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Let's say in a reaction, unimolecular (SN1), we have $\ce{CH3-CH(CH3)-CH2+}$ carbocation, and a nucleophile $\ce{Nu-}$. Generally the carbocation is first rearranged and then there is attack of $\ce{Nu-}$, but can't it be that a very strong nucleophile, strongly attracted towards the cation, attacks the cation befor it rearrange to give kinetically controlled product (KCP) (at least at lower temperatures) ?

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You are asking what would happen if Something like 1-bromo-2-methylpropane were to undergo an $\mathrm{S_N1}$ reaction. The simple answer is: Your premise won’t happen.

The first step in an $\mathrm{S_N1}$ reaction is the formation of a carbocation by the leaving group leaving. A general equation for this would be:

$$\ce{R3C-Br <=> R3C+ + Br-}$$

Where this reaction has its equilibrium and thus what concentration of the carbocation can be expected, depends on the relative stabilities of reactant and carbocation (the bromide is always the same so it doesn’t matter). Now primary carbocations are so unstable (in a thermodynamic sense) that thermodynamics predicts an equilibrium far to the left for primary haloalkanes:

$$\ce{RCH2-Br <<=> RCH2+ + Br-}$$

(This reaction arrow doesn’t do it justice. It should be a humungous left-pointing arrow and a minute right-pointing one.)

For tertiary or otherwise stabilised carbocations, we can expect a significant concentration of the right-hand side of the equation at equilibrium. Thus, we have a potential carbocation concentration that can potentially react with another nucleophile in an $\mathrm{S_N1}$ reaction. But primary haloalkanes will not even allow the carbocation to form, or in a different view, will pull back the bromide to regenerate the haloalkane before any nucleophile has a chance of attacking the carbocation.

That’s why primary haloalkanes predominantly react via the $\mathrm{S_N2}$ mechanism: There simply isn’t any significant amount of carbocation present for $\mathrm{S_N1}$ or any type of Wagner-Meerwein rearrangement.

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I think you're missing the point.

Consider the SN1 mechanism shown in Wikipedia. The overall reaction is:

$\ce{(CH3)_3C-Br + 2H2O -> (CH3)_3C-OH + H3O^+ + Br^-}$

There is no way that 3 molecules are going to simultaneous collide and make this reaction work. The "SN1 reaction" breaks this particular type of reaction into its intermediate steps.

If the nucleophile did some other sort of reaction, then it would be some other reaction mechanism, not a SN1 reaction. For instance there is a SN2 mechanism as well.

So chemistry isn't just about doing overall mass balanced equations. It is trying to show you how you got from point A to point B.

Multiple Reactions

It is possible that multiple reactions can happen. Chemists can change the reaction conditions, or use a catalyst, to change the kinetics (relative reaction rates) to favor a particular reaction.

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  • $\begingroup$ Alright you seem to have gained a friend here since I came here from the VLQ review. Note to self: Another instance of unjustified flagging $\endgroup$ – M.A.R. Nov 6 '15 at 20:27
  • $\begingroup$ @MaxW what I am asking is that in SN1 after the leaving group separate out, and the nucleophile is about to attack the cation, with a possibility of rearrangement (of cation), can the nucleophile attack the cation befor it rearranges ? $\endgroup$ – Chinmay Chandak Nov 7 '15 at 3:14
  • $\begingroup$ Look at this reaction. en.wikipedia.org/wiki/SN1_reaction#Stereochemistry Assuming 4 different groups, is what you want an "optically active" product? $\endgroup$ – MaxW Nov 7 '15 at 3:24
  • $\begingroup$ @MaxW i am not talking about optical activity. I am asking that can CH3−CH(CH3)−CH2-Nu be a product of SN1 reaction. (Nu = nucleophile ) $\endgroup$ – Chinmay Chandak Nov 7 '15 at 3:36
  • $\begingroup$ Let's assume the carbon reacting has four different groups so that it is stereoactive. The "rearrangement" of the carbocation results in a "planar" carbocation which can be attacked from either side. This would make racemic mixture. // If the carbocation kept its pyramid form then the product would be stereoactive too. Isn't that what you are getting at? This would be a special case of a SN1 reaction. // If this were a SN2 reaction then the product would be the opposite of the original optically active reactant. $\endgroup$ – MaxW Nov 7 '15 at 3:46

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