An SN2 reaction converts (R)-2-bromobutane to a product using cyanide (CN⁻) as the nucleophile. Complete inversion of spatial arrangement occurs at C2. Which statement about the product's R/S designation is correct?
AThe product must be (S)-2-cyanobutane, since SN2 inversion always converts R to S
BThe product must be (R)-2-cyanobutane, since inversion does not change the R/S designation
CThe product is a racemic mixture of R and S, since inversion is not stereospecific
DThe R/S designation of the product cannot be determined from inversion alone; it requires assigning CIP priorities to the new substituent arrangement
Spatial inversion (the umbrella flip) always occurs in SN2, but whether the R/S letter changes depends on CIP priority reassignment. If the incoming group (CN) has a different priority rank than the departing group (Br), the priority order of the four substituents may reshuffle, potentially keeping the same R/S label despite complete geometric inversion. The safe approach is always to draw the three-dimensional structure after reaction and assign priorities directly. The common misconception (option A) assumes inversion automatically flips the R/S label, which is only sometimes true.
Question 2 Multiple Choice
Why do tertiary alkyl halides essentially never undergo SN2 reactions, while methyl halides react fastest of all?
ATertiary substrates form more stable carbocations, making them prefer SN1 for thermodynamic reasons
BThe three alkyl groups surrounding the electrophilic carbon in tertiary substrates physically block the nucleophile from approaching the backside of the C–X bond
CThe C–X bond is significantly stronger in tertiary substrates, requiring prohibitively high activation energy
DTertiary carbons lack a suitable σ* antibonding orbital for the nucleophile to attack
SN2 requires backside attack — the nucleophile must approach along the C–X bond axis from the face opposite the leaving group. In a tertiary substrate, three bulky alkyl groups surround the carbon and physically block access to this approach trajectory. In a methyl substrate, three small hydrogen atoms leave the backside open. This is a purely steric (geometric) argument about transition-state accessibility, not about bond strength or carbocation stability. Option A describes why tertiary substrates prefer SN1, which is a separate mechanism — the SN2 failure is a steric issue, not a thermodynamic preference.
Question 3 True / False
In every SN2 reaction at a stereocenter, the spatial arrangement of substituents around the reacting carbon is completely inverted — every group moves to the opposite face — regardless of which nucleophile or leaving group is involved.
TTrue
FFalse
Answer: True
Walden inversion is a mechanistic inevitability of the SN2 pathway, not a property of specific reactants. The backside attack geometry — mandated by orbital symmetry (nucleophile lone pair into C–LG σ*) — forces the three remaining substituents to pass through a trigonal planar arrangement and emerge on the opposite face. This happens 100% of the time in every SN2 event, which is why SN2 is stereospecific: it produces a single enantiomer rather than a mixture.
Question 4 True / False
If an SN2 reaction causes complete spatial inversion at a chiral carbon, the product is expected to have the opposite R/S configuration label (R→S or S→R).
TTrue
FFalse
Answer: False
This is the most common misconception about Walden inversion. The spatial arrangement always inverts, but the R/S label is assigned by CIP priority rules — and those rules can reshuffle when a new group replaces the leaving group. If the incoming nucleophile has the same CIP priority rank as the leaving group it replaces, priorities are unchanged and the letter flips. But if the nucleophile has a different rank, the new priority ordering may yield the same letter as the starting material even though every group is now on the opposite face. R and S are labels applied to spatial arrangements, not properties of those arrangements themselves.
Question 5 Short Answer
Explain why the requirement for backside attack in SN2 reactions creates a direct connection between reaction rate and the steric environment around the electrophilic carbon.
Think about your answer, then reveal below.
Model answer: SN2 requires the nucleophile to approach along the axis of the C–leaving group bond, from the face directly opposite the leaving group. Any substituents on the electrophilic carbon occupy space near this backside approach trajectory. More and larger substituents (primary → secondary → tertiary) increasingly block the nucleophile's path, raising the transition-state energy and slowing (or preventing) the reaction. This is entirely a geometric argument about the trigonal bipyramidal transition state: the rate drops with steric bulk not because the bond is harder to break, but because the transition state geometry becomes inaccessible.
This link between geometry and reactivity is what makes SN2 a predictable, stereochemically controlled reaction. The same geometric constraint that causes the umbrella-flip inversion is what makes tertiary substrates unreactive — both effects trace back to the requirement that the nucleophile approach from the backside. Knowing this, you can predict both the stereochemical outcome (complete inversion) and the substrate scope (methyl > primary > secondary >> tertiary) from one structural principle.