Questions: SN1 Substitution Reactions

Question 1

A tertiary alkyl bromide reacts in aqueous ethanol with no added nucleophile. Why does SN1 proceed faster here than an SN2 reaction would?

• A. The tertiary substrate has more surface area for nucleophilic attack
• B. The tertiary carbocation intermediate is stabilized by hyperconjugation and inductive donation from three alkyl groups
• C. Aqueous ethanol is a strong nucleophile that accelerates bimolecular attack
• D. Tertiary substrates have lower activation energy for backside attack

Answer: (B)

SN1 rate depends only on forming the carbocation intermediate (unimolecular, rate-limiting step). Three alkyl groups stabilize the positive charge through hyperconjugation and inductive electron donation, lowering the activation energy for ionization. SN2, by contrast, is sterically blocked at tertiary carbons, so the bulkier the substrate, the slower the SN2 — not faster.

Question 2

An SN1 reaction at a pure stereocenter usually produces a perfectly racemic (50/50) mixture of enantiomers.

• T. True
• F. False

Answer: False

Complete racemization is an idealization. In practice, the departing leaving group briefly shields one face of the nascent carbocation before fully dissociating into the solvent shell. During this fleeting ion pair stage, nucleophilic attack from the retention face is slightly blocked, producing a modest excess of the inverted product. The result is partial inversion — mostly racemic but not perfectly 50/50.

Question 3

Why do polar protic solvents (e.g., water, ethanol) favor SN1 reactions over polar aprotic solvents (e.g., acetone, DMSO)?

Model answer: Polar protic solvents stabilize both the developing carbocation and the departing anionic leaving group through hydrogen bonding and ion-dipole interactions, lowering the activation energy for the ionization step.

SN1 depends on forming two charged species (a carbocation and a leaving group anion) from a neutral substrate. Polar protic solvents solvate both ions — their OH groups hydrogen-bond to anions and their dipoles stabilize cations — effectively lowering the energy of the transition state and intermediate. Polar aprotic solvents lack the hydrogen-bond donor ability needed to stabilize the anion as well, so they do not assist ionization as effectively.