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Which of the following will be more basic, hydrazine $\ce{H2N-NH2}$ or ammonia $\ce{NH3}$?

My chemistry teacher said that $\ce{NH3}$ is more basic since after giving $\ce{H+}$ to hydrazine results in $\ce{H3N+-NH2}$, and accommodation of the lone pair of $\ce{-NH2}$ is not possible by $\ce{-N+H3}$, since it doesn't have the space. So instead, $\ce{-NH2}$ will pull electrons from it making it unstable. Ammonia has no such problem so it must be more basic.

I am not so pleased with this argument. In $\ce{H3N+-NH2}$, although the lone pair cannot be accommodated, but the positive charge present on its sides , to an extent, should neutralize the intensity of the lone pair, making it somewhat stable. And also, not to forget, hydrazine has two spots where we can get the electrons, therefore, its ambident nature should also support it's basicity.

I need a bit of clarity on this.

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Obligatory link to Evans' pKa table.

Ammonia is more basic than hydrazine, by about one order of magnitude. This is expected, because the -NH2 group is more electronegative than -H or -CH3.

The second lone pair is not involved in the acid-base reaction, it does not point towards the -NH4+ group. You can, however, force two lone pairs into close proximity. This destabilizes the unprotonated form. The keyword is "proton sponge".

1,8-Bis(dimethylamino)naphthalene has a pKa of 12.3, it's one of the strongest known amine bases. Compare that to the pKa of aniline, which is something like 4.5.

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    $\begingroup$ Ok, I get yours and my teachers point, but please elaborate on why I am wrong. $\endgroup$ – Rohinb97 Oct 2 '14 at 7:46
  • $\begingroup$ How exactly were enol pKas determined? $\endgroup$ – Dissenter Feb 7 '15 at 0:08
  • $\begingroup$ What about the alpha effect? Will that not enhance the basicity of hydrazine? $\endgroup$ – Swedish Architect Feb 11 '17 at 19:05
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I think these could be the reason-

For the first part,

...positive charge present on its sides , to an extent, should neutralize the intensity of the lone pair, making it somewhat stable

This is not possible because $\ce{NH_3^+}$(no vacant orbital) doesn't have any space at all to get involved with the lone pair.

...and the second part

...hydrazine has two spots where we can get the electrons, therefore, its ambident nature should also support it's basicity.

Abel already answered that at one time only one $\ce{-NH_2}$ takes part when we determine basicity and the second $\ce{-NH_2}$ plays no role.

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  • $\begingroup$ To clarify the first part, I am not saying that the electrons will jump to the protonated nitrogen. I'm saying that the presence of a positive charge near the electrons will try to reduce its intensity and make it somewhat stable. For the second point you made, more number of nucleophilic sites would mean more chances of attack of an $H^+$, which adds to the basicity of Hydrazine. $\endgroup$ – Rohinb97 Apr 1 '15 at 8:46
  • $\begingroup$ You shouldn't compare the basicity of Hydrazine as a molecule. Just because it has two basic sites, it will not be more basic. It is akin to saying that just because Sulphuric acid has two acidic hydrogens, it is a stronger acid than Perchloro-acid, which is untrue. Remember, in any case, there will be only ONE protonation at a time. You should compare either Ka1, or Ka2, with the corresponding values for Ammonia. $\endgroup$ – Lexicon Apr 1 '15 at 14:17
  • $\begingroup$ I'm just saying that the probability of attack, and did not mean that it decreases it's $pK_b$ value. $\endgroup$ – Rohinb97 Apr 1 '15 at 23:13

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