Alcohols, Phenols and Ethers NEET PYQReal NEET & AIPMT 2005–2020 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
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Alcohols, Phenols and Ethers previous year questions
34 questions. Tap an option to check your answer.
Q1NEET 2020 · Q100
Reaction between acetone and methylmagnesium chloride followed by hydrolysis will give :
Not quite — the answer is C.
CH₃MgCl adds to the carbonyl carbon of (CH₃)₂C=O, and hydrolysis of the alkoxide gives (CH₃)₃C–OH, tert-butyl alcohol. A ketone with a Grignard reagent always gives a tertiary alcohol; isopropyl alcohol would come from reduction of acetone instead.
Q2NEET 2020 · Q135
Anisole on cleavage with HI gives :
Not quite — the answer is A.
HI protonates the ether oxygen and I⁻ attacks the methyl carbon by SN2, because the O–aryl bond has partial double-bond character and cannot be cleaved; so anisole gives phenol and methyl iodide. Iodobenzene is not formed, and anisole has a methyl group, not an ethyl group.
Q3NEET 2019 · Q47
The structure of intermediate A in the following reaction, is: C₆H₅–CH(CH₃)₂ (cumene) —O₂→ A —H⁺/H₂O→ C₆H₅OH (phenol) + CH₃COCH₃ (acetone)
Not quite — the answer is B.
Air oxidation of cumene attacks the tertiary benzylic C–H, giving cumene hydroperoxide C₆H₅C(CH₃)₂OOH, which rearranges in acid to phenol and acetone. Isopropyl phenyl ether is not formed by O₂ and would not give acetone on hydrolysis.
Q4NEET 2019 · Q69
The compound that is most difficult to protonate is:
Not quite — the answer is D.
In phenol the oxygen lone pair is delocalised into the benzene ring, so it is least available to accept a proton. In water, methanol and dimethyl ether the lone pairs are localised, and the electron-donating methyl groups make the ether easiest to protonate.
Q5NEET 2018 · Q52
Identify the major products P, Q and R in the following sequence of reactions: Benzene + CH₃CH₂CH₂Cl —(Anhydrous AlCl₃)→ P; P —(i) O₂, (ii) H₃O⁺/Δ→ Q + R
Not quite — the answer is D.
In Friedel–Crafts alkylation the n-propyl carbocation rearranges to the more stable isopropyl cation, giving cumene. Air oxidation gives cumene hydroperoxide, which on acid hydrolysis gives phenol and acetone (not propan-2-ol).
Q6NEET 2018 · Q60
The compound A on treatment with Na gives B, and with PCl₅ gives C. B and C react together to give diethyl ether. A, B and C are in the order
Not quite — the answer is D.
Ethanol with Na gives sodium ethoxide; with PCl₅ it gives chloroethane. Williamson synthesis C₂H₅ONa + C₂H₅Cl → C₂H₅OC₂H₅. Na does not convert ethanol to ethane, and the order of B and C matters.
Q7NEET 2018 · Q66
In the reaction: phenol + CHCl₃ + NaOH → sodium salt of o-hydroxybenzaldehyde (o-formyl sodium phenoxide, O⁻Na⁺ with –CHO at the ortho position), the electrophile involved is
Not quite — the answer is D.
In the Reimer–Tiemann reaction, OH⁻ removes a proton from CHCl₃ and the ⁻CCl₃ loses Cl⁻ to give dichlorocarbene :CCl₂, an electron-deficient electrophile that attacks the phenoxide ring at the ortho position. The dichloromethyl anion is a nucleophile, not an electrophile.
Q8NEET 2018 · Q68
Compound A, C₈H₁₀O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell. A and Y are respectively
Not quite — the answer is A.
The yellow precipitate with antiseptic smell is iodoform, formed by NaOI (from I₂ + NaOH). It requires a CH₃CH(OH)– or CH₃CO– unit; 1-phenylethanol has it, while 2-phenylethanol, p-methylbenzyl alcohol and the dimethylphenol do not.
Q9NEET 2017 · Q13
Which one is the most acidic compound?
Not quite — the answer is D.
Electron-withdrawing –NO₂ groups at ortho and para positions stabilise the phenoxide ion by resonance, so acidity rises with the number of nitro groups; 2,4,6-trinitrophenol (pKa about 0.4) is the strongest acid. The methyl group of p-cresol is electron-donating and makes it weaker than phenol.
Q10NEET 2017 · Q18
The heating of phenyl-methyl ethers with HI produces.
Not quite — the answer is C.
HI protonates anisole and I⁻ attacks the methyl carbon by SN2, giving phenol and CH₃I. Iodobenzene does not form because the aryl C–O bond has partial double-bond character and SN2 attack on an sp² ring carbon does not occur.
Q11NEET 2017 (Gujarati paper) · Q11
Of the following alcohols, the one that would react fastest with conc. HCl and anhydrous ZnCl₂ is:
Not quite — the answer is C.
The Lucas reagent reacts through a carbocation, so the tertiary alcohol 2-methylpropan-2-ol, which forms the most stable tertiary carbocation, reacts at once. Secondary butan-2-ol reacts in minutes, and the primary alcohols barely react at room temperature.
Q12NEET 2016 Phase 1 · Q63
Which of the following reagents would distinguish cis-cyclopenta-1,2-diol from the trans-isomer?
Not quite — the answer is A.
In the cis-1,2-diol both OH groups are on the same face of the ring, so acetone can bridge them to form a five-membered cyclic ketal (acetonide); the trans-diol cannot reach both groups. Ozone acts on C=C bonds, which neither diol has.
Q13NEET 2016 Phase 1 · Q82
The reaction Cyclopentanol (C₅H₉–OH) --NaH--> C₅H₉–O⁻ Na⁺ --Me–I--> C₅H₉–O–Me can be classified as:
Not quite — the answer is A.
NaH converts cyclopentanol into its sodium alkoxide, which then displaces iodide from methyl iodide by an SN2 reaction to give methoxycyclopentane, an ether. Forming an ether from an alkoxide and an alkyl halide is the Williamson ether synthesis; no alcohol is formed and no water is eliminated.
Q14AIPMT 2015
The reaction (CH₃)₃C–ONa + CH₃CH₂Cl → (CH₃)₃C–O–CH₂CH₃ + NaCl is called:
Not quite — the answer is B.
An alkoxide reacting with a primary alkyl halide by SN2 to give an ether is the Williamson synthesis. Gattermann–Koch and Etard reactions make aromatic aldehydes, not ethers.
Q15AIPMT 2014
Which of the following will not be soluble in sodium hydrogen carbonate?
Not quite — the answer is B.
Only acids stronger than carbonic acid release CO₂ from NaHCO₃ and dissolve. Benzoic acid, benzenesulphonic acid and picric acid (2,4,6-trinitrophenol) are strong enough, but o-nitrophenol, with only one nitro group and intramolecular H-bonding, is weaker than H₂CO₃.
Q16AIPMT 2014
Identify Z in the sequence of reactions: CH₃CH₂CH=CH₂ —HBr/H₂O₂→ Y —C₂H₅ONa→ Z
Not quite — the answer is D.
With peroxide, HBr adds anti-Markovnikov to but-1-ene giving 1-bromobutane (Y). Williamson synthesis with sodium ethoxide then gives CH₃(CH₂)₃–O–CH₂CH₃; the branched sec-butyl ether would come from Markovnikov addition.
Q17AIPMT 2014
Among the following sets of reactants which one produces anisole?
Not quite — the answer is A.
Phenol with NaOH forms sodium phenoxide, which reacts with CH₃I by Williamson synthesis to give C₆H₅OCH₃ (anisole). Neutral FeCl₃ only gives a colour test, and the other reagent sets give ketones or alcohols.
Q18NEET 2013
Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?
Not quite — the answer is C.
With tert-butyl methyl ether, HI cleavage goes by SN1 through the stable tert-butyl carbocation, giving tert-butyl iodide and methyl alcohol. For the primary and secondary alkyl methyl ethers the reaction is SN2, where I⁻ attacks the less hindered methyl group, giving methyl iodide and the other alcohol.
Q19AIPMT 2012 Mains
Which of the following compounds can be used as antifreeze in automobile radiators ?
Not quite — the answer is B.
Ethylene glycol (ethane-1,2-diol) is the standard radiator antifreeze: it lowers the freezing point of water and, unlike methyl or ethyl alcohol, has a high boiling point so it does not evaporate away.
Q20AIPMT 2012 Prelims
In the following reaction: (CH₃)₃C–CH=CH₂ —(H₂O/H⁺)→ A (major product) + B (minor product) The major product is
Not quite — the answer is B.
Protonation of the alkene gives the secondary carbocation (CH₃)₃C–C⁺H–CH₃, which undergoes a 1,2-methyl shift to the more stable tertiary carbocation (CH₃)₂C⁺–CH(CH₃)₂. Water attacks this to give 2,3-dimethylbutan-2-ol, (CH₃)₂C(OH)–CH(CH₃)₂. The unrearranged alcohol (CH₃)₃C–CH(OH)–CH₃ is the minor product.
Q21AIPMT 2012 Prelims
In the following sequence of reactions CH₃–Br —(KCN)→ A —(H₃O⁺)→ B —(LiAlH₄, ether)→ C, the end product (C) is
Not quite — the answer is A.
CH₃Br + KCN gives CH₃CN; acid hydrolysis converts it to CH₃COOH; LiAlH₄ then reduces the acid to CH₃CH₂OH, ethyl alcohol, as the end product. Acetaldehyde is not isolated because LiAlH₄ reduces carboxylic acids all the way to primary alcohols.
Q22AIPMT 2010 Mains · Q96
Among the following four compounds a. Phenol b. Methyl phenol c. Metanitrophenol d. Paranitrophenol The acidity order is
Not quite — the answer is A.
Electron-withdrawing –NO₂ increases acidity of phenol most from the para position (−R and −I), less from meta (−I only), while –CH₃ donates electrons and lowers acidity. So para-nitrophenol > meta-nitrophenol > phenol > methylphenol. Putting meta above para ignores the resonance effect.
Q23AIPMT 2010 Mains · Q118
Following compounds are given a. CH₃CH₂OH b. CH₃COCH₃ c. CH₃CH(OH)CH₃ d. CH₃OH Which of the above compound(s), on being warmed with iodine solution and NaOH, will give iodoform?
Not quite — the answer is C.
The iodoform test is given by compounds with CH₃CO– or CH₃CH(OH)– groups: ethanol, acetone and 2-propanol all qualify. Methanol has no such group and gives no iodoform.
Q24AIPMT 2010 Prelims · Q92
Which one of the following compounds has the most acidic nature?
Not quite — the answer is C.
Phenol is far more acidic (pKa ≈ 10) than any alcohol because the phenoxide ion is resonance-stabilised by the ring. Benzyl alcohol, cyclohexanol and dicyclohexylmethanol give alkoxide ions with no such delocalisation.
Q25AIPMT 2010 Prelims · Q93
Given are cyclohexanol (I), acetic acid (II), 2, 4, 6-trinitrophenol (III) and phenol (IV). In these the order of decreasing acidic character will be
Not quite — the answer is B.
Picric acid (2,4,6-trinitrophenol, pKa ≈ 0.4) is the strongest because three –NO₂ groups stabilise the phenoxide; acetic acid (pKa ≈ 4.8) is next, then phenol (≈ 10), then cyclohexanol (≈ 16). Putting phenol above acetic acid is the common trap.
Q26AIPMT 2009 Prelims · Q160
Consider the following reaction: ethanol —(PBr₃)→ X —(alc. KOH)→ Y —((i) H₂SO₄, room temperature; (ii) H₂O, heat)→ Z The product Z is:
Not quite — the answer is C.
PBr₃ converts ethanol to bromoethane (X); alcoholic KOH eliminates HBr to give ethene (Y). Cold conc. H₂SO₄ adds across the double bond giving ethyl hydrogen sulphate, which on boiling with water is hydrolysed back to ethanol (Z). Ethyl hydrogen sulphate is only the intermediate.
Q27AIPMT 2009 Prelims · Q179
HOH₂C–CH₂OH on heating with periodic acid gives:
Not quite — the answer is C.
Periodic acid cleaves the C–C bond between the two –OH bearing carbons of a 1,2-diol. Each CH₂OH becomes H₂C=O, so ethylene glycol gives two molecules of formaldehyde. Glyoxal would need oxidation without C–C cleavage.
Q28AIPMT 2009 Prelims · Q188
Consider the following reaction: Phenol —(Zn dust)→ X —(CH₃Cl, anhydrous AlCl₃)→ Y —(alkaline KMnO₄)→ Z The product Z is:
Not quite — the answer is B.
Distilling phenol with zinc dust removes the –OH, giving benzene (X). Friedel–Crafts methylation gives toluene (Y). Alkaline KMnO₄ oxidises the whole side chain to –COOH, giving benzoic acid (Z), not just benzaldehyde.
Q29AIPMT 2008 Prelims · Q94
Which one of the following is most reactive towards electrophilic attack?
Not quite — the answer is D.
The –OH group of phenol donates its lone pair into the ring (+R effect), which strongly activates the ring towards electrophiles. –CH₂OH is only weakly activating, –Cl deactivates by its –I effect, and –NO₂ strongly deactivates.
Q30AIPMT 2007 Prelims · Q97
In the reaction: (CH₃)₂CH–CH₂–O–CH₂–CH₃ + HI —Heated→ Which of the following compounds will be formed?
Not quite — the answer is C.
With two primary alkyl groups, I⁻ attacks the less hindered carbon by SN2 after protonation of the ether oxygen. The ethyl carbon is less hindered than the isobutyl carbon, so the products are CH₃CH₂I and (CH₃)₂CHCH₂OH.
Q31AIPMT 2006 Prelims · Q66
The general molecular formula, which represents the homologous series of alkanols is :
Not quite — the answer is D.
Alkanols are CₙH₂ₙ₊₁OH, which written as a molecular formula is CₙH₂ₙ₊₂O (e.g. CH₄O for methanol, C₂H₆O for ethanol). CₙH₂ₙO fits aldehydes and ketones, and CₙH₂ₙO₂ fits acids and esters.
Q32AIPMT 2006 Prelims · Q73
The major organic product in the reaction, CH₃–O–CH(CH₃)₂ + HI → Product is :
Not quite — the answer is D.
HI protonates the ether oxygen and I⁻ attacks by SN2 at the less hindered carbon, which is the methyl group. The products are CH₃I and isopropyl alcohol, (CH₃)₂CHOH. Attack at the secondary isopropyl carbon, giving CH₃OH and isopropyl iodide, is slower under these conditions.
Q33AIPMT 2006 Prelims · Q86
Ethylene oxide when treated with Grignard reagent yields :
Not quite — the answer is D.
The carbanion of RMgX attacks a ring carbon of ethylene oxide and opens the strained ring, giving RCH₂CH₂OMgX. Hydrolysis yields RCH₂CH₂OH, a primary alcohol with two more carbons than R. Secondary and tertiary alcohols come from aldehydes and ketones with Grignard reagents.
Q34AIPMT 2005 Prelims · Q65
Which one of the following compounds is most acidic ?
Not quite — the answer is B.
Phenols are more acidic than alcohols because the phenoxide ion is resonance-stabilised. The –NO₂ group withdraws electrons and further stabilises the phenoxide, so o-nitrophenol is most acidic; the –CH₃ of o-cresol lowers acidity.
How often Alcohols, Phenols and Ethers came up
Questions from this chapter in each paper on this site.
- NEET 2020: 2
- NEET 2019: 2
- NEET 2018: 4
- NEET 2017: 2
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 1: 2
- AIPMT 2015: 1
- AIPMT 2014: 3
- NEET 2013: 1
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 2
- AIPMT 2010 Mains: 2
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 3
- AIPMT 2008 Prelims: 1
- AIPMT 2007 Prelims: 1
- AIPMT 2006 Prelims: 3
- AIPMT 2005 Prelims: 1
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