Does the tetrahedral structure of the alkanes contribute to their lower reactivity? I thought that because a tetrahedral structure suggests ${sp^3}$ hybridization, it should contribute to its lower reactivity. Is this right?

  • $\begingroup$ The lower reactivity is attributed to the nonpolar C-H bonds that are present in alkanes. They are very stable and typically do not want to react... unless you do some exotic organometallic chemistry. That is a whole another topic, though. $\endgroup$
    Commented Nov 21, 2016 at 9:28
  • $\begingroup$ @TOLA3HPPA can you give me example of "exotic" organometallic please? I'm inquisitive :) $\endgroup$
    – ParaH2
    Commented Dec 22, 2016 at 22:02

2 Answers 2


Does the tetrahedral structure of the alkanes contribute to their lower reactivity?

Hybridization is at the center of the answer.

Due to poor P-P overlap, it only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about a double bond involving ${sp^2}$ hybridized carbons, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond involving ${sp^3}$ hybridized carbon atoms with highly directional orbitals that result in strong bonds. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane. Alternatively one can view ethene as a two-membered ring (ref. 1), no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring. It is easy to imagine that such a ring system would contain a significant amount of strain (ref. 2). Either way (and by the way these two views of ethene are equivalent and complementary) - using a poorer overlap argument or a higher strain energy argument, we see why alkenes (and even more so alkynes) would generally be more reactive than alkanes.

From this argument we would expect that if we were to somehow increase the strain in an alkane, then we should increase its reactivity. A good example of such behavior can be found in cyclopropane. Placing 3 "tetrahedral" carbon atoms into a ring dramatically increases the strain energy in the system (it changes the carbon hybridization too, see ref. 3) and the reactivity increases dramatically. Indeed, cyclopropane adds bromine much like an alkene (ref. 3)

Since someone above commented that ${sp^3}$ $\ce{C-H}$ bonds are "very stable and typically do not want to react". Let me just point out that $\ce{C-H}$ bond energies vary as follows

\begin{array}{|c|c|c|c|} \hline \ce{C-H} & \text{Bond Strength} \\ \text{bond type} & \text{(kcal/mol)} \\ \hline \ce{sp^3 C-H} & 101 \\ \hline \ce{sp^2 C-H} & 111 \\ \hline \ce{sp C-H} & 133 \\ \hline \end{array}

These bond dissociation energies (ref. 4) don't fit the observed reactivity pattern. They indicate that an ${sp^3}$ bond is actually easier to break than an ${sp^2}$ or ${sp}$ $\ce{C-H}$ bond


The carbon-hydrogen bonds are only very slightly polar; therefore, there are no portions of the molecules that carry any significant amount of positive or negative charge that can attract other molecules or ions. The carbon-carbon bonds are not polar at all and so these do not contribute to any charges on the molecule at all.

To my knowledge, the relative unreactivity of alkanes compared to other hydrocarbons such as alkenes or alkynes is due to the high amount of saturation across the carbon atoms.

I have heard little of the hybridization of the carbon atoms contributing to the reactivity if the alkanes. I was under the impression that the trends in reactivity were as I have described above.

  • $\begingroup$ You might want to amend your post to address the OP's query, especially regarding (a) tetrahedral geometry and; (b) hybridization in such geometry, both in the context of reactivity. $\endgroup$ Commented Nov 22, 2016 at 21:05
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    $\begingroup$ Thank you for the comment, I have updated the post. Was my mistake, I neglected to directly address the question by accident. Many thanks. $\endgroup$ Commented Nov 22, 2016 at 21:14

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