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When potassium ethoxide reacts with 2-bromopentane, an elimination reaction occurs. Which products are formed?

1. CH3CH(Br)CH2CH2CH3 → OC2H5 → CH3CH=CHCH2CH3 + CH2=CHCH2CH2CH3
2. CH3CH=CHCH2CH3 + CH2=CHCH2CH2CH3
3. CH3CH=CHCH2CH3 + CH2=CHCH2CH2CH3
4. CH3CH=CHCH2CH3 + CH2=CHCH2CH2CH3

Correct answer: CH3CH=CHCH2CH3 + CH2=CHCH2CH2CH3

Solution

Potassium ethoxide, being a strong base, promotes an E2 elimination reaction in 2-bromopentane, leading to the formation of two alkenes: pent-2-ene and pent-1-ene. The major product is pent-2-ene due to Zaitsev's rule.

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