An alkene “A” on reaction with O3 and Zn + H2O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene “A” gives “B” as the major product. The structure of product “B” is :
- H3C-CH2- Cl | CH3 |C-CH3
- H3C- Cl |CH- CH3 | CH3 |CH
- Cl-CH2-CH2- CH3 | CH3 |CH
- H3C-CH2- CH2Cl |CH-CH3
Answer
(1) H3C-CH2- Cl | CH3 |C-CH3
Solution
Ozonolysis products: propanone (CH3)2C=O + ethanal O=CH-CH3
⇒ Joining the carbonyl carbons by a double bond: A = 2-methylbut-2-ene, H3C- CH3 |C=CH-CH3
As per Markovnikov's rule, H+ adds to the =CH– carbon to give the more stable 3° carbocation, and Cl- then adds to it:
B = 2-chloro-2-methylbutane, H3C-CH2- Cl | CH3 |C-CH3
Answer: (1)