Based upon the results of the experiments carried out, rank (slowest to fastest)
ID: 508421 • Letter: B
Question
Based upon the results of the experiments carried out, rank (slowest to fastest)
the following compounds in order of reactivity with respect to an SN1 reaction.
Explain your reasoning for the ranking you propose. The compounds are
1-bromo-3-methylbutane, 1-bromo-2,2-dimethylpropane and 2-bromo-2-methylbutane.
I know the tertiary 2-bromo--2-methylbutane acts faster than the primary 1-bromo-3-methylbutane,
but I am just confused about which would react faster the 2-bromo-2-methylbutane or the neopentyl bromide?
Is it based on solvent alone then or tertiary dominates?
Please thoroughly explain reasoning, thank you
Explanation / Answer
2-bromo-2-methyl butane acts faster than primary 1-bromo-2,2-dimethyl propane and 1-bromo-3-methyl butane. Because tertiary carbocations are more stable than primary carbocation. In SN1 reaction tertiary alkyl halides are more reacive.
Neopentylbromide contains no beta hydrogen. Hence no elimination is possible. Although it is primary halide, it does not readily undergo SN2 reaction because of great steric hindrance in the transition state of the reaction. And since primary alkyl halides do not form carbocations (primary carbocation is unstable). Neither SN1 nor elimination (E1) reaction is possible. This this alkyl halide is essentially inert and no reaction is predicted.
Polar protic solvents like water, alcohol, formic acid, acetic acid are good for SN1 reaction. Because polar protic solvent solvate carbocation by ion dipole interaction and leaving group is solvated by hydrogen bonding. So SN1 reactions are favoured by polar protic solvents. But the formation of carbocation is dominating the solvent effect. In SN1 solvent stabilizes the carbocation only.