Class 12 Chemistry Chapter 7: Alcohols, Phenols and Ethers worksheet with answer key

Answer: (b) 2-methylpropan-1-ol

In hydroboration boron adds to the sp² carbon carrying more hydrogens, the =CH₂ end; oxidation replaces boron by –OH: (CH₃)₂CH–CH₂OH, 2-methylpropan-1-ol. This looks like anti-Markovnikov addition of water. (a) is the acid-catalysed hydration product; (c) and (d) have a different carbon skeleton.

Answer: (c) benzene-1,3-diol

Catechol is benzene-1,2-diol, resorcinol benzene-1,3-diol, and hydroquinone (quinol) benzene-1,4-diol. 3-Methylphenol is m-cresol.

Answer: (b) butane-1,2-diol

For a polyhydric alcohol the 'e' of the alkane is kept and 'di' goes before 'ol'; the chain is numbered to give the –OH groups the lowest locants: butane-1,2-diol. (a) drops the 'e'; (c) numbers from the wrong end; (d) treats –OH as a prefix.

Answer: (c) 2-hydroxy-5-methylbenzaldehyde

This is the Reimer–Tiemann reaction: –CHCl₂ enters ortho to the –O⁻ group and the substituted benzal chloride is hydrolysed by the alkali to –CHO. The para position is already held by CH₃, so –CHO goes ortho to –OH. Numbering from the CHO carbon: CHO at C-1, OH at C-2, and CH₃ (para to OH) at C-5. (b) is the Kolbe product (CO₂, not CHCl₃); (a) loses the –OH; (d) puts CH₃ meta to OH.

Answer: (b) CH₃–CH₂–C(CH₃)(OH)–CH₂–CH₃

In CH₃CH₂C(CH₃)(OH)CH₂CH₃ the –OH is on a carbon bonded to three other carbons: tertiary. (a) and (d) have –OH on a –CH₂– carbon (primary); in (c) the –OH carbon carries one hydrogen and two carbons (secondary).

(i) Methanol is oxidised first to methanal and then to methanoic acid, which may cause blindness and death. (ii) Ethanol competes for the enzyme that oxidises the aldehyde to the acid and swamps it, so less methanoic acid forms and the kidneys have time to excrete the methanol.

• methanal, then methanoic acid (1)
• ethanol swamps the enzyme; methanol excreted by kidneys (1)

C₆H₅OH + Zn → C₆H₆ + ZnO. In phenol the C–O bond has partial double bond character (the lone pair on oxygen is conjugated with the ring) and the carbon is sp² hybridised; such a carbon cannot undergo the nucleophilic substitution by Br⁻ that cleavage would need. C–O cleavage of this kind belongs to alcohols.

• C₆H₅OH + Zn → C₆H₆ + ZnO (1)
• partial double bond, sp² carbon resists nucleophilic substitution (1)

(i) Phenol-2,4-disulphonic acid; with concentrated nitric acid it gives 2,4,6-trinitrophenol (direct nitration of phenol gives a poor yield). (ii) The three –NO₂ groups withdraw electrons from the ring and the oxygen, which makes release of the hydrogen ion easy.

• phenol-2,4-disulphonic acid (1)
• three electron withdrawing –NO₂ groups ease release of H⁺ (1)

1. The alcohol is protonated: (CH₃)₃C–OH + H⁺ → (CH₃)₃C–OH₂⁺. 2. Water leaves, giving the carbocation (CH₃)₃C⁺; this is the slowest, rate-determining step. 3. A proton is lost from a CH₃ group: (CH₃)₂C=CH₂ + H⁺, and the acid is regenerated. Because step 2 decides the rate and the tertiary carbocation is the most stable, it forms easily; ethanol would have to form a much less stable primary carbocation, so it needs harsher conditions.

• step 1: protonation of the alcohol (1)
• step 2: (CH₃)₃C⁺ formed, rate determining (1)
• step 3: loss of H⁺ to 2-methylpropene; tertiary carbocation most stable, so easier (1)

Any two of: C₆H₅COCH₃ + C₂H₅MgBr; C₆H₅COCH₂CH₃ + CH₃MgBr; CH₃COCH₂CH₃ + C₆H₅MgBr. For example: C₆H₅COCH₃ + C₂H₅MgBr → C₆H₅–C(CH₃)(C₂H₅)–OMgBr; then + H₂O → C₆H₅–C(CH₃)(OH)–CH₂CH₃ + Mg(OH)Br. A ketone is needed because the product is a tertiary alcohol.

• first correct ketone + Grignard pair (1)
• second correct pair (1)
• equation with adduct and hydrolysis (1)

(i) 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂: ethanol gives up the proton of its O–H. (ii) C₂H₅OH + H⁺ → C₂H₅OH₂⁺ (with a strong acid such as H₂SO₄): the unshared pairs on oxygen accept a proton. (iii) Ethanol is a poorer proton donor than water, so its conjugate base, the ethoxide ion, is a better proton acceptor than hydroxide.

• acid behaviour, with sodium (1)
• base behaviour, protonation to C₂H₅OH₂⁺ (1)
• ethanol poorer proton donor than water, so ethoxide stronger base (1)