Alcohols, Phenols and Ethers: Application Problems (1)

easy • 3 min read
Tags Alcohols Phenols Ethers

Question

Compound A (C3H8OC_3H_8O) gives a positive Lucas test within 55 minutes. It reacts with PCCPCC to give a ketone. Identify A, write the structure, and give the IUPAC name. Also explain why phenol is more acidic than alcohols.

Solution — Step by Step

C3H8OC_3H_8O: degree of unsaturation =(2×3+2−8)/2=0= (2 \times 3 + 2 - 8)/2 = 0. So fully saturated, no rings or double bonds. Possible structures: 1-propanol, 2-propanol, methoxyethane (an ether).

Lucas reagent (ZnCl2/HClZnCl_2/HCl) distinguishes alcohols by class. Tertiary react instantly; secondary in 55-1010 minutes; primary react only on heating.

A reacts in 55 minutes — secondary alcohol.

PCCPCC oxidises alcohols to aldehydes (primary) or ketones (secondary). Since A gives a ketone, A is secondary.

Both clues agree: A is a secondary alcohol with C3H8OC_3H_8O — that’s CH3CH(OH)CH3CH_3CH(OH)CH_3, propan-2-ol (isopropanol).

 CH3−CH(OH)−CH3CH_3-CH(OH)-CH_3

IUPAC name: propan-2-ol.

Phenol’s pKa≈10pK_a \approx 10; ethanol’s pKa≈16pK_a \approx 16. Phenol is more acidic because:

  • The phenoxide ion (Ph−O−Ph-O^-) is stabilised by resonance — the negative charge delocalises into the ring (ortho and para positions).
  • The alkoxide (R−O−R-O^-) has no such delocalisation; the negative charge sits entirely on oxygen.

Greater stabilisation of the conjugate base means the parent acid is more easily ionised — more acidic.

Final answer: A is propan-2-ol, CH3CH(OH)CH3CH_3CH(OH)CH_3. Phenol is more acidic due to resonance stabilisation of phenoxide.

Why This Works

Lucas + PCC together pin down the alcohol class. C3H8OC_3H_8O has only one secondary alcohol structure, so the answer is unique.

For acidity, conjugate base stability is the key — Le Chatelier’s principle applied to the ionisation equilibrium. Resonance into the ring is what makes phenol special among −OH-OH compounds.

Alternative Method

NMR / IR spectroscopy could distinguish the isomers — primary alcohols show −CH2−OH-CH_2-OH split patterns, secondary show −CH−OH-CH-OH with a different multiplet. But for JEE Main MCQs, the chemical tests are faster.

Lucas test order to remember: tertiary < 5 sec, secondary 5-10 min, primary slow/heat needed. The reagent is ZnCl2ZnCl_2 in concentrated HCl. Tertiary carbocations form instantly; primary do not form readily under these conditions.

Students sometimes claim ethers (CH3OCH2CH3CH_3OCH_2CH_3, methoxyethane) react with Lucas reagent. Ethers do not have an −OH-OH group, so they don’t react with Lucas. If A reacts with Lucas, it cannot be the ether isomer.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Acid catalyzed dehydration of alcohols — Zaitsev product formation

medium

Alcohols, Phenols and Ethers: Conceptual Doubts Cleared (8)

medium

Alcohols, Phenols and Ethers: Diagram-Based Questions (5)

medium

Alcohols, Phenols and Ethers: Edge Cases and Subtle Traps (3)

hard

Alcohols, Phenols and Ethers: Exam-Pattern Drill (2)

medium