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In a conjugated ketone or aldehyde, nucleophiles could attack the double bond or the carbonyl carbon. I have two examples that differ in the location of initial attack, and I am wondering what determines the outcome.

I came across this question:

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The answer is given as 2nd options. I don't know why it is the case so I dug deeper and came across this post and inferred that nucleophiles prefer to attack the conjugated double bonds and also the carbonyl position.

But however this question:

enter image description here

I thought the hydride ion will attack the conjugated double bond, but looks like it does not affect the double bond instead attacks the carbonyl position to get the product in 1st option. So why it is the case? Is it based on the fact that carbocation formed in the benzylic position is more stable than the carbocation in the carbonyl position in the first question, which leads to the attack of nucleophile towards the benzylic carbocation ?

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    $\begingroup$ Well, "normal" Grignard would also only affect carbonyl. It's the first reaction that I'd wonder about, not the second. $\endgroup$
    – Mithoron
    Oct 23, 2023 at 12:31
  • $\begingroup$ @Mithoron Hi !. I thought it is based on the fact that benzylic carbocation is more stable than the one that would form at the carbonyl position. But in the links I have posted, the Lithium Aluminium Hydride attacks the conjugated double bond, in cinnamaldehyde. So it is an exception case only for cinnamaldehyde ? $\endgroup$ Oct 23, 2023 at 13:40
  • $\begingroup$ @Mithoron chemistry.stackexchange.com/questions/87138/… This one $\endgroup$ Oct 23, 2023 at 14:23
  • $\begingroup$ Well, that certainly looks unusual. $\endgroup$
    – Mithoron
    Oct 23, 2023 at 14:37
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    $\begingroup$ No, it's not. More like carbanion, if anything. $\endgroup$
    – Mithoron
    Oct 23, 2023 at 15:17

2 Answers 2

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[Mithoron in the comments] Well, "normal" Grignard would also only affect carbonyl. It's the first reaction that I'd wonder about, not the second.

Without the copper catalyst, the Grignard would attack the carbonyl. With it, it attacks the conjugated carbon-carbon double bond instead.

See Wikipedia article, which gives a 1941 reference (Kharash) referring to the copper-catalyzed version as a 1,4 addition.

A modern review article shows the current applications, and explains that you first obtain an 1,4 adduct (with the metal bound to oxygen) which then is quenched to get the aldehyde back.

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This is an example of the competition between conjugate (Michael) and direct addition to a conjugated carbonyl, which is itself an example of thermodynamic vs kinetic control.

You can read more about this in Clayden chapter 22, but in short,

The conjugate addition is the thermodynamic product and its formation is irreversible.

The direct addition is the kinetic product and its formation is reversible.

The conjugate addition can be favoured by:

  • Heating-as it is kinetically disfavoured, heating reduces the kinetic barrier to conjugate addition.

  • Using a soft nucelophile favours the conjugate addition, a harder nucleophile favours the direct addition.

Soft nucleophiles are ones good at attacking sp3 carbons, but not always sp2 carbons, even though in this case they would attack the sp2 alkene. They involve being polarisable, large, having a delocalised charge and having a high energy HOMO (not all of these characteristics are needed but they all help make a nucleiphile soft). Examples of soft nucleophiles include Thiolates(-SH), organnocopper reagents and deprotonated aromatic heterocycles.

Hard nucleophiles are small and densely charged, and examples include hydrides and gringards or organolithium derivatives.

There is no better example than the ones in your textbook. In the first example the copper is provided to transmetallate the gringard into an organocopper to favour the conjugate addition. Without the copper, a grigard would do a direct addition.

In the second example, the hydride is a hard nucleophile and so favours the direct addition.

Note that some nucleophiles are in between hard and soft and so if the stoichiometry allows, they will do both. A nice example is a soft hydride, like sodium borohydride.

There are chemoselective ways to reduce either the ketone or the conjugate double bond without reducing the other. Luche he reduction with NaBH4 and a catalytic Titanium (III) chloride would be used to reduce only the carbonyl. K-selectride would reduce only the double bond.

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  • $\begingroup$ Thank you for the Nice answer. ! :-) $\endgroup$ Oct 25, 2023 at 12:45

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