Question

All possible isomers of C$_3$H$_7$Br when reacted with aq. KOH. Find out total number of optically active products (without rearrangement).

Hint:

When looking for optically active products in nucleophilic substitution reactions, check for the formation of chiral centers in the products. Products with a plane of symmetry or no chiral centers are optically inactive.

Answer 1

Correct Answer: 6

Step 1: Understanding the reaction.
When C$_3$H$_7$Br reacts with aqueous KOH, the reaction is typically a nucleophilic substitution, where the bromine atom is replaced by a hydroxyl group. This is a typical alcohol formation reaction.
Step 2: Identifying the isomers.
The compound C$_3$H$_7$Br can exist as several structural isomers. These isomers are: - 1-Bromopropane - 2-Bromopropane - Isopropyl bromide (propyl bromide)
Step 3: Optically active products.
When these isomers react with KOH, they can form alcohols. The formation of optically active products requires the formation of a chiral center. For example: - In the case of 1-bromopropane, the product is propan-2-ol, which is optically active. - In the case of 2-bromopropane, the product is also optically active. - The reaction of isopropyl bromide leads to a non-chiral product (propan-2-ol is achiral), which does not form optically active products.
Step 4: Conclusion.
Thus, the total number of optically active products is 6.