The hydrolysis of optically active 2-bromobutane with aqueous $N a O H$ result in the formation of :

(+) butan-2-ol

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(-) butan-2-ol

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(

\pm

) butan-1-ol

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(

\pm

) butan-2-ol

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Solution

The correct option is A ( $\pm$ ) butan-2-ol
The hydrolysis of optically active $2 - b r o m o b u t a n e$ with aqueous $N a O H$ will result in the formation of both $(+)$ and $(-)$ $b u t a n - 2 - o l$ as the reaction will proceed via $S N^{1}$ mechanism.

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Similar questions

Arrange the following compounds in increasing order of boiling point. Propan - 1 - ol, butan - 1 - ol, butan - 2 - ol, pentan - 1 - ol
(a) Propan-1-ol, butan-2-ol, butan - 1 - ol, pentan-1-o|
(b) Propan-1-ol, butan - 1 - ol, butan-2-ol, pentan-1-ol
(c) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol
(d) Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol

Q. Arrange the following compounds in increasing order of boiling point.

Propan - 1 - ol, butan - 1 -ol, butan - 2- ol , pentan -1 -ol

Q. Arrange the following compounds in increasing order of boiling points . Propan-1-ol , butan-1-ol , butan-2-ol, pentan-1-ol.

Q. Butan-2-ol and pent-2-ol are:

Q. Butan-2-ol and pentan-2-ol are:

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

The correct option is A (±) butan-2-olThe hydrolysis of optically active 2−bromobutane with aqueous NaOH will result in the formation of both (+) and (−) butan−2−ol as the reaction will proceed via SN1 mechanism.

The hydrolysis of optically active 2-bromobutane with aqueous $N a O H$ result in the formation of :

The correct option is A ( $\pm$ ) butan-2-ol
The hydrolysis of optically active $2 - b r o m o b u t a n e$ with aqueous $N a O H$ will result in the formation of both $(+)$ and $(-)$ $b u t a n - 2 - o l$ as the reaction will proceed via $S N^{1}$ mechanism.