What might have been the product ratios observed in this experiment if an \"apro

Question

What might have been the product ratios observed in this experiment if an "aprotic solvent" had been used instead of water in the following reaction CH3CH2CH2CH2-OH + H2SO4 --> CH3CH2CH2CH2-OH2(+) Cl(-) + Br(-) --> CH3CH2CH2CH2-Cl + CH3CH2CH2CH2-Br + H2O I understand that water is polar and forms a compound with the intermediate which slows down the reaction and an aprotic solvent is less polar, but would someone expand on this?

Explanation / Answer

H2SO4 protonates the alcohol. Another alcohol attacks and displaces H2O, giving protonated ether. Ether loses proton. CH3CH2CH2CH2OH + H2SO4 ===> CH3CH2CH2CH2-OH2(+) ===> CH3CH2CH=CH2 + H2O H2SO4 protonates the alcohol. Electrons in C-H bond adjacent to CH2-OH2(+) shift over to displace H2O, yielding olefin.

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