CEB1013 ORGANIC CHEMISTRY

THE BASIC:
Resonance, Isomer & Hybridization

Course Learning Outcome:

CLO1: Describe the electronic structure, bonding and shape of the various
functional groups and extrapolate these to describe the origins of reactivity
of organic compounds

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LEARNING OUTCOME
By end of this lecture, student should be able to

1. Explain stability of organic compounds through their resonance structures.

2. Classify each pair of compounds as constitutional isomers, stereoisomers,
identical molecules
3. Determine the hybridization and geometry around every atom.
4. Compare bond length and bond strength for certain bonds.
5. Determine the shape around any atom bonded to two other atoms based on
the bond angle

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RESONANCE
For many molecules and ions, no single Lewis structure provides a truly accurate
representation

You have to show the electron movement (using arrow) and the formal charges for
every atoms.
▪ Curved arrow: a symbol used to show the redistribution of valence electrons
▪ In using curved arrows, there are only two allowed types of electron
redistribution: from a bond to an adjacent atom AND from an atom to an
adjacent bond
▪ Electron pushing is a survival skill in organic chemistry!!! learn it well!

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STABILITY OF RESONANCE
Rule 1: filled valence shells
▪ Structures in which all atoms have filled valence shells contribute more than
those with one or more unfilled valence shells

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STABILITY OF RESONANCE (C ONT.)

Rule 2: Maximum number of covalent bonds

▪ Structures with a greater number of covalent bonds contribute more than those
with fewer covalent bonds

+ •• +
CH3 O C H CH3 O C H
•• ••

H H
Greater contribution Lesser contribution
(8 covalent bonds) (7 covalent bonds)

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STABILITY OF RESONANCE (C ONT.)

Rule 3: Least separation of unlike charge

▪ Structures with separation of unlike charges contribute less than those with no
charge separation

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STABILITY OF RESONANCE (C ONT.)

Rule 4: Negative charge on the more electronegative atom

▪ Structures that carry a negative charge on the more electronegative atom
contribute more than those with the negative charge on the less electronegative
atom

O O O
(1) (2)
C C C
H3 C CH3 H3 C CH3 H3 C CH3
(a) (b) (c)
Lesser Greater Should not
contribution contribution be drawn

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STABILITY OF RESONANCE (C
ONT.)

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CLASS ACTIVITY 1.3
1. Draw all reasonable resonance structures for each species.

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CLASS ACTIVITY 1.3
2. Keeping the same atomic connections and moving only electrons, write a more
stable Lewis structure for each of the following. Be sure to specify formal
charges, if any, in the new structure.

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ISOMER
▪ Isomers are different molecules having the same molecular formula.

▪ Ethanol and dimethyl ether are constitutional isomers because they have the
same molecular formula, but the connectivity of their atoms is different.
▪ For example, ethanol has one C – C bond and one O – H bond, whereas dimethyl
ether has two C – O bonds.

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ISOMER (C
ONT.)

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ISOMER (C ONT.)

▪ Stereoisomer: Isomers with the same connectivity but a different orientation of

their atoms in space.

▪ The cis isomer has two groups on the same side of the ring.
▪ The trans isomer has two groups on opposite sides of the ring.

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CLASS ACTIVITY 1.4
3. With reference to compound A and B drawn below, label each compound as an
isomer, a resonance structure, or neither.

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CLASS ACTIVITY 1.4
4. Classify each pair of compounds as constitutional isomers, stereoisomers,
identical molecules, or not isomers of each other.

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HYBRIDIZATION

Visualizing Electron Orbitals

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HYBRIDIZATION – HYDROGENS
▪ What orbitals do the first- and second-row atoms use to form bonds?
▪ Let’s begin with hydrogen and then examine the orbitals used for bonding by
atoms in the second row.

▪ A σ bond concentrates electron density on the axis that joins two nuclei. All
single bonds are σ bonds.

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HYBRIDIZATION – CH4

▪ This lowest energy arrangement of electrons for an atom is called its ground
state.

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HYBRIDIZATION – CH4 (C ONT.)

▪ In this description, carbon should form only two bonds because it has only two
unpaired valence electrons, and CH2 should be a stable molecule.
▪ In reality, however, CH2 is a highly reactive species that cannot be isolated under
typical laboratory conditions. In CH2, carbon would not have an octet of
electrons.

Then, how?
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HYBRIDIZATION – CH4 (C ONT.)

▪ Chemists have proposed that atoms like carbon do not use pure s and pure p
orbitals in forming bonds. Instead, atoms use a set of new orbitals called hybrid
orbitals.
▪ The mathematical process by which these orbitals are formed is called
hybridization.

▪ These hybrid orbitals are called sp3 hybrids because they are formed from one s
orbital and three p orbitals.

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HYBRIDIZATION – CH4 (C ONT.)

▪ What do these new hybrid orbitals look like?

▪ The four hybrid orbitals form four equivalent bonds. We can now explain the
observed bonding in CH4.
▪ Each bond in CH4 is formed by overlap of an sp3 hybrid orbital of carbon with a
1s orbital of hydrogen.
▪ These four bonds point to the corners of a tetrahedron.

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CLASS ACTIVITY 1.5
Draw the hybrid orbitals for the following compounds

1. NH3

2. H2O

3. BF3

4. H2C=CH2

5. HC≡CH

▪ One 2s orbital and three 2p orbitals form four sp3 hybrid orbitals.
▪ One 2s orbital and two 2p orbitals form three sp2 hybrid orbitals.
▪ One 2s orbital and one 2p orbital form two sp hybrid orbitals.

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CLASS ACTIVITY 1.5 - ANSWER

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CLASS ACTIVITY 1.5 - ANSWER

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HYBRIDIZATION – SUMMARY

▪ The σ bond is formed from the end-on overlap of Csp2 – Osp2.

▪ The π bond is formed from the side-by-side overlap of C2p – O2p.

▪ The O atom has three sp2 hybrid orbitals.

▪ One is used for the σ bond of the double bond.
▪ The remaining two sp2 hybrids are occupied by the lone pairs.
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BOND LENGTH AND BOND STRENGTH

You should NOT remember the values,

but MUST understand it…
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BOND LENGTH AND BOND STRENGTH (C ONT.)

▪ Bond length is the average distance between the centers of two bonded nuclei.

▪ As the number of electrons between two nuclei increases, bonds become shorter
and stronger.
▪ Thus, triple bonds are shorter and stronger than double bonds, which are shorter
and stronger than single bonds.

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BOND LENGTH AND BOND STRENGTH (C ONT.)

▪ A Comparison of Carbon–Hydrogen Bonds

The length and strength of a C – H bond vary slightly depending on the hybridization
of the carbon atom.

Uhm… why?
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BOND LENGTH AND BOND STRENGTH (C ONT.)

▪ To understand why this is so, we must look at the atomic orbitals used to form
each type of hybrid orbital.
▪ A single 2s orbital is always used, but the number of 2p orbitals varies with the
type of hybridization. A quantity called percent s-character indicates the fraction
of a hybrid orbital due to the 2s orbital used to form it.

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BOND LENGTH AND BOND STRENGTH (C ONT.)

▪ Learn the general trends. Often knowing such trends is more useful than
learning a set of exact numbers, because we are usually interested in
comparisons rather than absolute values.

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BOND LENGTH AND BOND STRENGTH (C ONT.)

▪ Bond length decreases across a row of the periodic table as the size of the atom
decreases.

▪ Bond length increases down a column of the periodic table as the size of an atom
increases.

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BOND ANGLE
▪ Bond angle determines the shape around any atom bonded to two other atoms.
To determine the bond angle and shape around a given atom, we must first
determine how many groups surround the atom.
▪ A group is either an atom or a lone pair of electrons. Then we use the valence
shell electron pair repulsion (VSEPR) theory to determine the shape.
▪ VSEPR is based on the fact that electron pairs repel each other; thus: the most
stable arrangement keeps these groups as far away from each other as possible.

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BOND ANGLE (C
ONT.)

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BOND ANGLE (C
ONT.)

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BOND ANGLE (C
ONT.)

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CLASS ACTIVITY 1.6
Determine the geometry around all second-row elements in each compound.

1. NH2

2. H2O

3. CO2

4. NH4+
5. Predict the geometry around each indicated atom.

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WHAT HAVE YOU LEARNED?

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CEB1013 ORGANIC CHEMISTRY

THE BASIC:
Electron Configuration, Chemical
Structures & Formal Charge
Course Learning Outcome:
CLO1: Describe the electronic structure, bonding and shape of the various
functional groups and extrapolate these to describe the origins of reactivity
of organic compounds

1
LEARNING OUTCOME
By end of this lecture, student should be able to

1. Write electron configuration for any given atom.

2. Draw chemical structures of organic compounds and its derivatives
using several methods.
3. Calculate formal charges of each atom in the chemical compounds,
particularly organic compounds.

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PERIODIC TABLE

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ELECTRON CONFIGURATION
How the electron configures themselves in electron cloud?

Bohr diagrams indicate how many electrons fill each principal shell

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ELECTRON CONFIGURATION (C ONT.)

How the electron configures themselves in electron cloud?

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ISOTOPES – THE “ TWIN’
▪ Each of two or more forms of the same element that contain equal numbers of
protons but different numbers of neutrons in their nuclei, and hence differ in
relative atomic mass but not in chemical properties; in particular, a radioactive
form of an element.

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VALENCE ELECTRONS
▪ The outermost electrons.
▪ The presence of valence electrons can determine the element's chemical
properties and whether it may bond with other elements.

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CHEMICAL BONDS

What does the electron configuration

have to do with chemical bonds?

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IONIC BONDS
THE OCTET RULE
▪ This rule states that the elements lose or gain electrons in order to achieve
noble gas configurations.
▪ Atoms tend to combine in such a way that they have eight electrons in their
valence shells, giving them the same electron configuration as (the nearest)
Noble gas.

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IONIC BONDS (C ONT.)

“The Giver”

▪ To reach noble gas configuration, sodium has to either gain 7 electrons OR lost
one electron.
▪ It is easier for sodium to release one electron than to add 7 electrons to its
orbital.

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IONIC BONDS (C ONT.)

“The Receiver”

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IONIC BONDS (C ONT.)

▪ Sodium and fluorine undergoing ionic bonds to form sodium fluoride.

▪ Sodium loses its outer electron to give it a stable electron configuration, and
this electron enters the fluorine atom exothermically.

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IONIC BONDS (C ONT.)

ELECTRONEGATIVITY
▪ Electronegativity is a measure of the ability of an atom to attract electrons – the
higher the electronegativity, the higher its ability to attract electrons.

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COVALENT BONDS
Does carbon give OR receive electrons?

Carbon atoms have the ability to form huge network…

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COVALENT BONDS (C ONT.)

How the electron configures themselves in electron cloud?

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COVALENT BONDS (C ONT.)

▪ Carbon like to “share” electrons – known as covalent bond.

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COVALENT BONDS (C ONT.)

▪ Covalent bonds form by sharing of electrons between atoms of similar

electronegativities to achieve the configuration of noble gas.

1. Duplet rule, only for covalent bond involving hydrogen.

2. Octet rule, for (almost) other atoms.

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EXCEPTION TO OCTET RULE
▪ Molecules containing atoms of Group 3A elements, particularly boron and
aluminum

6 electrons in the

:
: F: valence shells of boron : Cl :
and aluminum
: :

: :
:F B : Cl Al
:F: : Cl :
:

:
Boron trifluoride Aluminum chloride

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EXCEPTION TO OCTET RULE (C ONT.)

▪ Molecules Atoms of third-period elements have 3d orbitals and may expand their
valence shells to contain more than 8 electrons
▪ Phosphorus may have up to 10

:
: Cl : :O:

: : : :

: : : :
: Cl Cl :

: :

:
CH3 -P- CH3 P H- O-P- O-H

:
CH3 : Cl Cl : O-H

:
Trimethyl- Phosphorus Phosphoric
phosphine pentachloride acid

▪ sulfur, another third-period element, forms compounds in which its valence shell
contains 8, 10, or 12 electrons

: O: : O:
:

H-S- H CH 3 -S-CH 3 H-O- S-O-H

:

:
:

:O : Compounds that do NOT follow

Did you
know?
octet rule are very reactive. It will
Hydrogen Dimethyl Sulfuric
react to form more stable
sulfide sulfoxide acid
compound that follow octet rule

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LEWIS STRUCTURE - ATOM
Lewis dot structure: The symbol of an element represents the nucleus and all inner
shell electrons dots represent valence electrons

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LEWIS STRUCTURE - BOND
Atoms bond together so that each atom acquires an electron configuration the
same as that of the noble gas nearest it in atomic number

Hydrogen do NOT follow octet rule

Did you
know?
due to its small size (and its orbital).
Instead of octet, it goes duplet.

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LEWIS STRUCTURE – BOND (C ONT.)

The number of shared pairs

▪ one shared pair forms a single bond
▪ two shared pairs form a double bond
▪ three shared pairs form a triple bond

Did you know?

In neutral molecules
▪ hydrogen has one bond
▪ carbon has 4 bonds and no lone pairs
▪ nitrogen has 3 bonds and 1 lone pair
▪ oxygen has 2 bonds and 2 lone pairs
▪ halogens have 1 bond and 3 lone pairs

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LEWIS STRUCTURE – BOND (C ONT.)

For more complicated molecules and molecular ions, it is helpful to follow the
step-by-step procedure outlined here:
1. Determine the total number of valence (outer shell) electrons. For cations,
subtract one electron for each positive charge. For anions, add one electron
for each negative charge.
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around
a central atom. (Generally, the least electronegative element should be
placed in the center.) Connect each atom to the central atom with a single
bond (one electron pair).
3. Distribute the remaining electrons as lone pairs on the terminal atoms
(except hydrogen), completing an octet around each atom.
4. Place all remaining electrons on the central atom.
5. Rearrange the electrons of the outer atoms to make multiple bonds with the
central atom in order to obtain octets wherever possible.

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STRUCTURAL FORMULA OF ORGANIC COMPOUNDS

Condensed
VS

Skeletal
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CONDENSED STRUCTURES
▪ Condensed structures are most often used for compounds having a chain of
atoms bonded together, rather than a ring.
▪ The following conventions are used:
1. All of the atoms are drawn in, but the two-electron bond lines are generally
omitted.
2. Atoms are usually drawn next to the atoms to which they are bonded.
3. Parentheses are used around similar groups bonded to the same atom.
4. Lone pairs are omitted.

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CONDENSED STRUCTURES (C
ONT.)

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CONDENSED STRUCTURES (C ONT.)

▪ Condensed structures containing heteroatom

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SKELETAL STRUCTURES
▪ Skeletal structures are used for organic compounds containing both rings and
chains of atoms.
▪ Three important rules are used to draw them:
1. Assume there is a carbon atom at the junction of any two lines or at the end
of any line.
2. Assume there are enough hydrogens around each carbon to make it
tetravalent.
3. Draw in all heteroatoms and the hydrogens directly bonded to them.

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SKELETAL STRUCTURES (C
ONT.)

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SKELETAL STRUCTURES (C ONT.)

Take care in interpreting skeletal structures for positively and negatively charged
carbon atoms, because both the hydrogen atoms and the lone pairs are omitted.
Keep in the mind the following
▪ A charge on a carbon atom takes the place of one hydrogen atom.
▪ The charge determines the number of lone pairs. Negatively charged carbon
atoms have one lone pair and positively charged carbon atoms have none.

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SKELETAL STRUCTURES (C ONT.)

Skeletal structures often leave out lone pairs on heteroatoms, but don't forget
about them.

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CLASS ACTIVITY 1.1
1. Convert each skeletal structure to a Lewis structure.

e. f. g.

h. i. j. k.

2. Write the condensed structures for compounds 1a to 1k.

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CLASS ACTIVITY 1.1
2. Convert each molecule into a skeletal structure

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FORMAL CHARGES
▪ Formal charge: the charge on an atom in a molecule or a polyatomic ion
▪ To derive formal charge
1. Write a correct Lewis structure for the molecule or ion
2. Assign each atom all its unshared (nonbonding) electrons and one-half its
shared (bonding) electrons
3. Compare this number with the number of valence electrons in the neutral,
unbonded atom

Or FC = Ve – (B+D)

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FORMAL CHARGES (C ONT.)

▪ Formal Charge Observed with Common Bonding Patterns for C, N, and O

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CLASS ACTIVITY 1.2
Calculate the formal charge for each structure.

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WHAT HAVE YOU LEARNED?

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