Previous year questions: Hydrocarbons

These ten questions come from the official sources used for Class 11 Chemistry. Six are from JEE Main 2026 papers, taken from six different shifts, and four are from the official NCERT Exemplar chapter on hydrocarbons. Each heading gives the paper and shift and the marks, and each answer is worked from the chemistry rather than copied from a key. Marks follow the book convention used for this book: 4 marks for a JEE Main item and 1 mark for an NCERT Exemplar multiple choice item.

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

One mole of an alkane requires 8 moles of oxygen for complete combustion. Find the sum of the number of carbon atoms and the number of hydrogen atoms in the alkane.

Answer: For an alkane the complete combustion equation is CnH2n+2 plus three n plus one by two molecules of oxygen, giving n carbon dioxide and n plus one water. Setting three n plus one by two equal to 8 gives three n plus one equal to 16, so n is 5 and the alkane is C5H12. The sum of the number of carbon atoms and hydrogen atoms is 5 plus 12, that is 17. The official key is 17.

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

The mass percentage ratio of carbon to hydrogen in a hydrocarbon is 12 to 1, and the molecule has two carbon atoms. Find the mass of carbon dioxide formed when 3.38 g of this hydrocarbon is completely burnt in oxygen.

Answer: With two carbon atoms the carbon mass is 24, and the mass ratio gives 2 g of hydrogen, so the formula is C2H2 with molar mass 26 g per mole. The amount of the hydrocarbon is 3.38 by 26, which is 0.13 mole. Each mole of C2H2 gives two moles of carbon dioxide, so the carbon dioxide formed is 0.26 mole. Multiplying by the molar mass 44 gives 11.44 g. The official key is 11.44.

Q3 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

An alkane requires 8 moles of oxygen for complete combustion, and on chlorination with chlorine in light it gives only one monochlorinated product. Find the total number of primary carbon atoms in the alkane.

Answer: The oxygen requirement fixes the alkane as C5H12, as in Q1. The condition that only one monochlorinated product is formed means that all the hydrogen atoms of the alkane are equivalent, which is true only for 2,2-dimethylpropane, the neopentane isomer. In neopentane the central carbon is quaternary and the four methyl carbons are primary, so the total number of primary carbon atoms is 4. The official key is 4.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

Arrange the following alkenes in the correct order of stability: A, 2-methylbut-2-ene; B, cis-but-2-ene; C, 2,3-dimethylbut-2-ene; D, prop-1-ene.

Answer: Alkene stability increases with the number of alkyl groups attached to the double bond carbons, because of hyperconjugation and the inductive effect. In C, 2,3-dimethylbut-2-ene, the double bond carries four methyl groups and there are twelve alpha hydrogen atoms. In A, 2-methylbut-2-ene, it carries three methyl groups with nine alpha hydrogens. In B, cis-but-2-ene, it carries two methyl groups with six alpha hydrogens, and in D, prop-1-ene, it carries one methyl group with three alpha hydrogens. The order is therefore C, then A, then B, then D. The official key matches this order, which is option A in the paper.

Q5 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

n-Butane on monochlorination under photochemical conditions gives an optically active compound P. P is further chlorinated to give dichloro compounds. Find the number of dichloro compounds, ignoring stereoisomers.

Answer: The monochlorination of n-butane gives 1-chlorobutane and 2-chlorobutane, and only 2-chlorobutane is optically active because its second carbon is a stereocentre. So P is 2-chlorobutane. Further chlorination can replace a hydrogen at the first carbon, at the second carbon, at the third carbon or at the fourth carbon. Substitution at the second carbon gives 2,2-dichlorobutane, substitution at the first carbon gives 1,2-dichlorobutane, at the third gives 2,3-dichlorobutane, and at the fourth gives 1,3-dichlorobutane after numbering from the nearer end. That gives four dichloro compounds ignoring stereoisomers, which is the official key.

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

Match the reactions with their mechanisms: A, Williamson synthesis; B, Friedel-Crafts reaction; C, bromination of vinyl benzene; D, chlorination of toluene in light. The mechanisms are free radical substitution, electrophilic addition, nucleophilic substitution and electrophilic substitution.

Answer: Williamson synthesis is the reaction of an alkoxide with an alkyl halide and it is a nucleophilic substitution, so A matches nucleophilic substitution. The Friedel-Crafts reaction is an electrophilic substitution on the aromatic ring, so B matches electrophilic substitution. Bromination of vinyl benzene adds bromine across the double bond of the side chain and is an electrophilic addition, so C matches electrophilic addition. Chlorination of toluene in light replaces a hydrogen of the methyl group by a chlorine atom, which is a free radical substitution, so D matches free radical substitution. The correct matching is A with nucleophilic substitution, B with electrophilic substitution, C with electrophilic addition and D with free radical substitution, which is the option printed as C in the paper. It helps to classify each reaction by the species that starts it: an alkoxide, an electrophile on the ring, an electrophile on the double bond and a radical from light.

Q7 (NCERT Exemplar, 1 mark)

Arrange the hydrogen halides in order of their decreasing reactivity with propene.

Answer: The reactivity decreases in the order HI, then HBr, then HCl. All three add through the same secondary carbocation intermediate, so the step that decides the rate is the breaking of the hydrogen halide bond. The bond energy is lowest for H-I, higher for H-Br and highest for H-Cl, so HI reacts fastest and HCl slowest. The correct option is HI greater than HBr greater than HCl. The same order is expected because the carbocation intermediate is identical in all three additions, so only the bond strength of the hydrogen halide decides the rate. This is also why the reaction with HI is the fastest and the reaction with HCl the slowest.

Q8 (NCERT Exemplar, 1 mark)

Which reaction of methane represents incomplete combustion?

Answer: Incomplete combustion is the reaction in which methane burns with a limited supply of oxygen and gives carbon black and water rather than carbon dioxide and water. The equation CH4 plus O2 giving carbon solid plus two water molecules shows this, and it is the correct option. The reactions giving methanol or formaldehyde are controlled oxidation reactions, and the reaction giving carbon dioxide and water is complete combustion.

Q9 (NCERT Exemplar, 1 mark)

Assertion: nitration of benzene with nitric acid requires concentrated sulphuric acid. Reason: the mixture of concentrated sulphuric acid and concentrated nitric acid produces the electrophile NO2+. Choose the correct option.

Answer: Both the assertion and the reason are correct, and the reason correctly explains the assertion. Concentrated sulphuric acid protonates nitric acid and helps remove water, which produces the nitronium ion, NO2+, and this electrophile attacks the benzene ring to give nitrobenzene. Without the sulphuric acid the nitronium ion is not formed efficiently, so the nitration is very slow.

Q10 (NCERT Exemplar, 1 mark)

Assertion: among isomeric pentanes, 2,2-dimethylpentane has the highest boiling point. Reason: branching does not affect the boiling point. Choose the correct option.

Answer: Both the assertion and the reason are wrong. 2,2-dimethylpentane is not even an isomer of pentane, because it contains seven carbon atoms, and among the isomeric pentanes the straight chain n-pentane has the highest boiling point. Branching does affect the boiling point, because a more branched molecule has a smaller surface area and weaker van der Waals forces, so the boiling point falls as branching increases. The correct choice is therefore that both statements are not correct.