Organic Chemistry

Section 1.4 Representation of Structure 15

In studying the molecules in this section, notice that when the atoms don’t bear a
formal charge or an unpaired electron, hydrogen and the halogens each have onecova-
lent bond, oxygen always has twocovalent bonds, nitrogen always has threecovalent
bonds, and carbon has fourcovalent bonds. Notice that (except for hydrogen) the sum
of the number of bonds and lone pairs is four: The halogens, with one bond, have three
lone pairs; oxygen, with two bonds, has two lone pairs; and nitrogen, with three bonds,
has one lone pair. Atoms that have more bonds or fewer bonds than the number re-
quired for a neutral atom will have either a formal charge or an unpaired electron.
These numbers are very important to remember when you are first drawing structures
of organic compounds because they provide a quick way to recognize when you have
made a mistake.

In the Lewis structures for and notice that each
atom has a complete octet (except hydrogen, which has a filled outer shell) and that
each atom has the appropriate formal charge. (In drawing the Lewis structure for a
compound that has two or more oxygen atoms, avoid oxygen–oxygen single bonds.
These are weak bonds, and few compounds have them.)

A pair of shared electrons can also be shown as a line between two atoms. Compare
the preceding structures with the following ones:

Suppose you are asked to draw a Lewis structure. In this example, we will use

1. Determine the total number of valence electrons (1 for H, 5 for N, and 6 for each
   ).


2. Use the number of valence electrons to form bonds and fill octets with lone-pair
   electrons.


3. If after all the electrons have been assigned, any atom (other than hydrogen) does
   not have a complete octet, use a lone pair to form a double bond.


4. Assign a formal charge to any atom whose number of valence electrons is not
   equal to the number of its lone-pair electrons plus one-half its bonding electrons.
   (None of the atoms in has a formal charge.)

H O N O

use a pair of electrons

to form a double bond

N does not have

a complete octet double bond

18 electrons have been assigned

H O N O

by using one of oxygen’s lone pairs

to form a double bond, N gets a

complete octet

HNO 2

O= 1 + 5 + 12 = 18

HNO 2.

C O H

O

H + − C H −−

O

O N H

O

H O O C

O

O NN

− −−

HCOH +

O

OCO

O

HONO NN

O

HHC

O

CH 2 O 2 , HNO 3 , CH 2 O, CO 32 - , N 2 ,

one bond one bond two bonds three bonds four bonds

H O N C

F Cl

I Br

H

hydride

ion

H

hydrogen

radical

− Br− Br BrBr ClCl

bromide

ion

H+

hydrogen
ion

bromine

radical

bromine chlorine