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In the following reactions: (i) CH3—CH=CH—CH3 → H+/heat → A (Major product) + B (Minor product) (ii) A → HBr, dark (in absence of peroxide) → C (Major product) + D (Minor product) The major products (A) and (C) are respectively:

1. CH3—CH=CH—CH3 and CH2—CH2—CH2—CH3
2. CH3—C=CH—CH3 and CH3—C—CH2—CH3
3. CH3—CH=CH—CH3 and CH3—CH2—CH2—CH3
4. CH3—C=CH—CH3 and CH3—CH2—CH2—CH3

Correct answer: CH3—C=CH—CH3 and CH3—CH2—CH2—CH3

Solution

In reaction (i), under acidic conditions and heat, 2-butene undergoes isomerization to form the more stable trans-isomer (A: CH3—C=CH—CH3). In reaction (ii), HBr adds to the double bond of A following Markovnikov's rule, yielding the major product (C: CH3—CH2—CH2—CH3).

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