Previous year questions: Organic Chemistry basics

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar. Each heading gives the paper and the marks. In statement questions, check each statement independently, and in the IUPAC priority question, compare the groups position by position.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

In IUPAC nomenclature, the correct order of decreasing priority of functional group is: (a) -CONH2, >C=O, -CHO, -NH2, -C≡C- (b) -CONH2, -COOCH3, -CHO, -NH2, -OH (c) -CONH2, -CHO, >C=O, -NH2, -C≡C- (d) -CONH2, -CHO, -CN, -NH2, -C≡C-

Answer: The IUPAC priority order puts the amide above the nitrile, the nitrile above the aldehyde, the aldehyde above the ketone, the ketone above the alcohol, the alcohol above the amine, and the amine above the alkyne. Option (a) puts the ketone before the aldehyde, which is wrong. Option (b) puts the amide above the ester, but an ester outranks an amide, and it also puts the amine above the alcohol, which is wrong. Option (d) puts the nitrile below the amine, but a nitrile outranks an amine and even an aldehyde, so it is wrong. Option (c) reads amide, aldehyde, ketone, amine, alkyne, which follows the priority order, so the answer is (c).

Hinglish: Priority order hai amide > nitrile > aldehyde > ketone > alcohol > amine > alkyne. Option (a) mein ketone aldehyde se pehle hai, galat. Option (b) mein amide ester se pehle hai, galat, kyunki ester ki priority zyada hai. Option (d) mein nitrile amine ke baad hai, galat. Option (c) ka order amide, aldehyde, ketone, amine, alkyne sahi hai, isliye answer (c).

Q2 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Given below are two statements. Statement I: 1,2,3-Trihydroxypropane can be separated from water by simple distillation. Statement II: An azeotropic mixture cannot be separated by fractional distillation. Choose the correct answer: (a) both true (b) both false (c) Statement I is true but Statement II is false (d) Statement I is false but Statement II is true.

Answer: Statement I is false. Glycerol, which is 1,2,3-trihydroxypropane, is a high boiling, viscous liquid that is miscible with water, and simple distillation would not separate it cleanly from water; a different method such as distillation under reduced pressure is needed. Statement II is true, because an azeotropic mixture boils at a constant temperature and its composition does not change on repeated distillation, so fractional distillation cannot separate it. The answer is (d).

Hinglish: 1,2,3-trihydroxypropane yani glycerol high boiling viscous liquid hai aur paani se miscible hai, isliye simple distillation se separation sahi nahi hai. Azeotropic mixture ka composition distill karne par nahi badalta, isliye fractional distillation se bhi separation nahi hoti. Answer (d).

Q3 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

Given below are two statements. Statement I: In a species where a positively charged CH group bridges a p-nitrophenyl ring and a p-methoxyphenyl ring, the carbocation is stabilised by the +R effect of the -OCH3 group. Statement II: In the corresponding carbanion, the carbanion is stabilised by the -R effect of the -NO2 group. Choose the correct answer: (a) both true (b) both false (c) Statement I is true but Statement II is false (d) Statement I is false but Statement II is true.

Answer: The methoxy group has a plus R effect and donates its lone pair into the ring, which stabilises the adjacent carbocation, so Statement I is true. The nitro group has a minus R effect and withdraws electron density from the ring, which stabilises the carbanion by spreading its negative charge, so Statement II is also true. Both statements are correct, so the answer is (a).

Hinglish: -OCH3 ka +R effect carbocation ko stabilise karta hai, isliye Statement I sahi. -NO2 ka -R effect carbanion ki negative charge ko delocalise karke use stabilise karta hai, isliye Statement II bhi sahi. Answer (a).

Q4 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Given below are two statements. Statement I: On the basis of inductive effect, the order of stability of alkyl carbanions is CH3 minus, then CH3CH2 minus, then (CH3)2CH minus, then (CH3)3C minus. Statement II: Allyl and benzyl carbanions are stabilised by the inductive effect and not by the resonance effect. Choose the correct answer: (a) both true (b) both false (c) Statement I is true but Statement II is false (d) Statement I is false but Statement II is true.

Answer: Alkyl groups have a plus I effect and push electron density toward the negatively charged carbon, which destabilises the carbanion. So the order with methyl most stable and tert butyl least stable is correct on the inductive basis, and Statement I is true. Statement II is false, because allyl and benzyl carbanions are stabilised mainly by resonance, not by the inductive effect. The answer is (c).

Hinglish: Alkyl groups +I effect se carbanion ko destabilise karte hain, isliye methyl sabse stable aur tert butyl sabse kam stable hai; Statement I sahi. Allyl aur benzyl carbanions resonance se stabilise hote hain, isliye Statement II galat. Answer (c).

Q5 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Match List I with List II. Purification technique: A. Simple distillation, B. Fractional distillation, C. Steam distillation, D. Distillation under reduced pressure. Used to separate: I. Steam volatile compound, II. Two liquids with a large difference in boiling points, III. Liquid decomposing at its boiling point, IV. Two liquids with close boiling points. Choose the correct answer.

Answer: Simple distillation separates two liquids with a large difference in boiling points, so A matches II. Fractional distillation separates liquids with close boiling points, so B matches IV. Steam distillation is used for steam volatile compounds, so C matches I. Distillation under reduced pressure is used for a liquid that decomposes at its boiling point, so D matches III. The matching is A-II, B-IV, C-I and D-III.

Hinglish: Simple distillation bade boiling point difference ke liye (A-II), fractional distillation close boiling points ke liye (B-IV), steam distillation steam volatile compound ke liye (C-I) aur reduced pressure distillation decompose hone wale liquid ke liye (D-III) hoti hai.

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

Given below are two statements. Statement I: A mixture of C12H22O11 (sugar) and NaCl can be separated by dissolving sugar in alcohol, due to differential solubility. Statement II: Rose essence from rose petals is separated by steam distillation due to its high volatility and insolubility in water. Choose the correct answer: (a) both true (b) both false (c) Statement I is true but Statement II is false (d) Statement I is false but Statement II is true.

Answer: Sugar dissolves in alcohol while sodium chloride does not, so alcohol can be used to separate the two by differential solubility, which makes Statement I true. Rose essence is a steam volatile compound that is immiscible with water, and steam distillation is the standard method for isolating it, so Statement II is also true. Both statements are correct, and the answer is (a).

Hinglish: Sugar alcohol mein dissolve ho jaata hai par NaCl nahi, isliye differential solubility se separation possible hai; Statement I sahi. Rose essence steam volatile aur paani mein insoluble hai, isliye steam distillation se nikalta hai; Statement II sahi. Answer (a).

Q7 (NCERT Exemplar, 1 mark)

Which of the following is the correct IUPAC name? (i) 3-Ethyl-4,4-dimethylheptane (ii) 4,4-Dimethyl-3-ethylheptane (iii) 5-Ethyl-4,4-dimethylheptane (iv) 4,4-Bis(methyl)-3-ethylheptane

Answer: The longest chain has seven carbons, so the root is heptane. Numbering must give the lowest locant set to the substituents; starting from the end nearer the ethyl group gives positions 3, 4 and 4, while starting from the other end gives 4, 4 and 5, so the first numbering is preferred. The substituents are listed alphabetically, and ethyl comes before methyl, so the name is 3-ethyl-4,4-dimethylheptane. Writing bis(methyl) instead of dimethyl is not the standard IUPAC form here, so the answer is (i).

Hinglish: Longest chain saat carbon ki hai, isliye heptane. Lowest locant set ke liye numbering us taraf se karo jisse 3,4,4 mile; ethyl alphabetically pehle aata hai, isliye 3-ethyl-4,4-dimethylheptane. Answer (i).

Q8 (NCERT Exemplar, 1 mark)

Electronegativity of carbon atoms depends on the state of hybridisation. In which of the following is the carbon marked with an asterisk most electronegative? (i) CH3-CH2-*CH2-CH3 (ii) CH3-*CH=CH-CH3 (iii) CH3-CH2-C≡*CH (iv) CH3-CH2-CH=*CH2

Answer: The more s character a hybrid orbital has, the more strongly it holds electrons, so electronegativity rises in the order sp three, then sp two, then sp. The carbon in option (iii) is part of a triple bond and is therefore sp hybridised, giving it the greatest s character and the highest electronegativity among the marked carbons. The answer is (iii).

Hinglish: Hybrid orbital mein s character jitna zyada, utni zyada electronegativity: sp^3 < sp^2 < sp. Option (iii) mein carbon triple bond mein hai, isliye sp hybridised hai aur sabse zyada electronegative hai. Answer (iii).

Q9 (NCERT Exemplar, 1 mark)

The fragrance of flowers is due to steam volatile organic compounds called essential oils. These are generally insoluble in water at room temperature but miscible with water vapour in the vapour phase. A suitable method for extracting these oils from flowers is: (i) distillation (ii) crystallisation (iii) distillation under reduced pressure (iv) steam distillation

Answer: The key property is that the oils are steam volatile and insoluble in water, so they can be carried over with steam and then separated from the water layer. Steam distillation uses exactly this property, and the mixture boils below the boiling point of water because the total vapour pressure is the sum of the two components. Crystallisation is for solids and the other distillation methods are for liquids that need a different separation principle. The answer is (iv).

Hinglish: Essential oils steam volatile hain aur paani mein insoluble hain, isliye steam ke saath carry hokar alag ho sakte hain. Steam distillation isi property par kaam karti hai. Crystallisation solids ke liye hoti hai. Answer (iv).

Q10 (NCERT Exemplar, 1 mark)

In which of the following compounds is the carbon marked with an asterisk expected to have the greatest positive charge? (i) *CH3-CH2-Cl (ii) *CH3-CH2-Mg+ Cl minus (iii) *CH3-CH2-Br (iv) *CH3-CH2-CH3

Answer: A positive charge on carbon is increased by an electron withdrawing group attached to it. Chlorine is more electronegative than bromine, so the carbon attached to chlorine carries a larger positive charge than the one attached to bromine. The Grignard carbon carries a partial negative charge because magnesium is electropositive, and the carbon in propane has no strong withdrawing group. The answer is (i).

Hinglish: Electron withdrawing group positive charge badhata hai, aur chlorine bromine se zyada electronegative hai, isliye Cl wale carbon par sabse zyada positive charge hoga. Grignard carbon par negative charge hota hai aur propane mein koi strong withdrawing group nahi hai. Answer (i).