1.

The hydrolysis of optically active 2- bromobutane with aqueous NaOH result in the formation of

Answer»

`(pm)` butan - 2- ol
`(pm)` butan - 1- ol
(-) butan - 2- ol
(+) butan - 2- ol

Solution :Usually `1^@` carbocation followes second order kinetics and `2^@and 3^@` carbocation follows first order kinetics.
The reaction between 2- tromobutonewith equal NaOH follows first order kinetic, i.e.,the rate of reaction depends upon the concentration of 2-brombutane (`S_N 1` MECHANISM).

Since, the 2-bromobutane is OPTICALLY active than the produt is a receive mixture. This is because carbocations are as intermadiate in `S_N1` REACTIONS. Since, carbonation being `Sp^2`-HYBRIDIZED is planar species, the attack of the nucleotide on it can occur from both the face with almost case giving a 50 : 50 mixture of two enantiomes.


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