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The example I am talking about is $\ce{CH2CHCH2^.}$.

To find the steric number and hence the hybridization, we use steric number = No. of lone pairs + number of sigma bonds. In this example, the last C atom has only one electron. Should we count it as a lone pair?

My teacher says that that electron will get delocalised so its hybridization will be $\mathrm{sp^2}$ and not $\mathrm{sp^3}$ (which I think is true). I didn't understand the delocalisation stuff. Is it true that here, there will be delocalisation and if not then what is happening here? If delocalisation is occurring here then please explain it too! Also please tell me in which cases is there delocalisation of electron(s)?

My teacher also said that had it been $\ce{CH3CH2CH2^.}$, then there would be no delocalisation of electrons. Why does this happen?

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  • $\begingroup$ The allyl radical has a resonance structure because the lone electron can combine with the pi bonding electron on the center carbon to make a double bond and leave the first carbon with a lone electron. Based on your question, you are implying that your teacher claims that the third carbon in the propyl radical a) does not have a delocalized electron AND b) therefore it will be $sp^3$ hybridized. I don't understand that so I won't post an answer, $\endgroup$ – Joseph Hirsch Dec 13 '16 at 16:17
  • $\begingroup$ but this may help: google.com/… $\endgroup$ – Joseph Hirsch Dec 13 '16 at 16:17
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Radicals are electron-deficient compounds. You should not treat a radical as a lone pair, as they are pretty different.

A good rule of thumb is to assume an orbital that contains the highest possible p-contribution for the single electron. The more s-contribution is mixed in, the ‘closer’ the single electron gets to the nucleus which is less favourable than having an electron pair in a similar orbital. A pure p-orbital fulfils the condition perfectly by having no s-contribution whatsoever. Thus, you should assume radicals to be in p-orbitals whenever possible.

This is similar to carbocations which have an orbital fully devoid of electrons. However, this empty orbital must always be a p-type orbital for electronic reasons. Radicals have the possibility of undergoing hybridisation and e.g. ending up in an $\mathrm{sp}^n$ orbital (examples: radicals on $\mathrm{sp^3}$ hybridised bridgehead atoms; phenyl radical).

Since we just established that the radical will occupy a p-type orbital and since there is a π bond nearby, we can delocalise. In resonance structure terms, this can be visualised in the following way:

$$\ce{H2C=CH-CH2^. <-> H2C^.-CH=CH2}$$

The radical is delocalised across two carbon atoms making the overall system more stable than if it weren’t. In molecular orbital terms, this is basically a stabilisation of the double bond at the expense of the radical-containing orbital which benefits the overall system. The delocalisation can, of course, only occur if there is a π bond nearby; the n-propyl radical does not feature one.

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  • $\begingroup$ Please examine my interpretation of paragraph 2. The lone electron in the radical goes into the p-orbital because it is less stable to have a single electron in an orbital with s-character as it is 'worse' to have single electrons close to the nucleus than to have electron pairs close to the nucleus? So can you clarify this example? In the methyl radical, is the geometry trigonal planar because it is more stable to have the lone electron in a pure p-orbital leaving open only 3$sp^2s$ for bonding, or does the stability of the t.p. geometry drive sp^2 hybridization? $\endgroup$ – Joseph Hirsch Dec 14 '16 at 2:29
  • $\begingroup$ @JosephHirsch The methyl radical should be $\mathrm{sp^2}$ in the absence of constraints, yes. While not instantly visible, radicals isomerise my a mechanism similar to amines, providing support for the view. $\endgroup$ – Jan Dec 14 '16 at 2:40
  • $\begingroup$ I understand it is $sp^2$. My question is, is it because the loan electron is more stable in a pure p-orbital or is it because the $sp^2$ allows a more stable geometry than $sp^3$? $\endgroup$ – Joseph Hirsch Dec 14 '16 at 2:47
  • $\begingroup$ @JosephHirsch The former. Geometries only play a role if they restrict a more stable electronic situation. E.g. geometry is the reason why water does not have $90^\circ$ angles (it would be too strained). $\endgroup$ – Jan Dec 14 '16 at 2:50
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I see the question has having the following parts:

1) Does the allyl radical have a delocalized electron?

2) Is this what makes it $sp^2$ instead of $sp^3$?

3) How does an unpaired electron factor into hybridization/how do you determine the hybridization state?

4) How can you tell if an electron is delocalized in a radical?

5) Why isn't the single electron delocalized in the propyl radical

6) Since the propyl radical does not have a delocalized electron, will the radical carbon be $sp^3$ hybridized

Answer 1) Yes enter image description here

2) No, the delocalization of the electron is not the determining factor in the hybridization of the carbon atoms in the allyl radical.

3) The unpaired electron is not equivalent to an electron pair for purposes of determining hybridization state. I believe that the hybridization state is driven by the steric stability of atoms, electron pairs and lone electrons around the hybridized atom, however the effects diminish as you move from attached atoms of different elements-->lone pairs-->unpaired electrons-->delocalized electrons. In the case of the allyl or propyl radical, the effects of a lone electron are not enough to favor $sp^3$ hybridization whether it is localized or delocalized. Therefore, in these molecules, you determine the hybridization state by the number of sigma bonds, but in molecules with lone pairs, the lone pairs may or may not alter the hybridization state depending on the stability gained and lost. Orbitals in Chlorine atoms in $\ce{Cl2}$ do not hybridize, but I believe that Chlorine is considered to be $sp^3$ hybridized in some compounds if the steric interactions of the electron pairs with the attached atom are great enough. Counting electron pairs and sigma bonds therefore is not a rule, but a guideline and ultimately it comes down to stability.

4) I think I would give a simple rule that as long as you can draw a Lewis diagram where a central carbon is attached to one atom with a lone electron and another atom by a double bond, the single electron will be delocalized. What you are doing is moving electrons ONLY by making and breaking Pi bonds.

5) Because the central carbon does not have a double bond that it can break to avail itself of an unpaired electron that can pair with the unpaired electron on the radical carbon.

6) No, in this case at least, neither a delocalized electron or unpaired electron will result in the $sp^3$ hybridization being more stable and energetically favorable. Now someone said that the propyl anion is still $sp^2$ but that is not what I am used to seeing. Usually carbanions are shown as being $sp^3$. The debate, I suspect will be whether it is $sp^3$ because the electron pair has enough effect to make the stability gained by a tetrahedral geometry pay off (my view) or if it is merely the function of non-geometry related "electronic factors".

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  • $\begingroup$ @jJan Sorry, the first comment is not obsolete as I did state that the Chlorine atom in HCl may be SP3 hybridized. It clearly is not. The correction is helpful and I am going to edit my answer to correct it. $\endgroup$ – Joseph Hirsch Dec 14 '16 at 1:52
  • $\begingroup$ The only compounds in which I would consider chlorine hybridised would be chloronium ions. $\endgroup$ – Jan Dec 14 '16 at 2:07

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