1 Before you start
A quick baseline. Your answers aren't graded now. You'll see the same questions at the end to measure what you've learned.
2 Valence: how many bonds each atom gets
You can now read a drawing. This module is about the rules that constrain what can be drawn at all: how many bonds each atom is allowed, what shape those bonds take, and how to work out a charge rather than guess it.
Essentially every drug is built from seven elements, and each has a preferred number of bonds. Learn these and most structures become self-checking.
- Carbon — four bonds, always. A carbon with three is either charged or a mistake.
- Nitrogen — three bonds and a lone pair. With four bonds it carries a positive charge.
- Oxygen — two bonds and two lone pairs. With one bond it is negative; with three it is positive.
- Sulfur — two, like oxygen, but it will also take four or six, which is how sulfoxides and sulfones exist.
- Phosphorus — three or five. Drug phosphates use five.
- Halogens — one bond each. Fluorine, chlorine, bromine, iodine.
That is the whole table, and it does more work than any other single thing you will memorise in this course. If a structure violates it, either you have misread the drawing or there is a charge you have not accounted for.
3 Formal charge, done by hand
The arithmetic is short. For any atom: formal charge = (valence electrons in the free atom) − (lone pair electrons) − (number of bonds).
Nitrogen has five valence electrons. In an amine it has three bonds and one lone pair, so 5 − 2 − 3 = 0, neutral. Protonate it and it has four bonds and no lone pair: 5 − 0 − 4 = +1. That is why a nitrogen with four bonds is always positive, with no exceptions worth remembering.
Oxygen has six. In an alcohol, two bonds and two lone pairs: 6 − 4 − 2 = 0. In a carboxylate, the singly-bonded oxygen has one bond and three lone pairs: 6 − 6 − 1 = −1. So an oxygen with one bond is negative.
The charge on a protonated amine
Salbutamol has a secondary amine. What is its formal charge at pH 7.4?
Step 1: study the solution
- At pH 7.4 an aliphatic amine (pKa near 10) is protonated
- Protonation adds a bond to the nitrogen and consumes the lone pair
- Nitrogen: 5 valence electrons, 0 lone pair electrons, 4 bonds
- 5 − 0 − 4 = +1
Step 2: fill the blanks
Now do the carboxylate of ibuprofen at the same pH.
Step 3: now you try
Deprotonation leaves an oxygen with one bond. Count its lone pairs.
A nitrogen in a structure is drawn with four bonds and no charge shown. What should you conclude?
4 Sigma, pi, and what a double bond actually is
Every single bond is a sigma bond: electron density lying along the axis between the two nuclei. It is cylindrically symmetric, which is why you can rotate around a single bond without breaking anything.
A double bond is one sigma bond plus one pi bond, and a pi bond is made from two p orbitals overlapping side by side, above and below the axis. Rotating would tear that overlap apart, so a double bond cannot rotate. That single fact is behind E/Z isomerism, behind the rigidity of aromatic rings, and behind the amide bond's behaviour, which gets its own treatment in module 6.
5 Hybridisation as bookkeeping
Hybridisation is not a physical process a molecule undergoes. It is a bookkeeping device that lets you predict geometry from a drawing, and that is all you need it for.
- sp³ — four sigma bonds, no pi. Tetrahedral, bond angles about 109°. A saturated carbon.
- sp² — three sigma bonds and one pi. Trigonal planar, angles about 120°. A carbonyl carbon, an alkene carbon, any aromatic ring atom.
- sp — two sigma bonds and two pi. Linear, 180°. A nitrile carbon, an alkyne.
The count that matters: add up the sigma bonds and lone pairs. Four means sp³, three means sp², two means sp. An sp³ carbon is a three-dimensional corner; an sp² carbon is part of a flat patch. Drug molecules that are almost all sp² are flat, and flat molecules stack, crystallise well and dissolve badly — which is unit 3.5's problem, and where the fraction of sp³ carbon in module 12 comes from.
Caffeine
Morphine
6 Lone pairs, and where they are
Lone pairs are not decoration either. They are what makes an atom a hydrogen bond acceptor, what makes a nitrogen basic, and what a metal ion coordinates to. Knowing where they are is knowing where the molecule can interact.
- Nitrogen with three bonds: one lone pair, available.
- Oxygen with two bonds: two lone pairs.
- A carbonyl oxygen: two lone pairs, and it is one of the best acceptors in drug chemistry.
The important qualification is that a lone pair can be spent. If it is delocalised into a pi system, it is no longer available to accept a hydrogen bond or to pick up a proton. That is the entire difference between an amine and an amide, and between pyridine and pyrrole. Both comparisons are coming, in module 3 and module 4 respectively — and both are the same idea.
7 Resonance: three cases that matter
Resonance means the real structure is not any one of the drawings; it is something in between, and the electrons are spread over more atoms than one drawing suggests. Three cases account for most of what you will meet.
The carboxylate. Both oxygens are equivalent. The negative charge is spread over both, each carrying half. This is why a carboxylic acid is so much more acidic than an alcohol: the anion it forms is stabilised, so it forms readily.
The amide. The nitrogen lone pair delocalises onto the carbonyl oxygen. The consequences are large and you will meet them repeatedly: the amide nitrogen is not basic, the C-N bond has partial double-bond character and so does not rotate freely, and the whole unit is flat.
The guanidinium. The positive charge is spread over three nitrogens, which stabilises it so thoroughly that the group stays protonated at every pH in the body. Arginine's side chain is a guanidinium, which is why arginine is the residue that so often grips a drug's carboxylate.
Metformin
Why is an amide nitrogen not basic, when an amine nitrogen is?
8 The octet rule and the exceptions you will actually see
Second-row elements — carbon, nitrogen, oxygen, fluorine — obey the octet rule and there is no useful exception. Below that row, atoms can exceed it.
In practice you will meet three groups drawn inconsistently across the literature, and it is worth knowing that the disagreement is about notation rather than chemistry. The sulfonamide and the sulfone are drawn either with two S=O double bonds or with S⁺–O⁻ pairs. The phosphate is drawn with a P=O or with P⁺–O⁻. The N-oxide is drawn as N⁺–O⁻ or, sloppily, as N=O. All of these describe the same molecule; different software will normalise them differently, and that is a real practical nuisance when you compare descriptor values between toolkits.
Celecoxib
9 Practice
Metformin
Ibuprofen
Take any structure from module 1 and find every lone pair in it. Which of them are available, and which have been spent on a pi system?