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Chapter 2: Problem 164

When propyne reacts with excess of chlorine water, the product obtained is (a) 1,1 -dichloro propan-2-one (b) 1,1 -dichloropropan-2-ol (c) 2,2 -dichloropropan- 1 -al (d) 1,2 -dichloropropane

Short Answer

Expert verified
Answer: 1,2-dichloropropane.

Step by step solution

01

Write the chemical equation for the reaction

Begin by writing the chemical equation for the reaction of propyne with chlorine water. We have: Propyne (C3H4) + Cl2 (aq) → Product
02

Determine the most reactive site

In propyne, the carbon-carbon triple bond is the most reactive site, as it can break and form new bonds with incoming atoms (in this case, chlorine). The triple bond consists of one sigma and two pi bonds. During the reaction, one of the pi bonds breaks to form new bonds with the incoming chlorine atom.
03

Apply electrophilic addition

Electrophilic addition occurs when an electrophile (a molecule attracted to electrons) interacts with a nucleophile (a molecule donating electrons). In this reaction, the chlorine molecule (Cl2) acts as the electrophile, and the carbon-carbon triple bond in propyne acts as the nucleophile. The bond between the chlorine atoms breaks, and one chlorine atom forms a bond with each carbon atom of the triple bond. This results in the addition of two chlorine atoms to the propyne molecule, converting it into a vicinal dihalide compound.
04

Identify the correct product

After the electrophilic addition reaction, the product obtained is 1,2-dichloropropane, which corresponds to option (d). The carbon-carbon triple bond converts into a single bond, and each carbon atom involved in the triple bond forms a new bond with a chlorine atom. So, the correct answer is (d) 1,2-dichloropropane.

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