Skip to main content
Chemistry LibreTexts

4.5.2: Other Polar Bonds

  • Page ID
    52314
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    We have seen that when hydrogen is covalently bonded to oxygen, nitrogen, or fluorine, the result is that the covalent bond is highly polarized and the majority of the electron density is located on the most electronegative atom. This means that the hydrogen atom has very little electron density remaining around it. Because hydrogen is such a small atom, the resulting positive charge density on the hydrogen atom is high. This leads to unusually strong attractions (H-bonds) with atoms that have lone pairs with which the positively charged hydrogen atom can interact. H-bonding is unique to molecules in which a hydrogen atom is covalently bonded to an oxygen, nitrogen, or fluorine atom. However, there are uneven charge distributions possible whenever two atoms with different electronegativities form a bond. Consider, for example, methanol (CH3OH). It has several different types of bonds with different distributions of charge in them. The familiar O–H bond in methanol is very much like the O–H bond found in water. That is, it is highly polarized and the hydrogen atom is a small, dense region of highly positive charge that can attract and will be attracted to regions of high electron density such as the lone pairs on oxygen. The methanol molecule also has a C–O bond and three C–H bonds. If we consider the differences in electronegativity we can predict the polarization of these bonds. Remember that carbon and hydrogen have quite similar electronegativities, and so the C–H bond is not very polarized. Carbon and oxygen, in contrast, are quite different in their electronegativities and the result is that the C–O bond is strongly polarized, with the δ+ located on the carbon atom and the negative end of the bond dipole on the oxygen atom. As we will see later this has implications for how methanol (and all C–O containing compounds) interact (and react) with other substances.

    An inspection of the Lewis structure can reveal (to the trained mind!) a huge amount about the structure and polarity of a molecule and taking that one step further we can make predictions about the properties of the compound. For example if we compare the relative boiling points of methanol (CH3OH, bp 65 °C) and ethane (CH3CH3, bp -88.6 °C) we see (just as you already predicted no doubt) that methanol has a much higher boiling point because it takes more energy to separate molecules of methanol. The question arises: is this because methanol can form an H-bond with itself? Can you draw a picture of how this happens? Or is it because of the C–O dipole? We can look at this idea a little more closely by comparing the boiling points of three compounds that have similar molecular weights (so that they experience similar London dispersion forces), but different types of bonds in them.

    If we classify the kinds of bonds as before we see that dimethyl ether has non-polar C–H bonds and polar C–O bonds. The C–O–C bond angle is about 104°. Because each atom (except for H) is surrounded by four centers of electron density, the molecule is not linear as pictured. (Why not?) The molecule as whole is polar but cannot form hydrogen bonds with itself because none of the hydrogen atoms have a significant δ+ as they would if they were bonded to an oxygen atom. We call the type of forces between dimethyl ether molecules, dipole–dipole forces. On the other hand, an ethanol molecule—which has exactly the same molecular weight and formula—can form hydrogen bonds with itself because it has an O–H bond, and so has a small partially positively charged hydrogen atom. This minor difference has a huge effect on boiling point: ethanol boils at 78 °C whereas dimethyl ether boils at –23 °C. Both of them are considerably higher than propane at -44 °C (remembering that absolute zero is –273.15 °C). From comparing these three similar compounds we can see that a simple dipole–dipole attraction increases the boiling point by 21 °C, and on top of that the H-bonding attraction in ethanol is worth another 99 °C, bringing the boiling point of ethanol to 78 °C.


    4.5.2: Other Polar Bonds is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?