Pent-1-ene with HCl gives:

The reaction of pent-1-ene with HCl (hydrochloric acid) results in the formation of 2-chloropentane.

Explanation:

When an alkene reacts with a hydrogen halide, such as HCl, the halogen atom (in this case, chlorine) adds to the alkene, resulting in the formation of an alkyl halide. This process is known as electrophilic addition.

Reaction mechanism:

The reaction proceeds through a carbocation intermediate. Here is the step-by-step mechanism for the addition of HCl to pent-1-ene:

Step 1: Formation of a carbocation
The pi bond in the alkene is attacked by the electrophilic hydrogen atom of HCl, resulting in the formation of a carbocation intermediate. In the case of pent-1-ene, the positive charge is formed on the carbon atom adjacent to the double bond.

Step 2: Attack of chloride ion
The chloride ion (Cl-) acts as a nucleophile and attacks the carbocation, forming a new bond between the carbon atom and the chlorine atom. This leads to the formation of the alkyl halide product.

Product formation:

In the case of pent-1-ene, the carbocation intermediate can undergo attack by the chloride ion at two different positions, resulting in the formation of two different products: 1-chloropentane and 2-chloropentane. However, due to the stability of the carbocation intermediate, the attack at the secondary carbon (2-position) is favored over the attack at the primary carbon (1-position).

Reason for the formation of 2-chloropentane:

The stability of the carbocation intermediate plays a crucial role in determining the regioselectivity of the reaction. The secondary carbocation formed during the reaction is more stable than the primary carbocation due to the presence of electron-donating alkyl groups. As a result, the chloride ion preferentially attacks the secondary carbon, leading to the formation of 2-chloropentane as the major product.

Therefore, the correct answer is option 'B' (2-chloropentane).