a) What would the product of the reaction be for an E2 mechanism

b) What would the products be for an E1 mechanism

enter image description here

The Br coming out of the page is the 79 isotope while the Br going into the page is the 81 isotope.

There is no mention of reaction conditions.

I am not totally sure what the effect of the isotope will have on bromine's ability to leave. I am guessing that the bromine-81 will leave since it is bigger and can stabilise the negative charge better.

I am also not sure how the products for the E1 and E2 mechanism will be different. Won't they be the same?

  • 7
    $\begingroup$ By any reasonable chemical definition of "size" the isotopes should be the same, so that can't be it. The first thing to think of when looking at isotopes is that the heavier isotope makes slightly stronger covalent bonds due to reduced zero-point energy of bond vibrations. For such high atomic masses, though, the difference must be extremely slight, so I don't know if that's the right path to go down. $\endgroup$ Feb 23 '16 at 9:45
  • $\begingroup$ @NicolauSakerNeto Thanks for that. I wasn't aware of the fact that heavier isotopes made stronger covalent bonds. This might be the way to go because I can't think of anything else that will effect its leaving ability. $\endgroup$
    – Nanoputian
    Feb 23 '16 at 11:11
  • 2
    $\begingroup$ This is commonly called a "kinetic isotope effect". The wikipedia article is decent.. $\endgroup$
    – Lighthart
    Feb 23 '16 at 14:45
  • 8
    $\begingroup$ I have a feeling that the isotopes are only there to distinguish between the two bromine atoms. Although there is an isotope effect, with the high mass bromine the reactivity difference will be negligible. Think about the difference in mechanism between E1 and E2. I would also guess that the stereochemistry is given to you, and that will play a critical role in an E2 reaction. If you provide the full structure (and reaction conditions, if possible) I can give a less speculative answer. $\endgroup$
    – jerepierre
    Feb 23 '16 at 17:34
  • 4
    $\begingroup$ By way of comparison, the kinetic isotope effect for the E2 elimination of $\ce{HCl}$ from (1-chloropropan-2-yl)benzene is only $k(\ce{^35Cl}):k(\ce{^37Cl}) = 1.0059$. The relative mass change for $\ce{^79Br}/\ce{^81Br}$ is even smaller. $\endgroup$
    – user7951
    Feb 23 '16 at 21:40

Under E1 conditions (typically hot, polar protic solvent), one of the bromines will leave as bromide to give a carbocation. Ignoring the isotope effect, there's about an equal propensity for either of the isotopic bromines to leave, giving about a 50/50 ratio of the two carbocations. Deprotonation at the beta-position gives an alkene. I've only shown the most substituted possible alkene since that will be the most stable and major product.

(Note, the carbocation could rearrange, but I'm ignoring that for this level of analysis.)

enter image description here

Under more strongly basic conditions an E2 reaction is more likely. There is a strong preference for an anti-periplanar arrangement of the leaving group, carbons where the alkene will form, and the deprotonated hydrogen. This stereochemical requirement is more easily visualized in the chair forms, which are not equivalent. The structure on the left is more stable since the methyl group is placed equatorially. In this conformation, there are two hydrogens that are anti-periplanar with the bromine-81 atom (green). A bulky base would react with the least hindered hydrogen (purple) and eliminate the bromine-81 to give the disubstituted alkene shown. A smaller base would be more likely to produce the most stable product, which results from deprotonating the hydrogen in blue.

Just for completeness I've shown the elimination product of the less stable chair conformation. In this case, there is only one anti-periplanar arrangement. If the orange hydrogen is deprotonated, then the bromine-79 atom can be eliminated to give the cyclohexene shown.

enter image description here

  • 1
    $\begingroup$ Thanks for the great answer. So basically, for elimination reactions in general, if the atom is heavy, different isotopes won't effect the end product. The end product will just depend on its stereochemistry, right ? While for smaller molecules the lighter isotope is more likely to leave. Am I right? $\endgroup$
    – Nanoputian
    Feb 24 '16 at 10:22
  • 2
    $\begingroup$ @Nanoputian The kinetic isotope effect is related to the ratio in the masses of the isotopes. The only atom that I would worry about difference in product distribution would be hydrogen. If isotopes are specified for heavy atoms, they are more useful as tags to determine mechanism. $\endgroup$
    – jerepierre
    Feb 24 '16 at 17:18

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