Chem210-1 F07 Ex3 p03a

Chem210-1 F07 Ex3 p03a - d Provide a step-by-step mechanism...

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4. When the compound 2-methyl propene is dissolved in methanol in the presence of fluoboric acid a new compound, "A", is formed. Compound "A" has a higher molecular weight than the starting alkene, and it has no fluorine in its structure. If ethanol is used as the solvent instead of methanol, a different product is observed. C CH 2 H 3 C H 3 C HBF 4 CH 3 OH compound "A" a) What is the structure of compound "A"? (3 points) b) Provide on the right the full Lewis structure (all atoms, bonds, lone pairs, formal charges) for the anion portion of fluoboric acid, HBF 4 - the counterion that's left when HBF 4 dissociates (3 pts) c) Why would HBF 4 be used here rather than some more common proton acid like HBr? (3 pts)
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Unformatted text preview: d) Provide a step-by-step mechanism for the formation of "A" from 2-methylpropene (6 points) O C CH 3 CH 3 CH 3 H 3 C F B F F F The anionic counterion, BF4-, cannot readily act as a nucelophile (neither B nor F can accomodate an additional bond), thus, only methanol is likely to attack the intermediate carbocation in the reaction. If, for example, H-Br had been used to protonate the alkene at least some of the Br-anions could compete with methanol for reaction with the carbocation. C CH 2 H 3 C H 3 C H BF 4 C CH 3 H 3 C H 3 C H O CH 3 H 3 C C CH 3 O CH 3 H H 3 C H 3 C C CH 3 O CH 3 H 3 C...
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