Computer Science, asked by mhetresanchita3, 6 months ago

explain the mechanical of alkaline hydrolysis reaction with aqueous KOH of tert- butyl bromide

Answers

Answered by masterrrrrrr
1

Answer:

The alkaline hydrolysis to tert-butyl bromide with aqueous alkali such as NaOH or KOH is as follows.

bromide

tert−butyl

​  

 

(CH  

3

​  

)  

3

​  

C

​  

−Br+NaOH(aq)→  

alcohol

tert−butyl

​  

 

(CH  

3

​  

)  

3

​  

C

​  

−OH+NaBr.

The rate of this reaction depends only on the concentration of the tert-butyl bromide and is independent of the concentration of alkali added.

Rate α[(CH  

3

​  

)  

3

​  

C−Br]

Rate =K[(CH  

3

​  

)  

3

​  

C−Br]

This is a first order reaction because rate of hydrolysis of (CH  

3

​  

)  

3

​  

−Br is independent of the concentration of alkali or OH  

 ions. This can be explained by two-step mechanism shown below. Each step is an elementary reaction with its own rate constant, step 1 proceeds much more slowly than step 2.

Step 1. (CH  

3

​  

)  

3

​  

C−Br  

k  

1

​  

 

​  

 

slow

​  

(CH  

3

​  

)C  

+

+Br  

 

Rate of reaction =k  

1

​  

[(CH  

3

​  

)  

3

​  

C−Br]

The first step consists of breaking of C-Br bond and it determines the rate of overall reaction. So, step 1 is called the rate-determining step. The rate determining step in this reaction involves only a single molecule, therefore, it is said to be unimolecular. Also, this type of mechanism is known as SN  

1

 mechanism(substitution, nucleophilic, unimolecular).

Step 2. (CH  

3

​  

)  

3

​  

C  

+

+OH  

 

k  

2

​  

 

​  

 

fast

​  

(CH  

3

​  

)C−OH

Rate of reaction=k  

2

​  

[(CH  

3

​  

)  

3

​  

C  

+

][OH  

]

The second step involves the attack of OH  

 ion. This is the fast step, since it is the bond formation step.

Energy profile diagram of SN  

1

 mechanism shows that rate of a reaction is independent of the concentration of nucleophile. The first step requires larger activation energy (ΔE  

1

​  

) than the second step (ΔE  

2

​  

). The first step to form carbocation determines the rate of overall reaction. The second step, which is the attack of nucleophile on carbocation is exothermic i.e., it is a lower energy transition state. The intermediate carbocation appears at a low point in the diagram. The conditions and reagents which favour the formation of carbocation will accelerate the SN  

1

 reaction. The energy difference between products and reactants is ΔH, i.e., Heat of reaction.

Explanation:

Answered by pds39937
1

Explanation:

The alkaline hydrolysis to tert-butyl bromide with aqueous alkali such as NaOH or KOH is as follows.

bromide

tert−butyl

(CH

3

)

3

C

−Br+NaOH(aq)→

alcohol

tert−butyl

(CH

3

)

3

C

−OH+NaBr.

The rate of this reaction depends only on the concentration of the tert-butyl bromide and is independent of the concentration of alkali added.

Rate α[(CH

3

)

3

C−Br]

Rate =K[(CH

3

)

3

C−Br]

This is a first order reaction because rate of hydrolysis of (CH

3

)

3

−Br is independent of the concentration of alkali or OH

ions. This can be explained by two-step mechanism shown below. Each step is an elementary reaction with its own rate constant, step 1 proceeds much more slowly than step 2.

Step 1. (CH

3

)

3

C−Br

k

1

slow

(CH

3

)C

+

+Br

Rate of reaction =k

1

[(CH

3

)

3

C−Br]

The first step consists of breaking of C-Br bond and it determines the rate of overall reaction. So, step 1 is called the rate-determining step. The rate determining step in this reaction involves only a single molecule, therefore, it is said to be unimolecular. Also, this type of mechanism is known as SN

1

mechanism(substitution, nucleophilic, unimolecular).

Step 2. (CH

3

)

3

C

+

+OH

k

2

fast

(CH

3

)C−OH

Rate of reaction=k

2

[(CH

3

)

3

C

+

][OH

]

The second step involves the attack of OH

ion. This is the fast step, since it is the bond formation step.

Energy profile diagram of SN

1

mechanism shows that rate of a reaction is independent of the concentration of nucleophile. The first step requires larger activation energy (ΔE

1

) than the second step (ΔE

2

). The first step to form carbocation determines the rate of overall reaction. The second step, which is the attack of nucleophile on carbocation is exothermic i.e., it is a lower energy transition state. The intermediate carbocation appears at a low point in the diagram. The conditions and reagents which favour the formation of carbocation will accelerate the SN

1

reaction. The energy difference between products and reactants is ΔH, i.e., Heat of reaction.

Similar questions