JEE Chemistry Alcohols Phenols and Ethers MCQs Set 25

Answer: (b) four

Explanation:
1. A positive iodoform test needs the \(\ce{CH3CH(OH)-}\) unit, i.e. \(\ce{CH3CH(OH)R}\) with R = \(\ce{C4H9}\).
2. There are four butyl groups (n-, iso-, sec- and tert-butyl).
3. The alcohols are hexan-2-ol, 4-methylpentan-2-ol, 3-methylpentan-2-ol and 3,3-dimethylbutan-2-ol, i.e. four.

Teacher's Note:
1. Count the isomers of the group R in CH3CH(OH)R. Ethanol is the only primary alcohol that gives the test.

Answer: (b) 2-methylbut-2-ene and 2-bromo-2-methylbutane

Explanation:
1. Dehydration gives a secondary carbocation which rearranges by a hydride shift to the more stable tertiary carbocation.
2. Loss of a proton gives the more substituted alkene (Zaitsev): A = 2-methylbut-2-ene.
3. HBr in the absence of peroxide adds by Markovnikov's rule, placing Br on the more substituted carbon: C = 2-bromo-2-methylbutane.

Teacher's Note:
1. Rearrangement to a more stable carbocation, then Zaitsev elimination, is the usual pattern for dehydration of secondary alcohols.

Answer: (a) III > II > IV > I

Explanation:
1. pKa values are about 0.4 (picric acid), 4.8 (acetic acid), 10 (phenol) and 16 (cyclohexanol).
2. A smaller pKa means a stronger acid.
3. Order: III > II > IV > I.

Teacher's Note:
1. Picric acid is a stronger acid than acetic acid because of three nitro groups.

Answer: (a) (iv) > (iii) > (i) > (ii)

Explanation:
1. p-Nitrophenol (pKa 7.2) is more acidic than m-nitrophenol (8.4) because the nitro group exerts both -I and -R effects from the para position.
2. Both are more acidic than phenol (10.0).
3. o-Methylphenol is the weakest because of the +I effect of the methyl group.

Teacher's Note:
1. Nitro > H > methyl for acidity; para nitro > meta nitro.

Answer: (a) 2-iodopropane

Explanation:
1. With limited HI glycerol gives allyl iodide.
2. With excess HI the allyl iodide adds HI by Markovnikov's rule to give 2-iodopropane (isopropyl iodide).

Teacher's Note:
1. Glycerol with HI first forms 1,2,3-triiodopropane, which eliminates iodine to allyl iodide; excess HI then gives isopropyl iodide.

Answer: (c) a-(ii), b-(iii), c-(i), d-(iv)

Explanation:
1. A primary amine gives the carbylamine test (foul-smelling isocyanide).
2. A terminal alkyne gives a white precipitate of silver acetylide with ammoniacal silver nitrate.
3. An ester undergoes alkaline hydrolysis (saponification).
4. A secondary alcohol gives turbidity with Lucas reagent in about five minutes.

Teacher's Note:
1. Match each functional group with its classical test before looking at the options.

Answer: (b) benzoic acid

Explanation:
1. Zinc dust reduces phenol to benzene (X).
2. Friedel-Crafts methylation gives toluene (Y).
3. Alkaline \(\ce{KMnO4}\) oxidises the side chain completely to a carboxyl group, and acidification gives benzoic acid (Z).

Teacher's Note:
1. Any alkyl side chain with an alpha hydrogen is oxidised to -COOH by hot alkaline permanganate.

Answer: (c) isobutyl alcohol and ethyl iodide

Explanation:
1. Both alkyl groups are primary, so the cleavage is \(\ce{S_N2}\).
2. Iodide attacks the less hindered group, the ethyl group, giving ethyl iodide and isobutyl alcohol.
3. (With excess HI the alcohol is further converted to the iodide, but the first products are those given in the option.)

Teacher's Note:
1. Iodide attacks the less crowded carbon in SN2 cleavage of ethers. The isobutyl group is more hindered because of the beta branching.

Answer: (a) \(\ce{CH3I}\) and \(\ce{(CH3)2CHOH}\)

Explanation:
1. The ether is protonated and iodide attacks the less hindered methyl group (\(\ce{S_N2}\)).
2. This gives methyl iodide and isopropyl alcohol.

Teacher's Note:
1. For ethers with primary and secondary alkyl groups the smaller group becomes the alkyl iodide.

Answer: (c) o-Nitrophenol

Explanation:
1. Phenols are more acidic than alcohols, so 2-chloroethanol is weaker than phenol.
2. The nitro group in o-nitrophenol withdraws electrons strongly and stabilises the phenoxide ion, while the methyl group in o-methylphenol destabilises it.
3. So o-nitrophenol is the most acidic.

Teacher's Note:
1. Rank by the stability of the conjugate base: phenoxides with electron-withdrawing groups are the most stable.

Answer: (a) Diethyl ether

Explanation:
1. Acetonitrile, acetamide and methyl acetate all contain polarised multiple bonds (C triple bond N or C=O) at which hydroxide can attack.
2. Diethyl ether has only C-O single bonds and no electrophilic carbon, so it resists nucleophilic attack by hydroxide.

Teacher's Note:
1. Ethers are unreactive towards bases and nucleophiles; they are cleaved only by strong acids such as HI.

Answer: (a) H-bonding in ethanol

Explanation:
1. Ethanol has an O-H bond and forms intermolecular hydrogen bonds.
2. Dimethyl ether has no hydrogen attached to oxygen, so it cannot form hydrogen bonds with itself.

Teacher's Note:
1. Boiling points of isomers are governed largely by hydrogen bonding.

Answer: (a) Aldehyde

Explanation:
1. Over hot copper a primary alcohol loses two hydrogen atoms to give an aldehyde.
2. Secondary alcohols give ketones, while tertiary alcohols dehydrate to alkenes.

Teacher's Note:
1. Remember the three outcomes: 1 degree gives aldehyde, 2 degree gives ketone, 3 degree gives alkene.

Answer: (d) reduction of CO with \(\ce{H2}\) over \(\ce{ZnO-Cr2O3}\)

Explanation:
1. Methanol is made by catalytic hydrogenation of carbon monoxide: \(\ce{CO + 2H2 -> CH3OH}\) at about 300 C and 200-300 atm with a \(\ce{ZnO-Cr2O3}\) catalyst.
2. The other methods are not industrial routes.

Teacher's Note:
1. Water gas or synthesis gas is the feedstock for methanol manufacture.

Answer: (c) 5

Explanation:
1. The primary alcohols are pentan-1-ol, 2-methylbutan-1-ol, 3-methylbutan-1-ol and 2,2-dimethylpropan-1-ol.
2. Only 2-methylbutan-1-ol has a chiral centre, so it counts twice (two enantiomers).
3. Total = 1 + 2 + 1 + 1 = 5.

Teacher's Note:
1. Do not forget optical isomers when the question says including stereoisomers.

Answer: (b) 3

Explanation:
1. Molar mass of \(\ce{C3H8O3}\) = 92 g/mol, so 0.092 g = 0.001 mol.
2. Methane formed = 67.2 mL / 22400 mL/mol = 0.003 mol.
3. Each active hydrogen releases one mole of methane, so the number of active H = 0.003 / 0.001 = 3 (the three OH groups of glycerol).

Teacher's Note:
1. Active hydrogen atoms (those on O, N or S) are counted by Zerewitinoff determination with \(\ce{CH3MgI}\).

Answer: (c) 5

Explanation:
1. Each acetylation replaces an H by \(\ce{COCH3}\), increasing the mass by 42.
2. Number of OH groups = (390 - 180)/42 = 5.
3. (This fits glucose, which has five OH groups.)

Teacher's Note:
1. Mass increase per acetyl group is 43 - 1 = 42.

Answer: (c) 5

Explanation:
1. Periodic acid cleaves the C-C bond between carbons carrying -OH or C=O groups.
2. The aldohexose has six oxygenated carbons in a row, so five C-C bonds are cleaved.
3. Therefore five moles of \(\ce{HIO4}\) are used.

Teacher's Note:
1. Number of HIO4 moles equals the number of C-C bonds between adjacent oxygen-bearing carbons.

Answer: (c) 3

Explanation:
1. The six carbons are cleaved at five C-C bonds.
2. The terminal carbons C1 and C6 become formaldehyde (X = 2).
3. The keto carbon C2 becomes \(\ce{CO2}\) (Z = 1).
4. The three middle CHOH carbons (C3, C4, C5) become formic acid, so Y = 3.

Teacher's Note:
1. In periodate cleavage, terminal CH2OH gives HCHO, each internal CHOH gives HCOOH and the carbonyl carbon of a keto group gives CO2.

Answer: (b) 4

Explanation:
1. \(\ce{HCl/ZnCl2}\), \(\ce{PCl3}\), \(\ce{PCl5}\) and \(\ce{SOCl2}\) all convert a primary alcohol to the alkyl chloride.
2. NaCl is not a reagent for replacing OH.
3. TsCl converts the alcohol only into a tosylate, not a chloride.
4. So four reagents work.

Teacher's Note:
1. Thionyl chloride is the best choice because the by-products \(\ce{SO2}\) and HCl are gases.

Answer: (b) 3

Explanation:
1. A positive test needs the \(\ce{CH3CH(OH)-}\) group (or ethanol).
2. Butan-2-ol, 1-phenylethanol and ethanol have it.
3. Pentan-3-ol, 1-phenylpropan-1-ol and benzyl alcohol do not, so three are positive.

Teacher's Note:
1. Iodoform test works for methyl carbinols (CH3CH(OH)R) and ethanol.