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

Question

The hydrolysis of optically active 2-bromobutane with aqueous NaOH result in the formation of (A) (+)-butan-2-ol (B) (-)-butan-2-ol (C) (±)-butan-1-ol (D) (±)-butan-2-ol

Tardigrade Admin · Accepted Answer

Trick In case of organic reactions, to differentiate between first order and second order kinetics, you should check the stability of carbocation. If carbocation is stable, then it will follow first order kinetics otherwise second order kinetics. Since, 2

^{\circ}

- carbocation is stable so given reaction follows first order kinetics.

Usually 1

^{\circ}

carbocation follows second order kinetics and 2

^{\circ}

or 3

^{\circ}

carbocation follows first order kinetics.

The reaction between 2-bromobutane with aqueous NaOH follows first order kinetics, i.e. the rate of the reaction depends upon the concentration of 2- bromobutane.

(S_{N} 1

mechanism)

Since, the 2-bromobutane is optically active than the product is a racemic mixture. This is because carbocations are as intermediates in

S_{N} 1

reactions Since, carbocation being

s p^{2}

-hybridised is planar (achiral) species, the attack of the nucleophile on it can occur from both the faces (front and rear) with almost equal case giving a 50:50 mixture of two enantiomers.

So, 2 bromobutane gives racemic of

(\pm)

butan-2-ol.

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

$(+)$ -butan- $2$ -ol

$(-)$ -butan- $2$ -ol

$(\pm)$ -butan- $1$ -ol

$(\pm)$ -butan- $2$ -ol

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

(+)-butan-2-ol

(−)-butan-2-ol

(±)-butan-1-ol

(±)-butan-2-ol

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

$(+)$ -butan- $2$ -ol

$(-)$ -butan- $2$ -ol

$(\pm)$ -butan- $1$ -ol

$(\pm)$ -butan- $2$ -ol