Question

2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is:
A. 1-Pentene
B. cis-2-pentene
C. trans-2-pentene
D. 2-Ethoxypentane

Answer 1

Hint: As we know that when alkyl halides react with the alcoholic KOH, then there will be an elimination reaction. It is found that the more substituted alkenes are more stable.

Complete answer:
- 2-Bromopentane is heated with potassium ethoxide in ethanol.
- As we know that when alkyl halide is reacted with ethoxide undergoes either elimination the substitution reaction like E1, E2 or SN1, SN2 reaction.
- In this reaction we can see that this reaction involves a strong nucleophile, that is ethoxide and a secondary alkyl halide. So, it is clear that this reaction is an E2 or SN1 reaction.
- It is found that this reaction takes place in a protic solvent, that is ethanol. As we know that ethoxide is an anion and it will form hydrogen bonds with the solvent. And further this makes the nucleophile less reactive. This is because the approach of the electrophile is hindered and this is found to block the substitution reaction.
\[\begin{align}
& C{{H}_{3}}-CHBr-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\text{ }+\text{ }{{C}_{2}}{{H}_{5}}-{{O}^{-}}{{K}^{+}}\to \\
& C{{H}_{3}}-CH=CH-C{{H}_{2}}-C{{H}_{3}}+KBr\text{ }+\text{ }{{C}_{2}}{{H}_{5}}-OH \\
\end{align}\]
- We can see here that the preferred mechanism is beta elimination that is by E2 mechanism and is found to occur through Zaitsev rule.
- The more substituted alkenes are more stable. We can see that as the bulky groups are placed far apart, therefore the trans isomer is preferred one. when 2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is trans-2-pentene.

Hence, we can conclude that the correct option is ©.

Note: - We should note here that trans-2-pentene is more symmetrical when compared to cis-isomers. Hence, the major product formed is trans-2-pentene.