Organic Chemistry Principles Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on General Organic Chemistry with step-by-step solutions covering electronic effects, IUPAC nomenclature priority, reaction mechanisms, isomerism and bond-length order.
A curated set of JEE Main 2026 previous-year questions on the basic principles of organic chemistry, each solved step by step so you can check both the final answer and the reasoning.
Solutions are AI-generated and pending review.
Solution
Electron-withdrawing power here is governed by the $-I$ (inductive) effect. The standard experimental order of $-I$ strength for these groups is:
$$-\text{NO}_2 \;>\; -\text{CN} \;>\; -\text{COOH} \;>\; -\text{I}$$The nitro group is the strongest electron-withdrawing group, followed by cyano, then carboxyl, and finally the iodo group is the weakest of the four.
Writing this in increasing order of electron-withdrawing power:
$$-\text{I} \;<\; -\text{COOH} \;<\; -\text{CN} \;<\; -\text{NO}_2$$Using the labels: $\;d < b < a < c$.
Answer: C
Solution
Statement I — check molecular formulas. Cyclohexanone is $\text{C}_6\text{H}_{10}\text{O}$. Each ring substituent replaces one H.
- 2,6-diethylcyclohexanone: add two $-\text{C}_2\text{H}_5$ groups $\Rightarrow \text{C}_6\text{H}_8\text{O}(\text{C}_2\text{H}_5)_2 = \text{C}_{10}\text{H}_{18}\text{O}$
- 6-methyl-2-n-propylcyclohexanone: add $-\text{CH}_3$ and $-\text{C}_3\text{H}_7 \Rightarrow \text{C}_6\text{H}_8\text{O}(\text{CH}_3)(\text{C}_3\text{H}_7) = \text{C}_{10}\text{H}_{18}\text{O}$
Same molecular formula, same functional group (ketone), but a different distribution of alkyl groups around the carbonyl — this is exactly metamerism. Statement I is TRUE.
Statement II — check for $\alpha$-hydrogens. Keto–enol tautomerism requires at least one $\alpha$-hydrogen. In 2,2,6,6-tetramethylcyclohexanone, both $\alpha$-carbons (C-2 and C-6) are fully substituted by methyl groups, so there are no $\alpha$-hydrogens. Hence no enol can form. Statement II is FALSE.
Answer: C
Solution
The IUPAC seniority order for principal characteristic groups (decreasing) is:
$$\text{acid} > \text{ester} > \text{amide} > \text{nitrile} > \text{aldehyde} > \text{ketone} > \text{alcohol} > \text{amine} > \text{alkyne}$$Testing each option against this order:
- Option A: amide $>$ ketone $>$ aldehyde — wrong, since aldehyde outranks ketone.
- Option D: amide $>$ aldehyde $>$ nitrile — wrong, since nitrile outranks aldehyde.
- Option C: $-\text{CONH}_2$ (amide) $>$ $-\text{CHO}$ (aldehyde) $>$ $>\!\text{C}=\text{O}$ (ketone) $>$ $-\text{NH}_2$ (amine) $>$ $-\text{C}\equiv\text{C}-$ (alkyne) — matches the correct seniority order exactly.
Answer: C
Solution
Match each mixture with the reagent that distinguishes its two components:
- A. Diethylamine (2°) + Ethylamine (1°): the carbylamine (isocyanide) test with $\text{CHCl}_3 + \text{KOH},\ \Delta$ is given only by the primary amine (foul-smelling isocyanide). Distinguishes them $\Rightarrow$ A–II.
- B. Acetaldehyde + Acetone: ammoniacal $\text{AgNO}_3$ (Tollens’ reagent) oxidises the aldehyde to give a silver mirror, but not the ketone $\Rightarrow$ B–IV.
- C. Ethanol + Phenol: neutral $\text{FeCl}_3$ gives a characteristic violet colour with phenol, not with ethanol $\Rightarrow$ C–III.
- D. Benzoic acid + Cinnamic acid: cinnamic acid has a C=C double bond and decolourises bromine water by addition; benzoic acid does not $\Rightarrow$ D–I.
Therefore: A–II, B–IV, C–III, D–I.
Answer: D
Solution
Classify each reaction by its mechanism:
- A. Williamson ether synthesis: an alkoxide ($\text{R}-\text{O}^-$) attacks an alkyl halide by an $S_N2$ pathway $\Rightarrow$ nucleophilic substitution (III).
- B. Friedel–Crafts reaction: an electrophile ($\text{R}^+$ or acylium) attacks the aromatic ring $\Rightarrow$ electrophilic substitution (IV).
- C. Bromination of vinylbenzene (styrene): $\text{Br}_2$ adds across the C=C double bond of the vinyl group $\Rightarrow$ electrophilic addition (I).
- D. Chlorination of toluene in light: $h\nu$ generates chlorine radicals that substitute at the benzylic $-\text{CH}_3$ (side chain) $\Rightarrow$ free radical substitution (II).
Therefore: A–III, B–IV, C–I, D–II.
Answer: C
Solution
Statement I. The methoxy group $-\text{OCH}_3$ has a lone pair on oxygen and shows a $+R$ (electron-donating resonance) effect. Through the para position it pushes electron density toward the benzylic carbon, delocalising and stabilising the positive charge of the carbocation. TRUE.
Statement II. The nitro group $-\text{NO}_2$ is a strong $-R$ (electron-withdrawing resonance) group. Through the para position it pulls electron density away from the benzylic carbon, delocalising and stabilising the negative charge of the carbanion. TRUE.
Both statements correctly match the substituent’s electronic effect to the charge it stabilises.
Answer: A
Solution
Identify the three carbon–oxygen bonds:
- $\alpha$: the carbonyl $\text{C}=\text{O}$ double bond.
- $\beta$: the single bond from the carbonyl carbon to the ester oxygen ($\text{C}-\text{O}$).
- $\gamma$: the single bond from the ester oxygen to the methyl group ($\text{O}-\text{CH}_3$).
The carbonyl $\text{C}=\text{O}$ ($\alpha$) is a genuine double bond, so it is the shortest.
The ester oxygen’s lone pair is delocalised into the carbonyl (resonance $\text{C}-\text{O}^+ = \text{C}$), giving the $\beta$ bond partial double-bond character, so it is shorter than a plain single bond.
The $\gamma$ bond ($\text{O}-\text{CH}_3$) is a normal single bond with the least double-bond character, so it is the longest.
Hence the increasing order of bond length is:
$$\alpha \;<\; \beta \;<\; \gamma$$Answer: B
Solution
Statement I. A carbanion carries a negative charge, so it is destabilised by electron-donating groups. Alkyl groups are $+I$ (electron donating), and each added alkyl group intensifies the negative charge, making the carbanion less stable. Thus more alkyl substitution $\Rightarrow$ lower stability:
$$CH_3^- > CH_3CH_2^- > (CH_3)_2CH^- > (CH_3)_3C^-$$This order is correct.
Statement II. Allyl and benzyl carbanions are primarily stabilised by resonance (delocalisation of the negative charge over the $\pi$-system / ring), not merely by an inductive effect. The statement claims the opposite, so it is incorrect.
Answer: C