Question

Predict the major product formed by 1,4 -addition of \(\mathrm{HCl}\) to cyclopentadiene.

Short answer

Answer: The major product formed by the 1,4-addition of HCl to cyclopentadiene is the one with hydrogen at C1 and chlorine at C4.

Step-by-step solution

Draw the structure of cyclopentadiene

First, let's draw the structure of cyclopentadiene. It is a 5-membered ring with two double bonds that are conjugated (alternating single and double bonds). The structure can be represented as follows: 

Identifying the reactive sites in cyclopentadiene and HCl

The double bonds in cyclopentadiene are the reactive sites where the 1,4-addition can take place. In HCl, the hydrogen atom acts as the electrophile and the chlorine atom acts as the nucleophile for this reaction, and they would be added across the reactive double bonds.

Predict the possible 1,4-addition products

Two possible 1,4-addition products can be formed when HCl is added to cyclopentadiene: one when the hydrogen atom attaches to C1 and the chlorine atom attaches to C4, and the other when the hydrogen atom attaches to C4 and the chlorine atom attaches to C1. Let's draw the structures of these two possible products: Product A (Hydrogen at C1, Chlorine at C4):  Product B (Hydrogen at C4, Chlorine at C1): 

Analyze the stability of the intermediates

To determine the major product, we need to analyze the stability of the carbocation intermediates formed during the reaction. The intermediate formed when HCl reacts with double bond C1=C2 is more stable than the intermediate formed when HCl reacts with double bond C4=C5 due to resonance effect. Thus, the hydrogen atom will be added preferentially to C1, and the chloride ion (nucleophile) will be added to C4.

Identify the major product

Based on the analysis of the intermediate stability, the major product will be the one with hydrogen at C1 and chlorine at C4. Therefore, Product A is the major product resulting from the 1,4-addition of HCl to cyclopentadiene. Here is the structure of the major product once again: