Radical reactions in the Pantone reactor. PG doctor in oceanography.
Radical reactions take place following the excitation of an electron from an atom passing the singlet state (or s2 s1) and the triplet state (T1) more stable by changing spin. This electron transmits its energy to other atoms to initiate reactions or returns to its initial state (s0) retransmitting heat or photon phosphorescence.
I'll call 'S' that atom 3S * when excited triplet state.
Type I reactions can take place between the S atom and a substrate HR where R = r-r-CH-CH2.
3S + RH -> S * + RH (direct transmission of energy)
3S + RH -> SH. + A. (abstraction of a hydrogen leading to the formation of radicals)
Type II reactions use an intermediary, eg oxygen, which occurs naturally in the form of di-radical .OO. which becomes singlet oxygen 1O2 *
3S O2 + -> + S * 1O2
1O2 * + RH -> ROOH (hydroperoxide)
From there a series of reactions can take place:
R. + O2 (.OO.) -> ROO.
ROO. + HS. -> ROOH + S
ROO. + ROOH -> RO. + RO.
RO. + HS. -> ROH (alcohol) + S
RO. + RH -> ROH + R.
RO. + O2 -> RO (ketone) + HO2.
RO. -réarrangement molecular types Mac Lafferty-> r-CHO (aldehyde) + r. cracking
RO. + O2 -> R-CO-CH3 (ketone) + r (alkene) + HO2 cracking
ROOH -énergie-> RO. + HO.
HO. + HO. -> H2O2 (hydrogen peroxide)
HO. + R -> ROH (alcohol)
HO2. -> O2 + H.
RO (ketone) -Energy + rearrangement molecularly> r-CO-CH3 (ketone shorter) + 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'm simplifying the C8H18 octane in this form H3C-CH2-CH2-CH2-CH2-CH2-CH2-CH3 or H3C- (CH2) 6-CH3.
The molecule is symmetrical so there 4 opportunities radical attack:
a) ° H2C- (CH2) 6-CH3
b) H3C- ° CH (CH2) 5-CH3
From there we will have training 4 peroxides:
a) ° OOCH2- (CH2) 6-CH3
b) H3C-HCOO ° - (CH2) 5-CH3
Tearing H ° on another molecule, this will cause the corresponding hydroperoxides:
a) HOOCH2- (CH2) 6-CH3
b) H3C-HCOOH- (CH2) 5-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)
or aldehyde and ketone 3:
a) OCH (CH2) 6-CH3
b) H3C-CO- (CH2) 5-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 it leads to cracking molecules C2 to C6. More unsaturated molecules will be excited more easily and better respond to radical reactions since C = C ° ° CC.
This also explains the rearrangement with ketones which are also as enols: CO-CH2- -hoc = CH