When ethyl bromide is added to potassium t-butoxide, the product is ethyl t-butyl ether.
`underset("ethyl bromide")(CH_3CH_2-Br)+underset("ptassium t-butoxide")((CH_3)_3C -OK)tounderset("ethyl t-butyl")((CH_3)_3C-O)-underset("ether")(CH_2CH_3)`
Which of the following statements is/are correct ?

Options

A. when the concentration of Ethyl bromide is doubled rate is also doubled.
B. when the concentration of potassium t-butoxide is tripled and the concentration of ethyl bromide is doubled rate will increases six times.
C. Elimination product dominates when temperature is raised.
D. when the concentration of potassium t-butoxide is tripled and the concentration of ethyl bromide is doubled rate will increase three times.

Answer

To analyze the reaction between ethyl bromide (CH₃CH₂Br) and potassium t-butoxide ((CH₃)₃COK) leading to the formation of ethyl t-butyl ether ((CH₃)₃C-O-CH₂CH₃), we need to evaluate the statements provided regarding the reaction kinetics and mechanisms.

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
The reaction between ethyl bromide and potassium t-butoxide is an example of an ether formation through an SN2 mechanism. In this reaction, the nucleophile (potassium t-butoxide) attacks the electrophilic carbon in ethyl bromide, leading to the formation of the ether.

2. **Write the Rate Law**:
The rate of the reaction can be expressed as:
\[ \text{Rate} = k [\text{ethyl bromide}]^1 [\text{potassium t-butoxide}]^1 \]
This indicates that the reaction is first-order with respect to both reactants.

3. **Evaluate Statement A**:
When the concentration of ethyl bromide is doubled, the rate of the reaction will also double. This is correct since the rate is directly proportional to the concentration of ethyl bromide.
- **Conclusion**: Statement A is correct.

4. **Evaluate Statement B**:
If both the concentration of ethyl bromide and potassium t-butoxide are doubled, the rate will increase by a factor of 6. This is calculated as follows:
\[ \text{New Rate} = k (2[\text{ethyl bromide}])(2[\text{potassium t-butoxide}]) = 4k[\text{ethyl bromide}][\text{potassium t-butoxide}] \]
Therefore, the rate will increase by a factor of 4, not 6.
- **Conclusion**: Statement B is incorrect.

5. **Evaluate Statement C**:
The statement regarding elimination reactions becoming dominant at higher temperatures is generally true. As temperature increases, the likelihood of elimination reactions (which can lead to the formation of alkenes) increases due to higher energy collisions.
- **Conclusion**: Statement C is correct.

6. **Evaluate Statement D**:
The statement about the concentration of potassium t-butoxide being tripled and ethyl bromide doubled leading to a rate increase of six times is incorrect. As calculated previously, the rate increases by a factor of 4 when both concentrations are doubled, and tripling one while doubling the other does not lead to a simple multiplication of the rate.
- **Conclusion**: Statement D is incorrect.

### Final Conclusion:
The correct statements are A and C.