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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. conversion

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?

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    $\begingroup$ 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 $\endgroup$ – Waylander May 4 at 18:14
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    $\begingroup$ I would also be inclined to reduce the amide with LiAlH4 $\endgroup$ – Waylander May 4 at 20:01
  • $\begingroup$ HBr is formed during the ring bromination of acetophenone. Be aware that acid catalyzed bromination of the methyl group may occur. $\endgroup$ – user55119 May 4 at 23:30

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