Animated Solution for Chemistry - Organic Chemistry: Which of the following compounds produces an optically inactive compound on hydrogenation?

Finding the Optically Inactive Product

Welcome to a classic organic chemistry puzzle! In this problem, we are presented with four distinct chiral alkenes. Our mission is to determine which of these molecules loses its optical activity after undergoing catalytic hydrogenation.

To solve this, we don't need to do complex math; we just need a sharp eye for molecular symmetry.

First, let's recall what hydrogenation actually does. When an alkene is treated with hydrogen gas ($H_2$) in the presence of a metal catalyst (like Palladium, Platinum, or Nickel), the $\pi$-bond of the alkene breaks. Two hydrogen atoms are added across the double bond, effectively converting the unsaturated alkene into a saturated alkane.

This reaction changes the identity of the groups attached to the carbon skeleton. For example, a vinyl group ($-CH=C{H}_2$) will be fully saturated to become an ethyl group ($-C{H}_{2}C{H}_3$).

For a molecule to be optically active, it must be chiral. In most simple organic molecules, chirality arises from the presence of a chiral center—a carbon atom bonded to four completely different groups.

If a reaction causes any two of these four groups to become identical, the molecule gains a plane of symmetry. It becomes superimposable on its mirror image, rendering it achiral and, consequently, optically inactive.

Let's mentally walk through the first three options to see what happens when we hydrogenate them:

- Option (a): The hydrogenation of the double bond results in an alkane where the central carbon is attached to a hydrogen atom, a methyl group, an ethyl group, and a propyl group. Since all four groups are distinct, the chiral center is preserved. - Option (b): Similar to (a), the product features a central carbon bonded to a hydrogen, a methyl, an ethyl, and a propyl group. It remains chiral. - Option (c): Once again, the hydrogenated product has four different groups (hydrogen, methyl, propyl, and ethyl) attached to the central carbon. It is still optically active.

Option (d)

Now, let's focus our attention on option (d), which is $3$-methylpent-$1$-ene.

Before the reaction, the central chiral carbon is bonded to: 1. A hydrogen atom ($-H$) 2. A methyl group ($-C{H}_3$) 3. An ethyl group ($-C{H}_{2}C{H}_3$) on the left 4. A vinyl group ($-CH=C{H}_2$) on the right

When we perform the hydrogenation, the vinyl group on the right gains two hydrogen atoms and transforms into an ethyl group ($-C{H}_{2}C{H}_3$).

Let's look at the central carbon now. On the left, we already had an ethyl group. And now, on the right, we have another ethyl group! Because these two groups are now identical, the central carbon is no longer a chiral center. The resulting molecule, $3$-methylpentane, possesses a plane of symmetry passing through the hydrogen, the methyl group, and the central carbon.

Because it is achiral, it is optically inactive. Therefore, option (d) is our correct answer.

In competitive exams, time is of the essence. Instead of drawing out every single product in detail, train your brain to visualize the groups attached to the chiral center. Ask yourself: "Will this reaction make any two groups identical?" If the answer is yes, you've instantly found your achiral product!