Major product of the following reaction is

The starting compound is allyl acrylate, which contains two different carbon-carbon double bonds.

Since:

$$\mathrm{2\ equivalents\ of\ HBr}$$

are used, both double bonds undergo electrophilic addition.

The double bond on the allyl side:

$$\mathrm{-O-CH_2-CH=CH_2}$$

undergoes normal Markovnikov addition.

The proton adds to the terminal carbon to generate the more stable secondary carbocation.

$$\mathrm{-CH=CH_2 \longrightarrow -C^+H-CH_3}$$

The bromide ion then attacks the carbocation.

Thus:

$$\mathrm{-O-CH_2-CH=CH_2 \longrightarrow -O-CH_2-CH(Br)-CH_3}$$

The double bond on the acrylate side:

$$\mathrm{CH_2=CH-C(=O)-}$$

is conjugated with the carbonyl group.

The electron-withdrawing:

$$\mathrm{C=O}$$

group alters the regiochemistry of addition.

Addition occurs such that bromine attaches to the terminal:

$$\mathrm{\beta}$$

carbon.

Thus:

$$\mathrm{CH_2=CH-C(=O)- \longrightarrow Br-CH_2-CH_2-C(=O)-}$$

Combining both additions, the final major product is:

$$\mathrm{Br-CH_2-CH_2-C(=O)-O-CH_2-CH(Br)-CH_3}$$