidation happens, the orange dichromate(VI) ion is reduced to green chromium(III) ions, so the mixture turns from orange to green.

Alcohol	Conditions	Product	Equation with [O]
Primary	Distil the product as it forms	Aldehyde	RCH₂OH + [O] → RCHO + H₂O
Primary	Heat under reflux with excess oxidising agent	Carboxylic acid	RCH₂OH + 2[O] → RCOOH + H₂O
Secondary	Heat	Ketone	R₂CHOH + [O] → R₂CO + H₂O
Tertiary	Heat	No reaction; stays orange	none

Why do the conditions matter for a primary alcohol? The aldehyde has no O–H group, so its molecules cannot hydrogen bond to each other. It boils at a lower temperature than the alcohol and the acid. If you distil as you go, the aldehyde leaves the flask before it meets more oxidising agent. Under reflux, the aldehyde vapour condenses and falls back into the flask, so it is oxidised further to the carboxylic acid.

Testing the products.

• Aldehyde: warm with Benedict’s or Fehling’s solution. The blue solution gives a red precipitate. A ketone gives no change.
• Carboxylic acid: add sodium carbonate or sodium hydrogencarbonate. Effervescence occurs; the gas is carbon dioxide, which turns limewater milky.

Worked example 5. Give the products, with equations, when 2-methylpropan-1-ol, (CH₃)₂CHCH₂OH, is oxidised (a) with distillation and (b) under reflux.

The –OH carbon carries two hydrogens, so the alcohol is primary.

(a) (CH₃)₂CHCH₂OH + [O] → (CH₃)₂CHCHO + H₂O. Product: 2-methylpropanal.

(b) (CH₃)₂CHCH₂OH + 2[O] → (CH₃)₂CHCOOH + H₂O. Product: 2-methylpropanoic acid.

10.19 Preparing and purifying a liquid organic product

Technique	Purpose	Points to get right
Heating under reflux	Heat a reaction for a long time without losing volatile substances	Condenser vertical, open at the top, water in at the bottom; anti-bumping granules; electric heater or water bath, as organic liquids are flammable
Extraction with a separating funnel	Separate two immiscible liquids, or move the product into a solvent	Invert and vent often; let the layers settle; run off the lower layer
Distillation	Separate a liquid from substances with different boiling temperatures	Thermometer bulb opposite the side-arm opening; keep the distillate that comes over near the product’s boiling temperature
Drying with an anhydrous salt	Remove traces of water from the organic layer	Add anhydrous magnesium sulfate or calcium chloride until some stays loose; the liquid goes from cloudy to clear; decant or filter
Boiling temperature determination	Check identity and purity	A pure liquid boils at a sharp, fixed temperature that matches data; an impure one boils over a range

A typical order for a crude liquid product is: react (often under reflux), distil out the crude product, wash it in a separating funnel, dry it, redistil, then measure the boiling temperature. Washing with sodium hydrogencarbonate solution removes acid; vent the funnel often, because carbon dioxide is released.

10.20 Core Practical 7: oxidising propan-1-ol

Making propanal. Warm propan-1-ol in a flask set up for distillation. Add the acidified dichromate(VI) slowly, so the alcohol stays in excess. Propanal distils over as it forms; collect it in a receiver cooled in ice, because it is very volatile. Test a sample with Benedict’s or Fehling’s solution.

Making propanoic acid. Use an excess of acidified dichromate(VI) and heat the mixture under reflux, so any propanal is oxidised further. Then rearrange the apparatus for distillation and collect the acid. Test a sample with sodium hydrogencarbonate.

CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O

CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O

Worked example 6. A student refluxes 6.00 cm³ of propan-1-ol (density 0.80 g cm⁻³) with excess oxidising agent and obtains 3.70 g of propanoic acid. Calculate the percentage yield. (Mr: propan-1-ol 60.0, propanoic acid 74.0.)

mass of propan-1-ol = 6.00 × 0.80 = 4.80 g
n(propan-1-ol) = 4.80 ÷ 60.0 = 0.0800 mol
theoretical n(acid) = 0.0800 mol (1 : 1)
theoretical mass = 0.0800 × 74.0 = 5.92 g
% yield = 3.70 ÷ 5.92 × 100 = 62.5%

Yields fall below 100% through incomplete oxidation and losses in transfer and distillation.

Common errors

• Classifying by the total number of carbons, or by branching elsewhere in the molecule, instead of looking only at the C–OH carbon.
• Writing “orange to green” for a tertiary alcohol. No oxidation happens, so it stays orange.
• Swapping the set-ups: distillation as it forms gives the aldehyde; reflux gives the acid.
• Testing for an aldehyde with sodium hydrogencarbonate, or for an acid with Benedict’s solution.
• Forgetting HCl and POCl₃ in the PCl₅ equation, or doing the –OH test on a wet sample.
• Missing one neighbouring carbon in elimination, or missing E and Z forms.

Next steps

Condense this into the revision notes, then try the practice questions.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11), Specification, Issue 1, September 2017, Pearson Education Limited. Topic 10: Organic Chemistry: Halogenoalkanes, Alcohols and Spectra, part 10C: Alcohols (outcomes 10.15 to 10.20).