
The slowest step of the reaction in a multi-step reaction is called rate-determining step.
A.True.
B.False.
Answer
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Hint: We know that the rate determining step is generally determined by the slowest step of the reaction in the overall rate of the reaction in chemical kinetics. It is also known as the rate limiting step. It tends to determine the rate or the speed of the reaction during which the reactants are able to convert into products.
Complete answer:
So the slow step is considered to be the least favorable step so it tends to take a while for the slightly unstable intermediate to form before it gets propelled for the formation of the products. Let us take the example of the $S{{N}_{1}}$ mechanism on the tertiary bromide. In this we see the bromide ion getting departing as the leaving group and then the poor or the slow nucleophile tends to attack the planar carbocation.
The process of attacking the cation of carbon is faster than forming it because the bromide tends to leave on its own when the nucleophile has been slow. The nucleophile has been slow because the bromide ion participates by itself that is in the first order process and so the bromine has little or no assistance for leaving the group faster. The other reason is the steric hindrance factor of the tertiary bromide is lessened when the formation of carbocation takes place as the planar molecule has been inherently less sterically hindered compared with tetrahedral conformation. So the slow step is the rate determining step.
Rate determining step is the elementary step in a reaction mechanism that can determine the observed rate of a reaction. The slowest step is the rate determining step in a reaction.
Therefore, the correct answer is option A.
Note:
Remember that the rate determining step is the slowest step which has been characterized by its high activation energy. The more is the activation energy the slower will be the speed or rate of reaction. The rate determining step is known as rate limiting step which determines the rate at which the ultimate product is formed or appeared.
Complete answer:
So the slow step is considered to be the least favorable step so it tends to take a while for the slightly unstable intermediate to form before it gets propelled for the formation of the products. Let us take the example of the $S{{N}_{1}}$ mechanism on the tertiary bromide. In this we see the bromide ion getting departing as the leaving group and then the poor or the slow nucleophile tends to attack the planar carbocation.
The process of attacking the cation of carbon is faster than forming it because the bromide tends to leave on its own when the nucleophile has been slow. The nucleophile has been slow because the bromide ion participates by itself that is in the first order process and so the bromine has little or no assistance for leaving the group faster. The other reason is the steric hindrance factor of the tertiary bromide is lessened when the formation of carbocation takes place as the planar molecule has been inherently less sterically hindered compared with tetrahedral conformation. So the slow step is the rate determining step.
Rate determining step is the elementary step in a reaction mechanism that can determine the observed rate of a reaction. The slowest step is the rate determining step in a reaction.
Therefore, the correct answer is option A.
Note:
Remember that the rate determining step is the slowest step which has been characterized by its high activation energy. The more is the activation energy the slower will be the speed or rate of reaction. The rate determining step is known as rate limiting step which determines the rate at which the ultimate product is formed or appeared.
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