CBSE Class 12 Chemistry Alcohols Phenols and Ethers NEET JEE Questions Set 03

Single Correct Answer

Question 1. 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 (NEET 2018)
(a) C₂H₅OH, C₂H₆, C₂H₅Cl
(b) C₂H₅OH, C₂H₅Cl, C₂H₅ONa
(c) C₂H₅Cl, C₂H₆, C₂H₅OH
(d) C₂H₅OH, C₂H₅ONa, C₂H₅Cl
Answer:
Ans. (d)
2C₂H₅OH (A) + 2Na → 2C₂H₅ONa (Sodium ethoxide, B) + H₂
C₂H₅OH + PCl₅ → C₂H₅Cl (C) + POCl₃ + HCl
C₂H₅ONa + C₂H₅Cl → C₂H₅OC₂H₅ (Diethyl ether) + NaCl

Teacher's Note:
1. Williamson synthesis needs an alkoxide (from alcohol + Na) and an alkyl halide (from alcohol + PCl₅).

Question 2. In the given reaction,
3-bromo-2,2-dimethylbutane \( \xrightarrow{\mathrm{C_2H_5OH}} \) 'A' (major product)
Product 'A' is (JEE Main 2021)
(a) 2-Ethoxy-2,3-dimethylbutane
(b) 2-Hydroxy-1,3-dimethylbutane
(c) 1-Ethoxy-3,3-dimethylbutane
(d) 2-Ethoxy-3,3-dimethylbutane
Answer:
Ans. (a)
(CH₃)₃C–CHBr–CH₃ (3-Bromo-2,2-dimethylbutane) \( \xrightarrow{\mathrm{C_2H_5OH}\ \text{(Ionization)}} \) (CH₃)₃C–C⁺H–CH₃ (2° Carbocation, less stable)
\( \xrightarrow{\text{1,2-Methyl shift}} \) (CH₃)₂C⁺–CH(CH₃)CH₃ \( \xrightarrow[-\mathrm{H^+}]{\mathrm{C_2H_5OH}} \) (CH₃)₂C(OC₂H₅)–CH(CH₃)–CH₃, 2-Ethoxy-2,3-dimethylbutane

Teacher's Note:
1. Ethanol is a weak nucleophile, so a 2° halide reacts by S_N1 (solvolysis); expect a methyl or hydride shift to a 3° cation.

Question 3. The reaction
Cyclopentanol (C₅H₉–OH) \( \xrightarrow{\text{NaH}} \) C₅H₉–O⁻Na⁺ \( \xrightarrow{\text{Me–I}} \) C₅H₉–O–Me
can be classified as (NEET Phase-I 2016)
(a) dehydration reaction
(b) Williamson alcohol synthesis reaction
(c) Williamson ether synthesis reaction
(d) alcohol formation reaction.
Answer:
Ans. (c)
C₅H₉–OH \( \xrightarrow{\text{NaH}} \) C₅H₉–O⁻Na⁺ \( \xrightarrow[-\mathrm{NaI}]{\text{Me–I}} \) C₅H₉–O–Me (Williamson ether synthesis)

Teacher's Note:
1. An alkoxide plus a primary (here methyl) halide by S_N2 is the Williamson ether synthesis; NaH is just used to make the alkoxide.

Question 4. Which of the following, upon treatment with tert-BuONa followed by addition of bromine water, fails to decolourise the colour of bromine ? (JEE Main 2017)
(a) C₆H₅–O–CH(Br)–CH₃
(b) C₆H₁₁–CO–CH(Br)–CH₃
(c) C₆H₁₁–O–CH₂–Br
(d) C₆H₅–CH₂–CH(Br)–C₆H₅
Answer:
Ans. (c)
The reactions of the given four compounds with tert-BuONa are shown below :
(a) C₆H₅–O–CH(Br)–CH₃ \( \xrightarrow{tert\text{-BuONa}} \) C₆H₅–O–CH=CH₂ (2° alkyl halides preferably undergo dehydrohalogenation to give alkenes; decolourises Br₂–H₂O due to unsaturation)
(b) C₆H₁₁–CO–CH(Br)–CH₃ \( \xrightarrow{tert\text{-BuONa}} \) C₆H₁₁–CO–CH=CH₂ (Dehydrohalogenation; decolourises Br₂–H₂O due to unsaturation)
(c) C₆H₁₁–O–CH₂–Br \( \xrightarrow{tert\text{-BuONa}} \) C₆H₁₁–O–CH₂–O^tBu (Williamson's synthesis, 1° alkyl halides undergo substitution; fails to decolourise Br₂–H₂O)
(d) C₆H₅–CH₂–CH(Br)–C₆H₅ \( \xrightarrow{tert\text{-BuONa}} \) C₆H₅–CH=CH–C₆H₅ (Dehydrohalogenation; decolourises Br₂–H₂O due to unsaturation)
Thus, option (c) is correct.

Teacher's Note:
1. Decolourising bromine water is a test for C=C; so the question is really which substrate undergoes substitution rather than elimination.

Question 5. C₆H₅OH + C₂H₅I \( \xrightarrow[\text{Anhyd. C}_2\text{H}_5\text{OH}]{^-\text{OC}_2\text{H}_5} \) gives
(a) C₆H₅OC₂H₅
(b) C₂H₅OC₂H₅
(c) C₆H₅OC₆H₅
(d) C₆H₅I
Answer:
Ans. (a)
C₂H₅O⁻ will abstract a proton from phenol to form phenoxide ion (C₆H₅O⁻), since phenol (pK_a ≈ 10) is a much stronger acid than ethanol (pK_a ≈ 16).
The phenoxide ion so formed attacks C₂H₅I by S_N2 reaction (Williamson synthesis) to form C₆H₅OC₂H₅.
C₆H₅O⁻ + C₂H₅I → C₆H₅OC₂H₅ + I⁻

Teacher's Note:
1. An acid–base reaction is faster than any substitution: ethoxide first converts phenol into phenoxide, and the phenoxide is then alkylated.

Question 6. Identify Z in the sequence of reactions :
CH₃CH₂CH=CH₂ \( \xrightarrow{\mathrm{HBr/H_2O_2}} \) Y \( \xrightarrow{\mathrm{C_2H_5ONa}} \) Z (AIPMT 2014)
(a) CH₃–(CH₂)₃–O–CH₂CH₃
(b) (CH₃)₂CH–O–CH₂CH₃
(c) CH₃–(CH₂)₄–O–CH₃
(d) CH₃CH₂–CH(CH₃)–O–CH₂CH₃
Answer:
Ans. (a)
CH₃CH₂CH=CH₂ \( \xrightarrow{\mathrm{HBr/H_2O_2}\ \text{(Anti-Mark. addn.)}} \) CH₃CH₂CH₂CH₂Br (Y) \( \xrightarrow[-\mathrm{NaBr}]{\mathrm{C_2H_5ONa}\ \text{(Williamson's synthesis)}} \) CH₃CH₂CH₂CH₂–O–CH₂CH₃

Teacher's Note:
1. HBr in the presence of peroxide adds anti-Markovnikov (only HBr shows this effect), giving the 1° bromide needed for a clean Williamson reaction.

Question 7. Among the following, the one which reacts most readily with ethanol is
(a) p-nitrobenzyl bromide
(b) p-chlorobenzyl chloride
(c) p-methoxybenzyl bromide
(d) p-methylbenzyl bromide
Answer:
Ans. (c)
Since benzyl carbocation is stabilized by resonance, therefore, reaction occurs by S_N1 mechanism. Further since electron-donating groups increase while electron-withdrawing groups decrease the stability of carbocations, therefore, p-methoxybenzyl bromide reacts most readily since it gives the most stable carbocation.
CH₃O–C₆H₄–C⁺H₂ ↔ CH₃–O⁺=C₆H₄=CH₂

Teacher's Note:
1. With a weak nucleophile like ethanol the rate depends on carbocation stability, so the strongest +R para substituent wins.

Question 8. Consider the reactions :
(i) (CH₃)₂CH–CH₂Br \( \xrightarrow{\mathrm{C_2H_5OH}} \) (CH₃)₂CH–CH₂OC₂H₅ + HBr
(ii) (CH₃)₂CH–CH₂Br \( \xrightarrow{\mathrm{C_2H_5O^-}} \) (CH₃)₂CH–CH₂OC₂H₅ + Br⁻
The mechanisms of reactions (i) and (ii) are respectively : (AIPMT Mains 2011)
(a) S_N1 and S_N2
(b) S_N1 and S_N1
(c) S_N2 and S_N2
(d) S_N2 and S_N1
Answer:
Ans. (c)
Since rearrangements do not occur in the given nucleophilic substitution reactions, therefore, carbocations are not the intermediates in these reactions. Thus, both the reactions occur by S_N2 mechanism, i.e., option (c) is correct.

Teacher's Note:
1. An isobutyl cation would rearrange to the tert-butyl cation; since the products keep the unrearranged skeleton, no carbocation was formed.

Question 9. In the reaction :
CH₃–CH(CH₃)–CH₂–O–CH₂–CH₃ + HI \( \xrightarrow{\text{Heated}} \) ......
Which of the following compounds will be formed ? (AIPMT 2007)
(a) CH₃–CH(CH₃)–CH₂OH + CH₃CH₂–I
(b) CH₃–CH(CH₃)–CH₂I + CH₃CH₂OH
(c) CH₃–CH(CH₃)–CH₃ + CH₃CH₂OH
(d) CH₃–CH(CH₃)–CH₂OH + CH₃CH₃
Answer:
Ans. (a)
Nucleophilic attack occurs on the smaller alkyl group.
CH₃–CH(CH₃)–CH₂–O–CH₂–CH₃ \( \xrightarrow{\mathrm{H^+}} \) CH₃–CH(CH₃)–CH₂–O⁺(H)–CH₂–CH₃ \( \xrightarrow{\mathrm{I^-}\ (\mathrm{S_N2})} \) CH₃–CH(CH₃)–CH₂OH + CH₃CH₂–I

Teacher's Note:
1. With two primary groups the iodide ion attacks the less hindered carbon in the S_N2 step, so the smaller alkyl group becomes the iodide.

Question 10. Heating of phenyl methyl ether with HI produces (NEET 2017)
(a) iodobenzene
(b) phenol
(c) benzene
(d) ethyl chloride
Answer:
Ans. (b)
C₆H₅–O–CH₃ \( \xrightarrow{\mathrm{H^+}} \) C₆H₅–O⁺(H)–CH₃ \( \xrightarrow{\mathrm{I^-}} \) C₆H₅–OH (Phenol) + CH₃I

Teacher's Note:
1. The aryl C–O bond is strengthened by resonance and an S_N2 attack on a ring carbon is impossible, so the alkyl group always ends up as the iodide.

Question 11. The major product of the following reaction is
m-CH₃O–C₆H₄–CH=CH₂ \( \xrightarrow[\text{heat}]{\text{Conc. HBr (excess)}} \) (JEE Main 2019)
(a) m-Br–C₆H₄–CH₂CH₂Br
(b) m-HO–C₆H₄–CH₂CH₂Br
(c) m-HO–C₆H₄–CH(Br)–CH₃
(d) m-Br–C₆H₄–CH(Br)–CH₃
Answer:
Ans. (c)
HBr brings about cleavage of the OCH₃ group and adds across the d.b. in accordance with Markovnikov's rule.
m-CH₃O–C₆H₄–CH=CH₂ \( \xrightarrow{2\mathrm{HBr}} \) m-HO–C₆H₄–CH(Br)–CH₃ + CH₃Br

Teacher's Note:
1. The aryl–O bond survives HBr (the methyl group leaves as CH₃Br), and HBr without peroxide adds to the vinyl group in the Markovnikov way.

Question 12. What are the products of the following reaction ?
C₆H₅OCH₂CH₂OH \( \xrightarrow[\text{heat}]{\text{excess HBr}} \) ?
(a) C₆H₅OH + BrCH₂CH₂Br
(b) C₆H₅OH + HOCH₂CH₂OH
(c) C₆H₅Br + HOCH₂CH₂OH
(d) C₆H₅OH + BrCH₂CH₂OH
Answer:
Ans. (a)
C₆H₅OCH₂CH₂OH \( \xrightarrow[\Delta]{\text{excess HBr}} \) C₆H₅OH + BrCH₂CH₂Br

Teacher's Note:
1. Excess HBr does two jobs here: it cleaves the alkyl–O bond of the ether and also converts the free alcoholic OH into C–Br.

Question 13. The major product formed in the following reaction is :
o-C₂H₅O–C₆H₄–CH(CH₃)–OCH₃ \( \xrightarrow[\text{Heat}]{\text{HI}} \) (JEE Main 2018)
(a) o-HO–C₆H₄–CH(OH)–CH₃
(b) o-I–C₆H₄–CH(I)–CH₃
(c) o-I–C₆H₄–CH(OH)–CH₃
(d) o-HO–C₆H₄–CH(I)–CH₃
Answer:
Ans. (d)
1° Alkyl phenyl ethers undergo S_N2 reaction giving the corresponding phenol and the alkyl iodide corresponding to the 1° alkyl group. In contrast, ethers containing one 2° alkyl group and one 1° alkyl group undergo S_N1 reaction giving alkyl iodide corresponding to the 2° alkyl group.
o-C₂H₅O–C₆H₄–CH(CH₃)OCH₃ \( \xrightarrow{\mathrm{HI}} \) (ArO⁺(H)–C₂H₅) \( \xrightarrow[(-\mathrm{CH_3CH_2I})]{\mathrm{I^-,\ S_N2}} \) o-HO–C₆H₄–CH(CH₃)OCH₃
\( \xrightarrow{\mathrm{HI}} \) o-HO–C₆H₄–CH(CH₃)–O⁺(H)CH₃ \( \xrightarrow[(-\mathrm{CH_3OH})]{\mathrm{S_N1}} \) o-HO–C₆H₄–C⁺H–CH₃ \( \xrightarrow{\mathrm{I^-}} \) o-HO–C₆H₄–CH(I)–CH₃
Thus, option (d) is correct.

Teacher's Note:
1. The benzylic 2° group leaves as a stabilised carbocation (S_N1) and gets the iodide, while the aryl–O bond is never broken.

Question 14. The major product [B] in the following reactions is
CH₃–CH₂–CH(CH₃)–CH₂–OCH₂–CH₃ \( \xrightarrow[\text{Heat}]{\text{HI}} \) [A] alcohol \( \xrightarrow[\Delta]{\mathrm{H_2SO_4}} \) [B] (JEE Main 2020)
(a) CH₂=CH₂
(b) CH₃–CH=C(CH₃)–CH₃
(c) CH₃–CH₂–C(CH₃)=CH₂
(d) CH₃–CH₂–CH=CH–CH₃
Answer:
Ans. (b)
CH₃CH₂–CH(CH₃)–CH₂–O–CH₂CH₃ \( \xrightarrow{\mathrm{H^+}} \) CH₃CH₂–CH(CH₃)–CH₂–O⁺(H)–CH₂CH₃ \( \xrightarrow{\mathrm{I^-}\ (-\mathrm{CH_3CH_2I})} \) CH₃CH₂–CH(CH₃)–CH₂OH [A]
\( \xrightarrow[-\mathrm{H_2O}]{\mathrm{H^+}} \) CH₃CH₂–CH(CH₃)–C⁺H₂ \( \xrightarrow{\text{1,2-Hydride shift}} \) CH₃CH₂–C⁺(CH₃)–CH₃ \( \xrightarrow[\text{(Saytzeff elimination)}]{-\mathrm{H^+}} \) CH₃CH=C(CH₃)–CH₃ [B]
Thus, option (b) is correct.

Teacher's Note:
1. HI cleaves the ether at the less hindered ethyl group; the 1° alcohol then dehydrates through a hydride shift to give the most substituted alkene.

Question 15. An organic compound 'A' (C₉H₁₀O) when treated with conc. HI undergoes cleavage to yield compounds 'B' and 'C'. 'B' gives yellow precipitate with AgNO₃ whereas 'C' tautomerises to 'D'. 'D' gives positive iodoform test. 'A' could be (JEE Main 2020)
(a) C₆H₅–O–CH₂–CH=CH₂
(b) C₆H₅–O–CH=CH–CH₃
(c) C₆H₅–CH₂–O–CH=CH₂
(d) p-H₃C–C₆H₄–O–CH=CH₂
Answer:
Ans. (c)
C₆H₅–CH₂–O–CH=CH₂, C₉H₁₀O (A) \( \xrightarrow[\Delta]{\mathrm{HI}} \) C₆H₅–CH₂I (B) + HO–CH=CH₂ (C) ⇄ CH₃–CHO (D) (Gives +ve iodoform test)
C₆H₅–CH₂I (B) \( \xrightarrow{\mathrm{AgNO_3}} \) AgI (Yellow ppt.)
Thus, option (c) is correct.

Teacher's Note:
1. Aryl iodides do not precipitate AgI, so B must be an alkyl or benzyl iodide; a vinyl alcohol (enol) always tautomerises to the aldehyde.

Question 16. The products in the following reactions are :
C₆H₅–O–CH₂–CH=CH₂ \( \xrightarrow[\text{373 K}]{\mathrm{HI}} \)
(a) C₆H₅–CH₂OH + CH₂=CHI
(b) C₆H₅–OH + HO–CH₂–CH=CH₂
(c) C₆H₅–OCH₂I + CH₃CHO
(d) C₆H₅–OH + I–CH₂–CH=CH₂
Answer:
Ans. (d)
C₆H₅–O–CH₂–CH=CH₂ \( \xrightarrow{\mathrm{HI}} \) C₆H₅–OH + CH₂=CHCH₂I
Option (d) is correct.

Teacher's Note:
1. In alkyl aryl ethers the O–aryl bond is never cleaved; the allyl group is easily attacked by I⁻ and becomes allyl iodide.

Question 17. Which of the following ethers is not cleaved by HI
(a) C₆H₅–CH₂–O–CH₃
(b) C₆H₅–O–C₆H₅
(c) C₆H₅–O–CH₂CH₃
(d) C₆H₅–O–CH₂–C₆H₄–CH₃ (p-)
Answer:
Ans. (b)
The C–O bond in phenols has some double bond character and hence is not cleaved by HI. In diphenyl ether both C–O bonds are of this type.

Teacher's Note:
1. An ether is cleaved by HI only if at least one alkyl (or benzyl) group is attached to O; diaryl ethers resist cleavage.

Question 18. The major product of the following reaction is
m-CH₃O–C₆H₄–CH(CN)–O–CH(CH₃)₂ \( \xrightarrow[\text{373 K}]{\text{HI (excess)}} \) (JEE Main 2020)
(a) m-HO–C₆H₄–CH(I)–CN
(b) m-I–C₆H₄–CH(OH)–CN
(c) m-HO–C₆H₄–CH(OH)–CN
(d) m-I–C₆H₄–CH(I)–CN
Answer:
Ans. (a)
m-CH₃O–C₆H₄–CH(CN)–O–CH(CH₃)₂ \( \xrightarrow{2\mathrm{H^+}} \) both ether oxygens are protonated; I⁻ attacks the smaller CH₃ group of the aryl methyl ether giving m-HO–C₆H₄– and CH₃I.
At the other ether oxygen, I⁻ attacks the benzylic carbon (carrying CN) and isopropyl alcohol, HO–CH(CH₃)₂, is released: m-HO–C₆H₄–CH(I)–CN + HO–CH(CH₃)₂

Teacher's Note:
1. The aryl–O bond is never broken by HI, so the ring keeps an OH; check which carbon on each ether oxygen is more easily attacked by I⁻.

Question 19. The major product C in the following reaction sequence will be
2-Methyl-2H-chromene (2-methyl-2H-1-benzopyran) \( \xrightarrow[\Delta]{\text{HBr (excess)}} \) A \( \xrightarrow[\mathrm{H^+}]{\text{KOH (alc.)}} \) B \( \xrightarrow{\mathrm{O_3,\ Zn/H_2O}} \) C (JEE Main 2020)
(a) o-HO–C₆H₄–CHO
(b) o-Br–C₆H₄–CHO
(c) o-Br–C₆H₄–CH₂CO₂H
(d) o-HO–C₆H₄–CO₂H
Answer:
Ans. (a)
2-Methyl-2H-chromene \( \xrightarrow{\text{HBr (Cleavage of ether)}} \) [o-HO–C₆H₄–CH=CH–CH(Br)–CH₃] \( \xrightarrow{\text{HBr (Addition of HBr to d.b.)}} \) o-HO–C₆H₄–CH(Br)–CH₂–CH(Br)–CH₃ (A)
\( \xrightarrow[-2\mathrm{HBr}]{\text{KOH (alc.)}} \) o-HO–C₆H₄–CH=CH–CH=CH₂ (B)
\( \xrightarrow[\text{(Ozonolysis)}]{\text{(i) O}_3\text{; (ii) Zn/H}_2\text{O}} \) o-HO–C₆H₄–CHO (C) + OHC–CHO + HCHO
Thus, option (a) is correct.

Teacher's Note:
1. The aryl–O bond survives every step, so the phenolic OH ends up in the final product; ozonolysis turns the ring-attached CH= into –CHO.

Question 20. In the following reaction sequences, structures of A and B respectively, will be
5-(Bromomethyl)chromane \( \xrightarrow[\Delta]{\mathrm{HBr}} \) A \( \xrightarrow[\text{Ether}]{\mathrm{Na}} \) B (Intramolecular Product) (JEE Main 2020)
(a) A = 2-(3-bromopropyl)-3-(bromomethyl)phenol and B = 2-methylindan-4-ol
(b) A = 2-(3-bromopropyl)-3-(bromomethyl)phenol and B = 5,6,7,8-tetrahydronaphthalen-1-ol
(c) A = 3-[2-bromo-6-(bromomethyl)phenyl]propan-1-ol and B = 5-(bromomethyl)chromane
(d) A = 3-[2-bromo-6-(bromomethyl)phenyl]propan-1-ol and B = 5-bromoisochromane
Answer:
Ans. (b)
5-(Bromomethyl)chromane \( \xrightarrow[\text{(Cleavage of ether)}]{\text{HBr}} \) 2-(3-bromopropyl)-3-(bromomethyl)phenol (A)
\( \xrightarrow[\text{(Intramolecular Wurtz reaction)}]{\text{Na/ether}} \) 5,6,7,8-tetrahydronaphthalen-1-ol (B)
Thus, option (b) is correct.

Teacher's Note:
1. HBr breaks the alkyl–O bond of the aryl alkyl ether, keeping OH on the ring; the two C–Br ends then join by Wurtz coupling to form a new six-membered ring.

Question 21. Which of the following compounds is resistant to nucleophilic attack by hydroxyl ions ?
(a) Methyl acetate
(b) Acetonitrile
(c) Acetamide
(d) Diethyl ether
Answer:
Ans. (d)
Diethyl ether itself being a Lewis base is not attacked by nucleophiles, i.e., OH⁻ ions while all others contain either a C=O or C≡N group which is easily attacked by OH⁻ ion.

Teacher's Note:
1. Ethers are inert to bases and nucleophiles because RO⁻ is a very poor leaving group; that is why ethers are used as solvents for Grignard reactions.

Question 22. Anisole \( \xrightarrow{(\mathrm{CH_3})_3\mathrm{CCl},\ \mathrm{AlCl_3}} \) \( \xrightarrow{\mathrm{Cl_2/FeCl_3}} \) \( \xrightarrow[\text{Heat}]{\mathrm{HBr}} \) X
The product X in the above series of reactions is
(a) 2-Bromo-4-tert-butylanisole
(b) 1-Bromo-2-chloro-4-tert-butylbenzene
(c) 1,2-Dibromo-4-tert-butylbenzene
(d) 2-Chloro-4-tert-butylphenol
Answer:
Ans. (d)
Anisole \( \xrightarrow{(\mathrm{CH_3})_3\mathrm{CCl},\ \mathrm{AlCl_3}\ \text{(F.C. alkylation)}} \) p-(CH₃)₃C–C₆H₄–OCH₃
\( \xrightarrow{\mathrm{Cl_2/FeCl_3}} \) 2-chloro-4-tert-butylanisole (methoxy group is a stronger o, p-directing group than C(CH₃)₃)
\( \xrightarrow{\text{HBr, Heat (Cleavage of ethers)}} \) 2-chloro-4-tert-butylphenol + CH₃Br

Teacher's Note:
1. When two groups compete, the stronger activator (–OCH₃) decides where the next substituent goes; HBr then removes only the methyl group.

One or More Correct Answers

Question 23. When phenol is treated with CHCl₃ and NaOH followed by acidification, salicylaldehyde is obtained. Which of the following species are involved in the above mentioned reaction as intermediate ?
(a) The cyclohexa-2,4-dien-1-one in which the ortho sp³ carbon carries H and –C⁻Cl₂
(b) o-HO–C₆H₄–CHCl₂
(c) The dienone in which the ortho sp³ carbon carries H and –CHCl(OH)
(d) o-⁻O–C₆H₄–CHCl₂
Answer:
Ans. (a, d)
(b) being a phenol (i.e., acidic) cannot exist in basic medium while (c) will undergo tautomerism to give aromatic compound, i.e., o-HO–C₆H₄–CH(OH)Cl.
Thus, species (a) and (d) are involved in the above reaction.

Teacher's Note:
1. In strongly basic solution a phenol exists only as phenoxide, so the intermediates carry O⁻; the dienone formed by :CCl₂ attack re-aromatises by proton transfer.

Question 24. In the following reaction, the product(s) formed is (are)
p-Cresol (4-CH₃–C₆H₄–OH) \( \xrightarrow[\mathrm{OH^-}]{\mathrm{CHCl_3}} \) ?
P: 2-hydroxy-5-methylbenzene-1,3-dicarbaldehyde; Q: 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-one; R: 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-ol; S: 2-hydroxy-5-methylbenzaldehyde (JEE Advanced 2013)
(a) P (major)
(b) Q (minor)
(c) R (minor)
(d) S (major)
Answer:
Ans. (b, d)
–OH group is o, p-directing. Since :CCl₂ is the reactive intermediate in R.T. reaction, it attacks the benzene ring at o, p-positions giving S (ortho) as the major product and Q (para) as the minor product.
At the para position the CH₃ group blocks re-aromatisation, so the –CHCl₂ group cannot be hydrolysed to –CHO and the dienone Q is isolated.

Teacher's Note:
1. When the para position of a phenol is occupied, the Reimer-Tiemann reaction can give a para-dichloromethyl dienone as a by-product.

Passage-Based Questions

Question 25. Comprehension 1: Riemer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehydes as depicted below.
p-CH₃–C₆H₄–OH (I) → [I] (Intermediate) → 2-Na⁺O⁻–5-CH₃–C₆H₃–CHO (II) \( \xrightarrow{\text{aq. HCl}} \) 2-HO–5-CH₃–C₆H₃–CHO (III) (IIT Paper II, 2007)
The structure of the intermediate [I] is
(a) 2-Na⁺O⁻–5-CH₃–C₆H₃–CH₂Cl
(b) 2-Na⁺O⁻–5-CH₃–C₆H₃–CHCl₂
(c) 2-Na⁺O⁻–5-CH₃–C₆H₃–CCl₃
(d) 2-Na⁺O⁻–5-CH₃–C₆H₃–CH₂OH
Answer:
Ans. (b)
The intermediate (I) in the R.T. reaction is (b) which on hydrolysis gives product (II).
Ar–CHCl₂ + 2OH⁻ → [Ar–CH(OH)₂] → Ar–CHO + H₂O

Teacher's Note:
1. The –CHCl₂ group (one H, two Cl) is at the aldehyde oxidation level; replacing both Cl by OH gives a gem-diol that loses water to form –CHO.

Matrix Match Questions

Teacher's Note:
1. Acidity order here: p-nitrophenol > o-nitrophenol > methanol > ethanol; only the ortho isomer, with intramolecular H-bonding, is steam volatile.

Teacher's Note:
1. For a Williamson synthesis, put the bulky group on the alkoxide and the small group on the halide; the reverse combination gives an alkene.

Integer and Numerical Answer Questions

Question 28. How many of the following ethers cannot be prepared by Williamson's synthesis ?
CH₃OCH₂CH₃, C₆H₅OCH₃, C₆H₅OCH₂CH₃, (C₆H₅)₂O, (CH₃)₃COCH₃, (CH₃)₃COCH₂CH₃, (CH₃)₃COC(CH₃)₃, (C₂H₅)₂O, C₆H₅CH₂OC₆H₅.
Answer:
Ans. 2
Williamson's synthesis requires that the alkyl halide should be 1° and alkoxide ion may be 1°, 2° or 3°. Thus, two ethers which cannot be prepared by Williamson's synthesis are : (C₆H₅)₂O, (CH₃)₃COC(CH₃)₃.

Teacher's Note:
1. Aryl halides do not undergo S_N2 and 3° halides eliminate, so an ether needing either of them on the halide side cannot be made.

Question 29. How many equivalents of CH₃MgBr are required to make 2-methylpropan-2-ol from ethyl ethanoate. (JEE Main 2021)
Answer:
Ans. 2
CH₃COOC₂H₅ (Ethyl ethanoate) \( \xrightarrow[-\mathrm{OC_2H_5MgBr}]{\mathrm{CH_3MgBr}} \) (CH₃)₂C=O \( \xrightarrow{\text{(i) CH}_3\text{MgBr; (ii) H}_3\text{O}^+} \) (CH₃)₂C(OH)CH₃, 2-Methylpropan-2-ol
Thus, 2-equivalents of CH₃MgBr are required.

Teacher's Note:
1. Esters always consume two equivalents of RMgX: the ketone formed after the first addition is more reactive than the ester and reacts at once.

Question 30. For the following question, enter the correct numerical value (in decimal notation), truncated/rounded-off to the second decimal place.
In the following reaction sequence, the amount of D (in g) formed from one mole of propyne is ...... .
CH₃C≡CH (1 mole) \( \xrightarrow[\text{330 K}]{\mathrm{H_2O/H_2SO_4}} \) A (80%) \( \xrightarrow{\text{(i) CH}_3\text{MgI; (ii) H}_3\text{O}^+} \) B (70%) \( \xrightarrow{\text{Na metal}} \) C (90%) \( \xrightarrow{\mathrm{CH_3\text{–}Br}} \) D (75%)
Answer:
Ans. 33.26
CH₃C≡CH (Propyne) \( \xrightarrow{\mathrm{H_2O/H_2SO_4,\ Hg^{2+}}} \) CH₃COCH₃, Acetone (A) (80%) \( \xrightarrow{\text{(i) CH}_3\text{MgI; (ii) H}_3\text{O}^+} \) (CH₃)₃C–OH, tert-Butyl alcohol (B) (70%)
\( \xrightarrow{\text{Na metal}} \) (CH₃)₃C–ONa, Sod. tert-butoxide (C) (90%) \( \xrightarrow[-\mathrm{NaBr}]{\mathrm{CH_3\text{–}Br}} \) (CH₃)₃C–OCH₃, tert-Butyl methyl ether (D) (75%)
Amount of D (in moles) \( = 1 \times \frac{80}{100} \times \frac{70}{100} \times \frac{90}{100} \times \frac{75}{100} = 0.378 \)
Mol. mass of tert-Butyl methyl ether (C₅H₁₂O) = 5 × 12 + 1 × 12 + 1 × 16 = 88 amu
∴ Amount of D (in g) = 0.378 × 88 = 33.264 = 33.26 g

Teacher's Note:
1. Overall yield in a multi-step synthesis is the product of the fractional yields of all steps; multiply carefully before converting moles to grams.

Assertion–Reason Questions

Question 31. Assertion: Phenol forms 2,4,6-tribromophenol on treatment with Br₂ in carbon disulphide at 273 K.
Reason: Bromine polarises in carbon disulphide. (AIIMS 2013)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. Phenol forms a mixture of p-bromophenol (major) and o-bromophenol (minor) on treatment with Br₂ in CS₂.
Correct reason. CS₂ suppresses the ionization of phenol thereby decreasing the electron-donating ability of the oxygen atom of the OH group to the benzene ring.

Teacher's Note:
1. The solvent decides the extent of bromination: bromine water gives 2,4,6-tribromophenol, while Br₂ in CS₂ or CHCl₃ at low temperature gives mainly monobromophenols.

Question 32. Assertion: A mixture of 2-nitrophenol and 4-nitrophenol can be separated by steam distillation.
Reason: 2-Nitrophenol is intramolecularly hydrogen bonded while 4-nitrophenol is intermolecularly hydrogen bonded.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Due to intramolecular H-bonding, 2-nitrophenol exists as discrete molecules and is volatile in steam, whereas 4-nitrophenol is associated through intermolecular H-bonding and is not steam volatile. Hence they can be separated by steam distillation.

Teacher's Note:
1. Intramolecular H-bonding lowers the boiling point and water solubility; intermolecular H-bonding raises both.

Question 33. Assertion: The ease of dehydration of the following alcohols is
phenol < cyclohexanol < cyclohex-3-en-1-ol < cyclohex-2-en-1-ol
Reason: Alcohols leading to conjugated alkenes are dehydrated to a greater extent. (AIIMS 2008)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
Phenol does not undergo dehydration. Cyclohexanol gives cyclohexene, cyclohex-3-en-1-ol gives the conjugated cyclohexa-1,3-diene, and cyclohex-2-en-1-ol forms a resonance-stabilised allylic carbocation which also gives the conjugated diene. The greater the stability of the carbocation and of the conjugated product, the easier the dehydration.

Teacher's Note:
1. Dehydration rate follows carbocation stability; an allylic alcohol dehydrates fastest because its cation is resonance-stabilised.

Question 34. Assertion: t-Butyl methyl ether is not prepared by the reaction of t-butyl bromide with sodium methoxide.
Reason: Sodium methoxide is a strong nucleophile. (AIIMS 2005)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (b)
Correct explanation. In presence of a strong base, i.e., sodium methoxide, tert-butyl bromide undergoes dehydrohalogenation to form isobutylene.

Teacher's Note:
1. CH₃O⁻ is both a good nucleophile and a strong base; with a 3° halide its basic character wins and elimination occurs.

Question 35. Assertion: tert-Butyl methyl ether on treatment with HI at 373 K gives a mixture of methyl iodide and tert-butyl alcohol.
Reason: The reaction occurs by S_N2 mechanism. (AIIMS 2015)
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. tert-Butyl methyl ether on treatment with HI at 373 K gives tert-butyl iodide and methyl alcohol.
Correct reason. The reaction occurs by S_N1 mechanism.

Teacher's Note:
1. When one group on the ether oxygen is 3°, cleavage goes through the stable 3° carbocation, so the 3° group becomes the iodide.

Question 36. Assertion: (CH₃)₃C–OCH₃ reacts with HBr to give (CH₃)₃C–Br and CH₃OH.
Reason: It follows S_N1 mechanism.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (a)
(CH₃)₃C–O⁺(H)–CH₃ breaks to give the stable (CH₃)₃C⁺ carbocation and CH₃OH (S_N1); Br⁻ then combines with (CH₃)₃C⁺ to give (CH₃)₃C–Br.

Teacher's Note:
1. Compare with methyl ethyl ether, where both groups are 1° and Br⁻ attacks the smaller group by S_N2.

Question 37. Assertion: Last traces of moisture can be removed from phenol by distilling with sodium metal.
Reason: Water is more acidic than phenol.
(a) If both assertion and reason are true, and reason is the true explanation of the assertion.
(b) If both assertion and reason are true, but reason is not the true explanation of the assertion.
(c) If assertion is true, but reason is false.
(d) If both assertion and reason are false.
Answer:
Ans. (d)
Correct assertion. Last traces of moisture cannot be removed from phenol by distilling with sodium metal.
Correct reason. Phenol being more acidic than water reacts with Na more readily than water.

Teacher's Note:
1. Sodium reacts with any acidic O–H; since phenol is a stronger acid than water, sodium would destroy the phenol itself.