Q.37. - Organic Chemistry

Question

Question: Methotrexate, a drug that inhibits the metabolism of folic acid, is used in the treatment of a variety of cancers and autoimmune disorders such as rheumatoid arthritis. (a) Give the hybridization of each N atom in methotrexate. (b) In what type of orbital does the lone pair of each N reside? (c) Explain why the bicyclic ring system that contains four N atoms is aromatic.

Step-by-Step Solution

Answer

Answer

 

a. The hydrogen atoms marked by an asterisk are $s{p}^2$ hybridized, whereas the other nitrogen atoms are $s{p}^3$ hybridized.  

b. The lone pair of the hydrogen atoms marked by an asterisk is located in the $s{p}^2$ orbital, and the lone pair of other nitrogen atoms are situated in the p orbital.

c. Methotrexate is cyclic, planar, conjugated, and obeys Huckel’s rule. Hence this molecule is aromatic.

Step 1: Methotrexate

Methotrexate is formerly familiar by another name which is noted as amethopterin. It can be utilized to cure several cancers like breast cancer and lung cancer.

Step 2: Aromatic heterocycles

Heterocycles comprising oxygen, nitrogen, or sulfur and possessing lone electron pairs are regarded to be aromatic. The lone pair can be a portion of the delocalized $\mathbf{\pi }$  system or can be localized on the heteroatom.

Step 3: Answers related to methotrexate

a. 

Methotrexate with labeled and unlabeled nitrogen atoms

 

The atoms denoted by an asterisk are surrounded by two groups and contain a lone pair. Hence the hybridization is $s{p}^2$ . The other nitrogen atoms are surrounded by three groups and have a lone pair of electrons. Hence the hybridization is $s{p}^3$ .   

 

 b. The hybridization of the nitrogen denoted by an asterisk is $s{p}^2$ the lone pair occupying one of the $s{p}^2$ orbital. The hybridization of the other nitrogen atoms is $s{p}^3$ the lone pair occupying the p orbital and taking part in delocalization.  

 

c. Methotrexate has a cyclic, planar, and completely conjugated system. The lone pair of electrons on nitrogen atoms situated in the bicyclic ring does not take part in the electron delocalization. It follows Huckel’s rule, and the number of pi electrons is $10\mathrm{\pi }(\left(4\times 2+2\mathrm{\pi }\right)$  electrons. Hence, this molecule is aromatic.