Question 1Level 2 · ExamSingle correct
Consider (I) butan-1-ol, (II) butan-2-ol, (III) 2-methylpropan-2-ol and (IV) ethoxyethane, all of molecular formula C4H10O. The correct decreasing order of their normal boiling points is ______. Exactly one option is correct.
AI > III > II > IV
BII > I > III > IV
CI > II > III > IV
DI > II > IV > III
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Answer: C
All four are isomers (same molar mass, similar dispersion forces), so intermolecular hydrogen bonding and molecular shape decide.
The three alcohols are hydrogen bonded through O–H; the ether has no O–H hydrogen and cannot hydrogen bond to itself, so it boils far lower (IV lowest, about 308 K).
Among the alcohols, branching makes the molecule more compact, lowers the surface area for van der Waals contact and hinders the approach of the O–H groups: unbranched primary I (about 391 K) > secondary II (about 373 K) > highly branched tertiary III (about 356 K).
Order: I > II > III > IV.
Answer: (C).
Difficulty: HardQuestion 2Level 2 · ExamOne or more correct
Which of the following equations or statements about the reactions of alcohols are correct? One or more options may be correct.
A2CH3CH2OH + 2Na → 2CH3CH2ONa + H2
B3CH3CH2OH + PCl3 → 3CH3CH2Cl + H3PO3
CPropane-1,2,3-triol reacts with excess PCl5 according to C3H5(OH)3 + 3PCl5 → C3H5Cl3 + 3POCl3 + 3HCl, so 1 mol of the triol releases 3 mol of HCl.
DEthanol reacts with sodium more vigorously than water does, because ethanol is the stronger acid.
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Answer: A, B, C
(A) Correct. Na displaces the acidic O–H hydrogen: 2 RO–H + 2 Na → 2 RO−Na+ + H2. Atoms: C4 H12 O2 Na2 on both sides.
(B) Correct. One PCl3 replaces three OH groups; the phosphorus ends up as phosphonic (phosphorous) acid H3PO3. Atoms: C6 H18 O3 P Cl3 on both sides.
(C) Correct. Each OH reacts with one PCl5: R–OH + PCl5 → R–Cl + POCl3 + HCl. The triol has three OH groups, so it needs 3 PCl5 and gives 1,2,3-trichloropropane, 3 POCl3 and 3 HCl. Atoms: C3 H8 O3 P3 Cl15 on both sides.
(D) Incorrect. Water is a slightly stronger acid than ethanol (the ethyl group's +I effect destabilises CH3CH2O−), so water reacts more vigorously with sodium.
Answer: (A), (B), (C).
Difficulty: HardQuestion 3Level 3 · AdvancedSingle correct
An alcohol X, C5H12O, gives turbidity immediately with Lucas reagent (conc. HCl + anhydrous ZnCl2) at room temperature. Heating X with conc. H2SO4 gives an alkene as the major product, and ozonolysis of this alkene (O3, then Zn/H2O) gives propanone and ethanal only. X is ______. Exactly one option is correct.
A2-methylbutan-2-ol, CH3CH2C(OH)(CH3)2
B3-methylbutan-2-ol, (CH3)2CHCH(OH)CH3
C2,2-dimethylpropan-1-ol, (CH3)3CCH2OH
D2-methylbutan-1-ol, CH3CH2CH(CH3)CH2OH
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Answer: A
Ozonolysis to propanone + ethanal means the alkene is (CH3)2C=CHCH3, 2-methylbut-2-ene.
Immediate turbidity with Lucas reagent at room temperature means a tertiary alcohol (fast SN1 via a tertiary carbocation).
(A) 2-Methylbutan-2-ol is tertiary: immediate turbidity. Acid dehydration gives the tertiary cation, which loses H+ mainly from the CH2 (Saytzeff) to give 2-methylbut-2-ene. Fits.
(B) 3-Methylbutan-2-ol also dehydrates to 2-methylbut-2-ene, but it is secondary: turbidity only after about 5 min.
(C) 2,2-Dimethylpropan-1-ol (primary, no turbidity at room temperature) gives 2-methylbut-2-ene only after a methyl shift.
(D) 2-Methylbutan-1-ol is primary: no turbidity at room temperature.
Answer: (A).
Difficulty: Very hard