Structure and Bonding
Hydrocarbons Organic chemistry is the study of the compounds of carbon,
principally the hydrocarbons and their derivatives. A hydrocarbon is a
compound that contains only carbon and hydrogen. Carbon and hydrogen can
unite to make compounds that contain long chains of carbon atoms, rings, and
multiple bonds between two carbon atoms. The following graphic shows models
of hydrocarbons. These compounds contain only C and H atoms.
H H H H
H C C C C H
H H H H
H H
H
C
H C
C H
H C
C H
C
H
H H H
H
H
H H H
C C C C C C H
H
H H
Note: Every C has four bonds (lines) and every H only one.
The number of hydrocarbons is enormous and justifies a branch of
chemistry called organic chemistry. Derivatives of hydrocarbons contain
heteroatoms—atoms that are not carbon or hydrogen. Most of the derivatives
we shall study contain oxygen or nitrogen atoms.
Alkanes Hydrocarbons that contain no rings or multiple bonds are called
alkanes. Unbranched alkanes are the basic building blocks of organic
molecules; they are named according to the number of carbon atoms in their
molecular structure.
Number of C atoms
1
2
3
4
5
6
7
8
9
10
Name
methane
ethane
propane
butane
pentane
hexane
heptane
octane
nonane
decane
Formula
CH4
CH3CH3
CH3CH2CH3
CH3CH2CH2CH3
CH3CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH2CH3
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
Note: Memorize these names and corresponding number of C atoms.
Lab 01, Fall 2012
1
Each name is composed of two parts, a stem or parent name and a suffix.
The parent name tells us the number of carbon atoms, and the suffix –ane tells us
that the compound is an alkane. For example, hexane tells us that six carbon
atoms are joined by single bonds. To draw a picture of hexane’s structure, join
six Cs with lines (these lines represent sigma bonds). Every C has four bonds, so
every C in the picture must have four lines.
C C C C C C
The molecular formula for hexane is C6H14, so fourteen H atoms are
bonded to the six carbon atoms. H can form only one bond, so an H is added to
the end of each line above to give the following structure.
H H H H H H
H C C C C C C H
H H H H H H
Every carbon atom requires two H atoms except for the end carbon atoms,
which require three hydrogen atoms. If n is the number of carbon atoms in an
alkane, then 2n + 2 is the number of hydrogen atoms. The general formula for
any alkane is CnH2n + 2. How many H atoms are required for n = 5? Check your
answer in the table above.
Obtain a model kit and make a model of hexane. Carbon atoms are black
spheres and hydrogen atoms are yellow spheres. Why do the black spheres have
four holes and the yellow spheres only one hole? Look at your model. Does the
chain of carbon atoms lie in a straight line or in a zigzag pattern?
Skeletal (Bond-Line) Structures The easiest way to show the structure
of hexane is with a skeletal structure. Skeletal structures take advantage of the
fact that carbon has four bonds and hydrogen one. So, all we need do to depict
hydrocarbons is to show the bonding of the carbon atoms. For example, hexane is
shown below.
1
3
2
5
4
6
The carbon atoms are located at the ends (1 and 6) and at bends (2, 3, 4 and
5). Because each carbon must have four bonds, there are three H atoms at C1 and
Lab 01, Fall 2012
2
C6 and two H atoms at C2, C3, C4 and C5. Note that every bond between two
carbon atoms is shown as a line. The key to drawing these structures is to know
that carbon has four bonds and hydrogen one. Examine your model of hexane and
note that the black carbon backbone resembles the structure shown above.
Cycloalkanes Join the two end carbons of your hexane model. The new
structure is a continuous loop or ring and is called cyclohexane. The prefix cyclomeans the following number of carbons (hex- = six) is in a ring, and the suffix
–ane means there are no multiple bonds in the ring compound. How does the
number of carbon atoms, n, compare in hexane and cyclohexane? Write a general
formula for cycloalkanes.__________ Three different bond-line depictions of
cyclohexane are shown below.
two dimensions
chair
boat
Draw skeletal structures of pentane and cyclopentane.
Sigma () and Pi () Bonds So far, all of the structures you have made
contain only single bonds. The first bond made between two carbon atoms is
called a bond and the second and third bonds are called  bonds. The symbol 
is the Greek sigma and  is pi.
H



H C C H
CC
H C C H

H
H
H
H
H
H
H
Alkenes Hydrocarbons that contain a carbon-to-carbon double bond (i.e., two
bonds, one  and one ) are called alkenes. Like alkanes, alkenes are named
with a stem that gives the number of carbon atoms but with the suffix –ene to
indicate a double bond and a number, if necessary, before the name to indicate
the location of the double bond in the carbon chain.
Lab 01, Fall 2012
3
Number of C atoms
2
3
4
5
6
7
8
9
10
Name
ethene
propene
1-butene
1-pentene
1-hexene
3-heptene
2-octene
4-nonene
1-decene
Formula
CH2=CH2
CH2=CHCH3
CH2=CHCH2CH3
CH2=CHCH2CH2CH3
CH2=CHCH2CH2CH2CH3
CH3CH2CH=CHCH2CH2CH3
CH3CH2CH2CH2CH2CH=CHCH3
CH3CH2CH2CH=CHCH2CH2CH2CH3
CH2=CHCH2CH2CH2CH2CH2CH2CH2CH3
Alkynes Hydrocarbons that contain a carbon-to-carbon triple bond (i.e., three
bonds, one  and two ) are called alkynes. Like alkenes, alkynes are named
with a stem that gives the number of carbon atoms and with a suffix –yne to
indicate a triple bond and a number, if necessary, before the name to indicate the
location of the triple bond in the carbon chain.
Number of C atoms
2
3
4
5
6
7
8
9
10
Name
ethyne
propyne
1-butyne
2-pentyne
3-hexyne
3-heptyne
2-octyne
3-nonyne
1-decyne
Formula
HC≡CH
HC≡CCH3
HC≡CCH2CH3
CH3CH2C≡CCH3
CH3CH2C≡CCH2CH3
CH3CH2C≡CCH2CH2CH3
CH3CH2CH2CH2CH2C≡CCH3
CH3CH2CH2CH2CH2C≡CCH2CH3
HC≡CCH2CH2CH2CH2CH2CH2CH2CH3
Shapes and Bond Angles in Organic Compounds
Tetrahedral Make a model of methane, including hydrogen atoms. We describe
the shape of the methane molecule as tetrahedral. A tetrahedral shape is one
in which a carbon atom is at the center of a four-sided regular pyramid. In
methane, the four hydrogen atoms are at the corners of the pyramid as shown
below.
Lab 01, Fall 2012
4
109.5o
Tilt your model of methane so you can observe the angle made by the
carbon atom and any two hydrogen atoms. This is the tetrahedral angle, which
is 109.5o.
Planar Make a model of ethene, including hydrogen atoms. We describe the
shape of the ethene molecule as planar. A planar shape is one in which six
atoms are coplanar (i.e., lie in the same plane).
120o
Lab 01, Fall 2012
5
Tilt your model of ethene so you can observe the angle made by one
carbon atom and its two hydrogen atoms. This is the planar angle, which is
120o.
Linear Make a model of ethyne, including hydrogen atoms. We describe the
shape of the ethyne molecule as linear. A linear shape is one in which four
bonded atoms lie in a straight line.
180o
Observe the angle made by any three atoms. This angle is the linear angle,
which is 180o.
It is important to be able to associate the three hydrocarbon families with
the kind of bonding (single, double or triple) and the geometry (shape and angles)
of organic molecules.
Family
Alkane
Alkene
Alkyne
Carbon
4 Single bonds
1 double bond
1 triple bond
Shape
Tetrahedral
Planar
Linear
Angle
109.5o
120o
180o
Orbitals in Carbon and Hydrogen
Carbon A carbon atom is made up of protons, neutrons and electrons. The total
number of protons (6) equals the total number of electrons (6), so the atom is
neutral or uncharged. The protons and neutrons are concentrated in the nucleus
Lab 01, Fall 2012
6
and do not participate in bonding. The 6 electrons are found outside the nucleus
in orbitals. Orbitals are three-dimensional spaces where electrons are found. The
electrons in the outermost main shell of carbon can take part in bonding with
other atoms; therefore, we are most interested in these electrons. Electrons in the
outermost main shell are called valence electrons. Carbon forms four bonds in
stable molecules. A carbon atom has four valence electrons, which are found in
four valence orbitals. When carbon atoms join to form molecules, some of the
orbitals may change shape (hybridize), but each carbon always has four orbitals.
Hybrid Orbitals Carbon can hybridize to make two hybrid orbitals, three hybrid
orbitals or four hybrid orbitals. Each bonded carbon atom will always have four
orbitals. As we shall see, these three kinds of hybridization are necessary for
carbon to make alkynes, alkenes and alkanes. Carbon always has four valence or
bonding orbitals. When carbon forms two hybrid orbitals, two orbitals remain
unhybridized. When carbon forms three hybrid orbitals, one orbital remains
unhybridized. When carbon forms four hybrid orbitals, no orbital remains
unhybridized. All hybrid orbitals are shaped like a teardrop. The following
graphic show a hybrid orbital.
A hybrid orbital
Unhybridized Orbitals Unhybridized orbitals in carbon are called p orbitals.
The shape of a p orbital is shown in the following graphic. A p orbital has two
lobes, and hybrid orbitals only one.
A p orbital
Lab 01, Fall 2012
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We can explain carbon-to-carbon single bonds, double bonds and triple
bonds by the use of the two kinds of orbitals shown above.
Hydrogen A hydrogen atom is made up of a proton and an electron (and
isotopes of hydrogen have neutrons). The sole electron is found in a spherical
orbital call an s orbital. A graphic of an s orbital is shown below.
An s orbital
Remember, an orbital is a three-dimensional space where electrons are found.
Because hydrogen has only one electron, this electron is in the outermost shell
and is a valence electron. Valence electrons participate in bonding.
Single Bonds, Double Bonds and Triple Bonds
For discussion purposes, we can consider covalent bonds to form
when two orbitals come together or overlap. The following rules will help you
understand the bonding in hydrocarbons.
1. Hydrogen always forms one single () bond.
2. Carbon always forms four bonds, which may be
a. four single () bonds.
b. a double bond ( and ) and two single () bonds.
c. a triple bond ( and two ) and one single () bond.
d. (rarely) two double bonds (two each  and ).
3. Sigma () bonds are formed from the overlap of hybrid or s orbitals.
4. Pi () bonds are formed from the overlap of p orbitals.
Single Bonds Carbon-to-carbon single bonds form when two hybrid orbitals
overlap end-to-end. Each hybrid orbital contains one electron, so when the
orbitals merge, the two electrons are found in the three-dimensional space on a
straight line between the two atoms. Think of the hybrid orbitals themselves as
electrons. The electrons have opposite spins and hold the atoms together like tiny
magnets attracting each other. The following graphic depicts the end-to-end
overlap of two hybrid orbitals from two carbon atoms forming a single bond,
Lab 01, Fall 2012
8
which is also a sigma bond. Each carbon has three other bonds that are indicated
by squiggly lines.
overlap
of hybrid
orbitals
carbon 1
 bond
carbon 2
Carbon-to-carbon single bond
Carbon to hydrogen single bonds form when a hybrid orbital of a carbon atom
overlaps with an s orbital from a hydrogen atom. The graphic below shows a
carbon atom forming one single (sigma) bond with hydrogen.
overlap
carbon
hydrogen
of hybrid
and s
orbitals
 bond
Carbon-to-hydrogen single bond
Double Bonds Carbon-to-carbon double bonds are composed of a  bond and
a  bond. The  bond forms on a straight line between the two carbon atoms as
above. The  bond cannot be in the same space (orbital) because an orbital holds
a maximum of two electrons. The  bond forms by the side-by-side overlap of
two p orbitals. Each p orbital has two lobes, which overlap above and below the
 bond. The following graphic shows the formation of a double bond. Each
carbon must have four bonds. The double bond accounts for two bonds to each
carbon, and the squiggly lines indicate the remaining two bonds to each carbon
atom.
Lab 01, Fall 2012
9
overlap of hybrid
orbitals ( bond)
and of p orbitals
( bond)
carbon 1
carbon 2
 bond
 bond
Carbon-to-carbon double bond
The figure below show the potential surfaces of two p orbitals coming
together to form a  bond. The underlying  bond is not shown.
Overlap of two p orbitals
Triple Bonds Carbon-to-carbon triple bonds are composed of a  bond and
two  bonds. The  bond forms on a straight line between the two carbon atoms
as in a single bond and double bond. The first  bond forms above and below the
 bond as in a double bond. The second  bond forms perpendicular to the first 
bond. All carbon-to-carbon  bonds form by the end-to-end overlap of hybrid
orbitals, and all carbon-to-carbon  bonds form by the side-by-side overlap of
Lab 01, Fall 2012
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two unhybridized p orbitals. The following picture shows how the two sets of p
orbitals overlap. The bond is not shown. Imagine both sets of p orbitals
centered on their respective carbon atoms.
xy plane
yz plane
Overlap of p orbitals
The graphic below shows the orbital arrangement in a carbon-carbon triple
bond.



A triple bond
Lab 01, Fall 2012
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Orbital Arrangements of Carbon in Single, Double and
Triple Bonds
Alkanes Consider the structure of ethane. We can determine what kinds of
orbitals overlap to make the bonds in ethane by examining its structure.
H H


H C C H

H H
ethane
Both carbon atoms are identical in ethane. One is shown in red for
discussion purposes. All four bonds to the red carbon are single () bonds. We
learned above that carbon always makes  bonds from hybrid orbitals. Thus, four
hybrid orbitals are required by the red carbon to form four single bonds. When a
carbon atom has four hybrid orbitals, we call that carbon atom sp3 hybridized,
and we call each of its four hybrid orbitals sp3-hybrid orbitals. A principle of
valence-bond theory is that hybrid orbitals around a given carbon orient as far
away from each other as possible. Therefore, the four hybrid orbitals orient
toward the corners of a regular tetrahedron. Thus, we can determine the shape of
a carbon atom by its hybridization. The orbitals of an sp3-hybridized carbon are
tetrahedral and result in methane having a tetrahedral shape as shown above.
Alkenes Consider the structure of ethene.
H
H
C
C
 
H
H
ethene
Lab 01, Fall 2012
12
The red carbon has four bonds and requires four orbitals, a p orbital for its
 bond and three hybrid orbitals for its three  bonds. A carbon atom with three
hybrid orbitals is called an sp2-hybridized carbon, and its three hybrid orbitals are
called sp2-hybrid orbitals. The three hybrid orbitals orient as far away from each
other as possible, in a plane at 120o angles to each other. The orientation of the
hybrid orbitals determines the shape around the carbon atom. The shape is planar.
Both carbon atoms of ethene are sp2 hybridized. Therefore, the six atoms of
ethyne are coplanar as seen above.
Alkynes Consider the structure of ethyne, which has the common name
acetylene.


H C C H

acetylene
The red carbon has four bonds and requires four orbitals, two p orbitals for
its two  bonds and two hybrid orbitals for its two  bonds. A carbon atom with
two hybrid orbitals is called an sp-hybridized carbon, and its two hybrid orbitals
are called sp-hybrid orbitals. The two hybrid orbitals orient as far away from each
other as possible, in a line 180o apart. The orientation of the hybrid orbitals
determines the shape around the carbon atom. The shape is linear. Both carbon
atoms of acetylene (ethyne) are sp hybridized. Therefore, the four atoms of
ethane are linear as seen above.
Summary: Carbon has four bonds, which may be  or. The first bond made
by carbon is always a  bond. The second bond of a double bond and the second
and third bonds of a triple bond are  bonds. Carbon makes  bonds with hybrid
orbitals and  bonds with p orbitals. The number of  bonds or hybrid orbitals
determines the shape around a carbon atom, because hybrid orbitals orient as far
apart as possible. An alkane carbon is sp3 hybridized, has four hybrid sp3 orbitals,
forms four  bonds 109.5o apart and is tetrahedral. An alkene carbon is sp2
hybridized, has three hybrid sp2 orbitals, forms three  bonds 120o apart plus
one bond and is planar. An alkyne carbon is sp hybridized, has two hybrid
orbitals, forms two  bonds 180o apart plus two  bonds and is linear.
Lab 01, Fall 2012
13
Branching Previously, we considered unbranched hydrocarbons. Many
hydrocarbons have branches in their structures. A branch, like a branch on a tree,
is a carbon chain that veers off the main or longest chain. Consider hexane shown
in the structure below.
H H H H H H
H C C C C C C H
H H H H H H
hexane
Hexane has no branches. Next, consider the structure below. Like hexane,
it has six carbon atoms in its main or longest chain. However, it also has a branch
at C3.
H H H H H H
1
the branch is
at C-3 not
C-4 because
3 is less than 4
H C
2
3
C C
H H
4
C
5
6
C C
H
longest continuous carbon
chain is six carbons or
hexane
H H H
H C
H
H
The parent name of a branched hydrocarbon is the longest continuous
chain of carbon atoms. Thus, the structure above is a derivative of hexane. The
side chain or branch contains one carbon atom and is a derivative of methane.
Thus, the name of the compound represented above is 3-methylhexane. The
methyl group contains one carbon and three hydrogen atoms (one less hydrogen
atom than in methane).
Partial Structures If we start with methane, CH4, and remove one of its H atoms,
we have CH3. CH3 is an incomplete or partial structure because carbon has only
three bonds, and carbon always has four bonds in complete structures.
H
Lab 01, Fall 2012
H
take away one H
H C H
H C
H
methane
H
methyl
14
The name suffix changes from -ane to -yl when one hydrogen atom is
removed from the end carbon atom of an alkane. Name the alkyl groups in the
following table.
CH3CH2
CH2CH2CH3
CH3CH2CH2CH2CH2
In the name of 3-methylhexane, the number 3 locates the position of the
methyl group on the longest chain. We count the number of carbons from each
end and give the smaller of the two numbers to the methyl group. In numbering
the structure above, we get 3 from the left end and 4 from the right end, so we use
3. In naming, a number is separated from a letter by a hyphen, so the name
becomes 3-methylhexane. Numbers are separated from other numbers by
commas (i.e., 3,3-dimethylhexane). Each branch on the longest chain is identified
by a name and a number.
Isomers Two compounds with the same molecular formula but different
structures are called isomers. The carbon-to-carbon skeleton in a hydrocarbon
shows the connectivity or constitution of the hydrocarbon. When two
hydrocarbon molecules with the same molecular formula differ in their
connectivity, we call them constitutional isomers. If they have different
formulas, they are not isomers but different compounds. You can draw the
structure of very simple hydrocarbons from their molecular formulas, but two or
more structures are possible for most formulas.
Draw the structures of CH4, C2H2, C2H6 and C2H4 and name them. These
compounds do not have isomers.
Draw the structures of at least three isomers of hexane. Hint: Start with
hexane, then make a pentane with a branch, butane with two branches, etc.
Summary of Hydrocarbons We have learned that hydrocarbons contain only
carbon and hydrogen and are comprised of alkanes, alkenes, and alkynes. They
can have single bonds, double bonds, triple bonds, or rings in their structures.
Single, double and triple bonds in hydrocarbons are made up of  bonds or 
bonds. During this course, we shall let the symbol  stand for the number of 
bonds in a structure and r stand for the number of rings in a structure. Shown
Lab 01, Fall 2012
15
below are the structures and formulas of five different organic compounds. In the
space below each structure, indicate how many bonds () and rings (r) are
present in each molecule. Then add the two numbers together to get the sum of 
bonds and rings in each structure.
C8H10
=
r=
r =
C6H12
=
r=
r =
C5H10
=
r=
r =
C5H12
=
r=
r =
C6H6
=
r=
r =
Derivation of DU Equation We call the sum of  bonds and rings in a
structure, the number of degrees of unsaturation (DU) in the compound. We see
that every time we make a new  bond or ring, starting with a hydrocarbon, we
must remove two hydrogen atoms. If we let h equal the number of hydrogen
atoms in an alkane, we can write h = 2n + 2, because 2n + 2 is the H subscript or
number of hydrogen atoms in the general formula of an alkane. Every time we
introduce a  bond or ring into an alkane, we reduce the number of hydrogen
atoms in the formula by two, so for any hydrocarbon h = 2n + 2 – 2 – 2r.
Rearrangement of terms gives the hydrocarbon equation below.
 + r = n + 1 - h/2
Equation 1
[Gross, R.A., Jr. Chem. Educator, [Online] 2003 8(1), 13-14; DOI 10.1333/s00897030646a,
“An Equation for Degrees of Unsaturation."]
Use the hydrocarbon equation to calculate DU ( + r) for each compound above
from its molecular formula. Then compare the calculated values to values
obtained by counting the  bonds and rings in the structures. We shall use the DU
equation throughout the course.
Derivation of a Total-Bonds Equation If we look at an alkane’s structure,
we can quickly see that the total number of bonds, B, which are all sigma bonds,
equals three times the number of carbon atoms, n, plus one. This relationship can
be expressed mathematically as shown below.
Lab 01, Fall 2012
16
B = 3n + 1
Equation 2
Satisfy yourself that this equation works by applying it to several alkanes.
Remember that alkanes have only sigma bonds, therefore, the number of pi bonds
equals zero, or  = 0.
We must make adjustments to the equation for compounds that have pi bonds and
rings. The total number of bonds B decreases by one for every pi bond or ring in
our compound. Thus, a general equation for the total number of bonds in a
hydrocarbon can be derived by subtracting Equation 1 from Equation 2.
B = 3n + 1 – ( + r) = 3n + 1 – (n + 1 – h/2)
B = 2n + h/2
Equation 3
[Gross, R.A., Jr. Chem. Educator, [Online] 2008 13(5), 284-286; DOI 10.1333/s0089708215a,
"Derivation and Use of a Total-Bonds Equation."]
Find the number of bonds B in methane and decane by Equations 2 and 3.
Find the number of bonds B in cyclopentane and in cyclohexene by Equation 3.
How many  bonds are in an acyclic (no rings) compound of formula C6H6?
Consider how organic molecules are joined together, and derive an equation for
the total number of bonds B in any covalent compound from its molecular
formula.
Lab 01, Fall 2012
17
Questions Structure and Bonding
Student No.___ Section____
Last name________________________ First name_________________________
2-pts each; all parts must be correct for credit.
1. Calculate  + r for C8H10. ans.________
2. Calculate B for C 21H30. ans._________
3. Find the formal charge on oxygen in: LiOCH2(CH3)2. Ans.______
4. The structure of the hydrocarbon shown below contains carbon-carbon single, double, and
triple bonds. In the spaces provided, show the hybridization, and geometry of the indicated
carbon atoms.
CH3CH2CH=CHCH2CH2C CCH3
hybridization
geometry
hybridization
geometry
hybridization
geometry
5. Circle the structures that represent polar compounds: CH4
CH2BrCl CH3F CHBr3 CCl4
6. Show the number of rings and  bonds in the following structures and the sum  + r.

r =
+r=

r =

r =

r =
+r=
+r=
+r=
7. A compound of formula C 20H36 has no rings. How many sigma bonds does it have? ans. _____
8. Indicate whether the following pair are isomers or are identical compounds. __________________
and
9. Draw skeletal structures of butane and octane (no credit for condensed or alternative structures).
10. What bond angle is associated with a methane carbon?
carbon?
an ethyne carbon?
Lab 01, Fall 2012
an ethene
18