Alcohols Phenols and Ethers Questions and Answers

Answer: B

The major product is (B), the acetonide of the diol.

2,2-Dimethoxypropane with an acid catalyst is the standard reagent for protecting a 1,2- or 1,3-diol as an isopropylidene acetal. The two CH₂OH groups of the starting diester form a 1,3-diol, and they close with the acetone unit into a six-membered 2,2-dimethyl-1,3-dioxane ring, releasing methanol. The two methyl ester groups are untouched, leaving dimethyl 2,2-dimethyl-1,3-dioxane-5,5-dicarboxylate.

(A) is a Meldrum's-acid-type ring formed from the carboxyl groups, which would need the free diacid. (C) and (D) are lactones formed from an ester and a hydroxyl group; acetal formation under these mild conditions does not convert the esters.

Answer: D

The key gives \(\mathrm{ClCH_{2}OH > BrCH_{2}OH > FCH_{2}OH > CH_{3}OH}\).

Acidity depends on how well the alkoxide ion XCH₂O^- is stabilised. A halogen on the carbon next to oxygen withdraws electron density inductively and spreads the negative charge, so every halomethanol is more acidic than methanol. That places CH₃OH last and rules out the two options that put it, or FCH₂OH below it, anywhere else.

Among the halogens, the gas phase removes solvation, and the ranking is no longer set by electronegativity alone: the polarisability of the halogen also helps disperse the charge. The key's order reflects chlorine combining a strong inductive pull with greater polarisability than fluorine, ahead of bromine, with fluorine last of the three. The option headed by FCH₂OH is the inductive-effect-only reading that the key does not follow.

Answer: C

The compounds are phenol (P), acetic anhydride (Q) and phenyl acetate (R), which is option (C).

A mono-substituted aromatic compound that decolourises bromine water, gives a purple-red colour with aqueous FeCl₃ and dissolves in aqueous NaOH is a phenol. Aniline does not give the last two. The reagent with two C=O stretching bands at about 1830 and 1750 cm⁻¹ is an acid anhydride, whose symmetric and asymmetric C=O stretches give a pair of bands. An acid chloride shows a single band near 1800 cm⁻¹. Acetylation of the phenolic OH with acetic anhydride and a base gives the ester, phenyl acetate.

Options (A) and (B) use acetyl chloride, which does not fit the two-band IR spectrum, and (B) starts with aniline. Option (D) has the right reagents but a ring C-acylated product, which is the Fries product and needs a Lewis acid, not a base.

Answer: A

The product P is the aryl vinyl ketone, 1-(4-methoxyphenyl)prop-2-en-1-one, which is option (A).

MnO₂ oxidises benzylic alcohols selectively, so the secondary benzylic OH becomes a ketone and the primary alcohol is left alone, giving ArCOCH₂CH₂OH. PBr₃ converts the primary alcohol to the bromide, ArCOCH₂CH₂Br. NaOEt then removes the acidic alpha hydrogen and expels bromide (E1cB elimination), forming the C=C of the vinyl ketone.

Option (B) would need oxidation of the primary alcohol and loss of the benzylic oxygen. Option (C) needs an alpha bromination that these reagents do not do, and (D) needs an alkyne. None can come from the three steps.

Answer: A

The IUPAC name is 3-Methtylbutoxybenzene (3-methylbutoxybenzene; the option prints an extra t).

The isopentyl group is \((CH_{3})_{2}CH-CH_{2}-CH_{2}-\). Numbered from the oxygen-bearing carbon, the methyl branch sits on C-3 of a four-carbon chain, so the alkoxy group is 3-methylbutoxy and the ether is named as a substituted benzene.

2-Methylbutoxy is the 'active amyl' group, a different isomer; 4-methylbutoxy is not a valid name (a methyl on C-4 just lengthens the chain to pentyloxy); and 1-methylbutoxy places the branch on the carbon attached to oxygen.

Answer: B

The reaction is the Reimer-Tiemann reaction.

Chloroform and NaOH generate dichlorocarbene, which attacks the phenoxide ion at the ortho position. Hydrolysis then gives salicylaldehyde (2-hydroxybenzaldehyde).

Kolbe's reaction uses \(CO_{2}\) to make salicylic acid; Gattermann-Koch formylates benzene with CO and HCl; the Stephen reaction reduces nitriles to aldehydes; and the Sandmeyer reaction replaces a diazonium group with a halide or cyanide.

Answer: D

Phenol is manufactured on a large scale from cumene.

Cumene (isopropylbenzene) is air-oxidised to cumene hydroperoxide, which is then cleaved with dilute acid to phenol and acetone. Most of the world's phenol is made this way, and the acetone is a valuable by-product.

Chlorobenzene (Dow process) and aniline (via diazonium salts) are older or laboratory routes, benzene is the starting point for cumene rather than the compound oxidised, and cyclohexane is not used to make phenol.

Answer: B

Phenol is converted into benzene by heating with zinc dust.

Zinc dust removes the oxygen of the phenolic group: C₆H₅OH + Zn → C₆H₆ + ZnO.

Sodium amalgam, LiAlH₄ and NaBH₄ are reducing agents for other groups (carbonyl compounds, for example) and do not remove a phenolic –OH. Cr₂O₃ is a catalyst for aromatisation, not a reagent for this conversion.

Answer: D

Intramolecular hydrogen bonding is present in o-nitrophenol.

In o-nitrophenol the –OH and –NO₂ groups are on adjacent carbons, so the hydroxyl hydrogen bonds to an oxygen of the nitro group within the same molecule, forming a six-membered ring (chelation). This is why o-nitrophenol is more volatile and less water-soluble than the para isomer.

In p-nitrophenol the groups are too far apart, so it forms intermolecular hydrogen bonds. Water, methanol and phenol also hydrogen bond between molecules, not within one.

Answer: A

Phenetole is Ethoxybenzene, C₆H₅OC₂H₅.

Phenetole is the common name of the ethyl phenyl ether: an ethoxy group (-OC₂H₅) attached to a benzene ring. Its IUPAC name is ethoxybenzene.

Methoxybenzene is anisole, the methyl phenyl ether, the usual confusion with phenetole. Methoxyethane is ethyl methyl ether, and 1-methoxypropane and 2-methoxypropane are methyl propyl and methyl isopropyl ethers; none of these contains a benzene ring.