JEE Chemistry Alcohols Phenols and Ethers MCQs Set 09

Answer: (c) 2-Methylpropan-2-ol

Explanation:
1. Lucas reagent converts the alcohol to an insoluble alkyl chloride through an SN1 path, so the speed depends on carbocation stability.
2. 2-Methylpropan-2-ol is tertiary and forms the stable tert-butyl carbocation, so turbidity appears immediately. Propan-2-ol takes 5 to 10 minutes and the primary alcohols give no turbidity at room temperature.

Teacher's Note:
1. Order of speed: tertiary, benzylic and allylic > secondary > primary.
2. Primary alcohols do not react at room temperature.

Answer: (a) PCC in CH2Cl2

Explanation:
1. Acidified KMnO4, excess dichromate and hot nitric acid are strong oxidants and take a primary alcohol through the aldehyde to the carboxylic acid.
2. Pyridinium chlorochromate in dichloromethane (no water) stops at the aldehyde stage and does not attack C=C bonds.

Teacher's Note:
1. PCC is used mainly for primary alcohols.
2. Water is needed for further oxidation to the acid, and PCC is used without it.

Answer: (d) Carboxylic acid

Explanation:
1. NaBH4 is a mild reagent that reduces aldehydes, ketones and acid chlorides.
2. LiAlH4 is stronger and also reduces carboxylic acids and esters, so the carboxylic acid is the answer.

Teacher's Note:
1. Neither reagent reduces an isolated C=C bond.
2. LiAlH4 must be used in dry ether; NaBH4 works in alcohol or water.

Answer: (b) Propan-1-ol

Explanation:
1. Hydroboration places boron on the less substituted carbon, the terminal CH2.
2. Oxidation replaces boron with OH, so the net result is anti-Markovnikov addition of water and the product is propan-1-ol.

Teacher's Note:
1. Acid-catalysed hydration of propene gives propan-2-ol instead.
2. Hydroboration-oxidation gives no rearrangement.

Answer: (b) 3,3-Dimethylbutan-2-ol

Explanation:
1. Oxymercuration-demercuration adds water in the Markovnikov sense, so OH goes to C-2.
2. No carbocation is formed, so the methyl group does not shift, and the product keeps the 3,3-dimethyl skeleton.
3. The product is 3,3-dimethylbutan-2-ol; acid-catalysed hydration would give the rearranged 2,3-dimethylbutan-2-ol.

Teacher's Note:
1. The mercurinium ion prevents rearrangement.
2. Compare with acid hydration, where a methyl shift occurs.

Answer: (d) H2O > CH3CH2OH > (CH3)2CHOH > (CH3)3COH

Explanation:
1. The acidity of an alcohol depends on how well its conjugate base (alkoxide) is solvated.
2. Bulky alkyl groups hinder solvation of the alkoxide and also push electrons toward oxygen, so acidity falls from water to primary, secondary and tertiary alcohols.

Teacher's Note:
1. Water is a slightly stronger acid than ordinary alcohols (methanol is a little stronger than water).
2. All alcohols are far stronger acids than terminal alkynes.

Answer: (a) the same configuration (retention)

Explanation:
1. Without pyridine the alcohol first forms a chlorosulphite, which then collapses with intramolecular delivery of chloride (SNi).
2. The chloride is delivered from the same side, so the configuration at the chiral carbon is retained.

Teacher's Note:
1. Adding pyridine changes the path to SN2 and gives inversion.
2. SNi is why SOCl2 is preferred over HX for sensitive chiral alcohols.

Answer: (c) But-2-ene

Explanation:
1. Dehydration follows Saytzeff's rule, giving the more substituted (more stable) alkene.
2. Butan-2-ol can lose H from C-1 (but-1-ene, minor) or from C-3 (but-2-ene, major).

Teacher's Note:
1. Saytzeff product = most substituted alkene.
2. But-2-ene exists as cis and trans forms, trans being the major.

Answer: (d) Diethyl ether

Explanation:
1. At about 140 °C with excess alcohol one molecule of protonated ethanol is attacked by another (SN2) and diethyl ether forms.
2. At 170 °C or above, with excess acid, ethene forms instead.

Teacher's Note:
1. Temperature decides ether versus alkene.
2. Ether formation is best for symmetrical ethers.

Answer: (b) Propanone

Explanation:
1. Hot copper dehydrogenates a secondary alcohol to a ketone and H2.
2. Propan-2-ol gives propanone (acetone).

Teacher's Note:
1. Primary alcohols give aldehydes and tertiary alcohols give alkenes under this treatment.
2. This reaction can be used to classify alcohols.

Answer: (a) Methyl iodide and ethanol

Explanation:
1. Both groups are primary, so the cleavage is SN2 and iodide attacks the smaller group.
2. The methyl group becomes CH3I and the ethyl group stays as ethanol (it later gives ethyl iodide with excess HI).

Teacher's Note:
1. The smaller alkyl group goes to the iodide.
2. With excess HI both groups end up as iodides.

Answer: (c) tert-Butyl iodide and methanol

Explanation:
1. A tertiary group makes the cleavage SN1: the protonated ether loses methanol to give the tert-butyl carbocation.
2. The carbocation captures iodide, giving tert-butyl iodide, and methanol is released.

Teacher's Note:
1. Rule: if one group is tertiary the tertiary group becomes the halide.
2. This is the exception to the 'smaller group becomes iodide' rule.

Answer: (a) Phenol and methyl iodide

Explanation:
1. Cleavage occurs at the alkyl-oxygen bond because the aryl-oxygen bond has partial double bond character.
2. Phenol and CH3I result. Phenol does not react further because an aromatic ring does not undergo nucleophilic substitution this way.

Teacher's Note:
1. Only HI is effective since iodide is a strong nucleophile.
2. Aryl-oxygen bond is not broken.

Answer: (d) Sodium tert-butoxide with methyl bromide

Explanation:
1. The Williamson reaction is an SN2 attack, so the alkyl halide should be primary (or methyl).
2. A tertiary halide would give elimination to isobutene with a strong base such as methoxide, so the halide must be CH3Br and the alkoxide the bulky one.

Teacher's Note:
1. Choose the less hindered halide.
2. Tertiary halide plus alkoxide gives alkene.

Answer: (c) shaking with fresh FeSO4 solution and adding KCNS to see a red colour

Explanation:
1. Ether peroxides oxidise Fe2+ to Fe3+.
2. Fe3+ forms a red complex with thiocyanate, so a red colour shows peroxide.

Teacher's Note:
1. Peroxides can be removed with aqueous KI or ferrous salt.
2. Never distil ether to dryness if peroxides may be present.

Answer: (b) Aqueous NaHCO3

Explanation:
1. Benzoic acid is strong enough to release CO2 from NaHCO3 (effervescence).
2. Phenol is a weaker acid than carbonic acid and does not react with NaHCO3. NaOH and Na dissolve both.

Teacher's Note:
1. Phenol turns FeCl3 violet but benzoic acid gives a buff precipitate, so FeCl3 can also be used.
2. Phenols dissolve in NaOH but not in NaHCO3.

Answer: (c) 2,4,6-Tribromophenol, a white precipitate

Explanation:
1. OH is a strongly activating o- and p-directing group, so the ring is brominated at all three activated positions even without a catalyst.
2. The product 2,4,6-tribromophenol separates as a white precipitate.

Teacher's Note:
1. With Br2 in CS2 at low temperature only o- and p-bromophenols form.
2. No Lewis acid is needed.

Answer: (a) Salicylaldehyde

Explanation:
1. Dichlorocarbene, :CCl2, formed from CHCl3 and base attacks the phenoxide ring, mainly at the ortho position.
2. Hydrolysis of the CHCl2 group gives CHO, so salicylaldehyde (o-hydroxybenzaldehyde) is obtained.

Teacher's Note:
1. Ortho product is favoured by chelation and by higher activation at ortho.
2. Kolbe reaction (with CO2) gives salicylic acid instead.

Answer: (d) Salicylic acid

Explanation:
1. In the Kolbe reaction CO2 acts as the electrophile.
2. With the small Na+ ion the ortho transition state is stabilised, so o-hydroxybenzoic acid (salicylic acid) is the major product.

Teacher's Note:
1. With potassium phenoxide the para acid becomes major.
2. Salicylic acid is used to make aspirin.

Answer: (b) Phenol and acetone

Explanation:
1. Acid protonates the hydroperoxide and a phenyl group migrates to oxygen with loss of water.
2. Hydrolysis of the resulting species gives phenol and acetone.

Teacher's Note:
1. Both products are valuable, which is why the process is economical.
2. The first step makes cumene from benzene and propene.

Answer: (b) Acrolein (propenal)

Explanation:
1. KHSO4 dehydrates glycerol, losing two molecules of water.
2. The product is CH2=CH-CHO, acrolein, which has a pungent smell.

Teacher's Note:
1. Glycerol with conc. HNO3 and H2SO4 gives nitroglycerine.
2. Glycerol with oxalic acid at 260 °C gives allyl alcohol.

Answer: (d) 1-Methoxypropan-2-ol

Explanation:
1. Under basic conditions the nucleophile CH3O- attacks the less hindered carbon of the epoxide by SN2 (the CH2 carbon).
2. That gives CH3OCH2CH(O-)CH3, and after protonation 1-methoxypropan-2-ol.

Teacher's Note:
1. Base-catalysed opening goes to the less substituted carbon.
2. Acid-catalysed opening goes to the more substituted carbon.

Answer: (a) 2-Methoxypropan-1-ol

Explanation:
1. Under acid conditions the epoxide oxygen is protonated and the ring opens with partial positive charge on the more substituted carbon.
2. Methanol attacks that more substituted (secondary) carbon, giving 2-methoxypropan-1-ol.

Teacher's Note:
1. Acid-catalysed opening is SN1-like.
2. A symmetrical epoxide gives the same product in both cases.

Answer: (c) Ethanal and methanal

Explanation:
1. HIO4 cleaves the bond between the two carbons carrying OH groups of a vicinal diol.
2. The CH(OH)CH3 end becomes CH3CHO and the CH2OH end becomes HCHO.

Teacher's Note:
1. Cis-diols react faster than trans-diols because a cyclic intermediate is needed.
2. Lead tetraacetate does the same job.

Answer: (d) Pinacolone (3,3-dimethylbutan-2-one)

Explanation:
1. One OH is protonated and lost as water to give a tertiary carbocation.
2. A methyl group migrates to the cationic centre, leaving a carbocation stabilised by the remaining OH.
3. Loss of H+ gives the ketone pinacolone, (CH3)3C-CO-CH3.

Teacher's Note:
1. This is the pinacol-pinacolone rearrangement.
2. Migratory order: H > phenyl > ethyl > methyl.