What are possible product obtained in the following reaction? (CH3)3−C−Br80%EtOH→20%H2O.
A
(CH3)2=CH2 via E1
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B
(CH3)3−C−OH via SN1
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C
(CH3)3C−O−Et via SN1
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D
(CH3)2=CH2 via E2
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Solution
The correct option is C(CH3)2=CH2 via E2 In Tertiary Alkyl Halides E2 reactions are favored over SN2 reactions. The greater the alkyl substitution, the faster the reaction, since in the Transiton stage, a double bond is formed partially. A greater substituted alkene is lower in energy. Hence the activating energy is reduced, making the reaction faster.