Final answer:
The most stable product formed when 2-bromo propyl is subjected to heat and NaOEt is propene option (a), due to a beta-elimination (E2 reaction) forming the alkene with a double bond between the alpha and beta carbons.
Step-by-step explanation:
The most stable product is formed when 2-bromo propyl is subjected to heat and NaOEt is most likely propene. This reaction is a type of elimination reaction known as a beta-elimination or E2 reaction.
2-Bromopropane, subjected to a strong base like NaOEt (sodium ethoxide) under heating conditions, will tend to lose a hydrogen atom from the beta carbon (the carbon adjacent to the carbon bearing the leaving group, which in this case is the bromine atom) and the bromine will leave, thus forming a double bond between the alpha and beta carbons.
This results in the formation of propene, the next largest alkene after ethene in the alkene series. Propene, also known as propylene, has three carbon atoms with a double bond between two of the carbon atoms.