The structure of the major product formed in the following reaction is :

The given substrate is $$2$$-bromobutane  $$\left(CH_3\,CH_2\,CHBr\,CH_3\right)$$ and the reagent is hot alcoholic $$KOH$$.

Step 1 : Nature of the reagent
Alcoholic $$KOH$$ behaves predominantly as a strong base. With alkyl halides it favours $$\beta$$-elimination (E2) far more than nucleophilic substitution, especially when a secondary (or tertiary) halide is present.

Step 2 : Identification of the β-hydrogens
In $$2$$-bromobutane the carbon bearing the halogen is C-2. Its two adjacent (β) carbons are   • C-1 carrying two hydrogens ($$\beta_1$$-H) and   • C-3 carrying one hydrogen ($$\beta_2$$-H). Removal of any one of these β-hydrogens together with the leaving group $$Br^-$$ will give an alkene.

Step 3 : Possible alkenes and Zaitsev (Saytzeff) rule
β-Elimination across C-1/C-2 gives $$CH_2=CH\,CH_2\,CH_3$$ (1-butene, monosubstituted).
β-Elimination across C-2/C-3 gives $$CH_3\,CH=CH\,CH_3$$ (2-butene, disubstituted).
Zaitsev rule: the more substituted alkene is formed preferentially because it is thermodynamically more stable. Hence 2-butene is the major product.

Step 4 : Geometrical (E/Z) outcome in an E2
The anti-periplanar arrangement required for a concerted E2 is achieved more easily when the two larger groups ($$CH_3$$ and $$CH_3$$) end up on opposite sides of the emerging $$C=C$$ bond. Therefore the trans (E) isomer of 2-butene is favoured over the cis (Z) form.

Major product
The principal product is trans-2-butene:

$$CH_3\,CH=CH\,CH_3$$  with $$CH_3$$ groups on opposite sides of the double bond.

Among the given structures, this corresponds to Option C.

Hence the correct answer is:
Option C which is: trans-2-butene ($$CH_3\,CH=CH\,CH_3$$).