$word_{count}$: 220

Step 1: Identify substrate class (methyl, 1 degree, 2 degree, 3 degree, allyl, benzyl, vinyl, aryl). Step 2: Identify reagent (strong vs weak nucleophile/base; bulky vs small). Step 3: Identify solvent (polar protic vs polar aprotic).

Methyl/1 degree + strong Nu + aprotic → SN2. If strong bulky base → E2.

2 degree — most competitive. Strong small Nu + aprotic → SN2 (major) + some E2. Strong bulky base → E2. Weak Nu + protic + heat → SN1/E1.

3 degree + strong base → E2 only (SN2 impossible). 3 degree + weak Nu + protic → SN1 (major) + E1. Heat shifts toward E1.

Allyl/Benzyl — both SN1 and SN2 are fast (resonance stabilisation).

Vinyl/Aryl → unreactive under normal conditions. Aryl requires SNAr (EWG activation) or benzyne (NaNH2).

Key tip for JEE: Most questions give a substrate + reagent + conditions. Follow the framework mechanically. Watch for rearranged products (= $\frac{SN1}{E1}$ with carbocation shift). Watch for stereochemistry clues (inversion = SN2, racemization = SN1).