Q.36. - Organic Chemistry

Question

Question: The purine heterocycle occurs commonly in the structure of DNA.

a. How is each N atom hybridized?

b. In what type of orbital does each lone pair on a N atom reside?

c. How many $\mathbf{\pi }$ electrons does purine contain?

d. Why is purine aromatic?

Step-by-Step Solution

Answer

Answer

a. The hybridization of each nitrogen in purine is $s{p}^2$ .

b. The unlabeled nitrogen atoms are $s{p}^2$ hybridized and have lone pairs in one of the $s{p}^2$ hybrid orbitals. The labeled nitrogen has its lone pair in a p orbital.  

c. Purine contains $10\mathrm{\pi }$ electrons.

d. Purine is aromatic as it is cyclic, planar, conjugated, and has $\left(4n+2\right)\mathrm{\pi }$  electrons.

Step 1: Purines

Purine is a compound comprising nitrogen as a heteroatom and falls under the category of heterocyclic compounds. These compounds form part of the DNA, and the different kinds of purine include adenine and guanine.

Step 2: Aromatic compounds

Several properties of aromatic compounds are given below:

• Each atom in the ring possesses a p orbital for the delocalization of electron density.
• Aromatic compounds are not subjected to the addition reactions of alkenes

Step 3: Purine heterocycle

a.

Purine heterocycle with labeled and unlabeled nitrogens

The nitrogen atom noted with an asterisk is surrounded by three groups and possesses a lone pair. This lone pair takes place in the delocalization of electrons. The other nitrogen atoms are surrounded by two groups and have one lone pair. Hence the hybridization of each nitrogen is $s{p}^2$ .

b. The three hydrogens that are not labeled with an asterisk are $s{p}^2$  hybridized, possessing lone pair of electrons in one of the $s{p}^2$ hybrid orbitals. The nitrogen marked with an asterisk contains lone pair of electrons in a p orbital.   

c. The purine molecule has $4\mathrm{\pi }$ bonds and a lone pair of electrons. According to Huckel’s rule, the number of pi electrons is given by $\left(4n+2\right)\mathrm{\pi }$ electrons, where n indicates a whole number. The value of $4n+2$ can be calculated as:

$$\begin{gathered} 4n+2 = 4\times 2+2 \\ =10\mathrm{\pi } \end{gathered}$$

 

d. For a compound to become aromatic, it must be cyclic, planar, completely conjugated, and contain $\left(4n+2\right)\mathrm{\pi }$  electrons. Purine is cyclic, planar, completely conjugated, and includes  $10\mathrm{\pi } \mathrm{i}.\mathrm{e}. \left(4\times 2+2\right)\mathrm{\pi }$ electrons. Hence, it is an aromatic molecule.