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Which of the following compounds give same SN1 and SN2 product?

a: 4‐chlorocyclohex‐1‐ene; b: 5‐(chloromethyl)cyclohexa‐1,3‐diene; c: 3‐chlorocyclohex‐1‐ene; d: 1‐chloro‐2‐methylcyclohexane

I think (b) should be the answer, but the given answer is (c). I think so as it is the only compound where carbon is not chiral. While in others racemic mixture in SN1 and enantiomer in SN2 are formed, in (b) there is no question about this. Please tell me if I got it right.

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    $\begingroup$ If B were to go for SN1 mechanism, rearrangement will take place to form a more stable carbocation. $\endgroup$
    – Aditya Dev
    Dec 25 '15 at 19:00
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    $\begingroup$ ‘where C is not chiral’ … which C? $\endgroup$
    – Jan
    Dec 25 '15 at 21:54
  • $\begingroup$ C means Carbon here. $\endgroup$
    – BEWARB
    Dec 27 '15 at 1:58
  • $\begingroup$ Yes, I was fully aware that C meant carbon. But you have many carbons. $\endgroup$
    – Jan
    Oct 27 '17 at 4:20
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For starters, none of your products show chirality in any way, so it does not matter whether stereoinformation is lost or whether you convert a racemic mixture into a racemic mixture where every single molecule if you had labelled it pre-reaction now had the opposite stereochemistry — in both cases, both reactants and products are racemic mixtures. So you cannot take the presence or absence of chiraliy to distinguish here.

What we can say, is that every compound can take part in an $\mathrm{S_N2}$ reaction (some faster, some slower) but not every one will easily take part in an $\mathrm{S_N1}$ reaction. Specificly, (a) and (d) would give a secondary carbocation intermediate while (b) would even give a primary carbocation. These cations are very unstable and would immediately undergo Wagner-Meerwein type rearrangements towards more stable — read tertiary or resonance stabilised — carbocations. Or, to quote one of my professors from uni:

Exactly one single unstabilised secondary carbocation exists, and that is the prop-2-ylium cation. (Because it cannot rearrange to a more stable one.)

Thus, only (c) will form a stable carbocation without any rearrangement happening: It would be an allyl cation which is well resonance stabilised. But now we have to dig a step deeper, because while all other reactants will certainly not give the same product in both cases, it could be that the correct answer is ‘none of these’, i.e. that somehow we can distinguish between products — because once we have the allyl cation, we could let the nucleophile attack on carbon 1 or carbon 3. Thankfully, though, both attacks give the same product here, because there are no other asymmetries present in the molecule, so we can transform one into the other by rotation.

And, never forget, the $\mathrm{S_N2}$ reaction could also follow the $\mathrm{S_N2'}$ mechanism — this, too, would switch the identities of the two carbons, so we can truly say that both mechanisms will lead to the same product.

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Those compounds in which 1,2 H-Shift is not possible to increase carbocation stability in $\mathrm{S_N1}$ reactions give same $\mathrm{S_N1}$ and $\mathrm{S_N2}$ product. It is because $\mathrm{S_N2}$ is a one step reaction and substitution takes place without rearrangement. Therefore answer would be (C)

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  • $\begingroup$ I fail to understand this answer. $\endgroup$
    – Jan
    Dec 25 '15 at 21:55
  • $\begingroup$ Can you tell which line you are not able to understand so I can clarify $\endgroup$
    – Vaibhav
    Dec 26 '15 at 1:34
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    $\begingroup$ I can’t understand how the first and second sentence go together. $\endgroup$
    – Jan
    Dec 26 '15 at 11:21
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I think compound c will react either via $\mathrm{S_N1}$ and $\mathrm{S_N2}$ and the product will be the same. Allylic carbon reacts by $\mathrm{S_N1}$ because the stability of carbocation and reacts via $\mathrm{S_N2}$ because its primary carbon.
Hope that helps.

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