CBSE Class 12 Chemistry Alcohols Phenols and Ethers NEET JEE Questions Set 01

Single Correct Answer

Question 1. In the following reaction,
(CH₃)₃C–CH=CH₂ \( \xrightarrow{\mathrm{H_2O/H^+}} \) A (Major product) + B (Minor product)
The major product is (AIPMT Prelim 2012)
(a) (CH₃)₂C(OH)–CH(CH₃)–CH₃
(b) HOCH₂–C(CH₃)₂–CH₂–CH₃
(c) (CH₃)₃C–CH(OH)–CH₃
(d) (CH₃)₃C–CH₂–CH₂OH
Answer:
Ans. (a)
(CH₃)₃C–CH=CH₂ \( \xrightarrow{\mathrm{H^+}} \) (CH₃)₃C–C⁺H–CH₃ (I) [2° carbocation, less stable]
\( \xrightarrow{\text{1,2-methyl shift}} \) (CH₃)₂C⁺–CH(CH₃)–CH₃ [3° carbocation, more stable] \( \xrightarrow{+\mathrm{H_2O},\ -\mathrm{H^+}} \) (CH₃)₂C(OH)–CH(CH₃)–CH₃ = A (Major)
Carbocation (I) gives the minor product (B): I \( \xrightarrow{+\mathrm{H_2O},\ -\mathrm{H^+}} \) (CH₃)₃C–CH(OH)–CH₃ = B (Minor product).

Teacher's Note:
1. Acid-catalysed hydration goes through a carbocation, so always check whether a 1,2-hydride or 1,2-methyl shift can give a more stable (3°) cation.

Question 2. Consider the following reaction,
ethanol \( \xrightarrow{\mathrm{PCl_5}} \) X \( \xrightarrow{\text{alc. KOH}} \) Y \( \xrightarrow{\text{(i) H}_2\text{SO}_4\text{, room temperature; (ii) H}_2\text{O, heat}} \) Z
The product Z is : (Karnataka CET 2016)
(a) C₂H₄
(b) CH₃CH₂OCH₂CH₃
(c) CH₃CH₂OSO₃H
(d) CH₃CH₂OH
Answer:
Ans. (d)
CH₃CH₂OH (Ethanol) \( \xrightarrow{\mathrm{PCl_5}} \) CH₃CH₂Cl (X) \( \xrightarrow[-\mathrm{HCl}]{\text{alc. KOH}} \) CH₂=CH₂ (Y)
CH₂=CH₂ \( \xrightarrow{\text{(i) H}_2\text{SO}_4\text{, r.t.; (ii) H}_2\text{O, heat}} \) CH₃CH₂OH, Ethanol (Z)

Teacher's Note:
1. Cold conc. H₂SO₄ adds to ethene to give ethyl hydrogen sulphate, which on boiling with water is hydrolysed back to ethanol.

Question 3. Acid-catalysed hydration of alkenes except ethene leads to the formation of
(a) primary alcohol
(b) secondary or tertiary alcohol
(c) mixture of primary and secondary alcohols
(d) mixture of secondary and tertiary alcohols
Answer:
Ans. (b)
Hydration of ethene gives a 1° alcohol, i.e., ethanol while all other alkenes give either 2° or 3° alcohols.
CH₂=CH₂ \( \xrightarrow{\text{(i) conc. H}_2\text{SO}_4\text{; (ii) H}_2\text{O, boil}} \) CH₃–CH₂OH (1° alcohol)
R–CH=CH₂ \( \xrightarrow{\text{(i) conc. H}_2\text{SO}_4\text{; (ii) H}_2\text{O, boil}} \) R–CHOH–CH₃ (2° alcohol)
R₂C=CH₂ \( \xrightarrow{\text{(i) conc. H}_2\text{SO}_4\text{; (ii) H}_2\text{O, boil}} \) R₂C(OH)–CH₃ (3° alcohol)

Teacher's Note:
1. Markovnikov addition puts the OH on the more substituted carbon, so only ethene can give a primary alcohol by acid-catalysed hydration.

Question 4. What is the IUPAC name of the organic compound formed in the following chemical reaction ?
Acetone \( \xrightarrow{\text{(i) C}_2\text{H}_5\text{MgBr, dry ether; (ii) H}_2\text{O/H}^+} \) Product (NEET 2021)
(a) 2-Methylbutan-2-ol
(b) 2-Methylpropan-2-ol
(c) Pentan-2-ol
(d) Pentan-3-ol
Answer:
Ans. (a)
(CH₃)₂C=O \( \xrightarrow{\text{(i) C}_2\text{H}_5\text{MgBr, dry ether}} \) (CH₃)₂C(OMgBr)C₂H₅
\( \xrightarrow{\text{(ii) H}_2\text{O/H}^+ \text{ (Hydrolysis)}} \) (CH₃)₂C(OH)CH₂CH₃, 2-Methylbutan-2-ol

Teacher's Note:
1. A ketone with a Grignard reagent always gives a 3° alcohol; number the chain so the carbon carrying OH gets the lowest locant.

Question 5. CH₃CH₂–C(OH)(Ph)–CH₃ cannot be prepared by (JEE Main 2019)
(a) PhCOCH₃ + CH₃CH₂MgX
(b) HCHO + PhCH(CH₃)CH₂X
(c) PhCOCH₂CH₃ + CH₃MgX
(d) CH₃CH₂COCH₃ + PhMgX
Answer:
Ans. (b)
Cleave the molecule at the three bonds to the carbinol carbon (–C₂H₅, –CH₃ and –Ph); the required reagents are:
cleavage of C–C₂H₅ bond: CH₃CH₂MgX + PhCOCH₃ ....... option (a)
cleavage of C–Ph bond: PhMgX + CH₃CH₂COCH₃ ....... option (d)
cleavage of C–CH₃ bond: CH₃CH₂COPh + CH₃MgX ....... option (c)
Thus, there is no option corresponding to reagents (b) and hence option (b) is correct.
Alternatively. Since addition of any Grignard reagent to formaldehyde always gives a 1° alcohol but the alcohol to be synthesized is a 3° alcohol, therefore, option (b) is correct.

Teacher's Note:
1. For a 3° alcohol, each of its three groups can come from the Grignard reagent while the other two come from the ketone.

Question 6. (CH₃)₂CH–CHO + CH₃MgBr \( \xrightarrow{\text{Ether}} \) A \( \xrightarrow{\mathrm{H_3O^+}} \) B
The IUPAC name of 'B' is
(a) 3-methylbutan-2-ol
(b) 2-methylbutan-3-ol
(c) 2-methylbutan-2-ol
(d) pentan-2-ol
Answer:
Ans. (a)
(CH₃)₂CH–CHO + CH₃MgBr \( \xrightarrow{\text{Ether}} \) (CH₃)₂CH–CH(OMgBr)–CH₃ (A)
\( \xrightarrow{\mathrm{H_3O^+}} \) (CH₃)₂CH–CH(OH)–CH₃, 3-Methylbutan-2-ol (B)

Teacher's Note:
1. Option (b) describes the same compound but with wrong numbering; the OH must get the lower locant.

Question 7. Ethyl ester \( \xrightarrow{\mathrm{CH_3MgBr\ (excess)}} \) P. The product P is
(a) (CH₃)₂C(OH)CH₃
(b) (C₂H₅)₂C(OH)CH₃
(c) (C₂H₅)₃C–OH
(d) (C₂H₅)₂C(OH)C₃H₇
Answer:
Ans. (a)
Since esters add two moles of the Grignard reagent to form 3° alcohols, therefore, two of the alkyl groups in 3° alcohol must be identical. Since the Grignard reagent is CH₃MgBr, therefore, the product must contain at least two methyl groups.
The only product which contains at least two (actually three) CH₃ groups is option (a). Thus,
CH₃COOC₂H₅ (Ethyl ester) \( \xrightarrow{\mathrm{CH_3MgBr}} \) [(CH₃)₂C(OMgBr)(OC₂H₅)] \( \xrightarrow{-\mathrm{Mg(OC_2H_5)Br}} \) CH₃COCH₃
\( \xrightarrow{\mathrm{CH_3MgBr}} \) (CH₃)₃C–OMgBr \( \xrightarrow[-\mathrm{Mg(OH)Br}]{\mathrm{H^+/H_2O}} \) (CH₃)₃C–OH

Teacher's Note:
1. An ester takes up two molecules of RMgX: the first converts it into a ketone, the second into a 3° alcohol with two identical R groups from the reagent.

Question 8. The major product (B) in the following sequence of reactions is
CH₃–C(CH(CH₃)₂)=CH–CH₂CH₃ \( \xrightarrow{\text{(i) B}_2\text{H}_6\text{; (ii) H}_2\text{O}_2\text{, OH}^-} \) (A) \( \xrightarrow[\Delta]{\text{Dil. H}_2\text{SO}_4} \) (B) (JEE Main 2020)
(a) CH₃–C(=C(CH₃)₂)–CH₂CH₂CH₃
(b) CH₂=C(CH(CH₃)₂)–CH₂CH₂CH₃
(c) CH₃–C(CH(CH₃)₂)=CHCH₂CH₃
(d) CH₃–CH(CH(CH₃)₂)–CH=CH–CH₃
Answer:
Ans. (a)
Hydroboration-oxidation gives alcohols corresponding to anti-Markovnikov's addition of H₂O to alkenes.
CH₃–C(CH(CH₃)₂)=CH–CH₂CH₃ \( \xrightarrow{\text{(i) B}_2\text{H}_6\text{; (ii) H}_2\text{O}_2\text{, OH}^-} \) CH₃–CH(CH(CH₃)₂)–CH(OH)–CH₂CH₃ (A)
\( \xrightarrow[-\mathrm{H_2O}]{\mathrm{H^+}} \) CH₃–CH(CH(CH₃)₂)–C⁺H–CH₂CH₃ \( \xrightarrow{\text{1,2-hydride shift}} \) CH₃–C⁺(CH(CH₃)₂)–CH₂CH₂CH₃
\( \xrightarrow[\text{(Saytzeff elimination)}]{-\mathrm{H^+}} \) CH₃–C(=C(CH₃)₂)–CH₂CH₂CH₃ (B)
Thus, option (a) is correct.

Teacher's Note:
1. Acid dehydration goes through a carbocation: let it rearrange to the 3° cation first, then apply Saytzeff's rule to get the most substituted alkene.

Question 9. CH₃CH₂CH=CH₂ \( \xrightarrow{\mathrm{B_2H_6};\ \mathrm{H_2O,\ H_2O_2,\ OH^-}} \) Z
What is 'Z' (NEET 2020)
(a) CH₃CH₂–CH(OH)–CH₃
(b) CH₃CH₂CH₂CHO
(c) CH₃CH₂CH₂CH₃
(d) CH₃CH₂CH₂CH₂OH
Answer:
Ans. (d)
Hydroboration-oxidation of terminal alkenes always give 1° alcohols.
6CH₃CH₂CH=CH₂ (But-1-ene) \( \xrightarrow{\mathrm{B_2H_6}\ \text{(Hydroboration)}} \) 2(CH₃CH₂CH₂CH₂)₃B \( \xrightarrow{\mathrm{H_2O,\ H_2O_2,\ OH^-}\ \text{(Oxidation)}} \) 6CH₃CH₂CH₂CH₂OH

Teacher's Note:
1. Boron attaches to the less substituted carbon of the double bond, so the OH ends up there: the net result is anti-Markovnikov addition of water.

Question 10. The major product of the following reaction is
(CH₃)₂C=CH–CH₂–CH₂–C(CH₃)=CH–CH₂OH \( \xrightarrow{\text{Dil. H}_2\text{SO}_4} \) (JEE Main 2020)
(a) (CH₃)₂C(OH)–CH₂CH₂CH₂–C(OH)(CH₃)–CH₂CH₂OH
(b) 2-(4-methylcyclohex-3-en-1-yl)propan-2-ol, i.e., a cyclohexene ring carrying CH₃ on a ring C=C carbon and –C(CH₃)₂OH at the ring carbon para to it
(c) 2-(3-methyl-2-vinylcyclopentyl)propan-2-ol, i.e., a cyclopentane ring carrying –CH₃, –CH=CH₂ and –C(CH₃)₂OH on three adjacent carbons
(d) (CH₃)₂CH–CH(OH)–CH₂CH₂–C(OH)(CH₃)–CH₂CH₂OH
Answer:
Ans. (b)
(CH₃)₂C=CH–CH₂CH₂–C(CH₃)=CH–CH₂OH \( \xrightarrow{\mathrm{H^+}} \) (CH₃)₂C=CH–CH₂CH₂–C(CH₃)=CH–CH₂–O⁺H₂ \( \xrightarrow{-\mathrm{H_2O}} \) allylic carbocation
The allylic carbocation is attacked by the other (trisubstituted) C=C bond to close a six-membered ring; this gives a 3° carbocation, (CH₃)₂C⁺– attached to the ring (4-methylcyclohex-3-en-1-yl).
Attack of H₂O on this 3° carbocation followed by loss of H⁺ gives 2-(4-methylcyclohex-3-en-1-yl)propan-2-ol (α-terpineol), i.e., option (b).

Teacher's Note:
1. With dilute acid an allylic alcohol ionises easily; when a second C=C is placed suitably, the cation is captured intramolecularly to form a ring before water adds.

Question 11. Reagent(s) which can be used to bring about the following transformation is (are)
→ (JEE Advanced 2016)
(a) LiAlH₄ in (C₂H₅)₂O
(b) BH₃ in THF
(c) NaBH₄ in C₂H₅OH
(d) Raney Ni/H₂ in THF
Answer:
Ans. (c)
NaBH₄ in C₂H₅OH neither reduces the acid and esters nor opens the epoxide ring. It reduces only the –CHO group to the 1° alcoholic group.
LiAlH₄ would also reduce the –COOH and ester groups (and can open the epoxide); BH₃ reduces –COOH readily; Raney Ni/H₂ can cleave the strained epoxide ring.

Teacher's Note:
1. NaBH₄ is the mild, selective reagent: it reduces aldehydes and ketones but leaves acids, esters and epoxides untouched.

Question 12. The major product of the following reaction is
Cyclopent-2-en-1-one \( \xrightarrow[\text{EtOH}]{\mathrm{NaBH_4}} \) (JEE Main 2019)
(a) Cyclopentanone
(b) Cyclopentane-1,3-diol
(c) Cyclopent-2-en-1-ol
(d) Cyclopentanol
Answer:
Ans. (c)
NaBH₄ reduces only the keto group and not the C=C double bond. Thus, option (c) is correct.

Teacher's Note:
1. NaBH₄ delivers hydride to the polar C=O group; an isolated or ring C=C bond is not reduced by it.

Question 13. The product of the following reaction is
Cyclopentylmethanol (C₅H₉–CH₂OH) \( \xrightarrow{\mathrm{H_3PO_4}} \) Product
(a) Methylenecyclopentane
(b) 1-Methylcyclopentene
(c) Cyclohexene
(d) 3-Methylcyclopentene
Answer:
Ans. (c)
C₅H₉–CH₂OH \( \xrightarrow[-\mathrm{H_2O}]{+\mathrm{H^+}} \) C₅H₉–C⁺H₂ (1° carbocation, less stable) \( \xrightarrow{\text{1,2-alkyl shift}} \) cyclohexyl carbocation (2° carbocation, more stable; ring expansion)
\( \xrightarrow{-\mathrm{H^+}} \) Cyclohexene

Teacher's Note:
1. A 1° carbocation next to a five-membered ring rearranges by ring expansion: it gains stability and relieves ring strain at the same time.

Question 14. Which is the final product (major) 'A' in the given reaction ?
1-(2-Methylcyclohexyl)ethan-1-ol (2-CH₃–C₆H₁₀–CH(OH)CH₃) \( \xrightarrow{\text{HCl}} \) 'A' (major product) (JEE Main 2021)
(a) 1-Chloro-1-ethyl-2-methylcyclohexane
(b) 1-Methyl-2-vinylcyclohexane (2-CH₃–C₆H₁₀–CH=CH₂)
(c) 1-(1-Chloroethyl)-2-methylcyclohexane (2-CH₃–C₆H₁₀–CHCl–CH₃)
(d) 1-Ethylidene-2-methylcyclohexane
Answer:
Ans. (a)
2-CH₃–C₆H₁₀–CH(OH)–CH₃ \( \xrightarrow[-\mathrm{H_2O}]{\mathrm{H^+}} \) 2-CH₃–C₆H₁₀–C⁺H–CH₃ (2° carbocation, less stable)
\( \xrightarrow{\text{1,2-hydride shift}} \) 3° carbocation on the ring carbon (1-ethyl-2-methylcyclohexyl cation) \( \xrightarrow{\mathrm{Cl^-}} \) 1-chloro-1-ethyl-2-methylcyclohexane
Thus, option (a) is correct.

Teacher's Note:
1. Alcohols react with HCl through carbocations, so the chloride appears where the most stable (3°) cation is formed after a hydride shift.

Question 15. Which of the following is not the product of dehydration of 2-cyclohexylbutan-2-ol, C₆H₁₁–C(OH)(CH₃)–CH₂CH₃ ? (RE-AIPMT 2015)
(a) 1-(Butan-2-yl)cyclohexene (C=C inside the ring)
(b) 2-Cyclohexylbut-2-ene, C₆H₁₁–C(CH₃)=CH–CH₃
(c) (Butan-2-ylidene)cyclohexane, C₆H₁₀=C(CH₃)–CH₂CH₃
(d) 2-Cyclohexylbut-1-ene, C₆H₁₁–C(=CH₂)–CH₂CH₃
Answer:
Ans. (a)
Dehydration is theoretically possible on all the three sides of carbon carrying the OH group giving alkenes (b), (c) and (d).
Only product (a) is not formed.

Teacher's Note:
1. In dehydration the new C=C bond always involves the carbon that carried the OH (unless a rearrangement occurs); list the H-bearing neighbours to find all the alkenes.

Question 16. The major product of the following reaction is
C₆H₅–CO–CH₂Br \( \xrightarrow[\text{MeOH, 25°C}]{\mathrm{NaBH_4}} \) (JEE Main 2019)
(a) C₆H₅–CH(OH)–CH₂OMe
(b) C₆H₅–C(OH)=CH–OMe
(c) 2-Phenyloxirane (styrene oxide)
(d) C₆H₅–CH(OH)–CH₂Br
Answer:
Ans. (c)
C₆H₅COCH₂Br \( \xrightarrow[\text{MeOH, 25°C}]{\mathrm{NaBH_4}} \) C₆H₅–CH(O⁻)–CH₂Br; the alkoxide oxygen attacks the neighbouring C–Br carbon (intramolecular S_N2)
\( \xrightarrow{-\mathrm{HBr}} \) 2-Phenyloxirane (an epoxide)

Teacher's Note:
1. An alkoxide formed next to a C–Br bond closes to an epoxide faster than any outside nucleophile can react; this is an intramolecular Williamson synthesis.

Question 17. Arrange the following compounds in increasing order of C–OH bond length : methanol, phenol, p-ethoxyphenol. (JEE Main 2020)
(a) phenol < methanol < p-ethoxyphenol
(b) methanol < p-ethoxyphenol < phenol
(c) phenol < p-ethoxyphenol < methanol
(d) methanol < phenol < p-ethoxyphenol
Answer:
Ans. (c)
In phenol, the lone pair on O is delocalised into the ring (C₆H₅–OH ↔ C₆H₅⁻=O⁺H structures). Due to resonance, C–OH bond has some double bond character and hence it has the shortest C–OH bond length.
In p-ethoxyphenol, the OC₂H₅ group also donates electrons to the ring by resonance; due to this, resonance by OH group is diminished to some extent, therefore C–OH bond length is little longer than in phenol.
In CH₃–OH, there is no resonance and hence C–OH bond length is the longest.
Thus, C–OH bond length increases in the order : phenol < p-ethoxyphenol < methanol, i.e., option (c) is correct.

Teacher's Note:
1. More double-bond character means a shorter bond; an electron-donating group para to OH competes with OH for resonance and lengthens the C–O bond slightly.

Question 18. The stable product of the following reaction is
\( \xrightarrow{\text{(i) H}_3\text{C–C}_6\text{H}_4\text{–SO}_2\text{Cl, Pyridine; (ii) NaCN, DMF}} \) (JEE Main 2022)
(a)
(b)
(c)
(d)
Answer:
Ans. (d)
Step (i): tosylation takes place at the O–H bond, so the configuration at the carbon carrying OH is retained: R–OH \( \xrightarrow[-\mathrm{HCl}]{\mathrm{TsCl,\ pyridine}} \) R–OTs.
Step (ii): CN⁻ (NaCN in DMF) displaces the tosylate group by nucleophilic substitution S_N2, with inversion of configuration at that carbon only; the other stereocentre is not affected.
So the product is the same skeleton with CN on a hashed bond in place of the wedged OH and the ethyl group still on a hashed bond, i.e., structure (d) [(2S,3R)-2,3-dimethylpentanenitrile from (2R,3R)-3-methylpentan-2-ol].

Teacher's Note:
1. TsCl does not touch the C–O bond, so it keeps the configuration; the inversion comes only from the S_N2 step.

Question 19. Reaction of Grignard reagent, C₂H₅MgBr with C₈H₈O followed by hydrolysis gives compound "A" which reacts instantly with Lucas reagent to give compound B, C₁₀H₁₃Cl. The compound B is (JEE Main 2021)
(a) C₆H₅–CH(CH₃)–CHCl–CH₃
(b) 2-(3-Chloropropyl)toluene, 2-CH₃–C₆H₄–CH₂CH₂CH₂Cl
(c) C₆H₅–CH(CH₃)–CH₂–CH₂Cl
(d) C₆H₅–C(Cl)(CH₃)–CH₂CH₃
Answer:
Ans. (d)
C₈H₈O \( \xrightarrow{\text{(i) C}_2\text{H}_5\text{MgBr; (ii) H}^+\text{/H}_2\text{O}} \) (A) \( \xrightarrow{\text{Lucas reagent}} \) C₁₀H₁₃Cl (B)
It appears that C₈H₈O is acetophenone. Thus,
C₆H₅COCH₃ (Acetophenone) \( \xrightarrow{\text{(i) C}_2\text{H}_5\text{MgBr; (ii) H}_3\text{O}^+} \) C₆H₅–C(OH)(CH₃)–CH₂CH₃ (A)
\( \xrightarrow{\text{Conc. HCl + ZnCl}_2 \text{ (Lucas reagent)}} \) C₆H₅–C(Cl)(CH₃)–CH₂CH₃, C₁₀H₁₃Cl (B)
Thus, option (d) is correct.

Teacher's Note:
1. An instant reaction with Lucas reagent signals a 3° (or benzylic) alcohol, so the product chloride has Cl on the carbon that carried OH.

Question 20. An organic compound A reacts with CH₃MgI to form an addition product which on hydrolysis forms the compound B. Compound B gives blue coloured salt in Victor-Meyer's test. The compounds A and B respectively are
(a) Acetaldehyde, tertiary butyl alcohol
(b) Acetaldehyde, ethyl alcohol
(c) Acetaldehyde, isopropyl alcohol
(d) Acetone, isopropyl alcohol
Answer:
Ans. (c)
CH₃CHO (Acetaldehyde) \( \xrightarrow{\text{(i) CH}_3\text{MgI; (ii) H}^+\text{/H}_2\text{O}} \) CH₃–CHOH–CH₃, Isopropyl alcohol (B) (2° Alcohol)
\( \xrightarrow{\text{Victor Meyer's test}} \) Blue colouration

Teacher's Note:
1. Victor Meyer's test: red for 1°, blue for 2° and colourless for 3° alcohols; a 2° alcohol from RMgX needs an aldehyde other than HCHO.

Question 21. An unknown alcohol is treated with the "Lucas reagent" to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism? (JEE Main 2013)
(a) tertiary alcohol by S_N2
(b) secondary alcohol by S_N1
(c) tertiary alcohol by S_N1
(d) secondary alcohol by S_N2
Answer:
Ans. (c)
3° Alcohols react fastest with Lucas reagent by S_N1 mechanism (i.e., carbocation intermediate) to give turbidity.

Teacher's Note:
1. Lucas test rates follow carbocation stability (3° > 2° > 1°), which is the signature of an S_N1 reaction.

Question 22. Iodoform can be prepared from all except (AIEEE 2012)
(a) isopropyl alcohol
(b) 3-methyl-2-butanone
(c) isobutyl alcohol
(d) ethyl methyl ketone
Answer:
Ans. (c)
Isobutyl alcohol does not contain the group CH₃CHOH– or CH₃CO– attached to C or H and hence does not give iodoform on treatment with I₂/NaOH.
Isopropyl alcohol: CH₃CHOH–CH₃; 3-Methyl-2-butanone: CH₃–CH(CH₃)–CO–CH₃; Ethyl methyl ketone: CH₃–CO–CH₂CH₃ all contain the required group, while isobutyl alcohol is (CH₃)₂CH–CH₂OH.

Teacher's Note:
1. Look for CH₃–CO– or CH₃–CH(OH)– in the molecule; a CH₂OH group at the end never gives the iodoform test.

One or More Correct Answers

Question 23. The correct combination of names for isomeric alcohols with molecular formula C₄H₁₀O is/are (JEE Advanced 2014)
(a) tert-butanol and 2-methylpropan-2-ol
(b) tert-butanol and 1,1-dimethylethan-1-ol
(c) n-butanol and butan-1-ol
(d) isobutyl alcohol and 2-methylpropan-1-ol
Answer:
Ans. (a, c, d)
Option (b) is wrong since 1,1-dimethylethan-1-ol is neither a common nor an IUPAC name. Thus, options (a, c, d) are correct.

Teacher's Note:
1. In IUPAC names the longest chain containing the carbon with OH is chosen, so a 'dimethylethanol' is always renamed as a propanol.

Matrix Match Questions

Teacher's Note:
1. Oxymercuration–demercuration gives Markovnikov alcohols without rearrangement; hydroboration–oxidation gives anti-Markovnikov alcohols by syn addition (H and OH on the same face, so CH₃ and OH end up trans).

Integer and Numerical Answer Questions

Question 25. The total number of structural isomeric alcohols having the molecular formula C₅H₁₂O is.
Answer:
Ans. 8
Eight (8) isomers are :
(i) CH₃CH₂CH₂CH₂CH₂OH (1°) ; (ii) CH₃CH₂CH₂–CHOH–CH₃ (2°) ; (iii) CH₃CH₂–CHOH–CH₂CH₃ (2°) ;
(iv) CH₃CH₂–CH(CH₃)–CH₂OH (1°) ; (v) CH₃–CH(CH₃)–CH₂CH₂OH (1°) ; (vi) CH₃–C(CH₃)(OH)–CH₂CH₃ (3°) ;
(vii) CH₃–C(CH₃)₂–CH₂OH (1°) ; (viii) CH₃–CH(CH₃)–CH(OH)–CH₃ (2°).

Teacher's Note:
1. Work chain by chain: pentane gives 3, 2-methylbutane gives 4 and 2,2-dimethylpropane gives 1 alcohol (stereoisomers are not counted as structural isomers).

Question 26. Among the following, the number of alcohols showing iodoform test is
CH₃OH, CH₃CH₂OH, (CH₃)₂CHOH, (CH₃)₃COH, CH₃CHOHCH₂CH₃, CH₃CH₂CH₂CH₂OH, CH₃CH₂CH₂OH, CH₃CHOHCH(CH₃)₂, (C₂H₅)₂CHOH.
Answer:
Ans. 4
Four (4) alcohols which contain the grouping CH₃CHOH– and hence give iodoform test are : CH₃CH₂OH, CH₃CHOHCH₃, CH₃CHOHCH₂CH₃, and CH₃CHOHCH(CH₃)₂.

Teacher's Note:
1. Ethanol is the only primary alcohol that gives the iodoform test, because its CH₂OH is attached to a CH₃ group (CH₃CH(OH)–H).

Question 27. The number of pentyl alcohols producing blue colouration in the Victor-Meyer's test is.
Answer:
Ans. 3
2° Alcohols give blue colouration in Victor-Meyer's test due to the formation of pseudonitrole. Refer to Ans. 81. The three 2° alcohols are : (ii), (iii) and (viii).

Teacher's Note:
1. Victor Meyer's test: 1° gives red (nitrolic acid), 2° gives blue (pseudonitrole), 3° gives no colour.

Assertion–Reason Questions

Question 28. Statement-1: m-Methoxyphenol is a stronger acid than p-methoxyphenol.
Statement-2: Methoxy group exerts +R-effect at both o- and p-positions.
(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. (b)
Correct explanation. Methoxy group exerts –I-effect at m-position and +R-effect at p-position thereby making m-methoxyphenol a stronger acid than p-methoxyphenol.

Teacher's Note:
1. At the meta position only the –I effect of OCH₃ operates, which increases acidity; Statement-2 alone does not explain why the meta isomer is stronger.

Question 29. Statement-1: 2-Pentanol and 3-pentanol cannot be distinguished by iodoform test.
Statement-2: 2-Pentanol and 3-pentanol both are secondary alcohols.
(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. (d)
Correct statement-1. 2-Pentanol can be distinguished from 3-pentanol by iodoform test.
2-Pentanol contains the CH₃CH(OH)– group and gives a yellow precipitate of CHI₃; 3-pentanol does not.

Teacher's Note:
1. The iodoform test depends on the CH₃CH(OH)– group, not on whether the alcohol is 1°, 2° or 3°.

Question 30. Statement-1: Equimolar mixture of conc. HCl and anhydrous zinc chloride is called Lucas reagent.
Statement-2: Lucas reagent can be used to distinguish between methanol and ethanol.
(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. Both methanol and ethanol are 1° alcohols and hence cannot be distinguished by Lucas reagent.

Teacher's Note:
1. Lucas reagent separates alcohols only by class (1°, 2°, 3°); two primary alcohols behave alike.

Question 31. Statement-1: Cleavage of anisole with HI at 373 K gives phenol and CH₃I.
Statement-2: Due to resonance O–C₆H₅ bond is stronger than O–CH₃ bond.
(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. (a)
Because of resonance, the O–C₆H₅ bond acquires partial double bond character and is stronger than the O–CH₃ bond. Hence the O–CH₃ bond breaks and I⁻ attacks the CH₃ group to give phenol and CH₃I.

Teacher's Note:
1. In aryl alkyl ethers it is always the O–alkyl bond that breaks, so the products are a phenol and an alkyl iodide.

Question 32. Statement-1: Anisole undergoes electrophilic substitution at o- and p-positions.
Statement-2: Anisole is less reactive than phenol towards electrophilic substitution reactions.
(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. (b)
Correct explanation. OCH₃ group is o, p-directing.
Statement-2 is also true (the CH₃ group makes the O lone pair less available than in phenol), but it does not explain the o, p-orientation.

Teacher's Note:
1. Orientation (where the electrophile goes) and reactivity (how fast it goes) are separate questions; a reason about one does not explain the other.

Question 33. Statement-1: The acidic strength of monosubstituted nitrophenol is higher than phenol because of electron-withdrawing nitro group.
Statement-2: o-Nitrophenol, m-nitrophenol and p-nitrophenol will have the same strength as they have one nitro group attached to the phenolic ring. (NEET 2022)
(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. Although o-, m- and p-nitrophenols have one –NO₂ group their effect is different at o, m- and p-positions. Their acidity character decreases in the order : p-nitrophenol > o-nitrophenol > m-nitrophenol.
Thus, Statement-1 is true but Statement-2 is false.

Teacher's Note:
1. The position of a substituent matters: –NO₂ stabilises phenoxide by resonance only from o- and p-positions, so the m-isomer is the weakest of the three.