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I encountered this question which asks how many possible radicals are formed when $\ce{CH3CH2C(CH3)3}$ is monosubstituted by $\ce{Br2}$. The answer given is 3 while I think it should be 4, reasoning that there is a "chiral" carbon if the radical is formed on carbon 3. While arguing with my tutor two questions came to my mind:

  1. What is the geometry of the $\ce{CH3\dot{\ce{C}}HC(CH3)3}$ radical?
  2. If the radical is trigonal pyramidal, does it transition quickly between the two possible states? If so are they considered two radicals?

My findings so far:

This answer on SE hints that the exact shape depends on the substituents on the alkyl radical.

Modern Physical Organic Chemistry states that all other (non-methyl) localized radicals are not planar.

Organic Reaction Mechanisms states that the geometry of free radicals is still controversial.

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First off, it was very astute of you to recognize that if the radical formed at carbon 3 (the methylene carbon in the starting compound) is pyramidal, then the radical would be chiral ( the lone radical electron serving as the 4th different substituent on that carbon) and 4 different radicals would be possible.

In a typical carbocation, the carbon bearing the positive charge is $\ce{sp^2}$ hybridized, meaning that the p-orbital at the cationic center is empty and the carbocation is planar. On the other hand, in a typical carbanion, the negatively charged carbon is pyramidal and approaches $\ce{sp^3}$ hybridization. This $\ce{sp^3}$-like orbital contains 2 electrons. Therefore, it would be reasonable to expect that a typical hydrocarbon radical, where the orbital holds 1 electron would be somewhere in between these two extremes and be slightly pyramidal.

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This is indeed the case and the degree of non-planarity is very sensitive to the substituents attached to the radical center. In the case of hydrocarbon radicals, such as those created from your starting material, the barrier to inversion from one pyramidal form to the other is extremely low, on the order of a few kcal/mole or so. Therefore, interconversion between the two enantiomeric forms of the radical would be extremely rapid at room temperature.

So yes, the $\ce{CH_3\dot{C}HC(CH3)3}$ radical is slightly pyramidal and it does transition rapidly between the two pyramidal enantiomers. Your answer of 4 radicals is correct, as long as you understand that 2 of them rapidly interconvert at room temperature.

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  • $\begingroup$ Indeed it should be less pyramidal than the carbanion. Chiral hydrogen-bearing carbon atoms adjacent to C=O groups will racemise slowly in basic solution through the carbanion intermediate (how much time they actually spend in the carbanion form will depend a lot on pH). One would expect racemization of the radical to be considerably more rapid than that. $\endgroup$ – Level River St Aug 12 '15 at 17:15
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    $\begingroup$ Be careful! It is not reasonable at all to assume that the alkyle radical is "something in between a carbanion and a carbocation". A quick look into the OC textbooks tells you that the methyl radical is in fact planar. The tBu radical on the other hand is pyramidal, but for steric(!) reasons, with an inversion barrier of a mere 1.2 kcal/mol. For the given example,the barrier is probably less, IF the thing is not in fact planar. And unless you have evidence for the contrary, i would assume the latter, and say "4" is wrong. $\endgroup$ – Karl Aug 12 '15 at 17:18
  • $\begingroup$ @KarlRatzsch Why is it not reasonable to assume that the alkyl radical is something in between a carbanion and a carbocation? I would agree that the barrier to planarity is very low in these cases. So low in fact that experiment (or theoretical calculations) cannot distinguish between planarity and a very small deviation towards pyramidal. OC textbooks are of little help in this case. Bottom line, there is an electron that can be stabilized by mixing in some s-character which will cause (slight) pyramidalization... $\endgroup$ – ron Aug 12 '15 at 18:14
  • $\begingroup$ @KarlRatzsch ...The 3 $\ce{C-H}$ bonds will give give up some s-character in order that the lone electron can exist in a lower energy orbital containing some s-character. $\endgroup$ – ron Aug 12 '15 at 18:15
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    $\begingroup$ I would also take some care with the categorising the radical in between a cation and an anion - the geometry certainly will be different from both cases. The radical has ungerade spin, while the others have gerade spin, which can often lead to significant deviations. However in computational chemistry, as a good practise kind of thing, it is often advised to calculate the singlet cation of the radical as a guess for the radical itself, since in most cases the behaviour is quite similar (no rule though). $\endgroup$ – Martin - マーチン Aug 12 '15 at 19:36
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As your books clearly indicate, nobody seems to have been able to find a chemical proof for the geometry of C-radicals. If it's not indeed planar, it equilibrates extremely quickly between the two pyramidal shapes. Unless there is some further steric constraint that keeps it in one conformation.

For the argument with your tutor: The correct answer is "two diffent primary radicals and one secondary radical". If someone asks for a number, and you have the slightest doubt, write two, three senctences explaining it. If you're lucky, the tutor might say "extra point for the one guy who noticed the question was stupid". If not, he'll still happily give you 8 out of ten. And if you're at a school where this is not the case: Get the hell out of there.

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Let's rephrase the question. How many different radicals can be formed from 2,2-dimethylbutane? Answer = 3 OR How many monobromo stereoisomers can be formed from 2,2-dimethylbutane? Answer = 4. Two primary bromides and a racemic secondary bromide.

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