I came across this question in my test:

This compound

enter image description here

is treated with $\ce{CH_3CH_2I}$ and then with $\ce{H_2SO_5}$ and heated. The product is:

enter image description here

B) $\ce{CH_2CH_2}$
and two other impossible options.

I first don't understand the role of $\ce{H_2SO_5}$ in this. I know about Hoffmann elimination of quaternary ammonium hydroxides but in that we use silver oxide to form the hydroxide from the iodide. And then heating is done for elimination.

Secondly the answer given is option B but I am not able to understand why is that. The second example in this page says otherwise Quaternary ammonium salts
Does this have anything to do with the $\ce{H_2SO_5}$?

My third doubt is a general one. What happens in general when different alkyl groups are attached to the the nitrogen in a quaternary ammonium hydroxide. Which one undergoes elimination first? Are there any rules to determine that?

  • $\begingroup$ @NilayGhosh How is that question related to this one $\endgroup$ Commented Jan 9, 2019 at 3:49
  • $\begingroup$ I think H2SO5 is used to oxidize Iodide so it could go away for sure and cleanly. R3NI might sublimate, but (R3N)(HSO4) likely cannot, and HI is a very strong acid unlikely to be displaced. In classical example Ag2O is used which isn't cheap and the products are distilled off. In this option hydrosulfate remains in reaction flask, which might be useful. $\endgroup$
    – permeakra
    Commented Jan 9, 2019 at 4:47
  • $\begingroup$ As for elimination, apparently Hoffman elimination favors products eliminating hydrogen from less stericaly hindered position. Usually it is used to judge the product from same radical as Hoffman elimination is usually used with trimetialalkyl salts, but there is no reason to not apply it between radicals as well. $\endgroup$
    – permeakra
    Commented Jan 9, 2019 at 4:49
  • $\begingroup$ The first question is how much EtI is the starting material treated with? If it is only 1eq then I think the product is the tertiary and the persulfate is oxidising that to the N-oxide then eliminating by a Cope elimination $\endgroup$
    – Waylander
    Commented Jan 9, 2019 at 7:48
  • $\begingroup$ @Waylander it is in excess $\endgroup$ Commented Jan 9, 2019 at 7:51

1 Answer 1


I believe this is an example of a Cope elimination mechanism here

The EtI is making the tertiary amine. The persulfate is taking the tertiary amine to the N-oxide. Heating this produces ethene and hydroxylamine of the starting amine.

This is consistent with the Cope as "The sterically demanding amine oxide function reacts preferentially with the more easily accessible hydrogens"

enter image description here


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