Problem

Solutions For An Introduction to Medicinal Chemistry Chapter 7 Problem 7Q

Step-by-Step Solution

Solution 1

A hydrogen bond is the bond formed between the hydrogen of one molecule and an electronegative atom of another molecule. Hydrogen involved in the bond formation is attached to the electronegative atom, which can be oxygen, nitrogen, and fluorine. Another effect involved is the mesomeric effect, which is the electron withdrawing or releasing capability of a molecule based on resonance structures.

It is symbolized by M. The mesomeric effect is negative (-M) when the molecule is an electron-withdrawing group. It attracts more electrons towards itself. The effect is positive (+M) when molecule is an electron releasing group. It has less affinity for electrons than the other group involved. Thus, it gives its electron to the other molecule.

In quinazoline, N-H (nitrogen-hydrogen) forms a strong hydrogen bond. In analog 1, H-N-R will form hydrogen bonds with H of water molecules. An R group in N-H-R tends to give electrons, reducing the electro negativity of N. The overall activity of N-H-R in hydrogen bond formation will decrease as compared to that in quinazoline.

Analog 2 has CN (cyanide) group, nitrile will show (-M) effect means it is an electron withdrawing group. N of CN will form hydrogen bonds with H2O. Presence of O in H2O creates a partial positive charge on the hydrogen atom.

It attracts the electron lone pair on the N in a nitrile and a strong hydrogen bond is formed. This bond is stronger than that involved in quinazoline. Therefore, analogue I is three times less active, whereas analogue II is 20 times more active than quinazoline.

Add your Solution
Textbook Solutions and Answers Search
Solutions For Problems in Chapter 7