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Explain the mechanism of S_N1 reaction taking 2-bromo-2-methyl propane (t-butyl bromide) |
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Answer» SOLUTION :Tertiary butyl bromide. When treated with aqueous KOH, tertiary butyl alcohol is formed. `(CH_(3))_(3)C-Br+KOH rarr (CH_(3))C-OH+KBr`. The `S_(N)`. REACTION mechanism involves two steps. step (i) : FORMATION of Carbonation. Tertiary butyl bromide undergoes ionization to form tertiary carbo cation of bromide ion. `CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")("C ")-Br overset(" Slow ")rarr CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")("C ")+Br^(-)` This is the slow step is hence is a rate determining step. Step - (ii) : Attack of nuclephile on carbocation. The NUCLEOPHILE `OH^(-)` attacks the positive centre of carbocation from any one of the direction forming the final product i.e., tertiary butyal alcohol. `CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")(C^(+))+OH^(-) overset(" Fast ")rarr CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")("C ")-OH` |
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