Oxygen-Containing Compounds Previous Year Questions (JEE Main 2026)

A curated set of JEE Main 2026 previous-year questions on oxygen-containing compounds (alcohols, phenols, ethers, and carbonyl compounds), each solved step by step so you can check both the final answer and the reasoning.

Solutions are AI-generated and pending review.

JEE Main 2026 · 6 Apr, Shift 1 Q6952782208

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)

Solution

Heating phenol with zinc dust reduces the C–OH bond, giving benzene as product X:

$$C_6H_5OH + Zn \longrightarrow C_6H_6 + ZnO$$

Moles of phenol taken: molar mass of phenol $= 6(12)+6(1)+16 = 94\ \text{g mol}^{-1}$

$$n_{\text{phenol}} = \dfrac{4.7}{94} = 0.05\ \text{mol}$$

The stoichiometry is $1:1$, so at 100% completion $0.05$ mol of benzene would form. At 60% completion:

$$n_X = 0.05 \times 0.60 = 0.03\ \text{mol} = 3 \times 10^{-2}\ \text{mol}$$

Answer: 3

JEE Main 2026 · 6 Apr, Shift 1

JEE Main 2026 · 4 Apr, Shift 2 Q695278441

The descending order of acidity among the following compounds is: A. phenol; B. 4-nitrophenol; C. 4-methoxyphenol; D. 4-nitrobenzoic acid; E. benzoic acid. Choose the correct answer from the options given below:

Solution

Two effects decide the order:

Carboxylic acids are far stronger than phenols (a carboxylate spreads negative charge over two equivalent oxygens), so D and E sit at the top.
Substituent effect: a $-NO_2$ group (electron-withdrawing) increases acidity, while $-OCH_3$ (electron-donating by resonance) decreases it.

Approximate $pK_a$ values (lower $pK_a$ = stronger acid):

Compound	Species	$pK_a$
D	4-nitrobenzoic acid	$\approx 3.4$
E	benzoic acid	$\approx 4.2$
B	4-nitrophenol	$\approx 7.1$
A	phenol	$\approx 10.0$
C	4-methoxyphenol	$\approx 10.2$

Descending acidity: $\mathbf{D > E > B > A > C}$.

Answer: D

A B > D > E > A > C
B D > B > E > A > C
C C > A > B > D > E
D D > E > B > A > C

JEE Main 2026 · 4 Apr, Shift 2

JEE Main 2026 · 2 Apr, Shift 2 Q691121216

An organic compound '$x$' where molar ratio of C, O and H are equal, on treatment with 50% KOH under reflux followed by acidification produced '$y$'. The most likely structure of '$y$' is : [Molar mass of '$x$' is 58 g mol$^{-1}$]

Solution

Since $C:O:H = 1:1:1$, the formula is $C_nH_nO_n$ with mass $29n$:

$$29n = 58 \;\Rightarrow\; n = 2 \;\Rightarrow\; C_2H_2O_2$$

This is glyoxal, $OHC\text{–}CHO$ — a dialdehyde with no $\alpha$-hydrogen.

With concentrated (50%) KOH, an aldehyde lacking $\alpha$-H undergoes the Cannizzaro reaction. Glyoxal undergoes an intramolecular Cannizzaro: one $-CHO$ is oxidised to $-COO^-$ while the other is reduced to $-CH_2OH$. After acidification:

$$OHC\text{–}CHO \xrightarrow[\text{then } H^+]{50\% \text{ KOH}} HOCH_2\text{–}COOH$$

Product $y$ is glycolic acid ($HOCH_2COOH$), matching option C.

Answer: C

A $CH_2=CH-\overset{O}{\overset{\
B }{C}}-OH$
C $CH_3-CH=CH-CH=O$
D $O=C(OH)-CH_2-OH$
E $CH_3-\overset{O}{\overset{\
F }{C}}-OH$

JEE Main 2026 · 2 Apr, Shift 2

JEE Main 2026 · 6 Apr, Shift 2 Q6911211264

'x' is the product which is obtained by the hydrolysis of prop-1-yne in the presence of mercuric sulphate under dilute acidic medium at 333 K. 'y' is the product which is obtained by the reaction of ethane nitrile with methyl magnesium bromide in dry ether followed by hydrolysis. IUPAC name of product obtained from 'x' and 'y' in the presence of barium hydroxide followed by heating is:

Solution

Find x: Markovnikov hydration of prop-1-yne ($CH_3\text{–}C\equiv CH$) with $HgSO_4/\text{dil. }H_2SO_4$ adds water at the internal carbon:

$$CH_3\text{–}C\equiv CH \xrightarrow{H_2O,\,HgSO_4} CH_3\text{–}\underset{O}{\overset{}{C}}\text{–}CH_3 \;=\; \textbf{acetone}$$

Find y: A nitrile with a Grignard reagent gives a ketone after hydrolysis:

$$CH_3C\equiv N + CH_3MgBr \rightarrow CH_3\text{–}C(=NMgBr)\text{–}CH_3 \xrightarrow{H_3O^+} CH_3\text{–}CO\text{–}CH_3 \;=\; \textbf{acetone}$$

So both x and y are acetone. With $Ba(OH)_2$ (base), acetone undergoes aldol condensation with itself; heating dehydrates the aldol:

$$2\,CH_3COCH_3 \xrightarrow{Ba(OH)_2} (CH_3)_2C(OH)\text{–}CH_2\text{–}COCH_3 \xrightarrow{\Delta,\,-H_2O} (CH_3)_2C{=}CH\text{–}COCH_3$$

The product is mesityl oxide. Numbering to give the ketone the lowest locant:

$$\underset{1}{CH_3}\text{–}\underset{2}{CO}\text{–}\underset{3}{CH}{=}\underset{4}{C}(CH_3)\text{–}\underset{5}{CH_3}$$

giving 4-methylpent-3-en-2-one.

Answer: 4 - Methylpent-3-en-2-one

A 2 - Methylpent-4-en-3-one
B 4 - Methylpent-3-en-2-one
C 4 - Methylpent-1-ene
D 2 - Methylpent-3-one

JEE Main 2026 · 6 Apr, Shift 2

JEE Main 2026 · 6 Apr, Shift 2 Q6911211265

An optically active alkyl bromide $C_4H_9Br$, reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form compound [D]. The functional group in compound D will be confirmed by:

Solution

Identify the starting bromide: an optically active $C_4H_9Br$ must have a stereocentre, so it is 2-bromobutane, $CH_3\text{–}CHBr\text{–}CH_2\text{–}CH_3$.

Trace the sequence:

[A] — ethanolic KOH → dehydrohalogenation (Saytzeff) → but-2-ene, $CH_3CH{=}CHCH_3$.
[B] — $Br_2$ addition → 2,3-dibromobutane.
[C] — ethanolic KOH then $NaNH_2$ → double dehydrohalogenation → but-2-yne, $CH_3\text{–}C\equiv C\text{–}CH_3$.
[D] — $H_2O/HgSO_4/\text{dil. }H_2SO_4$ (333 K) hydrates the internal alkyne → butan-2-one, $CH_3\text{–}CO\text{–}CH_2\text{–}CH_3$.

Butan-2-one is a methyl ketone ($CH_3CO\text{–}$ group), which gives a positive iodoform (haloform) test. Tollens’ (silver mirror) and Benedict’s are aldehyde tests; Lucas is an alcohol test — none apply here.

Answer: Haloform test

A Haloform test
B Lucas test
C Silver mirror test
D Benedict test

JEE Main 2026 · 6 Apr, Shift 2

JEE Main 2026 · 5 Apr, Shift 1 Q695278367

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 ________.

Solution

A → B: pentan-1-ol loses a molecule of water to give a pentene. Removal of $H_2O$ from an alcohol is acid-catalysed dehydration, so reagent D = conc. $H_2SO_4$ (or $H_3PO_4$) — not an oxidant like PCC. This immediately eliminates the two PCC options.

C → E (same reaction, excess D): $C$ is butane-1,3-diol, $HO\text{–}CH_2\text{–}CH_2\text{–}CH(OH)\text{–}CH_3$. With excess dehydrating acid, both $-OH$ groups are eliminated:

$$HO\text{–}CH_2\text{–}CH_2\text{–}CH(OH)\text{–}CH_3 \xrightarrow[-2\,H_2O]{\text{excess } H_2SO_4} CH_2{=}CH\text{–}CH{=}CH_2$$

giving the conjugated diene buta-1,3-diene. This matches option D.

Answer: D: Conc. $H_2SO_4$ or $H_3PO_4$; E: $CH_2=CH-CH=CH_2$

A D: Conc. H$_2$SO$_4$; E: $CH_2 = CH - CH(OH)CH_3$
B D: PCC; E: $HO - CH_2 - CH_2 - CH = CH_2$
C D: PCC; E: $CH_2 = CH - CH = CH_2$
D D: Conc. H$_2$SO$_4$ or H$_3$PO$_4$; E: $CH_2 = CH - CH = CH_2$

JEE Main 2026 · 5 Apr, Shift 1

JEE Main 2026 · 5 Apr, Shift 1 Q695278373

One mole of phenol is treated with dilute $HNO_3$ at 298 K to give a mixture of products. The mixture is separated by steam distillation. The steam volatile compound (X) is separated. The increase in percentage of oxygen in (X) with respect to phenol is ________ $\times 10^{-1}$ % (Given molar mass in g mol$^{-1}$ H:1, C:12, N:14, O:16)

Solution

Dilute $HNO_3$ nitrates phenol to a mixture of o-nitrophenol and p-nitrophenol. Only ortho-nitrophenol is steam volatile (intramolecular H-bonding lowers its boiling point; the para isomer forms intermolecular H-bonds and is non-volatile). So X = o-nitrophenol, $C_6H_4(OH)(NO_2)$.

% oxygen in phenol ($C_6H_6O$, $M = 94$):

$$\dfrac{16}{94}\times 100 = 17.02\%$$

% oxygen in o-nitrophenol ($C_6H_5NO_3$, $M = 72+5+14+48 = 139$; O mass $= 3\times16 = 48$):

$$\dfrac{48}{139}\times 100 = 34.53\%$$

Increase:

$$34.53 - 17.02 = 17.51\%$$

Expressed as $\times 10^{-1}\%$: $\;17.51 / 0.1 = 175.1 \approx \mathbf{175}$.

Answer: 175

JEE Main 2026 · 5 Apr, Shift 1

JEE Main 2026 · 5 Apr, Shift 2 Q691121516

Given below are two statements : **Statement I:** The condensation reaction between $CH_3-CH=O$ and $H_2N-NH-C(=O)-NH_2$ under optimum pH will produce $CH_3-CH=N-NH-C(=O)-NH_2$. **Statement II:** The molecule $Ph-CH(O-H)(O-CH_3)$ will generate $Ph-CH=O$ in the presence of dilute acid. In the light of the above statements, choose the correct answer from the options given below :

Solution

Statement I: Semicarbazide ($H_2N\text{–}NH\text{–}CO\text{–}NH_2$) is a nitrogen nucleophile that condenses with a carbonyl group, losing water to form a semicarbazone:

$$CH_3CHO + H_2N\text{–}NH\text{–}CO\text{–}NH_2 \xrightarrow{\text{opt. pH}} CH_3CH{=}N\text{–}NH\text{–}CO\text{–}NH_2 + H_2O$$

This is exactly the product written. True.

Statement II: $Ph\text{–}CH(OH)(OCH_3)$ has both an $-OH$ and an $-OR$ on the same carbon — it is a hemiacetal. Hemiacetals are unstable and, in dilute acid, revert to the parent aldehyde and alcohol:

$$Ph\text{–}CH(OH)(OCH_3) \xrightarrow{H^+} Ph\text{–}CHO + CH_3OH$$

So it does generate benzaldehyde. True.

Both statements are true.

Answer: Both Statement I and Statement II are true

A Both Statement I and Statement II are true
B Both Statement I and Statement II are false
C Statement I is true but Statement II is false
D Statement I is false but Statement II is true

JEE Main 2026 · 5 Apr, Shift 2

JEE Main 2026 · 8 Apr, Shift 2 Q691121597

'$x$' is the product which is obtained from benzene by reacting it with carbon monoxide and hydrogen chloride in the presence of cuprous chloride. '$y$' is the major product obtained from the benzene by reacting it with ethanoyl chloride in the presence of anhydrous $AlCl_3$. Product (major) obtained by heating $x$ and $y$ in the presence of alkali is $z$. Total number of $\pi$ (pi) electrons in $z$ is __________.

Solution

Identify x: benzene $+ CO + HCl$ with $CuCl$ is the Gattermann–Koch reaction → benzaldehyde, $C_6H_5CHO$ (no $\alpha$-H).

Identify y: benzene $+ CH_3COCl / \text{anhyd. }AlCl_3$ is Friedel–Crafts acylation → acetophenone, $C_6H_5COCH_3$ (has $\alpha$-H).

Form z: heating with alkali drives a cross aldol condensation — the $\alpha$-carbon of acetophenone attacks benzaldehyde, then dehydration gives chalcone:

$$C_6H_5CHO + C_6H_5COCH_3 \xrightarrow{\text{OH}^-,\,\Delta} C_6H_5\text{–}CH{=}CH\text{–}CO\text{–}C_6H_5$$

Count $\pi$ electrons in chalcone:

Source	$\pi$ bonds	$\pi$ electrons
Benzene ring 1	3	6
Benzene ring 2	3	6
$C=C$ (bridge)	1	2
$C=O$ (carbonyl)	1	2
Total	8	16

Answer: 16

JEE Main 2026 · 8 Apr, Shift 2

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