Edexcel International A Level Chemistry: Alcohols (YCH11) – Revision Notes

These notes condense Topic 10C: Alcohols of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017: outcomes 10.15 to 10.20, including Core Practical 7. The content is International AS (IAS), in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. Mechanisms for the 10.17 reactions are not required. If a point here is unfamiliar, go back to the alcohols study guide, which explains it step by step.

Links: course hub, printable checklist, alcohols practice questions, the related halogenoalkanes and mass spectra and IR topics, and the free diagnostics to find your gaps.

Naming and classes (10.15, 10.16)

• Longest chain containing the C–OH carbon; lowest number to the –OH carbon; suffix “-ol” with its locant (butan-1-ol).
• Two –OH groups: keep the “e”, add “-diol” (ethane-1,2-diol).
• Structural: CH₃CH₂CH₂OH. Displayed: every bond, including O–H. Skeletal: zigzag carbon chain, –OH written in full.
• Classify by the C–OH carbon only.
• General formula of the non-cyclic alcohols with one –OH: CₙH₂ₙ₊₁OH, so each extra CH₂ adds 14.0 to the Mr.

Class	Carbon neighbours of the –OH carbon	Hydrogens still on that carbon
Primary	1 (0 for methanol)	2
Secondary	2	1
Tertiary	3	0

Reaction summary (10.17, 10.18)

Reaction	Reagent and conditions	Product	Note
Combustion	Excess oxygen, ignite	CO₂ + H₂O	Balance O last
Chlorination	PCl₅ (solid), room temperature	Chloroalkane + POCl₃ + HCl	Steamy fumes of HCl
Bromination	KBr + 50% concentrated H₂SO₄, warm	Bromoalkane	HBr made in the flask
Iodination	Red phosphorus + iodine, warm	Iodoalkane	PI₃ made in the flask
Elimination	Concentrated H₃PO₄, heat	Alkene + H₂O	Check E-Z
Oxidation, 1°, distil	K₂Cr₂O₇ / dilute H₂SO₄, warm, distil as it forms	Aldehyde	Orange → green
Oxidation, 1°, reflux	Excess K₂Cr₂O₇ / dilute H₂SO₄, reflux	Carboxylic acid	Orange → green
Oxidation, 2°	K₂Cr₂O₇ / dilute H₂SO₄, heat	Ketone	Orange → green
Oxidation, 3°	K₂Cr₂O₇ / dilute H₂SO₄, heat	No reaction	Stays orange

Equations to know

Change	Equation
PCl₅	ROH + PCl₅ → RCl + POCl₃ + HCl
Making HBr	KBr + H₂SO₄ → KHSO₄ + HBr
Bromination	ROH + HBr → RBr + H₂O
Making PI₃	2P + 3I₂ → 2PI₃
Iodination	3ROH + PI₃ → 3RI + H₃PO₃
Elimination	CH₃CH₂CH₂OH → CH₃CH=CH₂ + H₂O
1° to aldehyde	RCH₂OH + [O] → RCHO + H₂O
1° to acid	RCH₂OH + 2[O] → RCOOH + H₂O
2° to ketone	R₂CHOH + [O] → R₂CO + H₂O

Count the [O]: one for aldehyde or ketone, two for carboxylic acid. Water appears once in each.

Method in steps: alkenes from elimination

1. Find the C–OH carbon.
2. Look at each adjacent carbon; keep only those with a hydrogen to lose.
3. Draw one C=C for each different neighbour (identical neighbours give the same alkene).
4. For each alkene, check whether both C=C carbons have two different groups. If so, give E and Z.
5. Count and name every product.

Method in steps: choosing the oxidation set-up

1. Classify the alcohol.
2. Tertiary: no reaction. Secondary: heat, ketone.
3. Primary and you want the aldehyde: alcohol in excess, add oxidising agent slowly, distil immediately, cool the receiver.
4. Primary and you want the acid: oxidising agent in excess, reflux, then distil the acid out.

Must-know distinctions

• Distil vs reflux. The aldehyde has no hydrogen bonding between its molecules, so it boils lowest and escapes if distilled as it forms. Reflux returns it to the flask, where it is oxidised to the acid.
• Aldehyde test vs acid test. Aldehyde: Benedict’s or Fehling’s solution, warm, blue to red precipitate. Acid: sodium carbonate or hydrogencarbonate, effervescence of CO₂ (limewater turns milky).
• PCl₅ test. Positive for any –OH, including carboxylic acids and water. Use a dry sample.
• 50% not 100% sulfuric acid. Fully concentrated acid oxidises HBr to bromine, wasting the HBr.
• Phosphoric not sulfuric acid for elimination. Concentrated sulfuric acid is a strong oxidising agent and causes more side reactions.

Techniques (10.19) in one line each

Technique	Key point
Reflux	Vertical condenser, open top; volatile liquids boil and condense back, so nothing escapes
Separating funnel	Immiscible layers; vent often; run off the lower layer; NaHCO₃ wash removes acid (vent the CO₂)
Distillation	Thermometer bulb at the side arm; collect the fraction near the expected boiling temperature
Drying	Anhydrous MgSO₄ or CaCl₂; add until some stays loose; cloudy liquid turns clear; decant
Boiling temperature	Pure: sharp and matches data. Impure: boils over a range

Typical order: react → distil off crude product → wash in separating funnel → dry → redistil → measure boiling temperature.

Core Practical 7 in brief (10.20)

• Propanal: warm propan-1-ol in distillation apparatus, add acidified dichromate(VI) dropwise, collect the distillate in an ice-cooled receiver, test with Benedict’s or Fehling’s.
• Propanoic acid: excess acidified dichromate(VI), reflux, then distil; test with NaHCO₃ or Na₂CO₃.
• Percentage yield: convert a liquid volume to mass with its density, then to moles; use the 1 : 1 ratio; actual ÷ theoretical × 100.

Small worked reminders

• Diol naming and classes. HOCH₂CH(OH)CH₂CH₃ is butane-1,2-diol. The –OH on C1 is primary; the –OH on C2 is secondary. Classify each –OH group separately.
• Only one neighbour with hydrogen. CH₃CH₂CH(CH₃)CH₂OH (2-methylbutan-1-ol) has the –OH on C1, and its only neighbour is C2. Elimination gives a single alkene, 2-methylbut-1-ene, with no E-Z isomers because C1 carries two hydrogens.
• Distilled product. 3-methylbutan-1-ol, distilled as it forms with acidified dichromate(VI), gives 3-methylbutanal: (CH₃)₂CHCH₂CH₂OH + [O] → (CH₃)₂CHCH₂CHO + H₂O.
• Volume to moles. For a liquid reactant, mass = volume × density, then moles = mass ÷ Mr. Do not skip the density step: treating 5.00 cm³ of propan-1-ol as 5.00 g overstates the moles, because its density is below 1 g cm⁻³.

Quick self-test

1. Name HOCH₂CH₂CH(CH₃)CH₂CH₃ and classify it.
2. Classify 1-methylcyclohexanol.
3. Write a balanced equation for the complete combustion of butan-1-ol.
4. Calculate the mass of oxygen needed to burn 1.85 g of butan-1-ol (Mr 74.0) completely.
5. Give the reagents and conditions to make 1-iodopropane from propan-1-ol.
6. State what you see when PCl₅ is added to dry ethanol, and name the gas.
7. Why must the sample be dry in the PCl₅ test?
8. Name every alkene formed when heptan-4-ol is heated with concentrated phosphoric acid.
9. Name the organic product when pentan-3-ol is heated with acidified potassium dichromate(VI), and give the colour change.
10. How would you tell butanal from butanoic acid with one test?
11. Calculate the atom economy for CH₃CH₂CH₂OH + HBr → CH₃CH₂CH₂Br + H₂O. (Mr: C₃H₇Br 122.9, H₂O 18.0)
12. 2.40 g of propan-1-ol (Mr 60.0) gives 0.928 g of propanal (Mr 58.0). Calculate the percentage yield.

Answers

1. 3-methylpentan-1-ol; primary, because only C2 is joined to the –OH carbon.
2. Tertiary: the ring carbon bearing –OH is bonded to three carbons (two ring carbons and the methyl).
3. C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O.
4. n = 1.85 ÷ 74.0 = 0.0250 mol; n(O₂) = 6 × 0.0250 = 0.150 mol; mass = 0.150 × 32.0 = 4.80 g.
5. Red phosphorus and iodine, warmed.
6. Steamy fumes; the gas is hydrogen chloride.
7. Water also reacts with PCl₅ to give HCl, so a wet sample gives a positive result whatever it is.
8. Both neighbours (C3 and C5) are equivalent, so one structure: hept-3-ene, as (E)-hept-3-ene and (Z)-hept-3-ene.
9. Pentan-3-one; orange to green.
10. Add sodium hydrogencarbonate: butanoic acid fizzes (CO₂); butanal does not. (Or Benedict’s: butanal gives a red precipitate, the acid does not.)
11. 122.9 ÷ (122.9 + 18.0) × 100 = 87.2%.
12. n(alcohol) = 2.40 ÷ 60.0 = 0.0400 mol; n(propanal) = 0.928 ÷ 58.0 = 0.0160 mol; yield = 0.0160 ÷ 0.0400 × 100 = 40.0%.

Where marks are usually lost

• Calling an alcohol secondary because the molecule is branched somewhere else; only the C–OH carbon counts.
• Giving a colour change for a tertiary alcohol with dichromate(VI). It stays orange.
• Writing “heat” for the aldehyde without saying the product is distilled off as it forms.
• Using one [O] for the carboxylic acid equation instead of two.
• Leaving POCl₃ or HCl out of the PCl₅ equation.
• Describing the PCl₅ result as “a gas is given off” without “steamy fumes” or a test for HCl.
• Naming only one product from elimination when there are two neighbouring carbons, or missing the E and Z forms.
• Saying the aldehyde boils lower because it has “weaker bonds”; the point is no hydrogen bonding between its molecules.
• Testing an acid with Benedict’s solution, or an aldehyde with sodium carbonate.
• Placing the distillation thermometer bulb in the liquid rather than at the side arm.

Next steps

Go back to the study guide for any answer you got wrong, then work through the practice questions under timed conditions.

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, Core Practical 7).