20
$\begingroup$

The boiling point of ammonia is −33 °C while that of $\ce{HCN}$ is 25 °C. In a recent AP (Advanced Placement) Chemistry test, a free response question asked why this is the case. Can someone shine light on this?

Based on Jan's answer:

  1. Although the $\ce{C-H}$ bond does not usually exhibit good hydrogen bond, $\ce{H-CN}$ is a special case in which the bond is polar enough to provide better hydrogen bonding than $\ce{NH3}$. I agree with Jan on this point.
  2. However, Jan's explanation is for the polarity of the $\ce{H-CN}$ bond is based on $\mathrm pK_\mathrm a$ (lower $\mathrm pK_\mathrm a$ suggests higher acidity and good hydrogen bonding). As one user pointed out, $\mathrm pK_\mathrm a$ is only defined in aqueous solution, so stating pK$_a$ as an evidence is not suitable for the system I have which is a pure $\ce{HCN}$ liquid. In such system, $\ce{HCN}$ does not significantly dissociate. A suggestion was made on using proton affinity as the evidence for the acidity of $\ce{HCN}$.

I perceive that the answers (reasons and evidence) are mostly targeted at the chemical process of dissociation of $\ce{HCN}$. Should this question be addressed more in terms of a physical process? I mean there is little dissociation of pure-liquid $\ce{HCN}$.

What I have been thinking about is a MO picture in which the cyano group withdraws electron density from the hydrogen atom, in a similar way to how acetate group withdraws electron from the hydrogen in acetic acid. What is your thought on this?

|improve this question|||||
$\endgroup$
  • 1
    $\begingroup$ Your MO attempt is certainly not wrong, but it would only justify a low pKa, or gas phase proton affinity, respectively. It's for one molecule only. You can extend the description to molecular clusters, but treating them with a MO only scheme will certainly give you the wrong answer. $\endgroup$ – Martin - マーチン♦ May 26 '15 at 4:37
  • $\begingroup$ You made a good point about one molecule. I didn't realize this sooner. However, could you tell me what you think about Jan's explanation? $\endgroup$ – Huy Nguyen May 26 '15 at 5:00
  • $\begingroup$ I think it raises an interesting point, but I would not consider it complete. I guess in liquid phase, there are a couple of more equilibria a play in HCN compared to NH3. I really don't like the premise of the whole question, the molecules are not really similar. $\endgroup$ – Martin - マーチン♦ May 27 '15 at 6:51
14
$\begingroup$

The enthalpy of vaporization of $\ce{HCN}$ is higher than for $\ce{NH3}$, which suggests that $\ce{HCN}$ molecules interact more strongly than $\ce{NH3}$ molecules. $\ce{C-H}$ bonds are not usually considered good hydrogen bond donors, but $\ce{HCN}$ is unusual. For example $\ce{HCN}$ has a $\mathrm pK_\mathrm a$ value of 9.2, indicating that the $\ce{CN}$ group is electron withdrawing and that it is a reasonably good hydrogen (bond) donor. This is likely due to the electronegativity of nitrogen, and also the high "s-content" of the sp-hybridized $\ce{CH}$ bond, which keeps the electron pair close to the nucleus.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Thanks for the answer. I think you're correct that C-H of $\ce{HCN}$ is a good hydrogen bond donor. As for the reason, I'd like to think that because the electron withdrawing ability of $\ce{CN-}$ is high, the hydrogen is electronically deficient, leading to an excellent electron acceptor. I would prefer your explanation if we're speaking of a reaction (where $\ce{CN-}$ is formed) $\endgroup$ – Huy Nguyen May 23 '15 at 9:16
  • 1
    $\begingroup$ There is a reasonably good and generally accepted (negative) correlation between hydrogen bond strength and pK$_a$ of the hydrogen donor $\endgroup$ – Jan Jensen May 23 '15 at 12:16
  • $\begingroup$ You're correct on the correlation between pKa and the strength of hydrogen bonding. I, however, think the polarity of $\ce{HCN}$ should be explained by MO theory and unlikely by the stabilization of the conjugate base which doesn't get produced anyway. Your argument is based on thermodynamic stabilization of a dissociation reaction, which occurs in the presence of a base. Note that there is purely $\ce{HCN}$ in the container here. $\endgroup$ – Huy Nguyen May 23 '15 at 16:15
  • $\begingroup$ @Huy If you wanted to have an explanation in terms of MO theory, you probably better include these points (and your thoughts and reasoning) into your question. $\endgroup$ – Martin - マーチン♦ May 25 '15 at 10:28
  • $\begingroup$ Please don't use MathJax like NH$_3$, use the mhchem package instead: $\ce{NH3}$ It might look okay for you, but it might break in other browsers, especially at line breaks. If you want to know more about mhchem, please have a look here and here. Please do not use markup in the title field, see here for details. $\endgroup$ – Martin - マーチン♦ May 27 '15 at 6:31
8
$\begingroup$

Supporting Jan's answer, actually: Consider the points below from the book Hydrogen Bonding: A Theoretical Perspective, p. 102.

"Moreover, the hydrogen in HCN is acidic enough that the molecule may act as an effective proton donor... When paired with NH$_3$, HCN acts as a proton donor..."
This agrees with Jan's note, on the pK$_a$ of HCN.

Your comment that "the hydrogen is electronically deficient, leading to an excellent electron acceptor" seems to be about hydrogen bonding, whereas there is evidence that HCN is more ionic in nature.

|improve this answer|||||
$\endgroup$
7
$\begingroup$

Besides the existing answers, which focus on the acidity of HCN, note that HCN is also a considerably larger molecule than NH3. Thus, even if the interactions between the molecules were qualitatively identical, one would still expect a higher boiling point for HCN on the basis of the size difference (and resulting stronger dispersion interactions; see comments) alone.

For an illustrative example, we can look at methylamine, CH3NH2, which resembles ammonia in most respects, except for having one of its hydrogens replaced by a bulky methyl group, making it similar in size to HCN. Its boiling point is −6.6 °C, well above the −33 °C for NH3.

The remaining ~32 K difference between the boiling points of HCN and CH3NH2 is then presumably explained by the stronger acidity of the HCN hydrogen, and thus the stronger hydrogen bonding between HCN molecules than for NH3 and CH3NH2.

|improve this answer|||||
$\endgroup$
  • 2
    $\begingroup$ @Jan Indeed, those have the same mass, but rather different boiling points. It's a good example, though it doesn't approach my question from the angle I had in mind, so forgive me for not being clear. The reason I'm a bit wary of the mass argument is this: how do we guarantee that, when comparing two molecules with different masses, the change in boiling point is not dominated by the accompanying change in surface area for interactions, rather than the change in mass? Is there a pair of molecules we could compare with different masses, but same polarity and same surface area? $\endgroup$ – Nicolau Saker Neto May 23 '15 at 14:35
  • 1
    $\begingroup$ @NicolauSakerNeto Then let me suggest diethyl ether and d10-diethyl ether, same difference in mass as between ammonia and HCN. The other properties, especially polarity should be very similar. $\endgroup$ – Jan May 23 '15 at 14:42
  • 2
    $\begingroup$ @NicolauSakerNeto And thinking about it more, the mass argument at school was only a proton argument, because more protons means more electrons (in a neutral molecule) which in turn means better polariseability and stronger van der Waals forces (or London interactions, as I’m told they’re called outside of Germany). $\endgroup$ – Jan May 23 '15 at 17:24
  • 2
    $\begingroup$ Just to clarify: gravitational attraction between molecules is much too small to have any measurable effect on the boiling point or any other chemical property. The correlation between molecular mass and boiling point is primarily due to an increase in dispersion interactions (the attractive part of the van der Waals or London forces). This force exist between each electron pair, so more electrons = more dispersion. The effect of isotopic substitution on intermolecular interactions comes mainly from change in vibrational frequencies. The sign is hard predict a priori. $\endgroup$ – Jan Jensen May 24 '15 at 8:20
  • 2
    $\begingroup$ @JanJensen: Naïvely, I would've expected mass to also contribute via the molecular velocity distribution (bigger molecules move slower, and so, intuitively, should have a harder time escaping the solution). But of course, thinking about it a bit more, the kinetic energy distribution is what really matters here, and that's independent of the molecular mass. Thanks, I learned something (or, at least, unlearned a wrong assumption) here. $\endgroup$ – Ilmari Karonen May 24 '15 at 13:07
2
$\begingroup$

Although $\ce{NH3}$ consist of hydrogen bonding $\ce{HCN}$ has a very stronger dipole-dipole interaction which makes its boiling point equivalent to alcohols.

See the diagram:

enter image description here

|improve this answer|||||
$\endgroup$
2
$\begingroup$

"Acidity" can't be responsible, since acidity refers to what happens in water solution. (I assume the user is referring to the $\mathrm{p}K_\mathrm{a}$ data.) I don't know how great the tendency of $\ce{HCN}$ is to ionize in the pure liquid, but I doubt it's significant. But if you have data, please cite it. Even for water, $K_\mathrm{w} = 10^{-14}$, its tendency to ionize does not contribute to its high boiling point.

|improve this answer|||||
$\endgroup$
  • $\begingroup$ Please think of a final answer. Include this answer. So far this is more like a comment. But thank you for saying what I wanted to say but no one tries to argue me wrong $\endgroup$ – Huy Nguyen May 24 '15 at 16:02
  • $\begingroup$ Though $pK_a$ is usually a measure of acidity in aqueous solution, there are measures which are independent of solvent, such as proton affinity. In that table, more positive means more exothermic, which means a stronger base and thus a weaker conjugate acid. Look at a few anions formed from $\ce{C-H}$ ionization, including cyanide, cyclopentadienide and trichloromethanide; cyanide tops them. Note also that the proton affinity of cyanide is almost the same as for phenlote. That is already remarkably acidic for a $\ce{C-H}$ bond $\endgroup$ – Nicolau Saker Neto May 24 '15 at 16:04
  • $\begingroup$ Although Jan's explanation is based on a popularly accepted correlation between acidity and C-H bond's polarity, this is NOT a causation. IMHO, the causation would be the MO picture in which the cyano group withdraws the electron cloud from the hydrogen, causing the bond to be polar and the hydrogen bond to be stronger. If you think I am wrong, please contribute to my knowledge. Otherwise, I will not learn anything from asking this question. $\endgroup$ – Huy Nguyen May 24 '15 at 22:35
  • $\begingroup$ @Huy If you are only stating your knowledge in the comments, I can understand why nobody takes the time to "argue" with you. $\endgroup$ – Martin - マーチン♦ May 25 '15 at 10:38
0
$\begingroup$

I think the crux of the question is not in hydrogen bonding. Recently, I did a question in the USA Chemistry Olympiad (Local Section) on boiling point of substances. I discovered that self-ionisation played a key role in determining boiling point and melting point. The answer to the question I attempted asking for the substance with the highest boiling point was pure sulfuric acid, which is apparently capable of self-ionisation to a large extent.

The reason why self-ionisation is able to raise the boiling point of the substance is because of the ionic interactions in the liquid of the resultant ions produced by the process. This strong ionic interactions increases the boiling point of the liquid, like in the case of sulfuric acid.

The same concept can be applied here. The hydrogen cyanide molecule being capable of self-ionisation, although not to such a significant extent, can produce ions in the liquid phase. This ionic interaction, although it may be very litte compared to sulfuric acid, does increase the boiling point of the substance.

In comparison, ammonia's self-ionisation is probably negligible and thus, the strongest interactions are only hydrogen bonding.

This post is a bit late but I hope this provides a new perspective.

|improve this answer|||||
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.