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A small amount of natural carbon is $\ce{^{14}C}$. This is radioactive and decays to $\ce{^{14}N}$. This is exploited in radiocarbon dating.

This $\ce{^{14}C}$ is taken up by living organisms. I know that natural biochemical processes can be sensitive to isotopes but I expect that the $\ce{^{14}C}$ can occur in any context where $\ce{^{12}C}$ could occur. (Please correct me if I am wrong here.)

So, $\ce{^{14}C}$ might occur as part of the backbone of a fatty acid:

$$. . . \ce{-CH2-{^{14}C}H2-CH2 -} . . .$$

When the $\ce{^{14}C}$ decays, this will become:

$$. . . \ce{-CH2-NH-CH2 -} . . .$$

(I assume one of the hydrogens would be lost.)

Are any of these unusual compounds stable or do they promptly react in some way?

Are they studied?

Do they have any interesting properties?

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  • $\begingroup$ I wonder if it can decay in-place i.e. in the middle of a compound the way you have depicted, without breaking the chain itself. The decay process is generally accompanied by a loss of heat (due to mass-defect) so the extra energy must manifest itself as kinetic energy of the daughter nuclei or heat, breaking the chemical bonds either way. $\endgroup$ – Gaurang Tandon Feb 11 '18 at 15:43
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    $\begingroup$ @GaurangTandon Will the result immediately break down: that's my question. I am reasonably sure that the $\ce{^{14}C}$ does not know or care that it is inside a fatty acid and hence will be liable to decay. I had in mind the $\ce{^{18}F}$-FDG used in PET scans though that is the reverse: an unusual molecule becomes a common one when the $\ce{^{18}F}$ decays to $\ce{^{18}O}$. $\endgroup$ – badjohn Feb 11 '18 at 16:24
  • $\begingroup$ Related: What happens to a radioactive carbon dioxide molecule when its carbon-14 atom decays? $\endgroup$ – user7951 Feb 11 '18 at 16:44
  • $\begingroup$ @Loong Thanks. "The decay energy of 14C is unusually low" but, nonetheless, it seems likely that the bonds will be broken. However, at least according to en.wikipedia.org/wiki/Fludeoxyglucose_(18F), the decay of $\ce{^{18}F}$ in $\ce{^{18}F}$-FDG does not break the molecule apart. $\endgroup$ – badjohn Feb 11 '18 at 17:00
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    $\begingroup$ @Karl So, this part is probably incorrect: "After 18F-FDG decays radioactively, however, its 2-fluorine is converted to 18O−, and after picking up a proton H+ from a hydronium ion in its aqueous environment, the molecule becomes glucose-6-phosphate labeled with harmless nonradioactive "heavy oxygen" in the hydroxyl at the C-2 position." $\endgroup$ – badjohn Feb 11 '18 at 17:13