Please rank the following compounds in DECREASING order of their reactivity toward the SN2 reaction with sodium ethoxide in ethanol.
a. methyl chloride (CH3Cl)
b. isopropylbromide ((CH3)2CHBr)
c. t-butyliodide (CH3)3Cl)
d. methyl iodide (CH3I)
e. ethyl chloride (CH3CH2Cl)
I have an answer, I'm just not sure if it is correct.
To rank the compounds in decreasing order of reactivity toward the SN2 reaction with sodium ethoxide in ethanol, we need to consider the factors that influence the reactivity of an alkyl halide in an SN2 reaction. These factors include the nature of the alkyl group, the nature of the leaving group, and steric hindrance.
1. Steric hindrance: The bulkier the alkyl group adjacent to the carbon atom with the leaving group, the slower the SN2 reaction will be, leading to lower reactivity.
2. Leaving group ability: The better the leaving group, the faster the SN2 reaction will be and the higher the reactivity will be.
3. Alkyl group nature: Alkyl groups can be electron-donating or electron-withdrawing. Electron-withdrawing groups increase the acidity of the carbon atom, making it more susceptible to nucleophilic attack. Therefore, alkyl halides with electron-withdrawing groups show higher reactivity.
Based on these factors, we can rank the compounds in decreasing order of reactivity toward SN2 reaction:
1. t-butyliodide ((CH3)3Cl) - This compound has the most steric hindrance due to the bulky t-butyl group, making it less reactive in SN2 reactions.
2. isopropylbromide ((CH3)2CHBr) - This compound has a smaller isopropyl group, resulting in less steric hindrance compared to t-butyliodide but still more than the remaining compounds.
3. ethyl chloride (CH3CH2Cl) - This compound has a smaller ethyl group, leading to less steric hindrance than isopropylbromide but more steric hindrance than the remaining compounds.
4. methyl chloride (CH3Cl) - This compound has the smallest alkyl group, resulting in the least steric hindrance among the remaining compounds.
5. methyl iodide (CH3I) - This compound has an electron-withdrawing iodine atom, making it more reactive than methyl chloride.
Therefore, the correct ranking of the compounds in decreasing order of reactivity toward SN2 reaction with sodium ethoxide in ethanol is:
t-butyliodide > isopropylbromide > ethyl chloride > methyl chloride > methyl iodide