Kindly help with this one.

Starting material has two carbons in the side chain (outside the aromatic ring), while the final product has four carbons. So we need to introduce two extra carbons. The best way is to use an acetylide ion, which can undergo an SN2 reaction with a primary alkyl halide.
First, we must create a good leaving group on the starting hydrocarbon:


Carry out free-radical bromination (Br₂, hν). This introduces Br on the more substituted (secondary) benzylic carbon.


Convert this secondary bromide to a primary bromide. Do this by:


E2 elimination to form the alkene, then


anti-Markovnikov addition of HBr (HBr, ROOR) to place Br on the terminal (primary) carbon.




Now we have a primary alkyl bromide:


Treat with an acetylide ion (e.g. Na⁺ −C≡CH). The acetylide does an SN2 attack on the primary bromide, adding two carbons and giving a terminal alkyne.


Finally, perform hydration of the alkyne (H₂O, H₂SO₄, HgSO₄). This gives the ketone after keto–enol tautomerism, which is the desired product.