Explanation:
To understand why the correct answer is 6, let's break down the reaction and analyze the possible substitution products.

Reaction:
Br2CH-CHBr2 + 2 NaCN → 2 NaBr + CN-CH-CH(CN)-CHBr2

Analysis:
In this reaction, the two bromine atoms (Br) in Br2CH-CHBr2 are substituted by two CN (cyanide) nucleophiles, resulting in the formation of the product CN-CH-CH(CN)-CHBr2.

To determine the number of different substitution products that are chiral, we need to consider the number of stereocenters in the product molecule. A stereocenter is an atom in a molecule bonded to four different groups, resulting in non-superimposable mirror images (chiral centers).

Possible Substitution Products:
In the product molecule CN-CH-CH(CN)-CHBr2, there are two carbon atoms bonded to four different groups (CN, H, Br, and another CN). These carbon atoms are potential stereocenters.

1. Carbon Atom #1:
The carbon atom in the middle (denoted as Carbon Atom #1) is bonded to two CN groups, one H atom, and one Br atom. Since all the groups bonded to it are different, Carbon Atom #1 is a stereocenter.

2. Carbon Atom #2:
The carbon atom on the right side (denoted as Carbon Atom #2) is bonded to two Br atoms, one CN group, and one Carbon Atom #1. Since all the groups bonded to it are different, Carbon Atom #2 is also a stereocenter.

Number of Possible Stereocenters:
Since there are two stereocenters in the product molecule, we need to consider all possible combinations of their configurations to determine the number of chiral substitution products.

Combinations:
To calculate the number of combinations, we use the formula 2^n, where n is the number of stereocenters.

In this case, n = 2 (two stereocenters)
So, the number of combinations = 2^2 = 4

Chiral Substitution Products:
However, not all combinations will result in chiral substitution products. We need to consider the symmetry of the molecule to eliminate any duplicates.

Chiral Product #1:
For the first combination, if both stereocenters have the same configuration (R or S), the molecule will be achiral. So, we need to consider different configurations for both stereocenters.

Chiral Product #2:
For the second combination, if the first stereocenter has configuration R and the second has configuration S, the molecule will be chiral.

Chiral Product #3:
For the third combination, if the first stereocenter has configuration S and the second has configuration R, the molecule will be chiral.

Chiral Product #4:
For the fourth combination, if both stereocenters have configuration R or both have configuration S, the molecule will be achiral.

Final Answer:
Thus, out of the 4 possible combinations, only 2 result in ch