Question

Model 3: Reactions of Bromine Radical with Various Alkanes For each reaction below, the most likely products are shown. H H H H Rxn I : Br -H H H H H H H H Br H Rxn IV C- H H CHy сн, H,C нC Rxn V BrH н.с CH -сн Critical Thinking Questions 5. Label the carbon radicals in Model 3 as methyl. primary, secondary, tertiary, allyl or benzyl. a) Which carbon radical in Model 3 is closest to having an octet? Explain your reasoning. (Recall that alkyl groups are electron donating.) The following is a list of radicals from highest to lowest potential encr this consistent with your conclusion in part a)? Explain H CH highest V.E. 30 or allyl benzyl x (halogen) Jowest V.E c) Which reaction in Model 3 is most likely? Explain 6. Atld curved half arrows to Rxn III in Model 3 so as to illustrate the mechanism of product formation. a) As in Model 2, draw the other sigma bond forming reaction that could have taken place in Rxn III, but did not. (Note: all 6 H's of ethane are equivalent.) b) Construct an explanation for why Rxn II in Model 3 happens instead of the reaction you drew in part a).

Answer 1

Since, there is no image of model 2, I couldn't answer the questions.

H 4 - •c- colt. t-u i-c-c- H- 4 J - I-J-I He -e - CH₂ - . 나 나 I Primary Freee readical scendatly I free radical) Tertiary [frice radical tree radical @ Terefiatey readical is closest to having an octet Reason? Alkyl groups such as methyl arte electron donating groups, so the elections will stay close to the free readical. Rxn & in model 3 is most likely. The carbon free readical formed in the reaction is a tertiary teadical. Tertiary readicals are more stable than primary and secondary. As free radicals. are eleefton deficient and it will be stable if more electron donating groups are attached to it, especicelly alkyl groups.