Animated Solution for Chemistry - Organic Chemistry: The total number of monohalogenated organic products in the following (including stereoisomers) reaction is ...... $A\stackrel{\text{(i) }{H}_2/Ni/\Delta \text{ (ii) }{X}_2/\Delta }{\to }$ (Simplest optically active alkene)

The journey begins with a fascinating puzzle: identifying the "simplest optically active alkene." For any organic molecule to exhibit optical activity, it must possess chirality, which usually means having at least one carbon atom bonded to four completely different groups.

If we start building alkenes from scratch, ethene, propene, and butene isomers simply don't have enough carbon atoms to create a chiral center. Even with five carbons, the best we can do is 3-methylbut-1-ene, where the third carbon is attached to a hydrogen, a vinyl group, and two identical methyl groups—making it achiral.

We must step up to six carbons. Enter 3-methylpent-1-ene ($C_6{H}_{12}$). In this molecule, the third carbon is bonded to a hydrogen atom, a methyl group, an ethyl group, and a vinyl group. Four distinct groups! This makes it the simplest acyclic alkene capable of optical activity.

Losing Chirality

The first reaction in our sequence is catalytic hydrogenation ($H_2/Ni/\Delta$). This is a classic reduction reaction where hydrogen gas adds across the carbon-carbon double bond, converting the alkene into an alkane.

When 3-methylpent-1-ene undergoes hydrogenation, the vinyl group ($-CH=C{H}_2$) is reduced to an ethyl group ($-C{H}_{2}C{H}_3$). The resulting molecule is 3-methylpentane.

Take a close look at 3-methylpentane. The third carbon is now bonded to a hydrogen, a methyl group, and two identical ethyl groups. Because two of the groups are the same, the molecule now possesses a plane of symmetry. It has lost its chirality and become an achiral (optically inactive) molecule.

Finding Unique Positions

Now comes the core of the problem: free radical monohalogenation ($X_2/\Delta$). We need to find every structurally distinct position where a halogen atom can substitute a hydrogen atom. In 3-methylpentane, there are four unique types of carbon environments:

1. The terminal primary carbons of the ethyl groups (C1 or C5). 2. The secondary carbons of the ethyl groups (C2 or C4). 3. The tertiary carbon in the center (C3). 4. The primary carbon of the methyl branch attached to C3.

Let's analyze the products formed at each of these positions and count their stereoisomers.

Counting Stereoisomers

Product P (Terminal Halogenation): If the halogen attaches to C1, we get 1-halo-3-methylpentane. The substitution breaks the symmetry, and C3 becomes chiral again! Since there is exactly one chiral center, this product exists as a pair of enantiomers. Count = 2

Product Q (Secondary Halogenation): If the halogen attaches to C2, we get 2-halo-3-methylpentane. This is where you must be careful! The substitution creates a new chiral center at C2, and it makes C3 chiral as well. With two distinct chiral centers and no plane of symmetry, the number of stereoisomers is $2^2=4$. Count = 4

Product R (Methyl Branch Halogenation): If the halogen attaches to the methyl group on C3, we get 3-(halomethyl)pentane. Here, C3 remains attached to two identical ethyl groups. Without a chiral center, this structure represents only a single molecule. Count = 1

Product S (Tertiary Halogenation): Finally, if the halogen attaches directly to C3, we get 3-halo-3-methylpentane. Once again, C3 is bonded to two identical ethyl groups, meaning no chirality is introduced. Count = 1

Summing them all up, the total number of monohalogenated organic products, including all stereoisomers, is $2+4+1+1=8$.