Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Amongst the following, the total number of compounds soluble in aqueous NaOH at room temperature is:
[Figure: Nine chemical structures labeled (I) through (IX): (I) benzoic acid, (II) 1-naphthol, (III) p-nitrophenol with N(CH3)2 group, (IV) hydroquinone, (V) salicylic acid, (VI) N,N-dimethylcyclohexylamine, (VII) 1-naphthoic acid, (VIII) tert-butylbenzene, (IX) 1-naphthol]
(1) 5
(2) 4
(3) 6
(4) 3
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Answer: Option (3)
Key idea: Acidity and solubility of organic compounds in aqueous NaOH - carboxylic acids and phenols form water-soluble sodium salts
Identify compounds that are acidic enough to react with NaOH: carboxylic acids (pKa ~ 4-5) and phenols (pKa ~ 10)
Consider compounds A, B and C with following structural formulae
$A = CH_3 - CH_2 - CH_2 - CH_2 - CH_2 - OH$
$B = CH_2 = CH - CH_2 - CH_2 - CH_3$
$C = HO - CH_2 - CH_2 - CH(OH) - CH_3$
For the conversion of B from A, reagent (D) required is _______ and structural formula of product (E) obtained when C undergoes same reaction using excess reagent (D) is _______.
Choose the correct answer from the options given below:
(1) D: Conc. H2SO4, E: $CH_2 = CH - CH(OH)CH_3$
(2) D: PCC, E: $HO - CH_2 - CH_2 - CH = CH_2$
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Answer: Option (1)
Key idea: Dehydration of alcohols using concentrated H₂SO₄
Identify that conversion of A (1-pentanol) to B (1-pentene) is a dehydration reaction requiring concentrated $H_2SO_4$ with heat
Reagent D = Concentrated $H_2SO_4$
Apply the same reagent (excess conc. $H_2SO_4$) to compound C: $HO-CH_2-CH_2-CH(OH)-CH_3$ (1,3-butanediol)
With excess acid, dehydration occurs at one of the OH groups. The primary OH at terminal position undergoes elimination
Product E formed is $CH_2=CH-CH(OH)CH_3$ (but-3-en-2-ol) with one OH group remaining
4.7 g of phenol is heated with Zn to give product X. If this reaction goes to 60% completion then the number of moles of compound X formed will be _______$\times 10^{-2}$. (Nearest Integer) (Given molar mass in g mol$^{-1}$: H:1, C:12, O:16)
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Answer: 3
Key idea: Reduction of phenol with zinc dust to form benzene; stoichiometry with percentage completion
Identify the reaction: Phenol $(C_6H_5OH)$ heated with $Zn$ gives benzene $(C_6H_6)$ as product $X$
Calculate molar mass of phenol: $M = 6(12) + 6(1) + 16 = 94$ g/mol
Consider the following reaction.
[Figure: Benzyl phenyl ether structure shown as C₆H₅-O-CH₂-C₆H₅] + HI → Product
**Statement I:** In the above reaction, product formed will be a mixture of benzyl alcohol and iodobenzene.
**Statement II:** In the above reaction, the -O-CH₂- bond is cleaved to give the product.
In the light of the above statements, choose the correct answer from the options given below:
(1) Both Statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
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Answer: Option (4)
Key idea: Ether cleavage with HI - regioselectivity based on carbocation stability
Benzyl phenyl ether reacts with HI: $C_6H_5-O-CH_2-C_6H_5 + HI$
The $O-CH_2$ bond cleaves (not $C_6H_5-O$) because benzyl cation $(^+CH_2C_6H_5)$ is stabilized by resonance, while phenyl cation is extremely unstable
Products formed: Phenol $(C_6H_5OH)$ and Benzyl iodide $(C_6H_5CH_2I)$
Statement I is FALSE (claims benzyl alcohol + iodobenzene, but actual products are phenol + benzyl iodide)
Statement II is TRUE (correctly identifies $O-CH_2$ bond cleavage)
Answer: Statement I is false but Statement II is true
FAQs
Frequently asked questions
How many questions come from Alcohols, Phenols and Ethers in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 5 questions on it, about 0.6 per shift.
Where can I practise Alcohols, Phenols and Ethers PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
