Animated Solution for Chemistry - Organic Chemistry: Amongst the following compounds, the optically active alkane having lowest molecular mass is

The quest for chirality is one of the most fascinating journeys in organic chemistry. It is the geometric property that gives molecules "handedness," much like our left and right hands. In this problem, we are tasked with finding the optically active alkane with the lowest molecular mass among four given options. Let's dive deep into the structural analysis of each molecule and uncover the hidden chiral center!

Before we jump into the options, we must establish the ground rules. What makes an alkane optically active?

For a simple hydrocarbon to exhibit optical activity, it must lack any plane or center of symmetry. In almost all basic cases, this implies the presence of at least one chiral carbon atom. A chiral carbon (or stereocenter) is an $s{p}^3$ hybridized carbon atom that is bonded to four completely distinct atoms or groups. If even two of the attached groups are identical, the molecule possesses a plane of symmetry and becomes achiral (optically inactive).

Our mission is clear: we need to inspect each option, locate the most substituted carbon atoms, and check if they are bonded to four different groups.

Let's systematically evaluate the given structures.

Option (a): n-butane The structure is $C{H}_3-C{H}_2-C{H}_2-C{H}_3$. If we look at the terminal carbons, they are attached to three identical hydrogen atoms. The internal carbons are attached to two identical hydrogen atoms. Since no carbon atom is bonded to four different groups, n-butane is completely achiral.

Option (b): 2-methylbutane The structure is $C{H}_3-C{H}_2-CH(C{H}_3)-C{H}_3$. Let's focus on the central $CH$ carbon, as it is the most substituted. It is bonded to: 1. A hydrogen atom ($-H$) 2. An ethyl group ($-C{H}_{2}C{H}_3$) 3. A methyl group ($-C{H}_3$) 4. Another methyl group ($-C{H}_3$)

Because it is attached to two identical methyl groups, it fails the chirality test. Thus, 2-methylbutane is achiral.

Option (d): 1-butyne The structure is $C{H}_3-C{H}_2-C\equiv CH$. Right away, we notice a triple bond. The carbons involved in the triple bond are $sp$ hybridized, meaning they have linear geometry and are only attached to two groups. A chiral center must be $s{p}^3$ hybridized (tetrahedral). Furthermore, the question specifically asks for an alkane, and this is an alkyne. We can safely eliminate this option.

Unveiling the Chiral Center

Now, let's turn our attention to Option (c): sec-butylcyclopropane.

Let's carefully trace the bonds of the central carbon atom. It is bonded to: 1. A methyl group ($-C{H}_3$) 2. A hydrogen atom ($-H$) 3. An ethyl group ($-C_2{H}_5$) 4. A cyclopropyl ring ($-C_3{H}_5$)

Take a moment to appreciate this structure. The methyl group, the ethyl group, and the cyclopropyl ring are all completely distinct hydrocarbon substituents. Along with the hydrogen atom, this central carbon is bonded to four entirely different groups!

Because it satisfies the strict condition of asymmetry, this carbon is a chiral center (often denoted with an asterisk, $C^{\ast }$). The presence of this chiral center, without any internal plane of symmetry, makes the entire molecule optically active.

We have successfully identified the optically active molecule. Option (c) is indeed the correct answer.

But let's push our understanding a bit further. Option (c) is a cycloalkane. What if the examiner asked for the lowest molecular mass optically active open-chain alkane?

To build an open-chain chiral alkane, the chiral carbon must be attached to the four smallest possible distinct alkyl groups (including hydrogen). These would be: - Hydrogen ($-H$) - Methyl ($-C{H}_3$) - Ethyl ($-C_2{H}_5$) - Propyl ($-C{H}_{2}C{H}_{2}C{H}_3$)

Combining these around a central carbon gives us 3-methylhexane ($C_7{H}_{16}$). This is a classic piece of trivia that frequently appears in competitive exams. Always remember to check the hybridization and the uniqueness of the four groups, and you will never fall into the chirality trap!