1. Read the substrate before anything else.
The carbon bearing the leaving group decides which of the four mechanisms are even possible.
Methyl → SN2 only 1° → SN2 2° → depends 3° → SN1/E1/E2 only 2. Methyl and primary carbons go SN2 only.
They cannot form a carbocation — a methyl or 1° cation is far too unstable — so SN1 and E1 are ruled out.
1-Bromobutane (1°) — backside attack, SN2 3. Tertiary carbons go SN1, E1, or E2 only.
A stable 3° carbocation forms readily, while steric crowding blocks the SN2 backside approach.
tert-Butyl bromide (3°) — SN2 blocked, cation easy 4. Secondary carbons are the swing case.
Any of the four mechanisms is possible, so the nucleophile/base, solvent, and temperature make the call.
2-Bromobutane (2°) — conditions decide 5. Allylic and benzylic halides also unlock SN1/E1.
Their carbocations are resonance-stabilized, so even at a 1°-looking carbon a cation pathway opens.
Allyl bromide — resonance-stabilized cation Benzyl bromide — resonance-stabilized cation
Summary
Substrate first · methyl/1° → SN2 only · 3° → SN1/E1/E2 only · 2° → swing case (conditions decide) · allylic/benzylic → SN1/E1 unlocked by resonance
Quiz yourself
Tap a question to reveal the answer — free, no login.
SN1 needs a carbocation intermediate, and a methyl or 1° cation is far too unstable to form, so only the concerted SN2 pathway is open.
SN2 is ruled out — the crowded 3° carbon blocks backside attack. Only SN1, E1, and E2 remain, favored by the easily formed stable 3° cation.
All four mechanisms are possible at a 2° carbon, so the substrate alone can't decide — the nucleophile/base strength, solvent, and temperature settle it.
Its cation is resonance-stabilized by the adjacent ring (or double bond, for allylic), making the carbocation stable enough for SN1/E1 to occur.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.