Question

2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is

• A. 2-ethoxypentane
• B. pentene-1
• C. trans-2-pentene
• D. cis-pentene-2

Answer 1

Answer: (C)

Potassium ethoxide is a strong base, and 2- bromopentane is a 2º bromide, so elimination raction predominates
$C{H}_{3}CH(Br)C{H}_{2}C{H}_{2}C{H}_3\stackrel{O{C}_2{H}_5^{-}}{\to }\underset{\text{Pentene - 2(major) trans}}{C{H}_{3}CH=CHC{H}_{2}C{H}_3}+\underset{\text{Pentene 1(min or) cis}}{C{H}_{2}CHC{H}_{2}C{H}_{2}C{H}_3}$ Since trans- alkene is more stable than cis.Thus trans-pentene -2 is the main product.