Q.An alkene 'A', on reaction with O3 and then 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 a) 1-chloro-3-methylbutane, Cl-CH2-CH2-CH(CH3)-CH3 b) 1-chloro-2-methylbutane, ClCH2-CH(CH3)-CH2-CH3 c) 2-chloro-2-methylbutane, H3C-CH2-C(CH3)(Cl)-CH3 d) 2-chloro-3-methylbutane, H3C-CH(Cl)-CH(CH3)-CH3

Step 1. Ozonolysis (O3, then Zn-H2O) of alkene A gives propanone (CH3-CO-CH3) and ethanal (CH3-CHO) in equimolar amounts -- so one alkene carbon must have carried two methyls (becoming propanone) and the other must have carried one methyl plus one hydrogen (becoming ethanal).

Step 2. Joining those two carbons back together at the former double bond reconstructs A as (CH3)2C=CH-CH3 -- 2-methylbut-2-ene.

Step 3. Addition of HCl to this unsymmetrical alkene follows Markovnikov's rule: the H adds to the carbon that already carries MORE hydrogens (the =CH-CH3 carbon, C3), while Cl adds to the MORE substituted carbon (the (CH3)2C= carbon, C2), since that pathway proceeds through the more stable, more substituted (tertiary) carbocation intermediate. …

Reconstruct alkene A from its two ozonolysis fragments first (matching each carbonyl product back to its own alkene carbon), then apply Markovnikov's rule (halogen to the more substitute …

• Applying anti-Markovnikov addition (Cl on the LESS substituted carbon) by mistake -- plain HCl addition (with no peroxide present) always follows Markovnikov's rule, gi …