[Mack]

Hi everyone, I am having a little trouble with deciding on which way this reaction will go. 2-iodo-2-methylpropane with NaN₃ in the solvent of DMF. I believe that no reaction will occur. Since it is tertiary no Sn2 and since NaN₃ is a decent nucleophile and the solvent is protic I figured that removes the Sn1 and E1. I figured E2 would also not occur either because NaN₃ is a better nucleophile than base. I was wondering if this logic was correct, thanks for any help

[antimatter101]

There should be some tertiary carbocation formed for the reaction to proceed. What is the concentration of H+?

[Altered State]

Hi everyone, I am having a little trouble with deciding on which way this reaction will go. 2-iodo-2-methylpropane with NaN₃ in the solvent of DMF. I believe that no reaction will occur. Since it is tertiary no Sn2 and since NaN₃ is a decent nucleophile and the solvent is protic I figured that removes the Sn1 and E1. I figured E2 would also not occur either because NaN₃ is a better nucleophile than base. I was wondering if this logic was correct, thanks for any help

As you said, Sn2 won't occur because your alkyl halide it is tertiary.

But you are wrong about unimolecular processes... Protic solvents do favor Sn1 or E1 reactions, they help stabilizing the charge of the ions that are formed. But keep in mind that this is not a very important factor in chosing a mechanism, generally both Sn1 and Sn2 can occur in both protic or aprotic solvents.
And obviously the fact that you have a strong nucleophile won't favor nothing to occur in your mixture, but totally the opposite.

You should know that if you have a good leaving group and a good nucleophile, reaction will occur in most cases, the facts to take in account you probably were given (http://jamesash.wpengine.netdna-cdn.com/wp-content/uploads/2012/08/compare-sn1sn2.png) are just to choose which mechanism adapts better to your reaction conditions.

Take a look: http://www.masterorganicchemistry.com/2012/08/08/comparing-the-sn1-and-sn2-reactions/

[Mack]

That is my bad I had a typo DMF is an aprotic solvent...so my problem still stands, I understand that it is the reaction that better adapts yet with so much conflicting its hard to say, but is why I considered E2??

[Altered State]

Like I said solvent characteristics are not determinant.

You have a tertiary alkyl halide, it will be Sn1 or E1 for sure.
You have strong nucleophile and no strong base, it will be Sn1.

[hinhthoi]

Actually there is not much clear cut on whether you should choose SN1 or E1 since both mechanism actually occur. The choice are different based on the condition of the reaction. Academic argument may prefer to choose SN1 because N3- is a better nucleophile than base. But there is no other information related to Pressure and temperature of the reaction. These 2 factors indeed affect the ratio of SN1/E1. Higher temperature and lower pressure prefer E1.

[orgopete]

Even though sodium azide is not a strong base, I still predict an E2 elimination. Hydrazoic acid has about the same acidity as acetic acid, so the result should be the same as treating with sodium acetate. Since the iodide is tertiary, an SN2 reaction will not occur due to the higher electron density donated by the methyl groups. That electron donation lengthen the C-I bond also. Lengthening the C-I bond increases the hyperconjugation of the C-H bonds (which are all equal). If they were not equal, then the more substituted would contribute more (more electron rich). This normally gives a Zaitsev product. In this case, they are all the same, so I predict an E2 elimination.

Because sodium azide is not as basic, I expect the reaction to be slower than a similar reaction in which a stronger base were present. None the less, I expect all other factors to be similarly applicable.

Now, since everything I've said is simply conjecture, I could be wrong. To answer this, it would be helpful if someone could look up this reaction (or similar) and post the results.