C — (CH₃)₃COH is 2-methyl-2-propanol (tert-butanol). The carbon bearing –OH is bonded to three other carbons → tertiary. A = 1°, B = 2°, D = 1°. Slide p.6; McMurry & Ballantine 8e Ch 17 §17.1
Question 2
The Lucas test uses ZnCl₂ in concentrated HCl. A tertiary alcohol gives a turbid (cloudy) solution:
B — 3° alcohols react immediately because the stable tertiary carbocation forms instantly. 2° requires 5–10 min; 1° does not react at RT (requires heat). Slide p.22–23; McMurry 8e §17.7
Question 3
Dehydration of 2-butanol at 170°C with H₂SO₄ produces mainly:
C — Zaitsev's rule: the more substituted (more stable) alkene is the major product. 2-butene (83%) vs 1-butene (17%). At 140°C ether forms; at 170°C alkene. Slide p.26; McMurry 8e §17.8
Question 4
Oxidation of a primary alcohol with excess KMnO₄ in acid yields:
D — 1° alcohol + KMnO₄/H⁺ → carboxylic acid (RCOOH). PCC stops at the aldehyde stage; K₂Cr₂O₄ and KMnO₄ go all the way. Ketones form from 2° alcohols; 3° alcohols do not oxidise. Slide p.29–31; McMurry 8e §17.10
Question 5
Which reagent oxidises a secondary alcohol to a ketone but cannot oxidise further?
A — PCC oxidises 1° → aldehyde (stops there) and 2° → ketone; it cannot oxidise ketones or carboxylic acids. NaBH₄ is a reducer. Slide p.29; McMurry 8e §17.10
Question 6
Phenol is a stronger acid than ethanol because:
B — Resonance stabilisation. The phenoxide anion (C₆H₅O⁻) delocalises the negative charge over the ring → lower energy → stronger acid. Acidity order: RCOOH > H₂CO₃ > phenol > H₂O > ROH. Slide p.38–39; McMurry 8e §16.4
Question 7
Adding FeCl₃ solution to an unknown compound gives a violet colour. The compound is most likely:
C — Phenol + FeCl₃ → violet complex. This is the classic phenol identification test. Alcohols, ethers, and thiols do not give this colour. Slide p.42; McMurry 8e §16.4
Question 8
Phenol reacts with excess Br₂ water to give a white precipitate. The product is:
D — 2,4,6-tribromophenol (white precipitate). –OH is a powerful o/p director; all three positions (2,4,6) are brominated rapidly without a Lewis acid catalyst. Used as a confirmatory test for phenol. Slide p.40; McMurry 8e §16.5
A — CH₃CH₂O⁻Na⁺ + CH₃CH₂CH₂Br → CH₃CH₂OCH₂CH₂CH₃ (ethyl propyl ether) + NaBr. This S₂ reaction requires a primary alkyl halide to avoid elimination. Slide p.58–59; McMurry 8e §17.11
Question 10
Diethyl ether stored in air forms explosive peroxides. The test to detect them uses:
B — KI/starch paper. Peroxides oxidise I⁻ → I₂, which turns starch blue. This is the standard safety test before distilling aged ethers. Slide p.52; McMurry 8e §17.11
Question 11
Excess HI cleaves diethyl ether. The products are:
C — With excess HI, both C–O bonds cleave: CH₃CH₂OCH₂CH₃ + 2 HI → 2 CH₃CH₂I + H₂O. With 1 equiv HI the intermediate is an alcohol + iodide. Slide p.54–57; McMurry 8e §17.11
Question 12
Which is the CORRECT statement about thiols (R–SH)?
D — 2 RSH + I₂ → RSSR + 2 HI (mild oxidation; I₂ = iodine). The reverse (RSSR → 2 RSH) uses Zn/H⁺ or other reducing agents. In biology: cysteine → cystine. Slide p.61–62; McMurry 8e §17.12
Question 13
The amino acid whose side chain oxidises to form a disulfide bond is:
A — Cysteine (R = –CH₂SH) is oxidised to cystine (two cysteine residues linked by –S–S–). Disulfide bonds stabilise tertiary protein structure. Serine and threonine carry –OH; tyrosine carries ArOH. Slide p.62; McMurry 8e §28.2
Question 14
Which statement about alcohols and water solubility is CORRECT?
B — –OH forms H-bonds with water, so C₁–C₄ alcohols are miscible. Beyond C₅, the hydrophobic hydrocarbon chain dominates and solubility drops sharply. Slide p.12–14; McMurry 8e §17.2
Question 15
TRUE or FALSE: There is no such thing as a quaternary alcohol. (2019 Past Paper III.7)
C — TRUE (the statement in the exam is TRUE). A quaternary carbon is bonded to four other carbons — it has no H and no room for –OH. Neopentyl alcohol [(CH₃)₃CCH₂OH] has a quaternary carbon but the –OH is on an adjacent primary carbon. 2019 Past Paper Part III Q7; Slide p.6
Question 16
IUPAC name of CH₃CH(OH)CH₂CH₃?
D — 2-butanol. Number from the end that gives the –OH the lowest locant: –OH on C2. “sec-butanol” (C) is a correct common name but not the IUPAC name. Slide p.8; McMurry 8e §17.1
Question 17
Reaction of alcohol with Na metal produces:
A — ROH + Na → RO⁻Na⁺ + ½ H₂. The alkoxide ion (RO⁻) is the product; the reaction is less vigorous than Na + water because alcohol is a weaker acid than water. Slide p.16; McMurry 8e §17.5
Question 18
What is the product of mild oxidation of 2-propanol with K₂Cr₂O₄?
B — 2-propanol is a secondary alcohol; K₂Cr₂O₄ oxidises it to acetone (propanone). The ketone cannot be oxidised further under these conditions. Slide p.30; McMurry 8e §17.10
Question 19
Phenol is less acidic than carbonic acid (H₂CO₃). This means phenol:
C — Classic property. Phenol dissolves in strong base (NaOH) to form sodium phenoxide. But because H₂CO₃ is a stronger acid than phenol, bubbling CO₂ through sodium phenoxide solution reprotoates phenol → white turbidity. This distinguishes phenol from carboxylic acids. Slide p.38–39; McMurry 8e §16.4
Question 20
Dehydration of ethanol at 140°C with H₂SO₄ produces mainly:
B — At 140°C, intermolecular dehydration gives diethyl ether (ethoxyethane). At 170°C, intramolecular dehydration gives ethene. Temperature controls the product! Slide p.24–26; McMurry 8e §17.8
1. Lucas reagent and Lucas test▼
A mixture of anhydrous ZnCl₂ in concentrated HCl. Used to distinguish 1°, 2°, and 3° alcohols by the time to form a turbid (cloudy) solution of the alkyl chloride. 3° = immediate; 2° = 5–10 min; 1° = no reaction at room temperature. Based on S⁼1 reactivity (carbocation stability). Slide p.22–23; McMurry 8e §17.7
2. Zaitsev's rule▼
In elimination reactions, the major product is the more-substituted (more stable) alkene. Applied to alcohol dehydration: 2-butanol gives mainly 2-butene (83%) rather than 1-butene (17%). Applies because the transition state for the major product resembles the more stable alkene. Slide p.26; McMurry 8e §17.8
3. Williamson ether synthesis▼
An S₂ reaction between an alkoxide ion (RO⁻) and a primary alkyl halide (R'X) to form an unsymmetrical ether: RO⁻ + R'X → R–O–R' + X⁻. Secondary or tertiary R'X gives elimination instead. Named after Alexander Williamson (1852). Slide p.58–59; McMurry 8e §17.11
4. Oxidation ladder of alcohols▼
1° ROH → RCHO (aldehyde, with PCC) → RCOOH (carboxylic acid, with KMnO₄ or K₂Cr₂O₄). 2° ROH → RCOR' (ketone). 3° ROH → no reaction (no H on the alpha carbon). The oxidation level increases as C–H bonds are replaced by C=O or C–O bonds. Slide p.29–31; McMurry 8e §17.10
5. Phenol▼
A compound with a hydroxyl group (–OH) directly attached to a benzene ring. Phenol is a stronger acid than aliphatic alcohols (pKₐ ~10) but weaker than carboxylic acids. It reacts with NaOH to form sodium phenoxide; CO₂ reprotoates phenoxide back to phenol. Identified by: FeCl₃ (violet), Br₂ water (2,4,6-tribromophenol white ppt). Slide p.33–42; McMurry 8e §16.4–16.5
6. Disulfide bond▼
A covalent –S–S– bond formed by mild oxidation of two thiol (–SH) groups: 2 RSH + I₂ → R–S–S–R + 2 HI. Reduced back to thiols by Zn/H⁺ or other reducing agents. In biology, cysteine residues in proteins form cystine linkages that stabilise tertiary and quaternary protein structure. Slide p.61–62; McMurry 8e §17.12
1. Describe the Lucas test. What are the results for 1°, 2°, and 3° alcohols and what is the mechanistic explanation?Essay
Reagent: Anhydrous ZnCl₂ in concentrated HCl (Lucas reagent).
Principle: Converts –OH to –Cl via S⁼1 (for 3° and 2°) or S⁼2 (for 1°). The alkyl chloride is insoluble in the aqueous reagent, causing turbidity.
3° alcohol: Turbid immediately at RT — stable 3° carbocation forms at once.
2° alcohol: Turbid in 5–10 min — 2° carbocation forms slowly.
1° alcohol: No reaction at RT; requires heat (or HBr) — no stable carbocation.
Mechanism (for 3°): ZnCl₂ acts as a Lewis acid coordinating to the –OH, facilitating departure as water to form the 3° carbocation; Cl⁻ attacks to give R–Cl.
Cite: Slide p.22–23; McMurry 8e §17.7
2. Explain the oxidation reactions of primary, secondary, and tertiary alcohols with appropriate reagents and products.Essay
+ KMnO₄/H⁺ or K₂Cr₂O₄/H₂SO₄ → carboxylic acid (RCOOH) (over-oxidation).
2° alcohol:
+ any chromium/permanganate oxidant → ketone (RCOR'). Ketones are resistant to further oxidation.
3° alcohol:
No reaction — no α-hydrogen on the carbon bearing –OH.
Cite: Slide p.29–31; McMurry 8e §17.10
3. Compare the acidity of phenol with water, alcohol, and carboxylic acids. How would you distinguish phenol from an aliphatic alcohol by chemical tests?Essay
Reason phenol > ROH: The phenoxide anion (C₆H₅O⁻) is stabilised by resonance — negative charge delocalized into the ring. Alkoxide (RO⁻) has no resonance stabilisation.
Chemical distinction:
FeCl₃ test: Phenol → violet; alcohol → no colour.
Br₂ water: Phenol → 2,4,6-tribromophenol white ppt; aliphatic alcohol → no immediate reaction.
Na₂CO₃: RCOOH reacts with Na₂CO₃ to evolve CO₂; phenol and alcohols do not (phenol is too weak).
Cite: Slide p.38–42; McMurry 8e §16.4
4. Describe the Williamson ether synthesis. Why must a primary alkyl halide be used? Give one example.Essay
Reaction: NaOR + R'X → R–O–R' + NaX
Mechanism: S₂ — the alkoxide (RO⁻) is a nucleophile; it attacks the back of the primary carbon bearing X.
Why primary R'X? Secondary and tertiary R'X undergo elimination rather than substitution when attacked by a strong base like RO⁻. Primary R'X has minimal steric hindrance and no α-H preference for elimination.
Reduction: RSSR + 2[H] (Zn/H⁺ or β-mercaptoethanol) → 2 RSH.
Biological significance:
Disulfide bonds (–S–S–) in proteins are covalent cross-links that stabilise tertiary (intramolecular) and quaternary (intermolecular) protein structure.
Example: insulin has 3 disulfide bonds critical for its active conformation.
Hair waving/perming: disulfide bonds in keratin are broken with reducing agents and reformed in a new shape by oxidation.