Animated Solution for Chemistry - Organic Chemistry: Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is (are) –

Welcome to a fascinating exploration of stereochemistry! In this problem, we are tasked with identifying which of the given unsaturated haloalkanes will yield an optically inactive product upon hydrogenation.

To solve this, we must first understand what hydrogenation does. Catalytic hydrogenation (using $H_2$ and a metal catalyst like $Pd$ or $Pt$) adds hydrogen atoms across a carbon-carbon double bond ($C=C$), converting it into a single bond ($C-C$).

However, the real trick lies in the stereochemistry. A molecule is optically active if it possesses a chiral center and lacks any internal plane of symmetry. If a reaction modifies the groups attached to a chiral center such that two groups become identical, the chiral center is destroyed, and the molecule becomes achiral (and thus, optically inactive). Let's analyze each option through this lens.

The Propenyl Shift

In option (A), the reactant is 4-bromo-2-hexene. The chiral center is located at carbon-4, which is attached to a hydrogen atom ($-H$), a bromine atom ($-Br$), a methyl group ($-C{H}_3$), and a propenyl group ($-CH=CH-C{H}_3$).

When we subject this molecule to hydrogenation, the double bond in the propenyl group is saturated, transforming it into a propyl group ($-C{H}_2-C{H}_2-C{H}_3$).

Let's re-evaluate the chiral center in the product, 2-bromohexane. It is now attached to: 1. $-H$ 2. $-Br$ 3. $-C{H}_3$ 4. $-C{H}_2-C{H}_2-C{H}_3$

Since all four groups remain distinct, the carbon atom retains its chirality. The product is chiral and optically active. Therefore, (A) is incorrect.

The Symmetry Reveal

Option (B) presents us with 3-bromo-1-pentene. Here, the chiral center at carbon-3 is attached to a hydrogen atom, a bromine atom, an ethyl group ($-C{H}_2-C{H}_3$), and a terminal vinyl group ($-CH=C{H}_2$).

Upon hydrogenation, the vinyl group absorbs $H_2$ and is converted into an ethyl group ($-C{H}_2-C{H}_3$).

Now, look closely at the product, 3-bromopentane. The central carbon is attached to: 1. $-H$ 2. $-Br$ 3. $-C{H}_2-C{H}_3$ (the original ethyl group) 4. $-C{H}_2-C{H}_3$ (the newly formed ethyl group)

Because two of the attached groups are now identical, the central carbon is no longer a chiral center! The molecule possesses an internal plane of symmetry passing right through the $H-C-Br$ bonds. This makes it a meso compound, which is achiral and optically inactive. Thus, (B) is a correct answer.

The Isopropyl Creation

In option (C), the reactant is 3-bromo-2-methyl-1-butene. The chiral center is attached to a hydrogen, a bromine, a methyl group, and an isopropenyl group ($-C(C{H}_3)=C{H}_2$).

Hydrogenation of the isopropenyl group yields an isopropyl group ($-CH(C{H}_3{)}_2$).

The resulting product is 2-bromo-3-methylbutane. Let's check the groups on the chiral center: 1. $-H$ 2. $-Br$ 3. $-C{H}_3$ 4. $-CH(C{H}_3{)}_2$

All four groups are completely different. The molecule remains chiral and optically active. So, (C) is incorrect.

The Diastereomeric Twin

Finally, let's look at option (D). If you observe carefully, the connectivity is identical to option (B); it is also 3-bromo-1-pentene, but with a different spatial arrangement at the chiral center (it is a diastereomer or enantiomer depending on the exact double bond geometry, but here it's simply a stereoisomer).

Just like in option (B), hydrogenating the vinyl group converts it into an ethyl group. The product is once again 3-bromopentane.

As we established earlier, 3-bromopentane has a plane of symmetry due to the two identical ethyl groups attached to the central carbon. It is achiral and optically inactive. Therefore, (D) is also a correct answer.

By carefully tracking how the substituents change during the reaction, we discovered that compounds (B) and (D) both yield the achiral molecule 3-bromopentane. This is a classic example of how a chemical reaction can destroy chirality by introducing symmetry. The correct options are (B) and (D).