Chapter 8: Q50P (page 365)

Which compound in each set is aromatic? Explain your choice.

Short Answer

Expert verified

(a) From the 1^st set, Cyclopropenyl cation is aromatic.

(b) From the 2^nd set, Cycloheptatrienyl cation is aromatic.

Step by step solution

Measurement of aromaticity

Aromaticity of a compound can be measured through theHückel’s rule, which is stated thatfor a planar, cyclic compound to be aromatic, its uninterrupted π cloud must contain (4n+ 2) π electrons, where nis any whole number. According to Hückel’s rule, then, an aromatic compound must have 2 (n= 0), 6 (n= 1), 10 (n= 2), 14 (n= 3), 18 (n= 4), and so on, π electrons.

Explanations of aromaticity of the compounds

(a) 1^st Compound is not aromatic because it does not have an uninterrupted ring of porbital-bearing atoms. One of its ring atoms is sp³ hybridized, and only sp² and sp carbons have porbitals. Therefore, cyclopropene does not fulfill the criterion for aromaticity.

2^nd Compound has all planar carbons(sp²) total 2 electrons a and to satisfy

Hückel’s rule of (4n+ 2) π electrons, n should be 0. So, the compound is aromatic.

3^rd Compound has 4 electrons (one from π bond and another from lone pair) but here n value can’t satisfyHückel’s rule of (4n+ 2) π electrons. So, the compound is not aromatic.

So, from the 1^st set, cyclopropenyl cation is the aromatic compound.

(b) 1^st Compound is not aromatic because it does not have an uninterrupted ring of porbital-bearing atoms. One of its ring atoms is sp³ hybridized, and only sp² and sp carbons have porbitals. Therefore, cycloheptatriene does not fulfill the criterion for aromaticity.

The 2^nd compound is aromatic as it has a planar structure, and 6 electrons which can be satisfied by Hückel’s rule of (4n+ 2) π electrons where n =1.

The 3^rd compound is not aromatic as it has 8 electrons which can’t be satisfied by Hückel’s rule of (4n+ 2) π electrons using any n value.

So, from the 2^nd set, cycloheptatrienyl cation is the aromatic compound.