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

Solution

Chemistry ma’am · 4 steps

Bromide leaves, fluorine stays, and the base pulls off the H next to fluorine.

1. Draw the chain. 3-bromo-2-fluoropentane is CH₃-CHF-CHBr-CH₂-CH₃. Fluorine sits on C2 and bromine on C3.
2. Pick the leaving group. The C-Br bond is much weaker than the C-F bond, and Br⁻ is a far better leaving group than F⁻. So the molecule loses HBr, and F stays in the product.
3. Choose the β-hydrogen. The carbons next to C3 are C2 and C4, so the H can come from either. The H on C2 is more acidic because the electronegative F beside it pulls electron density away (−I effect). Removing it also gives the more substituted alkene, CH₃-CF=CH-CH₂CH₃.

   Option	Lost	β-H taken from	Verdict
   (a)	HF	C1	F cannot leave in preference to Br
   (b)	HF	C3	F cannot leave in preference to Br
   (c)	HBr	C2, next to F	acidic H, trisubstituted alkene: major
   (d)	HBr	C4	less acidic H, disubstituted alkene: minor

4. So. The major product is 2-fluoropent-2-ene, CH₃CH₂CH=C(F)CH₃, which is option (c).

The idea to keep. In E2 the halogen with the weaker C–X bond leaves, so Br goes and F stays. A β-H next to an electronegative atom is more acidic, and pulling it off here also gives the more substituted alkene.

Answer from the NTA final answer key. Solution by PrepWiser.