The major product obtained on treatment of ​\[C{H_3}C{H_2}CH(F)C{H_3}\] with \[C{H_3}{O^ - }/C{H_3}OH\] is:
A. \[C{H_3}C{H_2}CH(OC{H_3})C{H_3}\]
B. \[C{H_3}CH = CHC{H_3}\]
C. \[C{H_3}C{H_2}CH = C{H_2}\]
D. \[C{H_3}C{H_2}C{H_2}C{H_2}OC{H_3}\]

Hint: Here, an alkyl halide is reacting with a methoxide ion (\[C{H_3}{O^ - }\] 9) in the presence of methanol (\[C{H_3}OH\]). The alkyl halide will undergo an elimination reaction called dehydrohalogenation. This elimination reaction will result in a mixture of products, only one of which will be the major product due to its stability.

Complete Step by Step Solution:
In the given question, we have been asked to predict the major product when 2-fluorobutane reacts with a methoxide ion (\[C{H_3}{O^ - }\] ) in the presence of methanol (\[C{H_3}OH\] ) as shown below:

Image: Reaction in question

The methoxide ion has lone pairs of electrons that it can donate to an electron-deficient carbon atom and thus, act as a nucleophile. However, it does not act like a nucleophile. Instead, the methoxide ion prefers to attack a hydrogen atom of 2-fluorobutane and abstract it as a proton (\[{H^ + }\]). Thus, the methoxide ion acts as a base.
If the methoxide ion had acted like a nucleophile, it would have caused a nucleophilic substitution reaction of 2-fluorobutane. Since it acts as a base instead, the reaction that would occur here would be an elimination reaction. Elimination reactions can be either “unimolecular” (the E1 mechanism) or “bimolecular” (the E2 mechanism). Methoxide ion is a strong base; therefore, it would prefer the bimolecular elimination pathway (E2).

The E2 elimination mechanism, in this case, would involve the methoxide ion abstracting a beta-hydrogen and the leaving of the fluorine atom (as fluoride \[{F^ - }\] ), in a single step. This reaction is called dehydrohalogenation.
The product of the reaction would be a mixture of two alkenes, but-1-ene and but-2-ene as shown below:

Image: Formation of but-2-ene


Image: Formation of but-1-ene

But-2-ene is called the Saytzeff product while but-1-ene is called the Hofmann product of the dehydrohalogenation reaction. Since the methoxide ion is less bulky, it is less sterically hindered. Less sterically hindered bases usually have the Saytzeff product as the major one and the Hofmann product the minor one. This is because the Saytzeff product in this reaction (but-2-ene) is more substituted than the Hofmann product (but-1-ene) which makes it more stable.
Thus, option B is correct.

Note: Students may be prone to mistaking the methoxide ion as a nucleophile, considering the reaction to be a nucleophilic substitution, and marking option A as their choice. They would be wrong if they did that. Nucleophilicity and basicity are not the same. Nucleophilicity refers to the ability of an electron-rich species to donate its lone pairs of electrons to electron-deficient species. Therefore, all nucleophiles are Lewis bases. But basicity refers to the ability of an electron-rich species to accept a proton. When an electron-rich species acts as an electron donor (i.e., a nucleophile), it causes substitution. When an electron-rich species acts as a proton acceptor (i.e., a base), it causes elimination. It is highly advised that students learn about the extent of nucleophilicities and basicities of different electron-rich species.