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In general soft nucleophiles do 1,4 conjugate addition to a carbonyl carbon under thermodynamic control. Will soft nucleophiles like organocuprates still preferentially undergo 1,4 addition if the β-carbon is more sterically hindered than the carbonyl carbon?

For example, if I consider the following case

2-cyclohexylideneacetaldehyde

should I just accept it as a rule of thumb that these "soft nucleophiles" undergo 1,4-nucleophilic addition to the carbonyl group, or does it vary depending on the structure and the situation?

In addition to the example above, what happens if there is a methyl group at C-2 of the ring?

2-(2-methylcyclohexylidene)acetaldehyde
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  • $\begingroup$ I don't see much steric hindrance in the example case. $\endgroup$
    – Martin - マーチン ♦
    May 19, 2021 at 22:47
  • $\begingroup$ Regarding your edit: The terms 'ortho', 'meta', and 'para' are only applicable to benzene rings. You have a cyclohexane ring here, those terms are not correct. $\endgroup$
    – orthocresol
    May 23, 2021 at 20:46
  • $\begingroup$ @orthocresol Though I recall "ortho product" and "para product / selectivity" when describing the outcome of asymetric dienes (e.g., methacrylate) with asymetric dienes (e.g., 2-methyl-buta-1,3-diene) in the course of Diels-Alder reactions, too. $\endgroup$
    – Buttonwood
    May 23, 2021 at 20:50
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    $\begingroup$ @Buttonwood Hmm, I recall Clayden uses these terms, but I think it's meant to be just a helpful mnemonic, not a serious scientific terminology. $\endgroup$
    – orthocresol
    May 23, 2021 at 21:00

1 Answer 1

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Your line of thought possibly is influenced by the Walden inversion which is characteristic for the nucleophilic substitution mechanism of $\mathrm{S_N2}$ where steric hindrance affects the rate when this umbrella-like structure flips upon entry of the nuclophile and departure of the nucleofuge around a formally $\mathrm{sp^3}$ hybridized center:

enter image description here

(credit)

The reaction shown by you however is an addition on to a carbon with molecular orbitals formally hybridized as $\mathrm{sp^2}$. Similar to the Bürgi–Dunitz trajectory for the entry of a nucleophile on to a $\ce{C=O}$ bond a) it is plausible that the angle of attack for the 1,4-addition b) equally is tilted, rather than an attack from the back.

enter image description here

(composed with a picture from here)

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    $\begingroup$ My impression (I never did much real-life organic chemistry) is that depending on the size of R and R', nucleophilic attack on the alkene (even via the B–D trajectory) can be quite tough: it probably depends on how big the nucleophile is, too. However, I can't say that I'm an expert on this, so had better leave it to the proper organic chemists :-) $\endgroup$
    – orthocresol
    May 23, 2021 at 21:22

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