JEE Chemistry Alcohols Phenols and Ethers MCQs Set 31

Answer: (a) 3,3-dimethylbutan-2-ol

Explanation:
1. Oxymercuration-demercuration follows Markovnikov addition of water and does not allow rearrangement.
2. The OH goes to C-2 of the alkene, giving \( \ce{(CH3)3C-CH(OH)-CH3} \).

Teacher's Note:
1. No carbocation rearrangement occurs in this method.
2. Hydroboration would instead give the anti-Markovnikov primary alcohol.

Answer: (d) Methylmagnesium iodide with phenol

Explanation:
1. The first three methylate the phenolic oxygen to give anisole.
2. A Grignard reagent is a strong base and simply removes the acidic phenolic H, giving methane and a magnesium phenoxide.

Teacher's Note:
1. Grignard reagents react with acidic hydrogens.
2. Williamson-type methylation of phenoxide works.

Answer: (a) Phenol gives benzyl alcohol

Explanation:
1. Hot copper dehydrogenates primary alcohols to aldehydes and secondary alcohols to ketones.
2. Tertiary alcohols have no alpha hydrogen and dehydrate to alkenes.
3. Phenol does not give benzyl alcohol.

Teacher's Note:
1. Copper at 573 K acts as dehydrogenation catalyst.
2. Phenol has no alkyl side chain to give benzyl alcohol.

Answer: (b) A, B and C

Explanation:
1. The first Grignard adds to the ester carbonyl to give the tetrahedral intermediate A.
2. A loses \( \ce{C2H5OMgBr} \) to give the ketone B (acetophenone).
3. A second Grignard adds to B to give the alkoxide C, which on hydrolysis gives the alcohol.

Teacher's Note:
1. Esters react with two moles of Grignard reagent.
2. All three species appear before the final hydrolysis.

Answer: (c) \( \ce{CH3COOCH2CH3} \)

Explanation:
1. Acylation of the ether cleaves the C-O bond.
2. Each ethyl group ends up attached to an acetyl group as ethyl acetate.

Teacher's Note:
1. Acid anhydrides or acid chlorides cleave ethers with a Lewis acid.
2. The product is an ester.

Answer: (a) ethane

Explanation:
1. Red P and HI is a strong reducing agent that cleaves ethers to alkanes.
2. One mole of diethyl ether gives two moles of ethane and one mole of water.

Teacher's Note:
1. Balance the carbon atoms to identify X.
2. HI with red phosphorus reduces any C-O bond.

Answer: (c) acetaldehyde, isopropyl alcohol

Explanation:
1. A blue colour in Victor Meyer's test shows that B is a secondary alcohol.
2. \( \ce{CH3CHO + CH3MgI} \) followed by hydrolysis gives \( \ce{CH3CH(OH)CH3} \), isopropyl alcohol.

Teacher's Note:
1. Blue means secondary, red means primary, colourless means tertiary.
2. Acetone with a Grignard would give a tertiary alcohol.

Answer: (c) \( \ce{CH3(CH2)3-O-CH2CH3} \)

Explanation:
1. With peroxide, HBr adds in anti-Markovnikov fashion to give 1-bromobutane (Y).
2. Sodium ethoxide then displaces bromide (Williamson synthesis) to give butyl ethyl ether.

Teacher's Note:
1. Peroxide effect reverses the orientation of HBr addition.
2. Williamson synthesis works best with a primary halide.

Answer: (b) \( \ce{CH3CH(OH)CH3} \) and \( \ce{CH3I} \)

Explanation:
1. Oxymercuration gives propan-2-ol (X), sodium isopropoxide (Y) and isopropyl methyl ether (Z).
2. HI cleaves a mixed ether by iodide attacking the smaller alkyl group in an SN2 reaction, giving \( \ce{CH3I} \) and isopropyl alcohol.

Teacher's Note:
1. Iodide attacks the less hindered methyl group.
2. Tertiary alkyl groups would instead go by SN1.

Answer: (c) \( \ce{HOCH2CH2CH2CH(Cl)CH3} \)

Explanation:
1. \( \ce{SOCl2} \) replaces the OH by Cl to give \( \ce{CH2=CHCH2CH2CHClCH3} \) (A).
2. Ozonolysis converts the terminal double bond to an aldehyde, \( \ce{OHCCH2CH2CHClCH3} \) (B).
3. \( \ce{NaBH4} \) reduces the CHO to CH2OH, giving C.

Teacher's Note:
1. Ozonolysis cuts the double bond and leaves C=O groups.
2. NaBH4 does not touch the C-Cl bond.

Answer: (c) \( \ce{CH2=C(CH3)-O-CH2CH3} \)

Explanation:
1. \( \ce{CH2=C(CH3)OCH2CH3} \) is an enol ether with formula \( \ce{C5H10O} \).
2. Its hydrolysis gives the enol of acetone (acetone) and ethanol.
3. Acetone and ethanol both contain the \( \ce{CH3CO-} \) or \( \ce{CH3CH(OH)-} \) group and give the iodoform test.

Teacher's Note:
1. Hydrolysis of an enol ether gives a carbonyl compound and an alcohol.
2. Option (a) would give propanal, which fails the iodoform test.

Answer: (b) loss of water to form the carbocation

Explanation:
1. Formation of the carbocation by loss of water needs bond breaking and is the slowest step.
2. Protonation of OH, hydride shift to a more stable cation and loss of H+ are all fast.

Teacher's Note:
1. The rate-determining step is the one that forms the high-energy carbocation.
2. Step 1 is a rapid acid-base reaction.

Answer: (b) \( \ce{CHCl3/NaOH} \)

Explanation:
1. Phenol with chloroform and aqueous NaOH undergoes the Reimer-Tiemann reaction.
2. The dichlorocarbene electrophile attacks the ortho position and hydrolysis gives salicylaldehyde.

Teacher's Note:
1. \( \ce{CCl4/NaOH} \) gives salicylic acid instead.
2. \( \ce{CO2} \) with NaOH is the Kolbe reaction.

Answer: (b) 1-(4-methoxyphenyl)ethanol

Explanation:
1. Dehydration proceeds through a benzylic carbocation.
2. The methoxy group donates electrons by resonance and stabilises the cation most, so this alcohol dehydrates fastest.

Teacher's Note:
1. Electron-donating groups favour carbocation formation.
2. A nitro group destabilises the carbocation.

Answer: (d) hydroquinone (benzene-1,4-diol)

Explanation:
1. Alkaline persulphate introduces an OH group para to the existing OH group.
2. The sulphate ester formed on hydrolysis gives hydroquinone.

Teacher's Note:
1. Elbs oxidation hydroxylates the para position.
2. Sulphonation needs conc. \( \ce{H2SO4} \), not persulphate.

Answer: (b) \( \ce{(CH3)3COC2H5} \)

Explanation:
1. Protonated tert-butyl alcohol loses water to give the stable tert-butyl cation.
2. Ethanol, being present in large amount as nucleophile, attacks the cation to give tert-butyl ethyl ether.

Teacher's Note:
1. Tertiary carbocations are attacked by the alcohol nucleophile.
2. Heating strongly would favour alkene.

Answer: (d) \( \ce{C2H2} \), \( \ce{H2O} \), phenol, \( \ce{H2CO3} \)

Explanation:
1. The pKa values are about 25 for ethyne, 15.7 for water, 10 for phenol and 6.4 for carbonic acid.
2. A lower pKa means a stronger acid, so the order of increasing acidity is as in option (d).

Teacher's Note:
1. Carbonic acid is stronger than phenol.
2. Phenol is stronger than water.

Answer: (b) \( \ce{CH3CH(OH)CH3} \)

Explanation:
1. Z is an alcohol, X is the alkyl chloride and Y is the alkene.
2. Hydration of Y must regenerate Z by Markovnikov addition.
3. Only propan-2-ol, via propene, fits: propene hydrates to propan-2-ol.

Teacher's Note:
1. Markovnikov hydration of propene gives a secondary alcohol.
2. Y is propene.

Answer: (d) a < c < b

Explanation:
1. The methyl group donates electrons and makes o-cresol less acidic than phenol.
2. Salicylic acid is a carboxylic acid and is much stronger than phenol.

Teacher's Note:
1. Electron-donating groups reduce acidity.
2. Carboxylic acids are stronger than phenols.

Answer: (d) hexane-3,4-diol

Explanation:
1. Periodic acid cleaves only vicinal diols, in which two OH groups are on adjacent carbons.
2. Hexane-3,4-diol breaks into two molecules of propanal.

Teacher's Note:
1. Vicinal means OH on adjacent carbons.
2. 1,3-, 2,4- and 1,6-diols resist cleavage.

Answer: (c) \( \ce{(CH3)2CHCH2OH} \) and \( \ce{CH3CH2I} \)

Explanation:
1. Both alkyl groups are primary, so iodide attacks the less hindered one by SN2.
2. The ethyl group gets the iodide, giving ethyl iodide and isobutyl alcohol.

Teacher's Note:
1. With two primary groups the smaller group takes the halogen.
2. Isobutyl is more hindered than ethyl.

Answer: (b) \( \ce{LiAlH4} \)

Explanation:
1. \( \ce{LiAlH4} \) reduces the carboxylic acid group to a primary alcohol without attacking an isolated C=C.
2. \( \ce{NaBH4} \) does not reduce acids, \( \ce{H2/Pt} \) would saturate the C=C and PCC would oxidise.

Teacher's Note:
1. LiAlH4 is a strong hydride reducing agent for acids.
2. Catalytic hydrogenation attacks alkenes.

Answer: (b) by reaction (ii)

Explanation:
1. In reaction (i) the HBr produced reverses the reaction and slows it.
2. In reaction (ii) silver oxide removes the bromide as insoluble AgBr, which drives the reaction forward.

Teacher's Note:
1. Removing the leaving group as a precipitate speeds up substitution.
2. Silver ion helps C-Br ionisation.

Answer: (d) \( \ce{H3PO3} \)

Explanation:
1. \( \ce{3C2H5OH + PBr3 -> 3C2H5Br + H3PO3} \).
2. The phosphorus stays in the +3 state and forms phosphorous acid.

Teacher's Note:
1. PBr3 does not oxidise phosphorus.
2. Balance the P, O and H atoms to find X.