Question

What occurs when two equiv of ammonia react with `"2-iodobutane"`?

Answer 1

From the context of the question, nucleophilic substitution occurs....

Explanation:

And so, we could write the stoichiometric equation as...

`H_3Cstackrel("*")CH(I)CH_2CH_3+2NH_3rarrH_3Cstackrel("*")CH(NH_3)CH_2CH_3+NH_4^(+)I^(-)darr`.

The starred carbon is chiral....were we to begin with a homochiral reactant, we would get a product with the OPPOSITE handedness given the stereochemical requirements of the `S_N2` reaction.

The second equivalent of ammonia in the reaction acts as the base to remove the one equivalent of `HI` formed....Because ammonia is a bit of a pfaff to use stoichiometrically, we could add ONE equivalent of `NH_4Cl` and ADD TWO equivalents of non-nucleophilic reagent, i.e. say `Et_3N`...we would then formulate the reaction as....

`H_3Cstackrel("*")CH(I)CH_2CH_3+NH_4Cl(s) + 2Et_3NrarrH_3Cstackrel("*")CH(NH_3)CH_2CH_3+Et_3NH^(+)Cl^(-)darr+Et_3NH^(+)I^(-)darr`

The added triethylamine, (i) deprotonates the ammonium salt, and then (ii) deprotonates the quaternized ammonia after it reacts with the carbon.... Do you follow?