Q.27 Which one of the following compounds is the simplest alkane that is optically active?

Option (C) is the correct answer: among the given structures, the simplest alkane that is optically active is the one shown in option (C).​

Introduction

In organic chemistry, many competitive exams ask which structure represents the simplest alkane that is optically active, i.e., the first alkane in increasing complexity that shows chirality and optical rotation. Understanding this requires identifying a chiral carbon atom correctly and comparing how many carbon atoms and substituents each option contains.​

Concept: When is an alkane optically active?

An alkane is optically active if:

• It contains at least one chiral (asymmetric) carbon atom.

• A chiral carbon is bonded to four different groups, so the molecule exists as non‑superimposable mirror images (enantiomers).

• For “simplest” alkane, the total carbon count and overall substitution pattern must be minimal among all optically active candidates.​

In the question, all options are skeletal formulas of alkanes; the task is to inspect each central carbon and ask whether it is attached to four different groups.

Option‑wise explanation

Option (A)

• The central branching carbon in (A) is attached to:

  • A methyl group on one side of the main chain.

  • Two identical alkyl segments on opposite sides of the chain (symmetrical environment).

  • One hydrogen atom.

• Because two of the attached groups are identical, this carbon is not chiral, and the alkane in option (A) is optically inactive.​

Option (B)

• In option (B), the supposed stereocenter carbon again connects to:

  • A methyl substituent.

  • Two equivalent alkyl chains extending in opposite directions.

  • One hydrogen.

• The presence of two identical alkyl groups makes this carbon achiral, so option (B) is also optically inactive and cannot be the required simplest alkane.​

Option (C) – Correct

• The branching carbon in option (C) is bonded to four different groups:

  • A methyl group.

  • An ethyl (or longer) alkyl segment on one side of the chain.

  • A different alkyl segment on the other side of the chain.

  • One hydrogen atom.

• Because all four substituents are different, this carbon is a true chiral center, making the molecule optically active.​

• Among the given choices, this structure has the lowest complexity (smallest chain length) while still containing a chiral carbon, so option (C) is the simplest alkane that is optically active.​

Option (D)

• The central carbon in option (D) is attached to:

  • Two identical alkyl groups that create a plane of symmetry.

  • One shorter alkyl chain.

  • One hydrogen atom.

• Because of the plane of symmetry and the pair of identical groups, there is no chiral center and the molecule is optically inactive, so option (D) is ruled out.​

Why option (C) is the simplest optically active alkane here

• Only option (C) provides a carbon atom bonded to four different substituents, fulfilling the condition for chirality and optical activity.​

• The other options either have symmetry or repeated substituents at the branching carbon, so they cannot rotate plane‑polarized light.

• External solutions for this specific multiple‑choice problem confirm that option (C) is the correct choice among the given structural formulas.​