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 :

Question

Which of the following does not obey the octet rule?

Answer 1

To solve this problem, we need to determine the structure of product "B" when alkene "A" reacts with HCl based on the given information.

Let's break down the problem step-by-step:

The alkene "A" undergoes ozonolysis (reaction with \( O_3 \) and \( Zn-H_2O \)) to give propanone and ethanol in equimolar amounts. This information allows us to deduce the possible structure of the alkene.
Ozonolysis is a process where the alkene is cleaved at the double bond, forming two carbonyl compounds. The given products are propanone (CH₃-C(=O)-CH₃) and ethanol (CH₃CH₂OH).
This suggests that the alkene "A" could be 2-methylbut-2-ene. When ozonolysis occurs, the double bond between the central carbon atoms cleaves to form two carbonyl compounds: propanone and ethanol, consistent with the products provided.
Next, we consider the reaction of alkene "A" with HCl. The addition of HCl to an alkene follows Markovnikov's rule, where the hydrogen atom adds to the carbon atom with the most hydrogen atoms, and the chlorine atom adds to the carbon atom with fewer hydrogen atoms.
Applying Markovnikov's Rule to 2-methylbut-2-ene, the chlorine atom will add to the more substituted carbon atom of the double bond, resulting in the formation of 2-chloro-3-methylbutane as the major product "B".

This is consistent with the given correct answer choice. Therefore, the structure of product "B" is 2-chloro-3-methylbutane.

In conclusion, the correct structure of product "B" formed by the addition of HCl to alkene "A" is 2-chloro-3-methylbutane.

Answer 2

To solve this problem, we need to determine the structure of product "B" when alkene "A" reacts with HCl based on the given information.

Let's break down the problem step-by-step:

The alkene "A" undergoes ozonolysis (reaction with \( O_3 \) and \( Zn-H_2O \)) to give propanone and ethanol in equimolar amounts. This information allows us to deduce the possible structure of the alkene.
Ozonolysis is a process where the alkene is cleaved at the double bond, forming two carbonyl compounds. The given products are propanone (CH₃-C(=O)-CH₃) and ethanol (CH₃CH₂OH).
This suggests that the alkene "A" could be 2-methylbut-2-ene. When ozonolysis occurs, the double bond between the central carbon atoms cleaves to form two carbonyl compounds: propanone and ethanol, consistent with the products provided.
Next, we consider the reaction of alkene "A" with HCl. The addition of HCl to an alkene follows Markovnikov's rule, where the hydrogen atom adds to the carbon atom with the most hydrogen atoms, and the chlorine atom adds to the carbon atom with fewer hydrogen atoms.
Applying Markovnikov's Rule to 2-methylbut-2-ene, the chlorine atom will add to the more substituted carbon atom of the double bond, resulting in the formation of 2-chloro-3-methylbutane as the major product "B".

This is consistent with the given correct answer choice. Therefore, the structure of product "B" is 2-chloro-3-methylbutane.

In conclusion, the correct structure of product "B" formed by the addition of HCl to alkene "A" is 2-chloro-3-methylbutane.

The Correct Option is C