An alkene on reductive ozonolysis (O3, then Zn/H2O) gives only propanone and propanal. Its reaction with HBr in the absence of peroxide gives mainly

Joining the carbonyl carbons of $\mathrm{(CH_3)_2C=O}$ and $\mathrm{OHC\!-\!CH_2CH_3}$ gives $\mathrm{(CH_3)_2C=CH\!-\!CH_2CH_3}$, 2-methylpent-2-ene. Ionic addition of HBr follows Markovnikov's rule through the more stable tertiary carbocation $\mathrm{(CH_3)_2C^+\!-\!CH_2CH_2CH_3}$, so $\mathrm{Br}$ ends up on C-2: 2-bromo-2-methylpentane. The secondary bromide (3-bromo-2-methylpentane) would form via the less stable secondary cation, and would be formed only if a hydride shift were possible, which would be uphill here.