S5 Chemistry – Unit 5: Alcohols and Ethers

5.1

Nomenclature and Classification

Alcohols — Definition Alcohols are organic compounds containing the hydroxyl functional group (–OH) bonded to a saturated carbon atom. General formula for monohydric alcohols: C_nH_2n+1OH (or R–OH). They are named using the suffix –ol.

IUPAC Nomenclature Rules for Alcohols

Find the longest chain containing the –OH group. Replace –e of the alkane with –ol.
Number the chain to give the –OH group the lowest locant.
State the locant of –OH before –ol: e.g. propan-1-ol, propan-2-ol.
Name and number other substituents alphabetically as usual.
Diols: suffix –diol; triols: –triol.

Classification by Degree of Substitution

Class	Definition	Example	IUPAC Name
Primary (1°)	C bearing –OH bonded to 1 other C (or 0 for methanol)	CH₃CH₂OH	Ethanol
Secondary (2°)	C bearing –OH bonded to 2 other C	CH₃CH(OH)CH₃	Propan-2-ol
Tertiary (3°)	C bearing –OH bonded to 3 other C	(CH₃)₃COH	2-methylpropan-2-ol

Common Alcohols

Name	Formula	Class	Common Name
Methanol	CH₃OH	Primary	Wood alcohol
Ethanol	CH₃CH₂OH	Primary	Alcohol (drinking)
Propan-1-ol	CH₃CH₂CH₂OH	Primary	–
Propan-2-ol	CH₃CH(OH)CH₃	Secondary	Isopropanol / rubbing alcohol
Butan-1-ol	CH₃(CH₂)₃OH	Primary	n-butanol
Butan-2-ol	CH₃CH(OH)CH₂CH₃	Secondary	–
2-methylpropan-2-ol	(CH₃)₃COH	Tertiary	tert-butanol
Ethane-1,2-diol	HOCH₂CH₂OH	Diol	Ethylene glycol
Propane-1,2,3-triol	HOCH₂CH(OH)CH₂OH	Triol	Glycerol

Example 1

Naming and Classifying Alcohols

Name and classify: (a) CH₃CH(OH)CH₂CH₃ (b) (CH₃)₂C(OH)CH₂CH₃

Longest chain containing –OH = 4 C (butane). –OH on C2 from left (lower than C3). → Butan-2-ol. C2 bonded to C1 and C3 → Secondary.

Longest chain = 4 C. –OH on C2. Two CH₃ on C2. → 2-methylbutan-2-ol. C2 bonded to C1, C3, and CH₃ (3 carbons) → Tertiary.

5.2

Physical Properties

Hydrogen Bonding

The –OH group can form intermolecular hydrogen bonds (O–H···O) between alcohol molecules. This makes alcohols have much higher boiling points than alkanes or ethers of similar molecular mass.

Example: ethanol (M_r = 46, B.P. = 78°C) vs propane (M_r = 44, B.P. = −42°C).

Solubility in Water

Short-chain alcohols (C₁–C₄) are completely miscible with water because the –OH group can form H-bonds with water molecules. As chain length increases, the non-polar hydrocarbon chain dominates and solubility decreases. Long-chain alcohols (C₆₊) are practically insoluble in water.

Alcohol	M_r	B.P. (°C)	Solubility in H₂O	State
Methanol	32	+65	Miscible	Liquid
Ethanol	46	+78	Miscible	Liquid
Propan-1-ol	60	+97	Miscible	Liquid
Butan-1-ol	74	+118	Slightly soluble	Liquid
Pentan-1-ol	88	+138	Slightly soluble	Liquid
Hexan-1-ol	102	+157	Insoluble	Liquid

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Comparing Boiling Points: 1° vs 2° vs 3° For the same molecular formula, boiling points follow: 1° > 2° > 3° alcohols — branching reduces surface area and hence van der Waals forces, lowering B.P. slightly. All three are much higher than the corresponding alkane due to H-bonding.

5.3

Preparation of Alcohols

Method 1: Hydrolysis of Halogenoalkanes

R-X + NaOH(aq) --heat/reflux--> R-OH + NaX e.g. CH3CH2Br + NaOH(aq) --> CH3CH2OH + NaBr (bromoethane) (ethanol)

Method 2: Hydration of Alkenes

Industrial method for ethanol: ethene reacts with steam over a phosphoric acid catalyst.

CH2=CH2 + H2O --conc. H3PO4, 300 degC, 60 atm--> CH3CH2OH For unsymmetrical alkenes, Markovnikov's rule applies: CH3CH=CH2 + H2O --H+--> CH3CH(OH)CH3 (major: propan-2-ol)

Method 3: Reduction of Carbonyl Compounds

Aldehyde + 2[H] --LiAlH4 or NaBH4--> Primary alcohol RCHO + 2[H] --> RCH2OH Ketone + 2[H] --LiAlH4 or NaBH4--> Secondary alcohol RCOR' + 2[H] --> RCH(OH)R'

LiAlH₄ (in dry ether) is a powerful reducing agent. NaBH₄ (in water or ethanol) is milder and more selective.

Method 4: Fermentation (see Section 5.4)

Ethanol is produced industrially and traditionally by fermentation of sugars using yeast enzymes.

Method 5: Reaction of Grignard Reagents with Carbonyl Compounds

A Grignard reagent (RMgX, prepared from R–X + Mg in dry ether) adds to aldehydes or ketones to give alcohols after hydrolysis:

RCHO + R'MgX --> [R'RCH-OMgX] --H2O/H+--> R'RCH-OH (aldehyde) (secondary alcohol) HCHO + RMgX --> RCH2OH (primary alcohol) RCOR' + R''MgX --> RR'R''COH (tertiary alcohol)

5.4

Fermentation

Fermentation Fermentation is the anaerobic (without oxygen) breakdown of sugars by enzymes in yeast to produce ethanol and carbon dioxide. It is the biological method for producing ethanol.

Overall Equation

C6H12O6 --yeast enzymes, 25-35 degC--> 2 C2H5OH + 2 CO2 (glucose) (ethanol) (carbon dioxide)

The reaction is carried out at 25–35°C (optimal enzyme activity). Above ~37°C, enzymes denature. Below 25°C, the reaction is too slow.

Conditions for Fermentation

Yeast (Saccharomyces cerevisiae) — contains the enzyme zymase
Temperature: 25–35°C (optimum ~30°C)
Anaerobic conditions (no oxygen — otherwise yeast uses aerobic respiration: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O)
pH: slightly acidic (~pH 4–6)
Aqueous medium (water as solvent)
Ethanol concentration cannot exceed ~14% — at higher concentrations yeast is killed

Source of Sugar

Sugar cane and sugar beet provide sucrose (C₁₂H₂₂O₁₁) which is first hydrolysed to glucose and fructose by invertase. Grain starch is first hydrolysed to glucose by amylase, then fermented. In developing countries, cassava, banana, and other starchy crops are used.

Industrial Ethanol: Fermentation vs Hydration of Ethene

Feature	Fermentation	Hydration of Ethene
Raw material	Sugars/starch (renewable)	Ethene from crude oil (non-renewable)
Temperature	~30°C (low energy)	~300°C (high energy)
Pressure	Atmospheric	~60 atm (high)
Catalyst	Yeast (biological)	H₃PO₄ (chemical)
Rate	Slow (batch process)	Fast (continuous process)
Purity	Low (requires distillation)	High (more pure product)
Sustainability	Renewable; carbon neutral	Fossil fuel dependent
Used in	Developing countries; biofuels	Industrial scale in developed countries

Example 2

Fermentation Calculation

Calculate the mass of ethanol produced when 180 g of glucose undergoes complete fermentation.

Equation: C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

M_r(glucose) = 180 g/mol → 180 g = 1 mol glucose.

1 mol glucose → 2 mol ethanol. M_r(ethanol) = 46 g/mol.

Mass of ethanol = 2 × 46 = 92 g.

5.5

Reactions of Alcohols

Reaction 1: Combustion

C2H5OH + 3O2 --> 2CO2 + 3H2O (complete combustion) General: CnH(2n+1)OH + (3n/2) O2 --> n CO2 + (n+1) H2O

Alcohols burn with a less luminous flame than alkenes/alkynes due to their lower C:H ratio.

Reaction 2: Reaction with Sodium Metal

Alcohols react with sodium metal to produce hydrogen gas and a sodium alkoxide. This shows alcohols are weakly acidic (though much weaker than water):

2 R-OH + 2 Na --> 2 R-O-Na+ + H2(g) e.g. 2 C2H5OH + 2 Na --> 2 C2H5ONa + H2 (ethanol) (sodium ethoxide)

The reaction is less vigorous than sodium with water (because the O–H bond in alcohol is slightly less polar than in water). Effervescence of H₂ is observed.

Reaction 3: Dehydration to Alkenes

R-CHOH-CH2R' --conc. H2SO4, ~170 degC (or Al2O3, heat)--> R-CH=CH-R' + H2O e.g. CH3CH2OH --conc. H2SO4, 170 degC--> CH2=CH2 + H2O (ethanol) (ethene)

At ~140°C (lower temperature), the ether is the major product instead (intermolecular dehydration).

Reaction 4: Conversion to Halogenoalkanes

R-OH + PCl5 --> R-Cl + POCl3 + HCl R-OH + SOCl2 --> R-Cl + SO2 + HCl R-OH + HBr --> R-Br + H2O (with conc. H2SO4 and NaBr)

(See Unit 4 for full details.)

Reaction 5: Oxidation

The oxidation products depend on whether the alcohol is primary, secondary, or tertiary. Oxidising agents used: acidified K₂Cr₂O₇ (orange → green) or acidified KMnO₄.

Alcohol Class	Oxidation Product (mild)	Oxidation Product (excess/reflux)
Primary (R–CH₂OH)	Aldehyde (RCHO) — distil off immediately	Carboxylic acid (RCOOH) — with excess oxidant/reflux
Secondary (R₂CHOH)	Ketone (RCOR') — cannot be further oxidised easily	Ketone (same product)
Tertiary (R₃COH)	Not oxidised (no H on C–OH)	Not oxidised under normal conditions

Primary: RCH2OH --[O]--> RCHO --[O]--> RCOOH (primary alcohol) (aldehyde) (carboxylic acid) Secondary: R2CHOH --[O]--> RCOR' (no further oxidation) (secondary) (ketone) Tertiary: R3COH --[O]--> no reaction under normal conditions

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Test: Distinguishing 1°, 2°, 3° Alcohols by Oxidation Add acidified K₂Cr₂O₇ (orange) and warm.
• Primary: orange → green (oxidised to aldehyde, then acid if excess oxidant)
• Secondary: orange → green (oxidised to ketone)
• Tertiary: no colour change (not oxidised)
To distinguish 1° from 2°: use Tollens’ reagent or Fehling’s (Unit 6) — only aldehydes (from 1°) give positive results.

Reaction 6: Esterification

Alcohols react with carboxylic acids in the presence of a concentrated H₂SO₄ catalyst to form esters and water (reversible reaction):

R-OH + R'COOH <==> R'COOR + H2O (alcohol) (carboxylic acid) (ester) (water) e.g. C2H5OH + CH3COOH <==> CH3COOC2H5 + H2O (ethanol) (ethanoic acid) (ethyl ethanoate)

This is a reversible, equilibrium reaction. Yield is improved by: using excess of one reactant, removing water (molecular sieves), or distilling off the ester as it forms.

Reaction 7: Intermolecular Dehydration → Ether

2 R-OH --conc. H2SO4, ~140 degC--> R-O-R + H2O e.g. 2 C2H5OH --conc. H2SO4, 140 degC--> C2H5-O-C2H5 + H2O (ethanol) (ethoxyethane / diethyl ether)

Example 3

Oxidation Products of Alcohols

Give the oxidation products of: (a) butan-1-ol with limited K₂Cr₂O₇/H⁺; (b) butan-1-ol with excess K₂Cr₂O₇/H⁺ under reflux; (c) butan-2-ol; (d) 2-methylpropan-2-ol.

Butan-1-ol is primary. Limited oxidant, distil off product → butanal (CH₃CH₂CH₂CHO).

Excess oxidant, reflux → butanoic acid (CH₃CH₂CH₂COOH).

Butan-2-ol is secondary → butan-2-one (CH₃COCH₂CH₃, a ketone). Orange → green.

2-methylpropan-2-ol is tertiary → no reaction. Dichromate remains orange.

Reaction 8: Iodoform (Triiodomethane) Test

Ethanol and all secondary alcohols with the structure CH₃CH(OH)R (methyl carbinols) give a yellow precipitate of CHI₃ (triiodomethane / iodoform) when warmed with iodine in alkaline solution (I₂/NaOH). This is a confirmatory test for ethanol and methyl ketones.

CH3CH(OH)R + I2/NaOH --> CHI3(s) + RCOONa (yellow ppt)

5.6

Ethers

Ethers — Definition Ethers are organic compounds with the general structure R–O–R', where two alkyl or aryl groups are joined by an oxygen atom. They are named as alkoxyalkanes (IUPAC) or as dialkyl ethers (common). General formula: C_nH_2n+2O (same as monohydric alcohols — they are isomers).

IUPAC Nomenclature of Ethers

The larger alkyl group is the parent chain; the smaller alkyl group + “oxy” is the prefix.

Structure	IUPAC Name	Common Name
CH₃–O–CH₃	Methoxymethane	Dimethyl ether
CH₃–O–C₂H₅	Methoxyethane	Methyl ethyl ether
C₂H₅–O–C₂H₅	Ethoxyethane	Diethyl ether
C₃H₇–O–CH₃	Methoxypropane	Methyl propyl ether

Physical Properties of Ethers

Ethers have no O–H group, so they cannot form H-bonds with each other. Their boiling points are similar to alkanes of comparable M_r and much lower than isomeric alcohols.

Example: ethoxyethane (C₄H₁₀O, B.P. 35°C) vs butan-1-ol (C₄H₁₀O, B.P. 118°C).

Ethers can accept H-bonds from water (O atom acts as H-bond acceptor), so short-chain ethers have slight solubility in water.

Chemical Properties of Ethers

Ethers are chemically very unreactive (stable C–O–C linkage). They do not react with sodium, oxidising agents, or bases under normal conditions. They are excellent inert solvents for organic reactions.

Important hazard: diethyl ether is extremely flammable (B.P. 35°C) and forms explosive peroxides on prolonged storage in air.

Alcohols vs Ethers (Isomers)

Alcohols and ethers with the same molecular formula are functional group isomers. Example: ethanol (CH₃CH₂OH) and methoxymethane (CH₃OCH₃) are both C₂H₆O.

Key Differences

Alcohols: react with Na (H₂ gas), oxidised by Cr₂O₇²⁻, form H-bonds, higher B.P. Ethers: do not react with Na, not oxidised, cannot H-bond with each other, lower B.P.

Example 4

Distinguishing Ethanol from Ethoxyethane

Describe two chemical tests to distinguish ethanol (CH₃CH₂OH) from ethoxyethane (C₂H₅OC₂H₅).

Test 1 — Sodium metal:
Ethanol: effervescence (H₂ gas), sodium dissolves.
Ethoxyethane: no reaction with sodium.

Test 2 — Acidified K₂Cr₂O₇:
Ethanol: orange solution turns green (oxidised to ethanal/ethanoic acid).
Ethoxyethane: no colour change (ethers not oxidised).

5.7

Uses of Alcohols and Ethers

Compound	Use	Reason
Ethanol	Alcoholic beverages	Produced by fermentation
Ethanol	Solvent (perfumes, medicines, paints)	Miscible with water and organic compounds
Ethanol	Biofuel (E10, E85 petrol blends)	Renewable; burns cleanly
Ethanol	Antiseptic and disinfectant (70% solution)	Denatures bacterial proteins
Methanol	Fuel (racing cars, camping stoves)	Burns cleanly; high octane
Methanol	Manufacture of methanal (formaldehyde) and acetic acid	Industrial feedstock
Methanol	Added to industrial ethanol (methylated spirits) as a denaturant	Makes ethanol unfit to drink (toxic)
Propan-2-ol	Rubbing alcohol; hand sanitiser	Antiseptic; evaporates quickly
Ethane-1,2-diol	Antifreeze in car radiators	Lowers freezing point of water
Ethane-1,2-diol	Manufacture of polyesters (PET)	Diol monomer for condensation polymerisation
Glycerol	Moisturiser in cosmetics and food	Hygroscopic; non-toxic
Glycerol	Manufacture of nitroglycerine (explosive); pharmaceuticals	Triol functional groups
Diethyl ether	Solvent for organic reactions (Grignard, LiAlH₄)	Inert; non-polar; low B.P.
Diethyl ether	Formerly used as anaesthetic	Now replaced due to flammability hazard

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✏️

Exercises

Name and classify each alcohol:
(a) CH₃CH₂CH(OH)CH₃ (b) (CH₃)₃COH (c) HOCH₂CH₂OH

(a) Butan-2-ol; secondary (C2 bonded to C1 and C3)
(b) 2-methylpropan-2-ol; tertiary (C2 bonded to three CH₃ groups)
(c) Ethane-1,2-diol; primary diol (each OH on a primary carbon)
Explain why ethanol (M_r = 46, B.P. = 78°C) has a much higher boiling point than propane (M_r = 44, B.P. = −42°C), despite having similar molecular masses.

Ethanol contains an –OH group and can form intermolecular hydrogen bonds (O–H···O) between molecules. These are much stronger than the van der Waals forces in propane. Significantly more thermal energy is required to overcome H-bonds, giving ethanol a much higher boiling point despite similar M_r.
Write the equation for the fermentation of glucose and state three conditions needed.

C₆H₁₂O₆ →^yeast 2 C₂H₅OH + 2 CO₂

Conditions: (1) Yeast (contains enzyme zymase); (2) Temperature 25–35°C; (3) Anaerobic conditions (no oxygen); (4) Slightly acidic pH (~4–6).
What are the products when propan-1-ol is oxidised with: (a) limited acidified K₂Cr₂O₇ (distil off product); (b) excess acidified K₂Cr₂O₇ under reflux?

Propan-1-ol is a primary alcohol.
(a) Limited oxidant, distil: → propanal (CH₃CH₂CHO) — an aldehyde. Orange → green.
(b) Excess oxidant, reflux: → propanoic acid (CH₃CH₂COOH). Orange → green.
Give the IUPAC name and the common name for C₂H₅–O–C₂H₅. Explain why this compound has a much lower boiling point (35°C) than butan-1-ol (118°C), even though both have the formula C₄H₁₀O.

IUPAC: ethoxyethane; Common: diethyl ether.

Butan-1-ol has an –OH group and forms strong intermolecular hydrogen bonds, requiring much more energy to vaporise. Ethoxyethane has no O–H group and cannot form H-bonds between its own molecules; it has only weaker van der Waals forces, giving it a much lower boiling point despite identical molecular formula.
A student reacts ethanol with ethanoic acid in the presence of concentrated H₂SO₄. (a) Name the type of reaction. (b) Write the equation. (c) Name the organic product. (d) How could the yield be increased?

(a) Esterification (Fischer esterification)
(b) C₂H₅OH + CH₃COOH ⇌^{conc. H₂SO₄} CH₃COOC₂H₅ + H₂O
(c) Ethyl ethanoate
(d) Use excess of one reactant (e.g. excess alcohol); remove water as it forms (molecular sieves or distillation); distil off ester as it forms to shift equilibrium to right (Le Chatelier's principle).

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Unit Test

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Instructions Total: 50 marks | Time: 50 minutes | Attempt all questions | Show all working.

Section A — Short Answer

30 marks

Q1 [4 marks]

Name and classify each as primary, secondary, or tertiary:
(a) CH₃OH (b) CH₃CH(OH)CH₂CH₃ (c) (CH₃)₃COH (d) HOCH₂CH(OH)CH₂OH

(a) Methanol; primary
(b) Butan-2-ol; secondary
(c) 2-methylpropan-2-ol; tertiary
(d) Propane-1,2,3-triol (glycerol); primary (C1 and C3) and secondary (C2)

Q2 [5 marks]

State all the conditions required for fermentation. Write the equation for the fermentation of glucose. Calculate the volume of CO₂ produced at STP when 90 g of glucose is fermented completely. (Molar volume at STP = 22.4 L/mol)

Conditions: yeast (enzyme zymase); temperature 25–35°C; anaerobic conditions (no O₂); slightly acidic pH (~4–6).

Equation: C₆H₁₂O₆ → 2 C₂H₅OH + 2 CO₂

Calculation:
M_r(glucose) = 180 g/mol → 90 g = 0.5 mol glucose
1 mol glucose → 2 mol CO₂→ 0.5 mol → 1.0 mol CO₂
Volume = 1.0 × 22.4 = 22.4 L

Q3 [6 marks]

Write equations for the following reactions of propan-1-ol:
(a) Combustion (complete) (b) Reaction with sodium (c) Oxidation with limited K₂Cr₂O₇/H⁺ (d) Oxidation with excess K₂Cr₂O₇/H⁺ under reflux (e) Dehydration with conc. H₂SO₄ at 170°C (f) Esterification with propanoic acid (name the ester)

(a) 2 C₃H₇OH + 9 O₂ → 6 CO₂ + 8 H₂O
(b) 2 CH₃CH₂CH₂OH + 2 Na → 2 CH₃CH₂CH₂ONa + H₂
(c) CH₃CH₂CH₂OH →^[O] CH₃CH₂CHO (propanal)
(d) CH₃CH₂CH₂OH →^{[O] excess} CH₃CH₂COOH (propanoic acid)
(e) CH₃CH₂CH₂OH →^{conc. H₂SO₄, 170°C} CH₃CH=CH₂ + H₂O (propene)
(f) C₃H₇OH + C₂H₅COOH ⇌^{conc. H₂SO₄} C₂H₅COOC₃H₇ + H₂O — propyl propanoate

Q4 [4 marks]

Give the IUPAC name and common name of C₂H₅OC₂H₅. Give two properties that make it a useful laboratory solvent, and one safety hazard associated with its use.

IUPAC: ethoxyethane; Common: diethyl ether

Properties as solvent: (any two) chemically inert (does not react with most reagents); low boiling point (35°C, easy to remove by evaporation); good solubility for non-polar and polar organic compounds; immiscible with water (useful for extraction).

Safety hazard: extremely flammable (B.P. 35°C; flash point −45°C); forms explosive peroxides on prolonged storage in air.

Q5 [5 marks]

Compare fermentation and hydration of ethene as industrial methods of producing ethanol. Discuss raw materials, conditions, rate, purity, and sustainability. Conclude which is preferable in a country that lacks oil but has abundant agricultural land.

Fermentation: Raw material = sugars/starch (renewable, agricultural); T = 25–35°C (low energy); atmospheric pressure; slow batch process; dilute product (~14%) requiring distillation; sustainable; carbon-neutral cycle.

Hydration: Raw material = ethene from crude oil (non-renewable); T = ~300°C (high energy); P = ~60 atm; fast continuous process; purer product; not sustainable long-term.

Conclusion: For a country without oil but with agricultural land, fermentation is preferable: it uses locally available renewable crops, requires less energy and infrastructure, and is economically viable without importing petroleum.

Q6 [6 marks]

Three unlabelled bottles contain ethanol, propan-2-ol, and 2-methylpropan-2-ol. Describe a sequence of chemical tests to identify each compound. State the reagent(s) used, observations, and reasoning for each step.

Test 1 — Acidified K₂Cr₂O₇ (warm):
All three give orange solution. Add and warm.
• Two samples turn green (ethanol and propan-2-ol are oxidised).
• One sample stays orange → 2-methylpropan-2-ol identified (tertiary, not oxidised).

Test 2 — Distinguish ethanol from propan-2-ol using iodoform test:
Add I₂/NaOH solution and warm to the two remaining samples.
• Yellow precipitate of CHI₃ → ethanol identified (has CH₃CH(OH)– structure).
• No yellow precipitate → propan-2-ol identified (wait — propan-2-ol also has CH₃CHOH structure → also gives iodoform!)

Revised Test 2 — Oxidation product test:
Oxidise both remaining with K₂Cr₂O₇/H⁺ and test the product with Tollens’ reagent (ammoniacal AgNO₃):
• Silver mirror → aldehyde produced → original alcohol was ethanol (primary → ethanal).
• No silver mirror → ketone produced → original alcohol was propan-2-ol (secondary → propanone).

Section B — Extended Response

20 marks

Q7 [10 marks]

(a) Describe all the chemical reactions of ethanol, writing equations for each. Include: combustion, reaction with sodium, dehydration, esterification, oxidation (to aldehyde and acid), and conversion to a halogenoalkane. [8 marks]

(b) Ethanol can be produced both by fermentation and by hydration of ethene. Write the equation for each process and state the conditions. [2 marks]

(a)
Combustion: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
With Na: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂
Dehydration (alkene): C₂H₅OH →^{conc. H₂SO₄, 170°C} CH₂=CH₂ + H₂O
Dehydration (ether): 2C₂H₅OH →^{conc. H₂SO₄, 140°C} C₂H₅OC₂H₅ + H₂O
Esterification: C₂H₅OH + CH₃COOH ⇌^{conc. H₂SO₄} CH₃COOC₂H₅ + H₂O
Oxidation to aldehyde: C₂H₅OH →^{[O] limited, distil} CH₃CHO (ethanal)
Oxidation to acid: C₂H₅OH →^{[O] excess, reflux} CH₃COOH (ethanoic acid)
Halogenoalkane: C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl

(b)
Fermentation: C₆H₁₂O₆ →^{yeast, 30°C, anaerobic} 2 C₂H₅OH + 2 CO₂
Hydration: CH₂=CH₂ + H₂O →^{conc. H₃PO₄, 300°C, 60 atm} C₂H₅OH

Q8 [10 marks]

(a) Using ethanol as your starting material and any other inorganic reagents, describe with equations how you would prepare: (i) ethanal; (ii) ethanoic acid; (iii) ethene; (iv) bromoethane; (v) ethyl ethanoate. State conditions for each. [8 marks]

(b) Ethane-1,2-diol is used as antifreeze. Explain, in terms of intermolecular forces, why it is effective at preventing water from freezing, and state one other use. [2 marks]

(a)(i) Ethanal: C₂H₅OH →^{acidified K₂Cr₂O₇, limited, distil off} CH₃CHO
(ii) Ethanoic acid: C₂H₅OH →^{excess acidified K₂Cr₂O₇, reflux} CH₃COOH
(iii) Ethene: C₂H₅OH →^{conc. H₂SO₄, 170°C} CH₂=CH₂ + H₂O
(iv) Bromoethane: C₂H₅OH + HBr →^{NaBr + conc. H₂SO₄, heat} C₂H₅Br + H₂O
(v) Ethyl ethanoate: C₂H₅OH + CH₃COOH ⇌^{conc. H₂SO₄} CH₃COOC₂H₅ + H₂O

(b) Ethane-1,2-diol (HOCH₂CH₂OH) has two –OH groups that form many hydrogen bonds with water molecules. This disrupts the regular H-bond network needed for ice crystal formation, lowering the freezing point of the mixture significantly below 0°C.
Other use: manufacture of PET (polyethylene terephthalate) polyester — reacts with terephthalic acid in condensation polymerisation.

← Unit 4: Halogenoalkanes Unit 6: Aldehydes & Ketones →