Organic Chemistry II, Please help! Free Radical Halogenation Problem Free-radica

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Organic Chemistry II, Please help!

Free Radical Halogenation Problem Free-radical halogenation reactions allow scientists to convert alkanes into haloalkanes. The great utility of this reaction stems from the fact that alkanes are extremely unreactive; they are nonpolar, nonacidic, and nonbasic. Haloalkanes, on the other hand, are relatively reactive; they undergo substitution and elimination reactions. Typically, free-radical halogenations are performed with the alkane as the solvent in the presence of molecular halogen 2) and light (hv) as the initiator. All of the mechanistic steps of the reaction can be classified as either initiation, propagation, or temination steps. Another reagent that can be used to affect alkane chlorination by a free-radical mechanism is t- butylhypochlorite. Since the O-CI bond is very weak, this reaction need not be run in the presence of light but instead can be themally activated Reaction 1 40C 65% 25% 10% . Which of the following most likely represents the initiation step of free-radical halogenation by t- butylhypochlorite? CI +(CHCH

Explanation / Answer

Solution :-

Q1) The answer is option C because the O-Cl bond is weak bons so it cleaves thermally to give the two radicals that is the Cl radical and other radical with oxygen having radical.

Q2) Answer is option C because the radical formed at the alpha carbon of the ring substituent is stabilized by the resonance of the pi electrons of the benzene ring so the answer is option C.

Q3) The answer is option C that is the racemic mixture can be formed because the radical formed can be give the retension of configuration or inversion of configuration so the product will have the 50 : 50 R and S conformer so it will be racemic mixture.

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