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Topic: Question about light absorption and homolysis mechanism.  (Read 644 times)

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Offline jhonn4460

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Question about light absorption and homolysis mechanism.
« on: June 25, 2021, 12:52:12 PM »
In this mechanism, the nonbonding electrons of the aryl azosulfone absorb a  photon  of visible light and are promoted to an antibonding pi orbital (n-π* transition), followed by homolysis of the adjacent sigma bonds. What i don't get is how exactly this promotion of electrons to antibonding pi orbitals causes the homolysis of adjacent sigma bonds. Could you point me to literature about this subject and/or explain how this works ?

Thank you.   

Offline Orcio_Dojek

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Re: Question about light absorption and homolysis mechanism.
« Reply #1 on: June 26, 2021, 05:06:30 AM »
Sulfone compounds are colourless, so absorption comes rather from -N=N- group.

Maybe this is similiar to the Sandmeyer reaction (formation of [CH3-C6H4-N≡N]+ and [SO2CH3]- and then single electron transfer).

Offline jhonn4460

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Re: Question about light absorption and homolysis mechanism.
« Reply #2 on: June 26, 2021, 11:17:25 AM »
Sulfone compounds are colourless, so absorption comes rather from -N=N- group.

Maybe this is similiar to the Sandmeyer reaction (formation of [CH3-C6H4-N≡N]+ and [SO2CH3]- and then single electron transfer).

They do absorb at this wavelenght, this  reaction is done under blue LED light. The radicals formed further react with dimethyl disulfide to yield aryl methyl sulfides as the desired product. This reaction is described in the article "Visible Light-Promoted Formation of C–B and C–S Bonds under
Metal- and Photocatalyst-Free Conditions" published on the Synthesis journal volume 51 pages 1243 - 1252 DOI:
10.1055/s-0037-1611648

Offline Corribus

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Re: Question about light absorption and homolysis mechanism.
« Reply #3 on: June 26, 2021, 06:52:30 PM »
Azo compounds may typically be colorless, but in this case the azo group, and probably the sulfate group, is conjugated to the aryl pi system. This narrows the benzyl pi-pi* energy gap, which pushes it closer to the visible region. More importantly, unlike, say, a vinyl substituent, the nitrogens that make up the azo group introduce non-bonding orbitals into the electronic system. This gives rise to very low-lying n->pi* transitions, which although weakly absorbing certainly extend into to visible region. These states could also be accessed with higher energy excitation through internal conversion of the coupled vibronic states.

Photochemistry following excitation of conjugated systems is usually complicated and discerning mechanisms via inspection can be difficult. Some insight can usually be gained by trying to figuring out where electron density lies in the photoexcited molecule. Excitation of n-pi* transitions (or relaxation to n-pi* states after excitation of higher lying states) undoubtedly creates a highly polarized system where electrons are shifted from the nonbonding nitrogen orbitals into the aryl-centric pi-system. Even so, some positions on the benzyl ring will probably be more electron dense than others, so theoretical treatments may help to understand where electron rich or deficient nuclei are, which can provide a window into what's going on with the photochemistry. As a first step, you may do some reading on the photoisomerization of the related model compound azobenzene, which I understand is still a matter of some debate.

https://en.wikipedia.org/wiki/Azobenzene

Keep in mind that the production of nitrogen gas in the reaction provides a thermodynamic driving force for photodissociation, as does the relatively weak N-S single bond. The n-->pi* states also hang around quite a bit longer than pi-->pi* states because their relaxation to the ground is symmetry forbidden (this is also why they absorption is weak). Meaning, there is more opportunity for photochemistry than in exclusively pi-pi* systems, where relaxation is orders of magnitude faster.

Hope that helps.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline jhonn4460

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Re: Question about light absorption and homolysis mechanism.
« Reply #4 on: June 26, 2021, 07:44:29 PM »
Azo compounds may typically be colorless, but in this case the azo group, and probably the sulfate group, is conjugated to the aryl pi system. This narrows the benzyl pi-pi* energy gap, which pushes it closer to the visible region. More importantly, unlike, say, a vinyl substituent, the nitrogens that make up the azo group introduce non-bonding orbitals into the electronic system. This gives rise to very low-lying n->pi* transitions, which although weakly absorbing certainly extend into to visible region. These states could also be accessed with higher energy excitation through internal conversion of the coupled vibronic states.

Photochemistry following excitation of conjugated systems is usually complicated and discerning mechanisms via inspection can be difficult. Some insight can usually be gained by trying to figuring out where electron density lies in the photoexcited molecule. Excitation of n-pi* transitions (or relaxation to n-pi* states after excitation of higher lying states) undoubtedly creates a highly polarized system where electrons are shifted from the nonbonding nitrogen orbitals into the aryl-centric pi-system. Even so, some positions on the benzyl ring will probably be more electron dense than others, so theoretical treatments may help to understand where electron rich or deficient nuclei are, which can provide a window into what's going on with the photochemistry. As a first step, you may do some reading on the photoisomerization of the related model compound azobenzene, which I understand is still a matter of some debate.

https://en.wikipedia.org/wiki/Azobenzene

Keep in mind that the production of nitrogen gas in the reaction provides a thermodynamic driving force for photodissociation, as does the relatively weak N-S single bond. The n-->pi* states also hang around quite a bit longer than pi-->pi* states because their relaxation to the ground is symmetry forbidden (this is also why they absorption is weak). Meaning, there is more opportunity for photochemistry than in exclusively pi-pi* systems, where relaxation is orders of magnitude faster.

Hope that helps.

This is very helpful. Thank you very much !

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