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I have researched three books: Brown's, Solomons' and Wade's. I could not find enough information about this.

My prof. says that secondary alkly halides undergo SN1 only under acidic environments.

We had this quiz. I got a terrible mark on one question.

I think these are the missing structures.

Now, I am wondering, did Sn2 occur here, if yes, at what rate? Was it more important than Sn1? What was the major pathway?

What is the major pathway for all secondary alkyl halides with acidic environment and with NONacidic environments?

Could anyone also just check my missing structures?

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    $\begingroup$ It is an SN1 reaction. Your major error was forgetting the double bond in the product. $\endgroup$ – jerepierre Nov 2 '14 at 0:36
  • $\begingroup$ Jerepierre, yes I forgot the double bonds on my exam. BUT, there is also the fact that there can be carbocation rearrangement. The double bond moves away. (I actually made a mistake on the uploaded images too. The double bond must be on the adjacent carbon.) $\endgroup$ – yolo123 Nov 2 '14 at 2:08
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Note that the starting material is drawn as a chiral molecule. Solvolysis of this starting material will generate a symmetric, achiral, allylic carbocation. Since a stabilized allylic carbocation is generated, the mechanism likely proceeds through an $\ce{S_{N}1}$ mechanism.

Both ends (there is positive charge delocalized at both ends of an allylic carbocation - draw the resonance structures and convince yourself) and both faces (top and bottom) of the allylic carbocation can capture the nucleophilic solvent molecule to restore neutrality as shown below. This results in the formation of two diastereomers of the 3,5-disubstituted cyclohexene - the cis and tran isomers. For each diastereomer, both enantiomers are also produced, as they must be, having passed through an achiral intermediate.

enter image description here

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  • $\begingroup$ Ron, could you please elaborate as to why SN2 is not major? $\endgroup$ – yolo123 Nov 2 '14 at 20:59
  • $\begingroup$ Normally solvolysis of a secondary halide can proceed through SN1 and\or SN2. Stabilizing the incipient ion further, as in this case through conjugation, shifts the balance towards SN1. Even if an SN2 component remains, it would be accompanied by an SN2' pathway as well. So again all 4 compounds would be formed, they just might have an excess of one enantiomer over the other. $\endgroup$ – ron Nov 2 '14 at 21:09

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