0
$\begingroup$

The book, Solomons' Organic Chemistry (for JEE Mains and Advance), contains the following question:

Hydrogen fluoride has a dipole moment of $\pu{1.82 D}$; its boiling point is $\pu{19.34 ^{\circ} C}$. Ethyl fluoride ($\ce{CH3CH2F}$) has an almost identical dipole moment and has a larger molecular weight, yet its boiling point is $\pu{-37.7 ^{\circ} C}$. Explain.

Now while explaining the (very) concept of boiling point and why it differs for different substances the book considers:

  1. Dipole-dipole forces

  2. London dispersion forces

  3. Molecular weight

But when I tried to explain the phenomenon stated in the question, I found it hard to explain the lower melting point. Following is my approach on the explanation:

  1. Since both have the same (similar) dipoles therefore there would be very negligible difference in their boiling points.

  2. London dispersion forces increase in magnitude as the surface area of the molecule increases. So contrary to what was said $\ce{CH3CH2F}$ should have stronger dispersion force interaction (due to larger surface area) and hence higher boiling point as compared to $\ce{HF}$ (given other properties were same).

  3. Given that $\ce{CH3CH2F}$ has higher mass than $\ce{HF}$, therefore it should lead to higher boiling point (as is stated in the question itself).

Now it can be clearly seen that the above stated arguments are against the phenomenon, so I believe that I am missing something in my explanation.

So

  • What is (if any) wrong/missing in my explanation?

  • Why is the boiling point of $\ce{CH3CH2F}$ lower than that of $\ce{HF}$?

$\endgroup$
  • $\begingroup$ Hint: Recheck how the dipole moment is defined. $\endgroup$ – Karl Mar 25 at 8:46
  • $\begingroup$ @Karl the said book only talked about the surface area while explaining such phenomenon. It says "and because the surface areas of larger molecules can be much greater, intermolecular dispersion attractions can also be much larger". $\endgroup$ – Johan Liebert Mar 25 at 8:55
  • 1
    $\begingroup$ they are in aqueous condition, then hydrogen bonding can be one of the reason in HF both hydrogen and fluorine are able to form hydrogen bond, where in ethyl fluoride only F which can form intermolecular bond. $\endgroup$ – Yuvraj Singh... Mar 25 at 9:23
5
$\begingroup$

The three reasons that you present are not sufficient to explain the high boiling point of pure $\ce{HF}$. There is a fourth reason, that you have not mentioned : the Hydrogen bond. Molecules $\ce{HF}$ are interacting with one another through $\ce{H}$-bonds, exactly like $\ce{H_2O}$ molecules attract one another because of $\ce{H}$-bonds. In the liquid state, $\ce{HF}$ behaves as if it is made of heavier molecules, something like $\ce{(HF)_n}$, with $\ce{n}$ being between about $4$ and $7$. The boiling point of $\ce{HF}$ and of $\ce{H_2O}$ is extremely high when comparing to substances of nearly the same molar weight, like $\ce{CH_4}$ or $\ce{Ne}$. The boiling point of $\ce{H_2O}$ and $\ce{HF}$ does not depend significantly on the van der Wals forces. $\ce{H}$-bonds are much more important than van der Wals forces.

And there are practically no $\ce{H}$-bonds in ethyl fluoride, because $\ce{H}$-bonds happens only when an $\ce{H}$ atom is attached to a strongly electronegative atom.

|improve this answer|||||
$\endgroup$
  • 3
    $\begingroup$ The last sentence is of course completely and utterly wrong. Just look up <CH-hydrogen bond>; but these are at least an order of magnitude weaker. Therefore they hardly play a role in the comparison of OP. $\endgroup$ – Martin - マーチン Mar 25 at 10:43
  • $\begingroup$ @ Martin. Please Martin ! Keep cool ! I perfectly know that some extremely weak H-bonds exist with C-H hydrogen bonds, and even between a lot of other X-H bonds. They can even be measured. But they are so weak that they do not have any visible effect on the boiling point. $\endgroup$ – Maurice Mar 25 at 17:29
  • 1
    $\begingroup$ Maurice, a technical note, adding a space between @ and the username like you did will not notify the person you're pinging. Just type @user and not @ user. $\endgroup$ – Guru Vishnu Mar 26 at 14:53

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.