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I have read that when the leaving group substituent is stuck in the equatorial position for some reason (like from a t-butyl group or a trans decalin), SN2 is either very slow or just doesn't happen at all because the rear attack is hindered by axial hydrogens/other substituents. I haven't been able to get a straight answer as to whether it can instead progress through SN1 at an acceptable rate, since the attached carbon will at least be secondary.

The only thing I can think of which would prevent this happening is the difficulty of forming a planar sp2 carbocation in a such a rigid system.

Any help is appreciated, thanks.

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I think it's difficult to say for sure without actually making experiments. It should depend a lot on the conformation of the reactant, the substituents, the leaving group and also on the solvent. Nevertheless, I would imagine that it should be possible in some cases if you have a good leaving group.

In the case of a substituted cis-decalin, for example (I suppose you actually meant to refer to cis because in the trans isomer both hydrogens are stuck in axial positions), the elimination of the leaving group must indeed induce some strain, but maybe not that much. The structure could adapt with small changes in angles all over the molecule and both cycles would keep chair-like conformations. A tertiary carbocation would also be quite favorable.

So yes, in my opinion it should be possible in some cases, specially if heating is employed or you have a good leaving group. However, I should reinforce the fact that it really depends on many factors. This type of system can get quite complex if you consider the influence of the solvent, the influence of the substituents and the possibility of side reactions, including eliminations reactions, which commonly compete with substitution (on the presence of bases).

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