Question

Carbons 1 and 4 of 1,3−cyclopentadiene are equivalent and give the same carbocation on protonation. Likewise, carbons 2 and 3 are equivalent. Write the structure of the carbocation formed by protonation of C−2 or C−3 to verify that it is not allylic and therefore not as stable as the one formed by protonation of C−1 or C−4.

Answer 1

Carbon 1 and 4 in 1,3−cyclopentadiene are equivalent efectively as you can see in the image below. When the molecule are protonated either in position 1 or 4, the double bond is placed in position 2,3.

When carbon 2 and 3 gets protonation respectively, the carbocation is not in the same position of the molecule.

Step-by-step explanation:

In the image below, you can see when 1,3−cyclopentadiene have protonation in the position 1, the first carbocation is placed in carbon 2. Then the pair of electrons from double bond turns to the carbon 2,3 having the carbocation in the carbon 4 of the 1,3−cyclopentadiene. When the molecule is protonated in the carbon 4, the carbocation is placed in carbon 3 and double bond in carbon 1 turns to position 2,3. This is characteristic from an allyl group.

On the other hand, when 1,3−cyclopentadiene is protonated whether in carbon 2 or 3, the double bond present either in carbon 1 or carbon 3 can't move due to the adjacent carbons are saturated.