question_answer

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

A) 2-Ethoxypentane

B) Pentene-1

C) cis-Pentene-2    

D) trans-Pentene-2

Correct Answer: D

Solution :

[d] Potassium ethoxide is a strong base, hence causes elimination reaction. The alkene formed is governed by Saytzeffrule "more substituted an alkene, higher will be its stability".  \[C{{H}_{3}}\overset{Br}{\mathop{\overset{|}{\mathop{C}}\,}}\,HC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\xrightarrow{^{-}O{{C}_{2}}{{H}_{5}}}\] \[C{{H}_{2}}=\underset{Pentene-1}{\mathop{CHC{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,+C{{H}_{3}}CH\] \[\underset{Pentene\,-2\,(more\,stable)}{\mathop{=CHC{{H}_{2}}C{{H}_{3}}}}\,\] Further pentene-2 shows geometrical isomerism, cis- and trans- in which trans-isomer having bulkieri groups away from each other is more stable than the cis-isomer.