CBSE Class 12 Chemistry Alcohols Phenols and Ethers NEET JEE Questions Set 02

Single Correct Answer

Question 1. 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 (NEET 2018)
(a) H₃C–C₆H₄–CH₂OH (p-) and I₂
(b) C₆H₅–CH₂CH₂OH and I₂
(c) C₆H₅–CH(OH)–CH₃ and I₂
(d) 2,4-dimethylphenol, (CH₃)₂C₆H₃–OH and I₂
Answer:
Ans. (c)
Since compound A, C₈H₁₀O on treatment with NaOI (produced by the action of I₂ on NaOH) gives a yellow ppt., (Iodoform test), therefore, compound A must contain –CHOH–CH₃ group, i.e., compound (A) must be a 2° alcohol as shown below :
C₆H₅–CH(OH)–CH₃, A (C₈H₁₀O) \( \xrightarrow{\text{NaOI (I}_2\text{/NaOH)}} \) C₆H₅–COONa + CHI₃ (Iodoform, yellow ppt.)
Thus option (c) is correct.

Teacher's Note:
1. NaOI is made from I₂ and NaOH; the yellow CHI₃ precipitate points to a CH₃CH(OH)– or CH₃CO– group.

Question 2. An organic compound neither reacts with neutral ferric chloride solution nor with Fehling solution. It, however, reacts with Grignard reagent and gives positive iodoform test. The compound is (JEE Main 2019)
(a) o-HO–C₆H₄–CO–CH₃ (2-hydroxyacetophenone)
(b) o-CH₃CH(OH)–C₆H₄–CO–C₂H₅
(c) o-CH₃O–C₆H₄–CHO (2-methoxybenzaldehyde)
(d) o-CH₃O–C₆H₄–CO–C₂H₅
Answer:
Ans. (b)
Since the given organic compound does not react with neutral FeCl₃, it cannot be a phenol. Therefore, option (a) is rejected. Since it does not react with Fehling solution, it cannot be an aldehyde, therefore, option (c) is also rejected. Further since it gives iodoform test, it must be a methyl ketone or a methyl carbinol. Since option (b) is a methyl carbinol and it also reacts with Grignard reagent, therefore, option (b) is correct.

Teacher's Note:
1. Use each test to strike out options: FeCl₃ detects phenols, Fehling's detects aliphatic aldehydes, and the iodoform test needs CH₃CO– or CH₃CH(OH)–.

Question 3. Identify the major products P, Q and R in the following sequence of reactions :
C₆H₆ + CH₃CH₂CH₂Cl \( \xrightarrow{\text{anhyd. AlCl}_3} \) P \( \xrightarrow{\text{(i) O}_2\text{; (ii) H}_3\text{O}^+/\Delta} \) Q + R (NEET 2018)
(a) P = C₆H₅–CH₂CH₂CH₃, Q = C₆H₅–CHO, R = CH₃CH₂OH
(b) P = C₆H₅–CH₂CH₂CH₃, Q = C₆H₅–CHO, R = C₆H₅–COOH
(c) P = C₆H₅–CH(CH₃)₂, Q = C₆H₅–OH, R = CH₃CH(OH)CH₃
(d) P = C₆H₅–CH(CH₃)₂, Q = C₆H₅–OH, R = CH₃COCH₃
Answer:
Ans. (d)
C₆H₆ (Benzene) + CH₃CH₂CH₂Cl \( \xrightarrow{\text{Anhyd. AlCl}_3 \text{ (F.C. alkylation)}} \) C₆H₅–CH(CH₃)₂, Isopropylbenzene (P) [the 1° carbocation rearranges to the 2° isopropyl carbocation]
\( \xrightarrow{\text{(i) O}_2\text{; (ii) H}_3\text{O}^+/\Delta} \) C₆H₅OH, Phenol (Q) + CH₃COCH₃, Acetone (R)
Thus, option (d) is correct.

Teacher's Note:
1. n-Propyl chloride gives isopropylbenzene (cumene) in Friedel-Crafts alkylation because the carbocation rearranges; cumene is then the starting material for phenol and acetone.

Question 4. The structure of intermediate A in the following reaction is
C₆H₅–CH(CH₃)₂ \( \xrightarrow{\mathrm{O_2}} \) A \( \xrightarrow[\mathrm{H_2O}]{\mathrm{H^+}} \) C₆H₅OH + CH₃COCH₃ (NEET 2019)
(a) C₆H₅–CH(CH₃)–CH₂–O–O–H
(b) C₆H₅–O–CH(CH₃)₂
(c) C₆H₅–C(CH₃)₂–O–O–H
(d) C₆H₅–O–O–CH(CH₃)₂
Answer:
Ans. (c)
Since tertiary H atom is easily abstracted by O₂ biradical, therefore, option (c) is correct.
C₆H₅–CH(CH₃)₂ \( \xrightarrow{\mathrm{O_2}} \) C₆H₅–C(CH₃)₂–O–O–H (cumene hydroperoxide)

Teacher's Note:
1. Air oxidation attacks the weakest C–H bond, the 3° benzylic one, so the –OOH group is placed on that carbon.

Question 5. The major product U in the following reaction is
C₆H₆ \( \xrightarrow[\text{High Pressure, heat}]{\mathrm{CH_2=CH\text{–}CH_3,\ H^+}} \) T \( \xrightarrow{\text{Radical Initiator, O}_2} \) U (JEE Advanced 2015)
(a) o-CH₃CH₂–C₆H₄–CH₂–O–O–H
(b) C₆H₅–C(CH₃)₂–O–O–H
(c) o-CH₂=CH–C₆H₄–CH₂–O–O–H
(d) C₆H₅–C(=CH₂)–CH₂–O–O–H
Answer:
Ans. (b)
C₆H₆ \( \xrightarrow[\text{high pressure (F.C. Alkylation)}]{\mathrm{CH_2=CH\text{–}CH_3,\ H^+}} \) C₆H₅–CH(CH₃)₂ (T) \( \xrightarrow{\text{Radical initiator, O}_2} \) C₆H₅–C(CH₃)₂–O–O–H (U)

Teacher's Note:
1. Propene and acid give the isopropyl cation, so T is cumene; its 3° benzylic C–H is the one converted into C–OOH.

Question 6. The compound which is most difficult to protonate is (NEET 2019)
(a) Ph–O–H
(b) H–O–H
(c) H₃C–O–H
(d) H₃C–O–CH₃
Answer:
Ans. (a)
Since the lone pair of electrons on the O atom in phenol is donated towards the benzene ring by resonance, therefore, it is not available for protonation. Thus, option (a) is correct.

Teacher's Note:
1. Protonation needs a free lone pair on oxygen; delocalisation into the ring makes phenol's oxygen the weakest base here.

Question 7. Arrange the following compounds in order of decreasing acidity.
p-ClC₆H₄OH (I), p-CH₃C₆H₄OH (II), p-NO₂C₆H₄OH (III), p-CH₃OC₆H₄OH (IV) (JEE Main 2013)
(a) IV > III > I > II
(b) II > IV > I > III
(c) I > II > III > IV
(d) III > I > II > IV
Answer:
Ans. (d)
Electron-withdrawing groups increase the acidity of phenols. Since –NO₂ group is a much more powerful electron-withdrawing group than Cl, therefore, p-nitrophenol (III) is a stronger acid than p-chlorophenol (I). Further, electron-donating groups decrease the acidity of phenols. Since –OCH₃ group is a more powerful electron-donating group than CH₃ group, therefore, p-methylphenol (II) is a stronger acid than p-methoxyphenol (IV). Combining the two trends together, the overall acidity of the four phenols decreases in the order : III > I > II > IV, i.e., option (d) is correct.

Teacher's Note:
1. Rank withdrawing groups first (strongest acid) and donating groups last; –NO₂ at the para position also stabilises the phenoxide ion by resonance.

Question 8. Which one is the most acidic compound ? (NEET 2017)
(a) Phenol
(b) p-Nitrophenol
(c) 2,4,6-Trinitrophenol
(d) p-Cresol (4-methylphenol)
Answer:
Ans. (c)
Due to the presence of three electron-withdrawing –NO₂ groups, picric acid is the strongest acid.

Teacher's Note:
1. Picric acid (2,4,6-trinitrophenol) has pK_a about 0.4, comparable to strong mineral acids, because each –NO₂ at o/p positions delocalises the negative charge.

Question 9. Strongest acid among the following is (AIIMS 2013)
(a) o-methoxyphenol
(b) p-methoxyphenol
(c) m-methoxyphenol
(d) phenol
Answer:
Ans. (c)
At m-position, OCH₃ group cannot exert its +R-effect. Instead, it exerts its –I-effect, and hence m-methoxyphenol is the strongest acid among phenol, o-, m- and p-methoxyphenols.

Teacher's Note:
1. –OCH₃ is electron-donating by resonance only at o- and p-positions; at the meta position only its electron-withdrawing inductive effect acts.

Question 10. Which of the following substituted phenol is the strongest acid ? (NEET 2020)
(a) p-Methoxyphenol
(b) p-Ethylphenol
(c) p-Methylphenol
(d) p-Nitrophenol
Answer:
Ans. (d)
Electron-withdrawing group such as –NO₂ increases the acidity of phenols.

Teacher's Note:
1. –OCH₃, –C₂H₅ and –CH₃ all donate electrons and make phenol weaker; only –NO₂ withdraws electrons and stabilises the phenoxide ion.

Question 11. Ortho-nitrophenol is less soluble in water than p- and m-nitrophenols because (AIEEE 2012)
(a) o-nitrophenol shows intramolecular H-bonding
(b) o-nitrophenol shows intermolecular H-bonding
(c) melting point of o-nitrophenol is lower than those of m- and p-isomers
(d) o-nitrophenol is more volatile in steam than those of m- and p-isomers
Answer:
Ans. (a)
Due to chelation, o-nitrophenol is less soluble in water than m- and p-isomers.

Teacher's Note:
1. When the OH is already H-bonded to the neighbouring NO₂ group (chelation), it cannot H-bond with water; the same fact explains its steam volatility.

Question 12. The structure of the compound that gives a tribromo derivative on treatment with bromine-water is
(a) 3-Methylphenol (m-cresol)
(b) C₆H₅–CH₂OH (benzyl alcohol)
(c) 2-Methylphenol (o-cresol)
(d) 4-Methylphenol (p-cresol)
Answer:
Ans. (a)
m-Cresol, C₇H₈O \( \xrightarrow[-3\mathrm{HBr}]{3\mathrm{Br_2}} \) 2,4,6-Tribromo-3-methylphenol, C₇H₅OBr₃
m-Cresol has two ortho- and one para- position free w.r.t. OH group and hence gives tribromo-derivative, C₇H₅OBr₃.

Teacher's Note:
1. Bromine water substitutes every free o- and p-position of a phenol; count how many of those three positions are unoccupied.

Question 13. The reaction of CH₃–CH=CH–C₆H₄–OH (p-) with HBr gives
(a) CH₃CHBrCH₂–C₆H₄–OH
(b) CH₃CH₂CHBr–C₆H₄–OH
(c) CH₃CHBrCH₂–C₆H₄–Br
(d) CH₃CH₂CHBr–C₆H₄–Br
Answer:
Ans. (b)
The addition of a proton at β-carbon gives a carbocation (I) which is stabilized by resonance because of the electron donating effect of the OH group. Subsequent addition of Br⁻ ion to α-carbon gives the addition product (II).
CH₃–CH=CH–C₆H₄–OH \( \xrightarrow{\mathrm{H^+}} \) [CH₃–CH₂–C⁺H–C₆H₄–OH (I) ↔ CH₃–CH₂–CH=C₆H₄=O⁺H] \( \xrightarrow{\mathrm{Br^-}} \) CH₃–CH₂–CHBr–C₆H₄–OH (II)

Teacher's Note:
1. The phenolic OH is not replaced by HBr; HBr only adds to the C=C so that the cation forms on the benzylic carbon, which the para-OH stabilises.

Question 14. The major product(s) of the following reaction is (are)
p-HO–C₆H₄–SO₃H \( \xrightarrow{\text{aqueous Br}_2 \text{ (3.0 equivalents)}} \) ?
P: 2,3,5-tribromo-4-hydroxybenzenesulphonic acid; Q: 2,4,6-tribromophenol; R: 3,4,5-tribromophenol; S: 2,3,6-tribromo-4-hydroxybenzenesulphonic acid (JEE Advanced 2013)
(a) P
(b) Q
(c) R
(d) S
Answer:
Ans. (b)
The OH group strongly activates the o- and p-positions. With aqueous Br₂, both ortho positions are brominated, and the –SO₃H group at the para position is replaced by Br (ipso substitution, desulphonation).
p-HO–C₆H₄–SO₃H + 3Br₂ → 2,4,6-tribromophenol + H₂SO₄ + 2HBr. Hence Q.

Teacher's Note:
1. An –SO₃H group para to OH is easily displaced by Br⁺; bromine never enters the meta positions of a phenol.

Question 15. Sodium phenoxide when heated with CO₂ under pressure at 125°C yields a product which on acetylation produces C,
C₆H₅ONa + CO₂ \( \xrightarrow[\text{5 atm}]{125^\circ\text{C}} \) B \( \xrightarrow[\mathrm{Ac_2O}]{\mathrm{H^+}} \) C
The product C would be (JEE Main 2014)
(a) m-CH₃COO–C₆H₄–COOH
(b) o-CH₃COO–C₆H₄–COOH
(c) 2,4-Diacetylphenol
(d) o-HO–C₆H₄–COCH₃
Answer:
Ans. (b)
Kolbe's reaction: C₆H₅ONa + CO₂ \( \xrightarrow[\text{5 atm}]{125^\circ\text{C}} \) o-HO–C₆H₄–COONa (sodium salicylate, B)
\( \xrightarrow{\mathrm{H^+}} \) salicylic acid \( \xrightarrow{\mathrm{Ac_2O}} \) o-CH₃COO–C₆H₄–COOH, acetylsalicylic acid (aspirin, C)

Teacher's Note:
1. Kolbe's reaction puts the –COOH ortho to OH; acetylation then occurs at the phenolic OH, not on the ring.

Question 16. Phenol on treatment with CO₂ in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with (CH₃CO)₂O in the presence of catalytic amount of H₂SO₄ produces : (JEE Main 2018)
(a) o-CH₃COO–C₆H₄–CO₂H
(b) p-CH₃COO–C₆H₄–CO₂H
(c) o-HO–C₆H₄–CO–O–CO–CH₃ (mixed anhydride)
(d) 4-CH₃COO–C₆H₃(CO₂H)₂-1,3 (4-acetoxybenzene-1,3-dicarboxylic acid)
Answer:
Ans. (a)
Phenol on treatment with CO₂ in presence of NaOH followed by acidification produces salicylic acid (X) (Kolbe's reaction).
Now (X) contains a –COOH group and OH group. Due to electron-withdrawing nature of the C=O part of the –COOH group, OH group of COOH is less nucleophilic than OH group of phenols. Therefore, on treatment of (X) with (CH₃CO)₂O, acetylation occurs at the OH group giving aspirin.
Phenol \( \xrightarrow{\text{CO}_2\text{/NaOH (Kolbe's reaction)}} \) [o-HO–C₆H₄–CO₂Na] \( \xrightarrow{\mathrm{H^+/H_2O}} \) o-HO–C₆H₄–COOH, Salicylic acid (X) \( \xrightarrow[\text{H}_2\text{SO}_4 \text{ (trace)}]{\text{(CH}_3\text{CO)}_2\text{O}} \) o-CH₃COO–C₆H₄–CO₂H, Aspirin
Thus, option (a) is correct.

Teacher's Note:
1. Salicylic acid has two OH groups but only the phenolic one is acetylated; the product is aspirin.

Question 17. Which of the following reagents may be used to distinguish between phenol and benzoic acid ? (AIIMS 2014)
(a) Neutral FeCl₃
(b) Aqueous NaOH
(c) Tollen's reagent
(d) Molisch reagent
Answer:
Ans. (a)
Neutral FeCl₃ gives a violet colour with phenol but not with benzoic acid.

Teacher's Note:
1. Both phenol and benzoic acid dissolve in aqueous NaOH, so NaOH cannot tell them apart; the violet/blue FeCl₃ colour is specific to the phenolic OH.

Question 18. In the reaction,
C₆H₅OH + CHCl₃ + NaOH → o-Na⁺O⁻–C₆H₄–CHO
the electrophile involved is (NEET 2018)
(a) dichloromethyl cation (C⁺HCl₂)
(b) formyl cation (C⁺HO)
(c) dichloromethyl anion (⁻CHCl₂)
(d) dichlorocarbene (:CCl₂)
Answer:
Ans. (d)
Dichlorocarbene. CHCl₃ + OH⁻ → ⁻CCl₃ + H₂O; ⁻CCl₃ → :CCl₂ + Cl⁻. The electron-deficient :CCl₂ attacks the ring of phenoxide ion at the ortho position (Reimer-Tiemann reaction).

Teacher's Note:
1. Dichlorocarbene is neutral but has only six valence electrons on carbon, which makes it the electrophile in the Reimer-Tiemann reaction.

Question 19. Identify the major products A and B for the below given reaction sequence
C₆H₆ \( \xrightarrow{\text{(i) CH}_3\text{–CHCl–CH}_3\text{, AlCl}_3\text{; (ii) O}_2\text{; (iii) H}^+\text{/H}_2\text{O}} \) P \( \xrightarrow{\text{Na}_2\text{Cr}_2\text{O}_7\text{/H}_2\text{SO}_4} \) A; P \( \xrightarrow{\text{Br}_2 \text{ in CS}_2} \) B (JEE Main 2022)
(a) A = phenol, B = p-bromophenol
(b) A = p-benzoquinone, B = p-bromophenol
(c) A = p-benzoquinone, B = 2,4,6-tribromophenol
(d) A = cyclohexa-2,4-dien-1-one, B = 2,4,6-tribromophenol
Answer:
Ans. (b)
C₆H₆ \( \xrightarrow{\text{(i) CH}_3\text{–CHCl–CH}_3\text{, AlCl}_3\text{; (ii) O}_2\text{; (iii) H}^+\text{/H}_2\text{O}} \) C₆H₅OH (P) + CH₃–CO–CH₃
P \( \xrightarrow{\text{Na}_2\text{Cr}_2\text{O}_7\text{/H}_2\text{SO}_4} \) p-Benzoquinone (A); P \( \xrightarrow{\text{Br}_2\text{/CS}_2} \) p-Bromophenol (B)
Thus, option (b) is correct.

Teacher's Note:
1. Phenol is oxidised by chromic acid to p-benzoquinone; in a non-polar solvent like CS₂ bromination stops at the mono stage, mainly para.

Question 20. Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br₂ to form product B. A and B are respectively : (JEE Main 2018)
(a) A = o-HO–C₆H₄–COOCH₃ (methyl salicylate) and B = methyl 5-bromo-2-hydroxybenzoate
(b) A = C₆H₅–O–CO–OCH₃ and B = m-Br–C₆H₄–O–CO–OCH₃
(c) A = C₆H₅–O–CO–OCH₃ and B = p-Br–C₆H₄–O–CO–OCH₃
(d) A = o-HO–C₆H₄–COOCH₃ and B = methyl 3-bromo-2-hydroxybenzoate
Answer:
Ans. (c)
Phenol reacts with methyl chloroformate in presence of NaOH to form compound (A).
C₆H₅OH + Cl–CO–OCH₃ (Methyl chloroformate) \( \xrightarrow[-\mathrm{NaCl}]{\mathrm{NaOH}} \) C₆H₅–O–CO–OCH₃ (A) \( \xrightarrow{\mathrm{Br_2}} \) p-Br–C₆H₄–O–CO–OCH₃ (B)
Due to resonance of lone pair of electrons on the O atom, OCOOCH₃ group is o, p-directing. Therefore, (A) on bromination gives product (B) in which Br enters at the unhindered p-position.
Thus, option (c) is correct.

Teacher's Note:
1. The phenoxide oxygen attacks the acyl carbon (O-acylation); an –O–CO–R group is still o,p-directing, though less activating than OH.

Question 21. The major product of the following reaction is
3-Methoxyphenol (m-MeO–C₆H₄–OH) \( \xrightarrow{\text{(i) ClCH}_2\text{CH}_2\text{COCl; (ii) AlCl}_3 \text{ (anhyd.)}} \) (JEE Main (online) 2018)
(a) 7-Methoxychroman-2-one (7-methoxy-3,4-dihydrocoumarin)
(b) 5-Methoxychroman-4-one
(c) 7-Methoxychroman-4-one
(d) 5-Methoxychroman-2-one (5-methoxy-3,4-dihydrocoumarin)
Answer:
Ans. (a)
Since acid chlorides are more reactive than alkyl halides, therefore, reaction occurs at the acid chloride giving product (X)
m-MeO–C₆H₄–OH \( \xrightarrow[-\mathrm{HCl}]{\mathrm{ClCH_2CH_2COCl}} \) m-MeO–C₆H₄–O–CO–CH₂CH₂Cl (X)
\( \xrightarrow[\text{F.C. alkylation (–HCl)}]{\text{Anhyd. AlCl}_3} \) the CH₂ group bonds to the ring carbon ortho to the ester oxygen and para to MeO, closing a six-membered lactone ring: 7-methoxychroman-2-one
Thus, option (a) is correct.

Teacher's Note:
1. First the more reactive acid chloride end acylates the phenol; then AlCl₃ makes the alkyl chloride end alkylate the ring intramolecularly, para to the strongly activating MeO.

Question 22. Identify A in the given reaction.
5-Hydroxy-4-(hydroxymethyl)-5,6,7,8-tetrahydronaphthalen-1-ol [a phenolic OH on the aromatic ring, a –CH₂OH group on the aromatic ring and a 2° alcoholic OH on the saturated ring] \( \xrightarrow{\mathrm{SOCl_2}} \) A (Major Product) (JEE Main 2021)
(a) Only the –CH₂OH changed to –CH₂Cl (phenolic OH and ring 2° OH unchanged)
(b) –CH₂OH changed to –CH₂Cl and the ring 2° OH changed to Cl (phenolic OH unchanged)
(c) Only the phenolic OH changed to Cl
(d) All three OH groups changed to Cl
Answer:
Ans. (b)
SOCl₂ converts alcoholic OH and not phenol OH group to Cl.
Thus, option (b) is correct.

Teacher's Note:
1. The C–O bond of a phenol has partial double-bond character and cannot be broken by SOCl₂, PCl₅ or HX; only alcoholic OH groups are replaced.

One or More Correct Answers

Question 23. Among the following, reaction(s) which gives (give) tert-butylbenzene as the major product is(are) (JEE Advanced 2016)
(a) C₆H₆ + (CH₃)₃C–Br, NaOC₂H₅
(b) C₆H₆ + (CH₃)₂CH–CH₂Cl, AlCl₃
(c) C₆H₆ + (CH₃)₂C=CH₂, H₂SO₄
(d) C₆H₆ + (CH₃)₂CH–CH₂OH, BF₃·OEt₂
Answer:
Ans. (b, c, d)
Only tert-butyl bromide in presence of NaOC₂H₅ undergoes dehydrohalogenation to get isobutylene: (a) (CH₃)₃C–Br \( \xrightarrow[-\mathrm{HBr}]{\mathrm{NaOC_2H_5}} \) (CH₃)₂C=CH₂
while all the remaining three reactions undergo F.C. alkylation to give tert-butylbenzene.
(b) (CH₃)₂CH–CH₂Cl \( \xrightarrow[-\mathrm{AlCl_4^-}]{\mathrm{AlCl_3}} \) (CH₃)₂CH–C⁺H₂ \( \xrightarrow{\text{1,2-hydride shift}} \) (CH₃)₃C⁺ (tert-butyl carbocation)
(c) (CH₃)₂C=CH₂ \( \xrightarrow{\mathrm{H^+}} \) (CH₃)₃C⁺ (tert-butyl carbocation)
(d) (CH₃)₂CH–CH₂OH \( \xrightarrow{\mathrm{BF_3\text{–}OEt_2}} \) (CH₃)₂CH–CH₂–O⁺(H)–B⁻F₃ \( \xrightarrow{-\mathrm{HO\text{–}BF_3}} \) (CH₃)₂CH–C⁺H₂ \( \xrightarrow{\text{1,2-hydride shift}} \) (CH₃)₃C⁺ (tert-butyl carbocation)
C₆H₆ + (CH₃)₃C⁺ \( \xrightarrow{-\mathrm{H^+}} \) C₆H₅–C(CH₃)₃ (tert-Butylbenzene)

Teacher's Note:
1. Any route that generates the tert-butyl cation in the presence of a Lewis or protic acid alkylates benzene; NaOC₂H₅ is a base and gives only elimination.

Passage-Based Questions

Question 24. 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)
Which one of the following reagents is used in the above reaction ?
(a) aq. NaOH + CH₃Cl
(b) aq. NaOH + CH₂Cl₂
(c) aq. NaOH + CHCl₃
(d) aq. NaOH + CCl₄
Answer:
Ans. (c)
For introduction of –CHO group, during Reimer-Tiemann reaction, a mixture of aq. NaOH + CHCl₃ is used, i.e., option (c) is correct.

Teacher's Note:
1. CHCl₃ gives –CHO (Reimer-Tiemann), while CCl₄ with NaOH gives –COOH (salicylic acid).

Matrix Match Questions

Teacher's Note:
1. Reimer-Tiemann with CHCl₃ gives an aldehyde, but with CCl₄ it gives a carboxylic acid; Kolbe's reaction uses CO₂ under pressure.

Teacher's Note:
1. Choose a test that only one member of the pair answers; tests that both pass (or both fail) cannot distinguish them.

Integer and Numerical Answer Questions

Question 27. How many of the structurally isomeric pentyl alcohols will produce immediate turbidity in Lucas test ?
Answer:
Ans. 1
Only 3° alcohols immediately give turbidity in the Lucas Test. Refer to Ans. 81, the only 3° alcohol is (vi), 2-methylbutan-2-ol.

Teacher's Note:
1. Immediate turbidity means 3°; 2° alcohols take about 5 minutes and 1° alcohols do not react at room temperature.

Question 28. How many of the following substances are more acidic than phenol ?
o-cresol, m-cresol, p-cresol, water, methyl alcohol, ethyl alcohol, 2,4-dimethylphenol, p-ethylphenol, dimethylcarbinol.
Answer:
Ans. 0
All alcohols and phenols having electron-donating groups are less acidic than phenol, i.e., all. In other words Ans. is zero.

Teacher's Note:
1. Water (pK_a about 15.7) and alcohols (pK_a about 16–18) are much weaker acids than phenol (pK_a about 10); alkyl groups on the ring also lower the acidity.

Question 29. The number of resonance structures of N is
2-Naphthol \( \xrightarrow{\mathrm{NaOH}} \) N (JEE Advanced 2015)
Answer:
Ans. 9
2-Naphthol \( \xrightarrow{\mathrm{NaOH}} \) sodium 2-naphthoxide (N)
N has nine resonating structures: three with the negative charge on O (one for each Kekulé arrangement of the two rings) and six with a C=O group and the negative charge on a ring carbon, i.e., on C-1 (two structures), C-3, C-4a, C-6 and C-8 (one each).
Total = 3 + 6 = 9.

Teacher's Note:
1. Count resonance structures by moving the charge only to alternate (ortho/para-type) carbons and then counting the different ways of placing the remaining double bonds.

Assertion–Reason Questions

Question 30. Statement-1: In Lucas test, primary, secondary and tertiary alcohols are distinguished on the basis of their reaction with conc. HCl + ZnCl₂, known as Lucas reagent.
Statement-2: Primary alcohols are most reactive and immediately produce turbidity at room temperature on reaction with Lucas reagent.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct statement-2. Tertiary alcohols are most reactive since they immediately produce turbidity at room temperature on reaction with Lucas reagent.

Teacher's Note:
1. The Lucas reaction goes through carbocations, so the order of reactivity is 3° > 2° > 1°.

Question 31. Statement-1: Ethyl phenyl ether can be prepared by Williamson ether synthesis.
Statement-2: Bromobenzene on reacting with sodium ethoxide gives ethyl phenyl ether. (JEE Main 2021)
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(b) Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.
(c) Statement-1 is True, Statement-2 is False.
(d) Statement-1 is False, Statement-2 is True.
Answer:
Ans. (c)
Correct statement-2. Sodium phenoxide on reacting with ethyl bromide gives ethyl phenyl ether.
Bromobenzene does not undergo nucleophilic substitution with sodium ethoxide because the C–Br bond has partial double bond character.

Teacher's Note:
1. For an alkyl aryl ether always take the phenoxide and the alkyl halide, never the aryl halide and the alkoxide.

Question 32. Assertion: Reaction of alcohols with SOCl₂ is catalysed by the presence of a tertiary amine (R₃N).
Reason: Tertiary amine promotes the reaction by reacting with the by product HCl. (AIIMS 2014)
(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)
R–OH + SOCl₂ \( \xrightarrow{\mathrm{R_3N}} \) R–Cl + SO₂ + HCl; the base R₃N (or pyridine) removes HCl as R₃NH⁺Cl⁻ and so drives the reaction forward.

Teacher's Note:
1. Pyridine or a 3° amine is added to SOCl₂ reactions to trap HCl; the other by-product, SO₂, escapes as a gas.

Question 33. Assertion: Hydroxy ketones are not directly used in Grignard reaction.
Reason: Grignard reagents react with hydroxyl group.
(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)
Grignard reagents are strong bases; they react with the acidic H of the –OH group (R–MgX + HO– → R–H + XMgO–) and are destroyed before they can add to the C=O group. Hence the OH group must be protected first.

Teacher's Note:
1. Any compound with an acidic H (OH, NH, COOH, terminal C≡CH, water) destroys a Grignard reagent, so such groups must be protected first.

Question 34. Assertion: p-F–C₆H₄–OH is less acidic than p-Cl–C₆H₄–OH.
Reason: –F exerts better + mesomeric effect than –Cl. (AIIMS 2011)
(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 effective 2p–2p overlap, F donates electrons to the ring (+R effect) much better than Cl (3p–2p overlap). This destabilises the phenoxide ion more in p-fluorophenol, so it is less acidic than p-chlorophenol.

Teacher's Note:
1. At the para position the +R effect of halogen competes with its –I effect; for F the +R effect is strong enough to make it the weakest acid among p-halophenols.

Question 35. Assertion: Phenoxide ion on treatment with active alkyl halide (e.g., CH₂=CH–CH₂Cl) gives two products, viz., O-substituted and C-substituted.
Reason: Phenoxide ion is an ambident nucleophile. (AIIMS 2011)
(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)
The negative charge of phenoxide ion is delocalised over O and the o- and p-carbon atoms. It can therefore attack an alkyl halide through O (giving allyl phenyl ether) or through ring carbon (giving o-allylphenol). Such a nucleophile is called ambident.

Teacher's Note:
1. An ambident nucleophile has two different atoms that can donate electrons; CN⁻, NO₂⁻ and phenoxide are common examples.