JEE Chemistry Alcohols Phenols and Ethers MCQs Set 07

Answer: (b) Intermolecular H-bonding in alcohols

Explanation:
1. Alcohols have an O–H bond, so their molecules are joined by intermolecular hydrogen bonds.
2. Ethers have no O–H bond and cannot form hydrogen bonds with each other.
3. Less energy is needed to separate ether molecules, so ethers boil lower and are more volatile.

Teacher's Note:
1. Hydrogen bonding raises boiling point.
2. Compare ethanol (78 C) with dimethyl ether (-24 C) of the same formula.

Answer: (c) CH₃CH₂C(CH₃)(OH)CH₂CH₃

Explanation:
1. Acid dehydration goes through a carbocation, so the alcohol that gives the most stable carbocation reacts fastest.
2. Option (c) is a tertiary alcohol and gives a tertiary carbocation; (a) and (d) are primary and (b) is secondary.
3. Order of ease of dehydration: tertiary > secondary > primary, so (c) is correct.

Teacher's Note:
1. Ease of dehydration follows 3° > 2° > 1°.
2. Identify the class of the carbon bearing OH first.

Answer: (d) C₆H₅CH(OH)CH₃

Explanation:
1. The iodoform test is given by compounds containing CH₃CH(OH)– (or CH₃CO–) attached to carbon or hydrogen.
2. Only C₆H₅CH(OH)CH₃ has the CH₃CH(OH)– unit. In (a) both sides of the carbinol carbon are ethyl groups, and (b) and (c) are primary alcohols other than ethanol.
3. So (d) gives yellow iodoform (CHI₃).

Teacher's Note:
1. Look for CH₃–CH(OH)– in the structure.
2. Ethanol is the only primary alcohol that gives the test.

Answer: (c) 2,4,6-Tribromophenol

Explanation:
1. The –OH group strongly activates the ring towards electrophilic substitution at the ortho and para positions.
2. In water, all three activated positions (2, 4 and 6) are brominated at once, giving a white precipitate of 2,4,6-tribromophenol.

Teacher's Note:
1. Bromine water on phenol gives a white precipitate.
2. In CS₂ at low temperature, mono-bromophenols form instead.

Answer: (b) CH₂=CH–CH₂OH

Explanation:
1. LiAlH₄ reduces the carboxylic acid group to a primary alcohol.
2. LiAlH₄ does not reduce an isolated C=C double bond, so the double bond is kept.
3. The product is allyl alcohol, CH₂=CH–CH₂OH.

Teacher's Note:
1. LiAlH₄ reduces polar C=O groups but usually not C=C.
2. Catalytic H₂ would reduce the double bond as well.

Answer: (b) CH₃CH(OH)CH₂CH₃

Explanation:
1. The Grignard reagent attacks an epoxide at the less hindered carbon, the CH₂ carbon.
2. The ring opens to give CH₃CH(OMgBr)CH₂CH₃.
3. Hydrolysis gives CH₃CH(OH)CH₂CH₃, butan-2-ol.

Teacher's Note:
1. In epoxide opening by strong nucleophiles, attack is at the less substituted carbon.
2. Methyl adds to CH₂, so the OH stays on the middle carbon.

Answer: (a) CH₃CHOHCH₃

Explanation:
1. A tertiary alcohol (CH₃)₃COH is made by adding CH₃MgI to a ketone, so B is acetone, CH₃COCH₃.
2. Acetone is formed by oxidation of a secondary alcohol, so A is propan-2-ol, CH₃CHOHCH₃.

Teacher's Note:
1. Grignard + ketone gives a tertiary alcohol.
2. Oxidation of a 2° alcohol gives a ketone.

Answer: (c) CH₃–CH=CH–CHO

Explanation:
1. PCC is a mild oxidising agent that oxidises primary alcohols only up to the aldehyde and does not touch the C=C bond.
2. So CH₃CH=CHCH₂OH gives CH₃CH=CHCHO (crotonaldehyde).

Teacher's Note:
1. PCC stops at the aldehyde stage.
2. It does not attack C=C bonds.

Answer: (b) o- and p-Nitrophenol

Explanation:
1. The –OH group directs incoming groups to the ortho and para positions.
2. Dilute HNO₃ at low temperature gives only mononitration, so a mixture of o- and p-nitrophenol forms.
3. Picric acid needs concentrated HNO₃ (with H₂SO₄).

Teacher's Note:
1. Dilute HNO₃ gives mono-nitrophenols; concentrated gives picric acid.
2. o- is steam volatile (intramolecular H-bond); p- is not.

Answer: (a) CH₃OH

Explanation:
1. Water gas or synthesis gas (CO + H₂) passed over a ZnO–Cr₂O₃ catalyst at 573–673 K and high pressure gives methanol.
2. CO + 2H₂ → CH₃OH.

Teacher's Note:
1. This is the industrial route to methanol.
2. The ratio is CO : H₂ = 1 : 2.

Answer: (d) CH₃C≡CH

Explanation:
1. Propanol dehydrates with conc. H₂SO₄ at 170 C to propene, X = CH₃CH=CH₂.
2. Br₂ adds to give 1,2-dibromopropane, Y = CH₃CHBrCH₂Br.
3. Alcoholic KOH followed by NaNH₂ removes two molecules of HBr to give propyne, Z = CH₃C≡CH.

Teacher's Note:
1. Vicinal dibromide plus strong base gives an alkyne.
2. 170 C favours elimination over ether formation.

Answer: (b) Iodoform test

Explanation:
1. Ethanol contains the CH₃CH(OH)– unit and gives yellow iodoform with I₂/NaOH.
2. Methanol does not give this test.
3. Both are fully soluble in water, and both react with sodium and do not react with HCl under ordinary conditions.

Teacher's Note:
1. Ethanol is the only primary alcohol giving the iodoform test.
2. Choose a test in which the two behave differently.

Answer: (b) C₆H₅OH and CH₃CH₂I

Explanation:
1. In an alkyl aryl ether, I⁻ attacks the alkyl carbon, because the aryl C–O bond is strong (partial double bond character) and sp₂ carbon does not undergo SN₂.
2. The products are phenol and ethyl iodide.
3. Phenol does not react further with HI to give iodobenzene.

Teacher's Note:
1. Aryl alkyl ethers always give phenol plus alkyl halide with HI.
2. The C(aryl)–O bond is not broken.

Answer: (a) Salicylic acid

Explanation:
1. This is the Kolbe–Schmitt reaction: sodium phenoxide reacts with CO₂ at about 400 K and 4–7 atm.
2. Carboxylation occurs mainly at the ortho position, and acidification gives salicylic acid.

Teacher's Note:
1. Kolbe–Schmitt gives ortho-hydroxybenzoic acid (salicylic acid).
2. Salicylic acid is the starting material for aspirin.

Answer: (b) 95.6% C₂H₅OH

Explanation:
1. Fractional distillation of fermented liquor gives rectified spirit, an azeotropic mixture of ethanol and water.
2. It contains 95.6% ethanol by mass and cannot be concentrated further by simple distillation.

Teacher's Note:
1. Rectified spirit is an azeotrope with 95.6% ethanol.
2. Absolute alcohol is 100% ethanol.

Answer: (d) Triphenyl phosphate

Explanation:
1. The C–OH bond in phenol has partial double bond character due to resonance, so the OH group is not easily replaced by Cl.
2. Phenol reacts with PCl₅ to give mainly triphenyl phosphate, with only a small amount of chlorobenzene.

Teacher's Note:
1. The phenolic C–O bond is strong.
2. Alcohols are converted to alkyl chlorides by PCl₅; phenols mostly are not.

Answer: (a) CH₃CH₂CH₂OH and CH₃CH₂Cl

Explanation:
1. The ether is protonated and chloride attacks the less hindered alkyl group by SN₂.
2. The ethyl group is less hindered than the propyl group, so ethyl chloride forms along with propan-1-ol.
3. So X = CH₃CH₂CH₂OH and Y = CH₃CH₂Cl.

Teacher's Note:
1. Primary alkyl groups: the smaller one is attacked.
2. With excess acid, the alcohol would also be converted to the halide.

Answer: (b) Phenol and phthalic anhydride

Explanation:
1. Two molecules of phenol condense with one molecule of phthalic anhydride in the presence of conc. H₂SO₄.
2. The product is phenolphthalein, an acid–base indicator.
3. Succinic acid, succinic anhydride and benzaldehyde do not form phenolphthalein.

Teacher's Note:
1. Phenol + phthalic anhydride (conc. H₂SO₄) gives phenolphthalein.
2. With resorcinol the same reaction gives fluorescein.

Answer: (c) m-Chlorophenol

Explanation:
1. Phenols are more acidic than alcohols because the phenoxide ion is stabilised by resonance, so (b) and (d) are weaker.
2. The electron-withdrawing chlorine (–I effect) further stabilises the phenoxide ion, so m-chlorophenol (pKa about 9.1) is stronger than phenol (pKa about 10).

Teacher's Note:
1. Electron-withdrawing groups increase the acidity of phenols.
2. Alcohols are weaker acids than phenols.

Answer: (c) Formaldehyde

Explanation:
1. Periodic acid cleaves the C–C bond between two adjacent OH-bearing carbons (Malaprade reaction).
2. HOCH₂–CH₂OH + HIO₄ gives 2 HCHO + HIO₃ + H₂O, so formaldehyde is formed.

Teacher's Note:
1. Periodic acid cleaves vicinal diols.
2. Each CH₂OH end becomes HCHO.