In-text Questions — Chains, Isomers and Homologous Series — Carbon and its Compounds

1. 12 marksNCERT Cl-10 Science, In-text Qs after §4.2.4, Q1

   How many structural isomers can you draw for pentane?

   Hint. Vary how branched the 5-carbon skeleton is — unbranched, one branch, or two branches.

   Step 1 — Start with the unbranched chain. n-pentane: CH₃-CH₂-CH₂-CH₂-CH₃, a straight 5-carbon chain.

   Step 2 — Introduce one branch. 2-methylbutane (isopentane): a 4-carbon chain with one -CH₃ branch on the second carbon.

   Step 3 — Introduce two branches on the shortest possible chain. 2,2-dimethylpropane (neopentane): a 3-carbon chain with two -CH₃ branches both on the central carbon, since that's the only way to place two branches and still have 5 carbons total.

   ✦ Answer: 3 structural isomers — n-pentane, 2-methylbutane, and 2,2-dimethylpropane.

   Where students slip. Missing 2,2-dimethylpropane because both branches sit on the same carbon — it's easy to assume branches must go on different carbons, but placing both methyl groups on the central carbon of propane gives a genuinely distinct, valid isomer.

2. 22 marksNCERT Cl-10 Science, In-text Qs after §4.2.4, Q2

   What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?

   Hint. One property is about how many bonds carbon forms; the other is about what it can bond to.

   Step 1 — Name the first property. Tetravalency — carbon has 4 valence electrons and forms 4 covalent bonds, giving huge scope for what it can attach to.

   Step 2 — Name the second property. Catenation — carbon's unique ability to bond to other carbon atoms, forming long chains, branched chains and rings, since this self-linking is what lets a small number of elements build an enormous variety of skeletons.

   ✦ Answer: Tetravalency and catenation.

   Where students slip. Naming only tetravalency — many elements form multiple bonds, but it's catenation (self-linking into chains and rings) combined with tetravalency that specifically explains carbon's enormous number of compounds.

3. 32 marksNCERT Cl-10 Science, In-text Qs after §4.2.4, Q3

   What will be the formula and electron dot structure of cyclopentane?

   Hint. Cyclopentane is the ring version of pentane — work out how many hydrogens each ring carbon can still hold.

   Step 1 — Set up the ring. Cyclopentane has 5 carbon atoms joined in a ring, each bonded by a single covalent bond to its two neighbours.

   Step 2 — Count remaining bonds per carbon. Each carbon uses 2 of its 4 bonds for the ring, leaving 2 bonds free for hydrogen — so each carbon in the ring carries exactly 2 hydrogen atoms.

   Step 3 — Write the formula. 5 carbons × 2 hydrogens each = 10 hydrogens, giving C₅H₁₀.

   ✦ Answer: C₅H₁₀ — a ring of 5 CH₂ units, each carbon singly bonded to its two ring neighbours and to 2 hydrogen atoms.

   Where students slip. Writing the open-chain pentane formula C₅H₁₂ — closing the chain into a ring uses up one extra bond per end carbon (compared to an open chain), which removes 2 hydrogens overall.

4. 44 marksNCERT Cl-10 Science, In-text Qs after §4.2.4, Q4

   Draw the structures for the following compounds. (i) Ethanoic acid (ii) Bromopentane (iii) Butanone (iv) Hexanal. Are structural isomers possible for bromopentane?

   Hint. For bromopentane, think about how many different carbon positions the bromine atom could sit on.

   Step 1 — (i) Ethanoic acid. CH₃-COOH — a methyl group attached to a carboxylic acid carbon (which is double-bonded to one oxygen and singly bonded to an -OH group).

   Step 2 — (ii) Bromopentane. CH₃-CH₂-CH₂-CH₂-CH₂-Br (1-bromopentane) is one structure — a 5-carbon chain with bromine on the end carbon.

   Step 3 — (iii) Butanone. CH₃-CO-CH₂-CH₃ — a 4-carbon chain with the C=O (ketone) group on the second carbon, not at the end.

   Step 4 — (iv) Hexanal. CH₃-CH₂-CH₂-CH₂-CH₂-CHO — a 6-carbon chain with the -CHO (aldehyde) group at the end.

   Step 5 — Isomers of bromopentane. Yes — bromine can sit on different carbons of the 5-carbon chain (1-, 2-, or 3-bromopentane), and the parent chain itself could be branched, so several structural isomers of bromopentane are possible.

   ✦ Answer: (i) CH₃-COOH (ii) CH₃-CH₂-CH₂-CH₂-CH₂-Br (iii) CH₃-CO-CH₂-CH₃ (iv) CH₃-CH₂-CH₂-CH₂-CH₂-CHO. Yes, structural isomers of bromopentane are possible.

   Where students slip. Placing the ketone C=O of butanone at the end of the chain, as if it were an aldehyde — a ketone's carbonyl group must sit on an internal carbon, which is exactly what separates butanone from an aldehyde of the same length.

5. 53 marksNCERT Cl-10 Science, In-text Qs after §4.2.4, Q5

   How would you name the following compounds? (i) CH₃—CH₂—Br (ii) H-C(=O)-H (methanal's structure) (iii) CH₃-CH₂-CH₂-CH₂-C≡C-H

   Hint. For (iii), count all six carbons in the chain before deciding where the triple bond sits.

   Step 1 — (i) CH₃-CH₂-Br. A 2-carbon chain with bromine on the end carbon — bromoethane.

   Step 2 — (ii) H-C(=O)-H. A single carbon double-bonded to oxygen with two hydrogens attached — this is methanal (formaldehyde), HCHO.

   Step 3 — (iii) CH₃-CH₂-CH₂-CH₂-C≡C-H. Counting all the carbons gives a 6-carbon chain, with a triple bond between the last two carbons (position 1 from that end) — hex-1-yne.

   ✦ Answer: (i) Bromoethane (ii) Methanal (iii) Hex-1-yne.

   Where students slip. Miscounting the carbons in (iii) and naming it as a shorter chain (e.g. pent-1-yne) — every CH₂ in the drawn zig-zag is a separate carbon, and this structure has six of them in total.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Science textbook, Reprint 2026-27 (jesc104.pdf) — five in-text question sets (13 questions total, not 18 as some older manifests claim) plus one end-of-chapter Exercise (15 questions, not 20). Unchanged by rationalisation.. Questions are referenced from the NCERT textbook for identification.