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Can a specific bond be broken by an engineered energy wave corresponding to the bond length?

For example, can we break the $\ce{C-N}$ bond and not the $\ce{C-H}$ in a hydrocarbon chain?

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    $\begingroup$ Arguably, the entire realm of chemistry is nothing but selective bond breaking. So yes, this can be done, though the light with specially selected frequency is a relatively rare tool to use. $\endgroup$ – Ivan Neretin May 7 '17 at 14:59
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    $\begingroup$ @IvanNeretin ...and selective bond-making. $\endgroup$ – stochastic13 May 7 '17 at 18:40
  • $\begingroup$ @SatwikPasani Yeah, right, that too. $\endgroup$ – Ivan Neretin May 7 '17 at 18:45
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In general, to break one bond vs another on demand in the presence of many others is a very difficult thing to achieve. Not possible at all by heating molecules thus there has been much effort to do so using lasers.

But this becomes very difficult also. This is because on exciting a molecule into an excited electronic state, such that it has enough energy to break a bond, and into the type of vibration involving the bond you wish to break, the energy then leaks out to other different vibrational normal modes and can then end up anywhere. So it does not just stay where the energy was initially placed.

This 'leaking' occurs due to vibrational anharmonicity which provides a 'coupling' with which one normal mode can swap energy with another. The process can be very fast and even if femtosecond laser pulses are used energy still moves all over the place and competes with accumulating energy into the 'victim' bond effectively defeating the process. The area is, however, one of very active study so progress in this area is to be anticipated.

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  • $\begingroup$ Thank you Porphyrin, do you happen to know which institutes lead the efforts or are active in these studies? $\endgroup$ – Mishal May 21 '17 at 16:36
  • $\begingroup$ Look for chemistry groups that use femtosecond laser spectroscopy and study 'coherent control'. There are also theoretical groups that study this. I'm sure that you will find several quite easily. $\endgroup$ – porphyrin May 21 '17 at 18:21

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