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Analyze the generalized rate data: RX ...

Analyze the generalized rate data:
`RX + M^(ɵ) rarr` Product
`|{:("Experiment",[RX] "Substrate",[M^(ɵ)] "Attaking species","Rate"),(I,0.10 M,0.10 M,1.2 xx 10^(-4)),(II,0.20 M,0.10 M,2.4 xx 10^(-4)),(III,0.10 M,0.20 M,2.4 xx 10^(-4)),(IV,0.20 M,0.20 M,4.8 xx 10^(-4)):}|`
For the reaction under conisderation, `3^(@)` alkyl has been found to be the most favourable alkyl group. Which of the following attacking species `(M^(ɵ))` will give the best yield in the reaction ?

A

`(CH_(3))_(2)CH-O^(ɵ)`

B

`(CH_(3))_(3)C-O^(ɵ)`

C

`overset(ɵ)(OH)`

D

`CH_(3)CH_(2)O^(ɵ)`

Text Solution

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To analyze the generalized rate data and determine which attacking species (M^(ɵ)) will give the best yield in the reaction, we can follow these steps: ### Step 1: Understand the Given Data We are provided with experimental data that includes concentrations of the substrate (RX) and the attacking species (M^(ɵ)), along with the corresponding reaction rates. The data is as follows: | Experiment | [RX] (Substrate) | [M^(ɵ)] (Attacking species) | Rate (mol/L·s) | |------------|------------------|-----------------------------|-----------------| | I | 0.10 M | 0.10 M | 1.2 x 10^(-4) | | II | 0.20 M | 0.10 M | 2.4 x 10^(-4) | | III | 0.10 M | 0.20 M | 2.4 x 10^(-4) | | IV | 0.20 M | 0.20 M | 4.8 x 10^(-4) | ### Step 2: Analyze the Reaction Order From the data, we can see that when the concentration of RX is doubled (from Experiment I to II), the rate also doubles. This suggests that the reaction is first-order with respect to RX. Similarly, when the concentration of M^(ɵ) is doubled (from Experiment III to IV), the rate also doubles, indicating that the reaction is first-order with respect to M^(ɵ). Thus, we can conclude that the overall reaction is second-order (first-order in RX and first-order in M^(ɵ)). ### Step 3: Identify the Mechanism The problem states that a tertiary alkyl group is the most favorable. In organic chemistry, reactions involving tertiary alkyl groups typically follow either an E2 or SN2 mechanism. Given that the reaction is second-order and involves a strong base (as implied by the preference for a tertiary alkyl), it is likely that this reaction follows the E2 mechanism, which is favored by strong bases. ### Step 4: Determine the Best Attacking Species E2 reactions are typically favored by strong bases. The order of basicity is crucial here. Strong bases will yield better results in E2 reactions. ### Step 5: Conclusion Based on the analysis, the best yield for the reaction will be achieved with the strongest base among the attacking species (M^(ɵ)). Thus, if we are given a list of potential attacking species, we should select the one that is the strongest base. ### Final Answer The attacking species (M^(ɵ)) that will give the best yield in the reaction is the strongest base available in the options provided. ---
To analyze the generalized rate data and determine which attacking species (M^(ɵ)) will give the best yield in the reaction, we can follow these steps: ### Step 1: Understand the Given Data We are provided with experimental data that includes concentrations of the substrate (RX) and the attacking species (M^(ɵ)), along with the corresponding reaction rates. The data is as follows: | Experiment | [RX] (Substrate) | [M^(ɵ)] (Attacking species) | Rate (mol/L·s) | |------------|------------------|-----------------------------|-----------------| | I | 0.10 M | 0.10 M | 1.2 x 10^(-4) | ...
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