1. Write the substrate clearly

The compound is 3-bromo-2-fluoropentane:

$\mathrm{CH_3-CH(F)-CH(Br)-CH_2-CH_3}$

In an E2 elimination, the leaving group is usually the better leaving group. Since

$\mathrm{Br^-} \text{ is a much better leaving group than } \mathrm{F^-},$

the elimination will occur mainly by removal of HBr, not HF.

So we consider elimination of a $\beta$-hydrogen adjacent to the carbon bearing Br.

2. Identify the $\alpha$-carbon and possible $\beta$-carbons

The carbon bearing Br is carbon-3. Thus:

• $\alpha$-carbon = $\mathrm{C_3}$
• possible $\beta$-carbons = adjacent carbons $\mathrm{C_2}$ and $\mathrm{C_4}$

So two eliminations are possible:

(i) Elimination from $\mathrm{C_2}$

Remove H from $\mathrm{C_2}$ and Br from $\mathrm{C_3}$:

$\mathrm{CH_3-C(F)=CH-CH_2-CH_3}$

This gives 2-fluoropent-2-ene.

(ii) Elimination from $\mathrm{C_4}$

Remove H from $\mathrm{C_4}$ and Br from $\mathrm{C_3}$:

$\mathrm{CH_3-CH(F)-CH=CH-CH_3}$

This gives 4-fluoropent-2-ene (or equivalently 2-fluoropent-3-ene depending on numbering convention).

3. Decide the major product

Ordinarily, E2 reactions often give the more substituted alkene (Saytzeff product). However, here one of the possible alkenes has fluorine directly attached to a double-bond carbon (a vinylic fluoride):

$\mathrm{CH_3-C(F)=CH-CH_2-CH_3}$

Fluorine is strongly electron-withdrawing by the $-I$ effect, which destabilizes this alkene relative to the alternative product. Therefore, the alkene with fluorine not directly on the double bond carbon in the same unfavorable way is preferred.

Hence the major product is:

$\mathrm{CH_3-CH(F)-CH=CH-CH_3}$

4. Conclusion

The major E2 elimination product of 3-bromo-2-fluoropentane is

$\boxed{\mathrm{CH_3-CH(F)-CH=CH-CH_3}}$

This corresponds to option C.

5. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So the derived answer agrees with the stored answer.