Answer:

When ethanolic KOH is reacted with 1,1-dichloropropane, it undergoes an elimination reaction known as dehydrohalogenation. This reaction results in the removal of a hydrogen atom and a chlorine atom from adjacent carbon atoms, forming a triple bond between the carbon atoms. The product formed in this reaction is propyne.

Explanation:

The reaction between ethanolic KOH and 1,1-dichloropropane proceeds through an E2 (bimolecular elimination) mechanism. Let's break down the process step by step:

1. Formation of an alkoxide ion:
- The ethanolic KOH (potassium hydroxide dissolved in ethanol) dissociates into potassium cation (K+) and hydroxide anion (OH-).
- The OH- then reacts with the 1,1-dichloropropane, abstracting a proton (H+) from one of the carbon atoms, resulting in the formation of an alkoxide ion (CH3CH2O-).

2. Transition state formation:
- The alkoxide ion acts as a strong base and abstracts a proton (H+) from the adjacent carbon atom, leading to the formation of a transition state.
- In this transition state, the leaving group (chlorine) and the hydrogen being abstracted are anti-periplanar, which means they are in a staggered conformation on opposite sides of the molecule.

3. Simultaneous bond formation and bond cleavage:
- In the transition state, the C-H bond is breaking, and the C-Cl bond is also breaking.
- At the same time, a new bond starts to form between the adjacent carbon atoms, resulting in the formation of a triple bond (C≡C) between the two carbon atoms.

4. Product formation:
- Finally, the leaving group (chlorine) departs with its pair of electrons, and the triple bond between the carbon atoms is formed.
- The product formed is propyne (CH3C≡CH), which is an alkyne.

Therefore, the correct answer is option 'B' - Propyne.