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I can understand that $\mathrm{S_N2}$ reactions prefer primary carbons, because of the steric hindrance that would be an obstacle.

But what is the exact reason that $\mathrm{S_N1}$ reactions prefer tertiary carbons ?

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The intermediate in an SN1 reaction is a carbocation. The stability of that intermediate determines how favorable that particular reaction is, so a more stable intermediate means a more favorable reaction.

Carbocations are quite unstable on their own, but the inductive effect of nearby C-C bonds provide some electron density to balance out the positive charge, making the carbocation intermediate more stable. Tertiary carbons have the largest number of adjacent C-C bonds, the largest inductive effect, the most stable carbocation intermediate, and are thus favored in SN1.

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  • $\begingroup$ Because this is about reaction rate, you should refer to transition state energies for the most correct response. $\endgroup$ – Zhe Oct 12 '17 at 16:43
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Some answers are already addressing the key issues, but I see some important missing elements. While cation stability is important, we're talking about reaction rates, so we should be using kinetic, rather than thermodynamic arguments. Here, I will invoke the Hammond postulate to convert a thermodynamic argument into a kinetic one.

First, we establish that the $S_{\mathrm{N}}1$ mechanism proceeds via an intermediate cation. This intermediate is, importantly, higher in energy than the reactants.

Second, the first step of the $S_{\mathrm{N}}1$ reaction is the slow step and therefore (given the first fact), also the rate determining step.

By the Hammond postulate, in an endothermic reaction, the transition state will be structurally and energetically similar to the reaction product. In this specific case, the transition state will resemble the cation. This means that we can expect the transition state for the rate determining step to be strongly affected by factors that can stabilize the intermediate cation.

Generally, we speak about hyperconjugative effects as the factors that most influence the stability of the cation. Specifically, with increasing substitution on the cation, adjacent $\ce{C-H}$ bonds are able to donate some of their electron density into the empty orbital of the cation, helping to stabilize it.

Therefore, $S_{\mathrm{N}}1$ reactions will be fastest for species that generate a tertiary cation, because those reactions go through cation-like transitions states that are lower in energy.

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SN1 reaction occurs in two steps.In the first step there occurs a slow cleavage in the polarised C-X bond to produce a carbocation and a halide ion.Generally SN1 reactions are carried out in polar protic solvents.The energy required for breaking the C-X bond is obtained through solvation of halide ion with the proton of the solvent.Tertiary alkyl halides undergo SN1 reaction very fast because of the high stability of tertiary carbocations.

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    $\begingroup$ This makes zero sense and is completely wrong. $\endgroup$ – orthocresol Jan 19 '17 at 8:27

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