Reaction of Ethene with Bromine in CCl_4 will result in

The correct answer is Option (2) → Vicinal- Dibromide

Alkenes react with bromine in an inert solvent like $\text{CCl}_4$ by electrophilic addition. The $\pi$ bond breaks and two bromine atoms add across the double bond on adjacent carbons.

Option 1: Gem-Dibromide

Geminal dihalides have both halogens on the same carbon ($-\text{CBr}_2-$). These usually form from the addition of $\text{HX}$ twice or from substitution reactions, not from $\text{Br}_2$ addition to an alkene. Ethene does not give this product under these conditions.

Option 2: Vicinal-Dibromide

Vicinal means on neighboring carbons. $\text{Br}_2$ adds across the $\text{C}=\text{C}$ via a bromonium ion intermediate, giving $\text{Br}$ atoms on each carbon of the former double bond. Ethene forms 1,2-dibromoethane, which is a classic test for unsaturation.

$\text{CH}_2 = \text{CH}_2 + \text{Br}_2 \overset{{\text{CCl}_4}}{\longrightarrow} \text{BrCH}_2\text{CH}_2\text{Br}$

Option 3: Vinylic-Dibromide

Vinylic halides have halogen directly attached to a double-bond carbon ($\text{C=C--Br}$). But in this reaction, the double bond is destroyed during addition, so no vinylic product forms.

Option 4: Allylic-Bromide

Allylic bromination occurs in the presence of light or NBS and involves substitution at the carbon next to a double bond. Here, reaction is addition in $\text{CCl}_4$, not radical substitution, so this product is not formed.