Radical reactions and cracking

Radical reactions in the Pantone reactor. PG doctor in oceanography.

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Radical reactions take place following the excitation of an electron of an atom which changes to the singlet state (s2 or s1) then to the triplet state (T1) which is more stable by changing its spin. This electron transmits its energy to other atoms to initiate reactions or returns to its initial state (s0) by re-emitting heat or a phosphorescence photon.

I'm going to call this atom 'S', 3S * when it is excited to the triplet state.

Type I reactions can take place between this S atom and an 'RH' substrate where R = r-CH-CH2-r.

3S * + RH -> S + RH * (direct energy transmission)

3S * + RH -> SH. + R. (tearing of a hydrogen leading to the formation of radicals)

Type II reactions use an intermediate, for example oxygen, which occurs naturally in the form of a .OO di-radical. which becomes singlet oxygen 1O2 *

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3S * + O2 -> S + 1O2 *
1O2 * + RH -> ROOH (hydroperoxide)

From there a series of reactions can take place:

R. + O2 (.OO.) -> ROO.

ROO. + SH. -> ROOH + S
ROO. + ROOH -> RO. + RO.

RO. + SH. -> ROH (alcohol) + S
RO. + RH -> ROH + R.
RO. + O2 -> RO (ketone) + HO2.

RO. –Mac Lafferty type molecular rearrangement–> r-CHO (aldehyde) + r. cracking

RO. + O2 -> r-CO-CH3 (ketone) + r (alkene) + HO2 cracking

ROOH –energy-> RO. + HO.

HO. + HO. -> H2O2 (hydrogen peroxide)
HO. + R. -> ROH (alcohol)

HO2. -> O2 + H.

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RO (ketone) –energy + molecular rearrangement-> r-CO-CH3 (shorter ketone) + r (alkene) cracking

As seen, these reactions are interleaved and a multitude of products can be generated, including ketones, alcohols, aldehydes, alkenes, of identical size or shorter than the starting molecule.

Example with octane (28/09/2005)

I schematize the octane C8H18 in this form H3C-CH2-CH2-CH2-CH2-CH2-CH2-CH3 or H3C- (CH2) 6-CH3.

The molecule is symmetrical so there are 4 possibilities of radical attack:

a) ° H2C- (CH2) 6-CH3
b) H3C- ° CH (CH2) 5-CH3
c) H3C-CH2-°CH-(CH2)4-CH3
d) H3C-(CH2)2-°CH-(CH2)3-CH3

From there we will have training 4 peroxides:

a) ° OOCH2- (CH2) 6-CH3
b) H3C-HCOO ° - (CH2) 5-CH3
c) H3C-CH2-HCOO°-(CH2)4-CH3
d) H3C-(CH2)2-HCOO°-(CH2)3-CH3

Tearing H ° on another molecule, this will cause the corresponding hydroperoxides:

a) HOOCH2- (CH2) 6-CH3
b) H3C-HCOOH- (CH2) 5-CH3
c) H3C-CH2-HCOOH-(CH2)4-CH3
d) H3C-(CH2)2-HCOOH-(CH2)3-CH3

Could lead to a primary alcohol and favored secondary 3 because ls radicals are more stable tertiary groups on secondary than on the primary:

a) HOCH2- (CH2) 6-CH3 (primary alcohol)
b) H3C-HCOH- (CH2) 5-CH3 (secondary alcohol)
c) H3C-CH2-HCOH- (CH2) 4-CH3 (secondary alcohol)
d) H3C- (CH2) 2-HCOH- (CH2) 3-CH3 (secondary alcohol)

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or aldehyde and ketone 3:

a) OCH (CH2) 6-CH3
b) H3C-CO- (CH2) 5-CH3
c) H3C-CH2-CO-(CH2)4-CH3
d) H3C-(CH2)2-CO-(CH2)3-CH3

By molecular rearrangement ketones can lead to shorter molecules:

b) H3C-CO- (CH2) 5-CH3 [C8] -> H3C-CO-CH3 [C3] + HC = CH- (CH2) 2-CH3 [C5]
c) H3C-CH2-CO- (CH2) 4-CH3 [C8] -> H3C-CH2-CO-CH3 [C4] + HC = CH-CH2-CH3 [C4]
d) H3C-(CH2)2-CO-(CH2)3-CH3 [C8] -> H2C=CH2 [C2]+ H3C-CO-(CH2)3-CH3 [C6]
d) H3C-(CH2)2-CO-(CH2)3-CH3 [C8] -> H2C=CH-CH3 [C3]+ H3C-CO-(CH2)2-CH3 [C5]

In short, this cracking leads to molecules from C2 to C6. In addition, the unsaturated molecules will be excited more easily and react better to radical reactions because C = C <=> ° CC °.

This also explains the rearrangement with ketones which are also in the form of enols: -CO-CH2- <=> -HOC = CH-

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