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This is a mechanism provided in a problem set in my ochem course.

As far as I know, double bond is electron rich, which is nucleophilic and attacks electrophilic species. In this case, why does one double bond attack another carbon in double bond?

IMO, wouldn't a SN2 reaction make more sense? enter image description here

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  • $\begingroup$ You need to consider the heteroatoms in the two molecules $\endgroup$
    – Waylander
    Commented Nov 16, 2022 at 7:37
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    $\begingroup$ SN2 and SN2' (SN2 prime) reaction of this allylic substrate leads to the formation of the same product.The attack of nucleophile to the double bond is explained using frontier orbitals.The attack of nucleophile to the double bond is called SN2 prime reaction. $\endgroup$
    – hkilic
    Commented Nov 16, 2022 at 8:17

1 Answer 1

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Several points about the question itself. First, the enamine of n-butyraldehyde is unquestionably of the (E)-configuration and its preparation does not require an acid catalyst. Mannich[1] prepared 1,1-dipiperidinobutane (1,1'-(butane-1,1-diyl)dipiperidine) from piperidine and n-butyraldehyde in the presence of potassium carbonate. Elimination to the enamine and piperidine occurred upon distillation. Secondly, the enamine employed in the allylation is that of propionaldehyde.

Whether or not $\ce{S_N2 or S_N2'}$ allylation occurs is not straightforward. N-Allylation may be kinetically favored over direct C-allylation. Stork, et al.[2] have noted that Cope rearrangement of the N-allylated product leads to a C-allylation aldehyde which would, if C-allylation were kinetically favored, lead to allylation at the more hindered end of the allylic electrophile. The following is from the Stork paper:

Blockquote

  1. C. Mannich and H. Davidsen, Chem. Ber., 1936, 69, 2106.
  2. G. Stork, A. Brizzolara, H. Landesman, J. Szmuszkovicz, and R. Terrell, J. Am. Chem. Soc. 1963, 85, 207–222.
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