2-Methyl propyl bromide reacts with C2H5O and gives 'A' whereas on reaction with C2H5OH it gives 'B'. The mechanism…

Step 1: Identify the substrate

2-Methylpropyl bromide = Isobutyl bromide

Structure: $\ce{(CH3)2CH-CH2Br}$

This is a primary halide (Br on primary carbon).

Step 2: Reaction with $\ce{C2H5O-}$ (ethoxide ion)

Reagent: Strong nucleophile (ethoxide ion)

Mechanism: $S_N2$ (primary halide + strong nucleophile)

$\ce{(CH3)2CH-CH2Br + C2H5O- -> (CH3)2CH-CH2-O-C2H5 + Br-}$

Product A = Isobutyl ethyl ether (2-methylpropyl ethyl ether)

Step 3: Reaction with $\ce{C2H5OH}$ (ethanol)

Reagent: Weak nucleophile (ethanol)

Mechanism: For primary halide with weak nucleophile:

• $S_N2$ is still possible but slow
• Or $S_N1$ if carbocation can rearrange

Carbocation rearrangement:

If $S_N1$ occurs:

1. $\ce{(CH3)2CH-CH2Br -> (CH3)2CH-CH2+ + Br-}$ (1° carbocation)
2. Hydride shift: $\ce{(CH3)2CH-CH2+ -> (CH3)3C+}$ (3° carbocation, more stable)
3. $\ce{(CH3)3C+ + C2H5OH -> (CH3)3C-O-C2H5}$

Product B = tert-Butyl ethyl ether (after rearrangement)

But wait, primary halides typically don't undergo $S_N1$ easily. Let me reconsider.

Actually, with weak nucleophile (ethanol), the reaction is very slow. If it does proceed via $S_N1$, rearrangement is likely.

Step 4: Match with options

Looking at option (a):

• $S_N2$ with $\ce{C2H5O-}$ → isobutyl ethyl ether ✓
• $S_N1$ with $\ce{C2H5OH}$ → tert-butyl ethyl ether (rearranged) ✓

Answer: (a)

Key Points:

• Strong nucleophile ($\ce{RO-}$) → $S_N2$ → no rearrangement
• Weak nucleophile ($\ce{ROH}$) → $S_N1$ (if possible) → rearrangement likely
• Carbocation rearrangement: 1° → 2° or 3° (more stable)
• Hydride shift: H moves with its bonding electrons
• Methyl shift: $\ce{CH3}$ moves with its bonding electrons