Organic conversion involving Grignard reagent and alkyl group with a carbonyl carbon on an benzene ring

I have been asked to synthesize the organic compound shown below using the following reagents and solvents.

$$\ce{Mg, Zn, Hg, FeBr3, AlCl3, Br2, LiAlH4, KMnO4}$$, dil. $$\ce{H2SO4}$$, conc. $$\ce{NH3}$$, conc. $$\ce{HCl, H2O, D2O}$$, ether, ethanol, $$\ce{CH3C(O)Cl, CH3C(O)CH3, C6H6}$$.

I have done it as shown below.

But in the answer given that the -$$\ce{C(O)CH3}$$ group attached to the benzene ring is reduced to -$$\ce{CH2CH3}$$ before introducing $$\ce{Mg}$$/dry ether. Then the $$\ce{Mg}$$/dry ether is introduced followed by $$\ce{D2O}$$. Then the alkyl group is oxidized and so on...

Is there is specific reason for doing that or is my method correct?

• Cannot form a Grignard in the presence of a ketone. Form an acetal of the ketone then do the Grignard and your scheme will work – Waylander May 4 '19 at 18:14
• I would also be inclined to reduce the amide with LiAlH4 – Waylander May 4 '19 at 20:01
• HBr is formed during the ring bromination of acetophenone. Be aware that acid catalyzed bromination of the methyl group may occur. – user55119 May 4 '19 at 23:30