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Alcohols, Phenols & Ethers

10 NEET-UG practice questions on Alcohols, Phenols & Ethers, part of Chemistry. The first 10 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

  1. Question 1 · Hard

    Read these statements on the acid-catalysed dehydration of alcohols. A. Ease of dehydration follows the order tertiary > secondary > primary. B. Ethanol with conc. H₂SO₄ at 443 K gives ethene as the main product. C. The slowest step is the protonation of the alcohol. D. Excess ethanol with conc. H₂SO₄ at 413 K gives ethoxyethane. Which of these are correct?

    1. AA, C and D only
    2. BB, C and D only
    3. CA and B only
    4. DA, B and D only
    Show answer

    Answer: D · A, B and D only

    Dehydration passes through a carbocation, so its ease follows carbocation stability, which makes A correct. Temperature decides the product of ethanol: ethene at 443 K (B), and the ether at 413 K with excess alcohol (D). C is wrong, because protonation is fast and loss of water to form the carbocation is the slowest step.

    • A: This includes C and leaves out B, which is the standard preparation of ethene.
    • B: This includes C and leaves out A, which is the standard reactivity order.
    • C: This leaves out D, though the ether does form at 413 K.

    Remember this

    In acid dehydration, carbocation formation is the slow step; lower temperature with excess alcohol favours the ether.

  2. Question 2 · Easy

    Ethane-1,2-diol (0.2 mol) reacts completely with excess sodium. What volume of H₂ forms at STP? (Molar volume = 22.4 L)

    1. A6.72 L
    2. B2.24 L
    3. C8.96 L
    4. D4.48 L
    Show answer

    Answer: D · 4.48 L

    Each -OH gives half a mole of H₂, so a diol gives 1 mol of H₂ per mole. So 0.2 mol of diol gives 0.2 mol of H₂, and 0.2 × 22.4 = 4.48 L.

    • A: This treats the compound as a triol, taking 0.3 mol of H₂.
    • B: This counts only one -OH group per molecule.
    • C: This forgets that hydrogen is diatomic, taking 0.4 mol of H₂.

    Remember this

    Moles of H₂ with sodium equal half the moles of -OH groups present.

  3. Question 3 · Medium

    tert-Butyl ethyl ether is to be made by Williamson synthesis. Which pair of reactants gives it in good yield?

    1. ACH₃CH₂ONa and CH₃CH₂Br
    2. B(CH₃)₃CBr and CH₃CH₂ONa
    3. C(CH₃)₃CONa and CH₃CH₂Br
    4. D(CH₃)₃CONa and (CH₃)₃CBr
    Show answer

    Answer: C · (CH₃)₃CONa and CH₃CH₂Br

    Williamson synthesis is SN2, so it needs an unhindered, primary alkyl halide. The bulky group must come in as the alkoxide, with CH₃CH₂Br as the substrate.

    • A: This pair gives diethyl ether, not the required ether.
    • B: A tertiary halide with a strong base eliminates to give 2-methylpropene.
    • D: This pair has a tertiary halide, which eliminates, and has no ethyl group at all.

    Remember this

    Keep the alkyl halide primary; a tertiary halide with an alkoxide undergoes elimination and gives the alkene.

  4. Question 4 · Easy

    With Lucas reagent at room temperature, which alcohol turns cloudy immediately?

    1. A2-Methylpropan-1-ol
    2. B2-Methylpropan-2-ol
    3. CButan-1-ol
    4. DButan-2-ol
    Show answer

    Answer: B · 2-Methylpropan-2-ol

    Lucas reagent is conc. HCl with ZnCl₂. Tertiary alcohols form the insoluble alkyl chloride at once by way of a stable carbocation. Secondary alcohols take some minutes and primary ones show no turbidity at room temperature.

    • A: This is a branched primary alcohol, and branching does not change its class.
    • C: This is a primary alcohol, which shows no turbidity at room temperature.
    • D: This is a secondary alcohol, which needs several minutes.

    Remember this

    Lucas turbidity: tertiary at once, secondary in minutes, primary not at room temperature.

  5. Question 5 · Medium

    Arrange these compounds in increasing order of acid strength. A. Ethanol B. Phenol C. 4-Methylphenol D. 4-Nitrophenol

    1. AA, B, C, D
    2. BA, C, B, D
    3. CD, B, C, A
    4. DC, A, B, D
    Show answer

    Answer: B · A, C, B, D

    Phenoxide ions are stabilised by resonance, so phenols are more acidic than ethanol. An electron-donating methyl lowers the acidity of phenol slightly, while an electron-withdrawing nitro at the para position raises it strongly.

    • A: This ranks 4-methylphenol above phenol, treating methyl as electron-withdrawing.
    • C: This reverses the whole order.
    • D: This puts 4-methylphenol below ethanol, ignoring phenoxide resonance.

    Remember this

    Electron-withdrawing groups strengthen a phenol as an acid; electron-donating groups weaken it; every phenol here beats ethanol.

  6. Question 6 · Medium

    A sample of 9.4 g of phenol reacts completely with excess bromine water. What mass of the precipitate forms? (C = 12, H = 1, O = 16, Br = 80)

    1. A17.3 g
    2. B9.4 g
    3. C25.2 g
    4. D33.1 g

    Answer and explanation in LearnFree account

  7. Question 7 · Hard

    Given below are two statements: Statement I: Anisole with HI gives phenol and methyl iodide, not iodobenzene and methanol. Statement II: tert-Butyl methyl ether with HI gives tert-butyl alcohol and methyl iodide. Choose the correct answer.

    1. AStatement I is incorrect but Statement II is correct
    2. BStatement I is correct but Statement II is incorrect
    3. CBoth Statement I and Statement II are incorrect
    4. DBoth Statement I and Statement II are correct

    Answer and explanation in LearnFree account

  8. Question 8 · Easy

    Phenol is shaken with bromine water. What is the white precipitate?

    1. A4-Bromophenol
    2. B2,4,6-Tribromophenol
    3. C2,4-Dibromophenol
    4. D3,5-Dibromophenol

    Answer and explanation in LearnFree account

  9. Question 9 · Medium

    Arrange the compounds drawn in order of increasing acid strength. A. Structure A B. Structure B C. Structure C D. Structure D

    Figure: MedExams
    Four hydroxy compounds labelled A to D: one simple alcohol and three phenols with different ring substituents.
    1. AB, A, D, C
    2. BB, A, C, D
    3. CA, B, D, C
    4. DB, D, A, C

    Answer and explanation in LearnFree account

  10. Question 10 · Medium

    X and Y in the figure are isomeric alcohols. Given below are two statements: Statement I: With Lucas reagent at room temperature, X turns turbid almost at once, while Y does not. Statement II: On heating with concentrated acid, X dehydrates under milder conditions than Y. Choose the correct answer.

    Figure: MedExams
    Two four-carbon alcohols labelled X and Y, one branched and one straight-chain.
    1. AStatement I is correct but Statement II is incorrect
    2. BBoth Statement I and Statement II are correct
    3. CStatement I is incorrect but Statement II is correct
    4. DBoth Statement I and Statement II are incorrect

    Answer and explanation in LearnFree account

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