write the mechanism of alkaline hydrolysis of an alkyl halide which involves formation of carbocation as an intermediate
Answers
Answer:
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.
Consider the action of aqueous sodium hydroxide or potassium hydroxide on a tertiary alkyl halide such as t-butyl bromide. has displaced the weaker nucleophile, and therefore, it is nucleophilic substitution (SN) reaction. The nucleophile ( : O ¯ H ) attacks the carbocation forming t-butyl alcohol.