When 2 -bromopentane is heated with potassium ethoxide, we get trans-pent-2-ene. Because when alkyl halides react in ethanol with potassium ethoxide, it give trans product as major product due to more symmetrical nature as compared to cis-isomer. The reaction occurs as follows:
CH3−
C
|
Br
H−CH2−CH2−CH3+C2HS−OK 2- bromopentane
Ethanol
───────▶
CH3−
CH=CH−CH2
trans-pent- 2 -ene
−CH3
(It shows elimination reaction.) Hence, trans −2− pentane is the main product and option (a) is the correct answer.