Question

An alkene \("A"\) on reaction with \(O_3\) and \(Zn - H_2{0}\) given propanone and ethanol in equimolar ratio. Addition of \(HCI\) to alkene \("A"\) gives. \("B"\)as the major product. The structure of product \("B"\) is : 

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is C

To solve this problem, we need to determine the alkene "A" that gives propanone and ethanol upon ozonolysis. We also need to find the structure of product "B" formed when "A" reacts with HCl as the major product.

1. **Ozonolysis Reaction**:

Ozonolysis of alkenes breaks the double bond, resulting in compounds with carbonyl groups. In this case, the products are propanone (C_3H_6O) and ethanol (C_2H_5OH). The structure that leads to these products must be deduced.

2. **Determine the structure of "A"**:

Propanone (acetone) is derived from a C_3 fragment, and ethanol is derived from a C_2 fragment. Therefore, the alkene "A" must be 2-methylpropene ((CH_3)_2C=CH_2), where ozonolysis cleaves the double bond to yield propanone and ethanol.

3. **Reaction with HCl**:

When "A" (2-methylpropene) reacts with HCl, we follow Markovnikov's rule, which states that the hydrogen atom will add to the carbon with the most hydrogens already attached. This produces a carbocation at the more stable tertiary position.

Thus, addition of HCl results in the formation of "B": 2-chloro-2-methylpropane.

The correct structure of product "B" is:

4. **Verification**:

• Ozonolysis of 2-methylpropene yields propanone and ethanol, matching the given conditions.
• Addition of HCl forms the stable tertiary carbocation, leading to the formation of 2-chloro-2-methylpropane as the major product.

This confirms that the structure of "B" is indeed correct.