The substitution reaction studied here with Chlorobutane and KOH is known to have a second order rate equation meaning the transition state in the slow step involves both nucleophile and electrophile. Knowing that, which of the following statements are true? Select all that apply. A) Adding more Chlorobutane (the electrophile) will not change the rate of the reaction. B) Adding more KOH (the nucleophile) will make the reaction go faster. Adding more Chlorobutane (the electrophile) will make the reaction go faster. (D) Adding more KOH (the nucleophile) will not change the rate of the reaction. E) Adding more KOH (the nucleophile) will make the reaction go slower.

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**Understanding Second-Order Reaction Rates in Chlorobutane and KOH Reactions**

The substitution reaction between chlorobutane and KOH is characterized by a second-order rate equation. This implies that both the nucleophile (KOH) and the electrophile (chlorobutane) play a role in the transition state during the slow step of the reaction. Given this information, consider which of the following statements are accurate:

- **Option A:** Adding more chlorobutane (the electrophile) will not change the rate of the reaction.
- **Option B:** Adding more KOH (the nucleophile) will make the reaction go faster. *(Correct)*
- **Option C:** Adding more chlorobutane (the electrophile) will make the reaction go faster. *(Correct)*
- **Option D:** Adding more KOH (the nucleophile) will not change the rate of the reaction.
- **Option E:** Adding more KOH (the nucleophile) will make the reaction go slower.

### Explanation

In second-order reactions, the rate is proportional to the product of the concentrations of two reactants. Therefore, increasing the concentration of either the nucleophile (KOH) or the electrophile (chlorobutane) will increase the reaction rate. Thus, options B and C are the correct answers.
Transcribed Image Text:**Understanding Second-Order Reaction Rates in Chlorobutane and KOH Reactions** The substitution reaction between chlorobutane and KOH is characterized by a second-order rate equation. This implies that both the nucleophile (KOH) and the electrophile (chlorobutane) play a role in the transition state during the slow step of the reaction. Given this information, consider which of the following statements are accurate: - **Option A:** Adding more chlorobutane (the electrophile) will not change the rate of the reaction. - **Option B:** Adding more KOH (the nucleophile) will make the reaction go faster. *(Correct)* - **Option C:** Adding more chlorobutane (the electrophile) will make the reaction go faster. *(Correct)* - **Option D:** Adding more KOH (the nucleophile) will not change the rate of the reaction. - **Option E:** Adding more KOH (the nucleophile) will make the reaction go slower. ### Explanation In second-order reactions, the rate is proportional to the product of the concentrations of two reactants. Therefore, increasing the concentration of either the nucleophile (KOH) or the electrophile (chlorobutane) will increase the reaction rate. Thus, options B and C are the correct answers.
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