B. 18] Given the following trend in bond strengths, how would you expect the sel

Question

B. 18] Given the following trend in bond strengths, how would you expect the selectivity of radical oxygenations to compare to the analogous chlorination and bromination reactions? Include a reaction coordinate diagram to support your answer. Bond Dissociation Epergies for Homolytic Cleavage (kcal/mol) H-OH H-cl H-Br Allylic C-OH Allylic C-H 119 103 HO-OH Cl- c Br-Br C-cl C-Br 51 t of hiattan 91 87 4 80 68 Rodical exigenetions wouid be tes setrctive thon brominahos bult husel ner e gerective than cao nnotions because its asociated DDB tndt wil) more like be analegows to cdleinatlon ducjs Ran cosdisgte

Explanation / Answer

Higher BDE represents that bond is so strong. Bond formation always favors towards more stable compounds consequently their bond dissociation energy also higher. Hence radical oxygenation is more selective than chlorination or bromination. In addition oxygen is more electronegative element than bromine and chlorine, this means it is highly capable of holding an electron (radical) in its valence shell. Moreover compare to Br & Cl, oxygen atom is a bit smaller in size also, this accounts for its higher reactivity in radical oxygenation.

Remember Carbocation & radical stability: Tertiary > secondary > primary

Carbanion stability: Tertiary < secondary < primary

the order of halogen reactivity towards radical mechanism is F2 > Cl2 > Br2 > I2.

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