An alkene on reaction with and given propanone and ethanol in equimolar ratio. Addition of to alkene gives. as the major product. The structure of product is :
1
2
3
4
Official Solution
Correct Option: (3)
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.
