Skip to content
NCERT Exemplar · Q57

Q.Assertion (A): Pent-1-ene and pent-2-ene are position isomers.
Reason (R): Position isomers differ in the position of functional group or a substituent.

(i) Both A and R are correct and R is the correct explanation of A.
(ii) Both A and R are correct but R is not the correct explanation of A.
(iii) Both A and R are not correct.
(iv) A is not correct but R is correct.
Tripura TbseShort· 1mImportance★★★★★
93% · 121/130 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 differ only in where a functional group or substituent sits on the same carbon skeleton. Pent-1-ene and pent-2-ene have their double bond at different positions along the same five-carbon chain, making them position isomers. Both statements are correct, and R explains A perfectly.

Understanding Position Isomerism

Structural isomers share the same molecular formula but differ in how their atoms are connected. Within structural isomerism, position isomers form a specific subclass: they have the same carbon skeleton and the same functional group, but that functional group occupies a different position on the chain.

Think of it as rearranging furniture in the same room. The room (carbon skeleton) stays identical, the furniture piece (functional group) is the same, but you've moved it to a different spot.

Analyzing the Assertion and Reason

Let me work through each statement systematically.

1. Examining the Reason (R) first

The reason states that position isomers differ in the position of a functional group or substituent. This is the textbook definition of position isomerism. For alkenes, the functional group is the C=C\ce{C=C} double bond. For alcohols, it's the −OH\ce{-OH} group. For haloalkanes, it's the halogen atom. The definition is accurate and complete.

2. Checking the Assertion (A)

Pent-1-ene has the structure:

CHX2=CH−CHX2−CHX2−CHX3\ce{CH2=CH-CH2-CH2-CH3}

The double bond sits between carbon-1 and carbon-2.

Pent-2-ene has the structure:

CHX3−CH=CH−CHX2−CHX3\ce{CH3-CH=CH-CH2-CH3}

The double bond sits between carbon-2 and carbon-3.

Both molecules have:

  • The same molecular formula: CX5HX10\ce{C5H10}
  • The same carbon skeleton: an unbranched five-carbon chain
  • The same functional group: one C=C\ce{C=C} double bond
  • Different positions for that double bond

This matches the definition of position isomers exactly.

3. Does R explain A? …

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.