The decreasing order of reactivity in terms of SN2 reaction with sodium ethoxide in ethanol for the given compounds can be determined based on the relative steric hindrance at the carbon atom bearing the leaving group. The general trend is that the more hindered the carbon atom, the lower the reactivity towards SN2 reactions.
Here is the decreasing order of reactivity:
Methyl iodide (CH3I): Methyl iodide is the most reactive compound among the given options. It has the least steric hindrance because it has only one carbon substituent, making it the most accessible for nucleophilic attack.
Ethyl chloride (C2H5Cl): Ethyl chloride is less reactive than methyl iodide due to the presence of an additional ethyl group, which increases the steric hindrance. However, it is still more reactive than the remaining compounds.
Isopropyl bromide (C3H7Br): Isopropyl bromide has more steric hindrance compared to ethyl chloride because it has a larger isopropyl group attached to the carbon atom bearing the leaving group.
Neopentyl bromide (C5H11Br): Neopentyl bromide is more hindered than isopropyl bromide because it has three bulky methyl groups attached to the carbon atom bearing the leaving group. The steric hindrance in this compound significantly reduces its reactivity towards SN2 reactions.
Propyl iodide (C3H7I): Propyl iodide is less reactive than neopentyl bromide due to the presence of only one iodine atom as a leaving group, but it is still more hindered than the previously mentioned compounds.
Therefore, the decreasing order of reactivity in terms of SN2 reaction with sodium ethoxide in ethanol is: Methyl iodide > Ethyl chloride > Isopropyl bromide > Neopentyl bromide > Propyl iodide.