Skip to content
NCERT Exemplar · Q19

Q.Consider the following four compounds:
I. CH3—CH2—CH2—CH2—CHO
II. CH3—CH2—CH2—CO—CH3
III. CH3—CH2—CO—CH2—CH3
IV. CH3—CH(CH3)—CH2—CHO
Which of the following pairs are position isomers? (Note: more than one of the given options may be correct.)

(i) I and II
(ii) II and III
(iii) II and IV
(iv) III and IV
Sikkim CbseMCQ· 1mImportance★★★★★
64% · 83/130 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Position isomers have the same carbon skeleton and the same functional group, differing only in the location of that group. Here, I and II are not position isomers (different functional groups), II and III are (both ketones, same skeleton), II and IV are not (different skeletons), and III and IV are not (different skeletons). The correct pair is (ii).

Let’s first get the concept straight. Position isomerism occurs when two compounds share the same molecular formula and the same functional group, but the functional group is attached at a different carbon atom along the same carbon chain. That means the carbon skeleton (the backbone) must be identical — no branching differences, no change in chain length. If the functional group itself changes, that’s functional group isomerism, not position isomerism. If the skeleton changes (e.g., straight chain vs. branched), that’s chain isomerism.

Now examine each compound:

  1. Compound I: CH₃—CH₂—CH₂—CH₂—CHO

    This is pentanal — a straight-chain aldehyde with the –CHO group at the end (carbon 1). Its molecular formula is C₅H₁₀O.

  2. Compound II: CH₃—CH₂—CH₂—CO—CH₃

    This is pentan-2-one — a straight-chain ketone with the carbonyl group at carbon 2. Same formula C₅H₁₀O, but the functional group is different (ketone vs. aldehyde). So I and II are functional group isomers, not position isomers. Option (i) is wrong.

  3. Compound III: CH₃—CH₂—CO—CH₂—CH₃

    This is pentan-3-one — a straight-chain ketone with the carbonyl at carbon 3. Compare II and III: both are straight-chain C₅ ketones, same skeleton, same functional group — only the carbonyl position differs (C2 vs. C3). That’s textbook position isomerism. Option (ii) is correct.

  4. Compound IV: CH₃—CH(CH₃)—CH₂—CHO

    This is 3-methylbutanal — a branched aldehyde. Its skeleton is not straight; it has a methyl branch at carbon 3. Compare II and IV: II is a straight-chain ketone, IV is a branched aldehyde — different skeletons and different functional groups. Not position isomers. Option (iii) is wrong. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.