What is the order of reactivity from fastest to slowest in an SN2 reaction for the following substrates?
a. t-butyl bromide
b. isopropyl bromide
c. neopentyl bromide
d. ethyl bromide
e. methyl bromide
So far have methyl bromide> ethyl bromide> neopentyl bromide> isopropyl bromide > t-butyl bromide
I know t-butyl is on a 3 prim carbon so that would be last
isopropyl is on a 2 prime carbon so thats second
but the methyl, ethyl, and neopentyl is where i am confused at since they are both primary
Your initial listing is correct... The rate depends upon the least stable carbon attached to the halogen for Sn2 reaction mechanisms.
MtBr with methyl carbon is the fastest substitution
EtBr and neopentyl-Br both have primary carbons attached to the Bromide
Isopropyl Br => secondary carbon and...
t-BuBr => tertiary carbon (prefers to follow the Sn1 pathway via ionization; i.e., is a 2-step process).
To determine the order of reactivity in an SN2 reaction, you need to consider the factors that affect the reaction rate. In an SN2 reaction, a nucleophile attacks the substrate and displaces the leaving group in a single step. The rate of the reaction depends on the steric hindrance and the nature of the leaving group.
In this case, the steric hindrance is the primary factor for determining the order of reactivity. Steric hindrance refers to the obstruction of the nucleophile's approach to the substrate due to bulky substituents. The more hindered the substrate is, the slower the reaction rate.
Let's examine each substrate and evaluate their steric hindrance:
a. t-butyl bromide: t-butyl group is highly hindered due to its three bulky methyl groups. Therefore, it is the slowest to react.
b. isopropyl bromide: isopropyl group has two methyl groups attached to the secondary carbon, which introduces some steric hindrance but less compared to t-butyl group. Therefore, it is the second slowest.
c. neopentyl bromide: neopentyl group has four methyl groups attached to the tertiary carbon. Although it is sterically hindered, it is still less hindered than t-butyl group. Therefore, it is faster than t-butyl and isopropyl bromides.
d. ethyl bromide: ethyl group is a primary carbon with only one methyl group attached. It is less hindered compared to neopentyl, isopropyl, and t-butyl groups. Therefore, it is faster than these three substrates.
e. methyl bromide: methyl group is the least hindered, as it has no additional substituents. Therefore, it is the fastest to react.
Based on the analysis above, the correct order of reactivity from fastest to slowest in an SN2 reaction for the given substrates would be:
methyl bromide > ethyl bromide > neopentyl bromide > isopropyl bromide > t-butyl bromide.