Which substrate would you expect to undergo SN2 substitution most readily with a strong nucleophile?
A(CH₃)₃CBr — tert-butyl bromide
BCH₃CH₂CH₂Br — n-propyl bromide (primary)
C(CH₃)₂CHBr — isopropyl bromide (secondary)
DCH₃Br — methyl bromide
SN2 reactivity is methyl > primary > secondary >> tertiary, because the nucleophile must approach the backside of the electrophilic carbon through a trigonal bipyramidal transition state. Methyl bromide (CH₃Br) has no alkyl substituents blocking the backside, making it the most accessible. Tert-butyl bromide has three methyl groups flanking the reactive carbon, creating severe steric crowding that essentially prevents backside attack — SN2 at tertiary centers is so slow as to be negligible.
Question 2 True / False
In an SN2 reaction at a stereocenter, inversion of the spatial arrangement of substituents generally results in a change from R to S configuration (or vice versa).
TTrue
FFalse
Answer: False
Backside attack always causes Walden inversion — the spatial arrangement of the three non-leaving substituents is flipped like an umbrella turning inside out. However, whether the R/S label changes depends on the CIP priority ranking of the incoming nucleophile relative to the other substituents. If the nucleophile has lower priority than the leaving group, the inversion of geometry coincides with an R→S or S→R change. But if the new substituent has higher priority than what it replaced, the spatial inversion can appear to leave the R/S designation unchanged, because the priority ranking that determines the label has also changed.
Question 3 Short Answer
Why does SN2 rate depend on the concentration of both the nucleophile and the substrate, while SN1 rate depends only on the substrate concentration?
Think about your answer, then reveal below.
Model answer: SN2 is a concerted, one-step mechanism: the nucleophile attacks and the leaving group departs simultaneously in a single transition state. Both the nucleophile and the substrate must collide to form this transition state, so the rate depends on both concentrations (rate = k[substrate][nucleophile]). SN1 involves a two-step mechanism where the rate-determining step is unimolecular ionization of the substrate to form a carbocation intermediate — the nucleophile is not involved in this slow step, so its concentration does not affect the rate.
The 'bimolecular' in SN2 refers to the kinetics: two species appear in the rate law because both must be present at the transition state. In SN1, the slow step is the spontaneous ionization of the C–LG bond; the nucleophile only participates in the fast second step after the carbocation has already formed. This mechanistic difference has practical consequences: increasing nucleophile concentration accelerates SN2 but has no effect on SN1.