Question

Propane molecule on chlorination under photochemical condition gives two di-chloro products, "x" and "y". Amongst "x" and "y", "x" is an optically active molecule. How many tri-chloro products (consider only structural isomers) will be obtained from "x" when it is further treated with chlorine under the photochemical condition?

• A. 3
• B. 2
• C. 5
• D. 4

Hint:

Optically active compounds are those that have chiral centers. In this case, the addition of chlorine at various positions leads to a variety of structural isomers.

Answer 1

The Correct Option is A

To solve this problem, we first need to understand what happens during the chlorination of propane under photochemical conditions and then analyze the structural isomers obtained from the further chlorination of the optically active dichloro derivative.

Step 1: Chlorination of Propane 

Propane (C₃H₈) on chlorination mainly forms two monochloro derivatives:

• 1-chloropropane: Chlorine substitutes at the terminal carbon.
• 2-chloropropane: Chlorine substitutes at the second carbon.

When these monochloro products undergo further chlorination, they produce dichloro derivatives. One such dichloro derivative is optically active, which occurs when both chlorine atoms attach to the central carbon, resulting in a chiral center. This is 2,2-dichloropropane.

Step 2: Further Chlorination of 2,2-Dichloropropane

2,2-Dichloropropane will undergo further chlorination to form several trichloro derivatives. The possible sites for additional chlorination are:

• Either of the two hydrogen atoms at the 1-position.
• The hydrogen atom at the 3-position.
• Further substitution at the central carbon is less likely due to existing chlorine atoms creating steric hindrance.

Step 3: Possible Trichloro Isomers

The three possible trichloro structural isomers are:

1. 1,2,2-Trichloropropane: Formed by chlorination at the 1-position.
2. 3,2,2-Trichloropropane: Formed by chlorination at the 3-position.
3. Further, consider another distinct option at the 1-position which remains structurally unique compared to existing di-substitution.

Hence, among the options given, the correct number of structural isomers that "x", 2,2-dichloropropane, can achieve by further chlorination is 3.

The correct answer is therefore:

3

Answer 2

To determine the number of trichloro derivatives from optically active 1,2-dichloropropane:

1. Identification of Compound "x":
The optically active dichloro product must be 1,2-dichloropropane (CH₃CHClCH₂Cl) as it contains a chiral center at C-2.

2. Possible Trichloro Products:
Chlorination can occur at three distinct positions:

1. Methyl group chlorination:
   CH₃CHClCH₂Cl → CCl₃CHClCH₂Cl
2. Chiral center chlorination:
   CH₃CHClCH₂Cl → CH₃CCl₂CH₂Cl
3. Terminal chloromethyl group:
   CH₃CHClCH₂Cl → CH₃CHClCHCl₂

3. Isomer Count:
These substitutions yield three distinct structural isomers. No additional unique structures are possible through single hydrogen substitutions on this compound.

The number of possible trichloro structural isomers is 3.