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I was just doing an organic chemistry problem from an online source recently, on the topic of conjugate additions (or 1,4-additions). Below is an image illustrating the problem.

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I chose the upper route while the solution manual claims that the answer is the lower route. My rationale for choosing the upper route was that perhaps the alkene is not nucleophilic enough to react feasibly in an electrophilic addition with $\ce {HCl}$. This is due to the presence of the electron-withdrawing carbonyl. Thus, it is important that we first get rid of the carbonyl by using the ethyl Grignard. Then, we can easily add the $\ce {HCl}$ for the electrophilic addition.

Upon reconsidering my reasoning, perhaps I was too narrow-minded in saying that the $\ce {HCl}$ must add in the fashion of electrophilic addition. Perhaps, conjugate addition could also take place. And also, by choosing the upper route, I cannot assure that my chlorine atom would end up in the desired position due considerations of the stability of the carbocation intermediate. Thus, I agree that my choice is incorrect.

The correct choice is the lower route and this does ensure that the $\ce {Cl}$ is in the right place in the final product, as it adds via a conjugate fashion. But how is the writer of the solution so sure that the $\ce {HCl}$ would add in a conjugate fashion, instead of via an electrophilic addition?

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Grignards unless modified with Cu(I) salts add 1,2 not 1,4 so the EtMgBr will add to the ketone to give the t-alcohol and leaving the double bond intact.

Consider then the HCl step - if it is strong enough to protonate the double bond leading to Cl addition then it will protonate the t-alcohol in preference leading to elimination or t-Cl formation. So the answer is to do the reaction on the double bond first then the Grignard addition. Protonation will occcur at the carbonyl oxygen but you have to remember that the double bond is in conjugation with the carbonyl system. This polarises the double bond, leading to Cl- attack at the end of the system.

The Grignard is more nucleophile than base so you will get majority addition to the ketone though some deprotonation/elimination may occur depending on the reaction conditions.

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  • $\begingroup$ Ok thanks for the insights! But I would like to know approximately how much of it would be electrophilic addition of HCl across the double bond and how much of it would be conjugation addition with Cl- $\endgroup$ – Tan Yong Boon Jul 4 '18 at 23:01
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    $\begingroup$ I doubt there would be any HCl addition across the double bond as that is not the site of protonation, the carbonyl oxygen is. You simply cannot think of the double bond in isolation, its reactivity is completely altered by the carbonyl in conjugation. If you are happy with the answer please accept it $\endgroup$ – Waylander Jul 5 '18 at 6:41

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