Hepta-1,6-dien-3-one is reacted with excess HBr to form an intermediate compound A . Compound A is then treated with 1 equivalent of NaI in acetone to yield the major product B . The structure of product B is:

Step-by-step solution

The reactant hepta-1,6-dien-3-one ( CH ₂= CH - C (= O )- CH ₂- CH ₂- CH = CH ₂ ) has two double bonds: an , -unsaturated ketone (C1=C2) and an isolated terminal alkene (C6=C7). Step 1: Addition of excess HBr . For the conjugated alkene ( CH ₂= CH - C (= O )- ), electrophilic addition places the proton at the -carbon to avoid placing a positive charge adjacent to the electron-withdrawing carbonyl group. The bromide ion attacks the -carbon, yielding a primary bromide: BrCH ₂- CH ₂- C (= O )- . For the isolated alkene