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2-Methylpropyl bromide (isobutyl bromide) undergoes two separate reactions: (i) with sodium ethoxide (C2H5O-) and (ii) with ethanol (C2H5OH). What are the mechanisms and the products A and B formed in reactions (i) and (ii) respectively?

1. SN2 in (i), A = isobutyl ethyl ether; SN1 in (ii) with rearrangement, B = tert-butyl ethyl ether
2. SN1 in (i), A = tert-butyl ethyl ether; SN1 in (ii), B = 2-butyl ethyl ether
3. SN1 in (i), A = tert-butyl ethyl ether; SN2 in (ii), B = isobutyl ethyl ether
4. SN2 in (i), A = 2-butyl ethyl ether; SN2 in (ii), B = isobutyl ethyl ether

Correct answer: SN2 in (i), A = isobutyl ethyl ether; SN1 in (ii) with rearrangement, B = tert-butyl ethyl ether

Solution

With the strong nucleophile ethoxide, SN2 occurs directly at the primary carbon giving isobutyl ethyl ether. With the weak nucleophile ethanol under protic conditions, SN1 proceeds via a hydride-shifted tertiary carbocation, giving tert-butyl ethyl ether.

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