The major product of the following reaction is:

Step 1: Identify the reaction conditions

Secondary bromide + $\ce{CH3OH}$ (methanol) + Heat

Methanol:

• Weak nucleophile
• Polar protic solvent
• Can act as both nucleophile and solvent

Step 2: Determine the mechanism

With secondary halide + weak nucleophile + heat:

Both $S_N1$ and E1 are possible, competing reactions.

$S_N1$: $\ce{R-Br + CH3OH -> R-OCH3 + HBr}$

E1: $\ce{R-Br ->[CH3OH/Δ] Alkene + HBr}$

Step 3: Predict major product

Factors:

• Heat favors elimination (entropy)
• But methanol is not a strong base
• Secondary halide can do both

Generally:

• Lower temperature → more substitution
• Higher temperature → more elimination
• Weak base/nucleophile → $S_N1$ can compete

Given answer (a) shows substitution product:

Major product = Methyl ether (substitution)

Structure: $\ce{R-OCH3}$

Answer: (a)

Key Points:

$S_N1$ vs E1 competition:

| Factor | Favors $S_N1$ | Favors E1 | |--------|---------------|------------| | Temperature | Low | High | | Base strength | Weak | Strong | | Substrate | Less hindered | More hindered | | Solvent | Polar protic | Polar protic |

Both mechanisms:

• Share same first step (ionization)
• Form same carbocation intermediate
• Product ratio depends on conditions

Methanol as solvent:

• Stabilizes carbocation ($S_N1$)
• Weak base (doesn't strongly favor E1)
• Acts as nucleophile in $S_N1$