This MCQ module is based on: Classification Preparation Alcohols Phenols
This assessment will be based on: Classification Preparation Alcohols Phenols
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Replace one hydrogen of water by an alkyl group and you arrive at an alcohol (R–OH); replace it by an aryl group and you have a phenol (Ar–OH). Slip an alkyl group between the two hydrogens of water and you get an ether (R–O–R'). Three closely related families, yet their chemistry diverges dramatically because of the carbon to which the oxygen is attached — sp³ in alcohols and ethers, sp² (aromatic) in phenols.
These oxygen-containing molecules are everywhere: ethanol in beverages and sanitisers, methanol as an industrial feedstock, glycerol in soaps and pharmaceuticals, phenol as an antiseptic (Lister, 1867) and the precursor to aspirin, bakelite and countless dyes; diethyl ether was the first surgical anaesthetic; anisole and eugenol give cloves and aniseed their scent. Learning their preparation and properties opens up a large slice of modern organic chemistry.
If the –OH is on an sp³ carbon, the alcohol is 1° (primary), 2° (secondary) or 3° (tertiary) depending on whether the –OH-bearing carbon is attached to 1, 2 or 3 other carbons. Two special sp³-OH cases are named separately:
If the –OH is on an sp² carbon of C=C the compound is a vinyl alcohol / enol (CH₂=CH–OH) — normally unstable and tautomerises to the carbonyl (acetaldehyde, CH₃CHO).
A phenol carries one or more –OH groups directly attached to the benzene ring. Common examples: phenol itself (C₆H₅OH), the three cresols (methylphenols), catechol (1,2-), resorcinol (1,3-) and hydroquinone (1,4-dihydroxybenzene).
An ether has the –O– oxygen sandwiched between two carbons: R–O–R'. When the two groups are identical (e.g., C₂H₅–O–C₂H₅) the ether is symmetrical; when they differ (e.g., CH₃–O–C₂H₅) it is unsymmetrical (a mixed ether).
IUPAC: take the longest carbon chain that contains the –OH carbon, drop the terminal "e" of the alkane and append "ol". Number the chain so that –OH gets the lowest locant. For common names, combine the alkyl group name with the word "alcohol".
| Structure | Common name | IUPAC name |
|---|---|---|
| CH₃OH | Methyl alcohol | Methanol |
| CH₃CH₂OH | Ethyl alcohol | Ethanol |
| (CH₃)₂CHOH | Isopropyl alcohol | Propan-2-ol |
| (CH₃)₃COH | tert-Butyl alcohol | 2-Methylpropan-2-ol |
| CH₂=CHCH₂OH | Allyl alcohol | Prop-2-en-1-ol |
| HOCH₂CH₂OH | Ethylene glycol | Ethane-1,2-diol |
| HOCH₂CH(OH)CH₂OH | Glycerol | Propane-1,2,3-triol |
The parent name is simply phenol. Substituents are indicated with locants numbering the –OH carbon as C-1, or with o-, m-, p- prefixes for disubstituted rings.
Common name: list the two alkyl/aryl groups alphabetically followed by "ether" (ethyl methyl ether, diethyl ether). IUPAC: the smaller R–O– group is treated as an alkoxy substituent on the larger alkane (methoxyethane, ethoxyethane). Anisole = methoxybenzene (C₆H₅OCH₃).
The oxygen in R–O–H is sp³-hybridised — two of its four sp³ orbitals hold lone pairs, and the other two form σ bonds to C and H. The C–O–H angle in methanol is about 108.9°, very close to water's 104.5° and the perfect tetrahedral angle 109.5°. The slight opening compared with water reflects lower lone-pair/lone-pair repulsion when an alkyl group replaces one H.
In phenol, the oxygen is bonded to an sp² aromatic carbon. An oxygen lone pair can delocalise into the ring, giving the C–O bond partial double-bond character. Consequences: the C(aryl)–O bond is shorter (~136 pm) than a C(sp³)–O bond (~143 pm), the –OH is harder to break from the ring (no C–O cleavage), and the ring becomes electron-rich at the ortho and para positions.
The ether oxygen is sp³-hybridised as in alcohols. The C–O–C angle in dimethyl ether is 111.7°, slightly larger than tetrahedral because the two bulky methyl groups push each other apart. C–O bond length ≈ 141 pm.
Alkene + H₂O in dilute acid proceeds via protonation to the more stable carbocation; water then captures the cation and loses a proton. The OH ends up on the more substituted carbon.
Diborane adds to the alkene so that boron attaches to the less hindered (terminal) carbon; the resulting trialkylborane is then oxidised by alkaline H₂O₂ to give the alcohol with OH on that same carbon — formally an anti-Markovnikov hydration. The net stereochemistry is syn-addition. This is the cleanest way to make primary alcohols from terminal alkenes.
Metal hydrides (LiAlH₄, NaBH₄) or catalytic hydrogenation (H₂/Ni, Pd or Pt) deliver hydride to the carbonyl carbon:
LiAlH₄ — a more powerful hydride than NaBH₄ — reduces acids and esters all the way to primary alcohols.
The magnesium-carbon bond of R–MgX is strongly polarised with δ⁻ on carbon; the carbanion-like R adds to the carbonyl carbon of an aldehyde, ketone, ester or CO₂. Aqueous workup then gives the alcohol.
| Carbonyl partner | Product after H₃O⁺ | Class of alcohol |
|---|---|---|
| HCHO (formaldehyde) | R–CH₂OH | Primary (1°) |
| R'CHO (other aldehyde) | R–CH(OH)–R' | Secondary (2°) |
| R'₂C=O (ketone) | R–C(OH)R'₂ | Tertiary (3°) |
| CO₂ (dry ice) | R–COOH | Carboxylic acid (not an alcohol) |
| R'COOR'' (ester) | R₂C(OH)–R' | Tertiary (uses 2 equiv RMgX) |
Chlorobenzene is heated with aqueous NaOH under severe conditions (623 K / 300 atm); acidification of the resulting sodium phenoxide gives phenol. The high temperature and pressure are needed because the C–Cl bond on the ring is strengthened by resonance and is reluctant to break.
Benzenesulphonic acid is obtained by sulphonating benzene with oleum. Fusing its sodium salt with solid NaOH and then acidifying the mixture gives phenol.
A primary aromatic amine is first converted to a diazonium salt (NaNO₂/HCl, 273–278 K). Warming the aqueous solution replaces –N₂⁺ by –OH, liberating nitrogen gas.
More than 90% of the world's phenol is made this way. Cumene (isopropylbenzene, from benzene + propene/H₃PO₄) is oxidised by air to cumene hydroperoxide, which rearranges in dilute acid to give phenol and acetone — both commercially valuable.
Name (CH₃)₂CH–CH(OH)–CH₃.
Step 1: longest chain through the –OH carbon = 4 C (butane).
Step 2: numbering from the end nearer the –OH: C-1 CH₃, C-2 CH(OH), C-3 CH(CH₃), C-4 CH₃. So the –OH is at C-2 and the methyl branch is at C-3.
Answer: 3-methylbutan-2-ol.
What is the major product when 2-methylpropene is treated with dilute H₂SO₄?
Protonation gives two possible cations: (CH₃)₃C⁺ (3°, stable) or (CH₃)₂CH–CH₂⁺ (1°, unstable). The 3° cation forms; water attacks it and deprotonation yields tert-butanol.
Predict the alcohol obtained from but-1-ene + B₂H₆, then alkaline H₂O₂.
Boron attaches to the terminal (less hindered) carbon; –OH replaces B in the oxidation step. The net effect is anti-Markovnikov water addition.
Starting from ethylmagnesium bromide, how would you make (i) propan-1-ol, (ii) butan-2-ol, (iii) 2-methylbutan-2-ol?
(i) React C₂H₅MgBr with HCHO, then H₃O⁺ → C₂H₅–CH₂OH (propan-1-ol).
(ii) React C₂H₅MgBr with CH₃CHO → CH₃–CH(OH)–C₂H₅ (butan-2-ol).
(iii) React C₂H₅MgBr with CH₃COCH₃ → (CH₃)₂C(OH)–C₂H₅ (2-methylbutan-2-ol).
Suggest a reagent that reduces only the C=O of hex-2-enal (CH₃CH₂CH₂CH=CHCHO) and leaves the C=C untouched.
NaBH₄ is selective for carbonyls and does not reduce isolated C=C. (LiAlH₄ would also leave the double bond alone but is harder to handle.) Product: hex-2-en-1-ol.
Convert aniline into phenol.
List the economic advantages of the cumene process over the Dow process.
(i) It operates at low pressure and moderate temperature, avoiding the 300 atm / 623 K conditions of the Dow process. (ii) Both co-products — phenol and acetone — have large markets, so the route has essentially no waste by-products. (iii) Cumene is easily prepared from cheap benzene and propene.
You have a bottle of methylmagnesium iodide (CH₃MgI) and access to four carbonyl compounds in the stockroom: (a) HCHO, (b) CH₃CHO, (c) (CH₃)₂C=O, (d) dry CO₂.
(a) HCHO: CH₃–CH₂–OH (ethanol, 1°).
(b) CH₃CHO: (CH₃)₂CH–OH (propan-2-ol, 2°).
(c) (CH₃)₂C=O: (CH₃)₃C–OH (2-methylpropan-2-ol, 3°).
(d) CO₂: CH₃COOH (ethanoic acid — carboxylic acid, not an alcohol).
Pick a target compound and a starting material; the tool recommends the best route.
1. Which reagent combination converts but-1-ene cleanly into butan-1-ol?
2. Classify 2-methylpropan-2-ol and justify.
3. True/False: Enols (C=C–OH) are generally more stable than the corresponding aldehyde/ketone tautomer.
4. Fill in the blank: The co-product obtained along with phenol in the cumene process is ________.
5. Give the structure of the alcohol produced when phenylmagnesium bromide (C₆H₅MgBr) reacts with propanal followed by H₃O⁺.
Assertion (A): The C–O bond in phenol is shorter than that in methanol.
Reason (R): In phenol, the oxygen lone pair is delocalised into the aromatic ring, giving partial double-bond character to the C–O linkage.
Assertion (A): Reaction of phenylmagnesium bromide with dry ice followed by H₃O⁺ gives benzyl alcohol.
Reason (R): Grignard reagents add to C=O of CO₂ to give primary alcohols.
Assertion (A): The Dow process for phenol requires temperature ≈ 623 K and pressure ≈ 300 atm.
Reason (R): The C(aryl)–Cl bond has partial double-bond character and is resistant to nucleophilic substitution.
Sit a full paper on what you have been studying, marked question by question.