UNIT VIII - BONDING
1
I. Introduction - What do we know about the interaction of different elements?
A. We know that elements with similar properties are in the same group (column) on the Periodic Table.
B. We also know that elements in the same group have the same number of valence electrons.
C. Therefore..........
S. How many valence electrons does an atom of this element have?
1) He 2
2) C 4
3) F 7
4) Na 1
5) Al 3
6) Ne 8
7) O 6
8) N 5
* Lewis Structures - shows valence electrons alone or bonding
For each of the following elements, give the number of valence electrons, the Lewis Structure, and the number of
single electrons and electron pairs:
T.
Na
1)
Na
3)
B
5)
N
7)
F
Mg
1 single
2)
Mg
3 singles
4)
C
3 singles, 1 pair
6)
O
1 single, 3 pairs
8)
Ne
3 singles
2)
Si
3 singles, 1 pair
4)
S
1 single, 3 pairs
6)
Ar
B
2 singles
C
N
4 singles
O
F
2 singles, 2 pairs
Ne
4 pairs
S.
Al
1)
Al
3)
P
5)
Cl
Si
P
S
Cl
He
4 singles
2 singles, 2 pairs
Ar
4 pairs
Cs
He
1 pair*
8)
Cs
1 single
* Note that He has the two dots paired because its valence shell is the first energy level (1s), which can
only hold up to two electrons in its one orbital.
7)
UNIT VIII - BONDING
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II. Introduction to bonding
A. Octet Rule - all atoms “want” 8 valence electrons - a complete shell
B. compound - two or more elements that are chemically combined - What does that mean??
C. chemical bond - force of attraction between two atoms - interaction is between valence electrons
III. Ionic Bond - formed between a metal and a nonmetal; involves a complete transfer of electrons
* results in two ions with full positive and negative charges
* these ions then attract each other due to electrostatic force
Examples:
1) sodium chloride (NaCl)
Na
-
+
Na
Cl
Na
Cl
Cl
sodium gives its electron
as a result separate
away to chlorine
ions are formed
2) magnesium chloride (MgCl2) – Mg has 2 valence electrons wants to lose 2
Cl has 7 valence electrons wants to gain 1; need 2 Cl’s to take the 2
electrons from Mg
-1
-1
+2
Mg
Cl
Cl
Give the Lewis Structure for:
T.
1) Al+3
2) F-1
+3
Al
Al
-1
to form a +3 ion, aluminum
F
gives away 3 electrons
S.
1) K+1
2) I-1
3) N-3
-1
+1
-1
I
K
4) potassium fluoride
N
5) sodium oxide
6) aluminum chloride
-1
-1
+3
K
-2
-1
+1
Cl
F
Na
Al
+1
+1
O
Na
-1
Cl
Cl
UNIT VIII - BONDING
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IV. Covalent Bond - formed between 2 nonmetals; involves sharing of electrons
* happens because nonmetals have high electronegativity
* formed by a sharing of single (unpaired) electrons
* therefore, each unpaired (or single) electron will be able to form one covalent bond (shared pair)
Give the Lewis Structure for the following molecules, then circle all shared pairs of electrons:
* also list the number of single, double, and triple bonds on each molecule
T.
F
xx
F
both fluorine atoms "want"
eight electrons, so they form
a shared pair with the two
singles
x
xx
F
xx
x
xx
xx
F
xx
shared pair = 1 covalent bond
3) O2
2) Cl2
single bond
Cl
Cl
double bond
Cl Cl
O
O
O
4) CH4
H
C
H
H
H
H
4 single bonds
C H
H
H
S.
1) HCl
2) NH3
H
N H
H Cl
H
3) N2
4) CO2
triple bond
N
N
O
C
O
O
UNIT VIII - BONDING
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V. Hybrid Orbital Theory
A. VSEPR Model - Valence Shell Electron Pair Repulsion Model
* basic idea is that all electron pairs (- charged) want to be as far away from each other as
possible
B. How does this work?
2s
1s
2p
1. Orbital Diagram for carbon:
* right now, the 2s electrons can’t bond, so we combine the 2s and the 3 2p’s to form
3
2sp
C
3
4 sp orbitals of equal energy:
* now we have 4 orbitals with single electrons that can form bonds, this is hybridization
3
* any element with 4, 5, 6, 7, or 8 valence electrons will exhibit sp hybrid orbitals when
bonding with other atoms
2s
1s
2p
2p
2sp
1s
2. Beryllium Compounds:
2 valence electrons
1s
2s
2p
1s
2
2sp
2p
3. Boron Compounds:
3 valence electrons
VI. Molecular Geometry - Rules:
Hybrid
Orbitals
Shape
Bond
Angle
Shared
Pairs
3
Unshared
Pairs
0
2
1
1
2
180o
sp
linear
sp2
120o
trigonal planar
sp3
109.5o
tetrahedron
Shared Pairs
4
3
2
1
Unshared Pairs
0
1
2
3
Shape
trigonal
planar
bent
linear
Shape
tetrahedron
pyramid
bent
linear
* If central atom has two shared pairs and zero unshared pairs,
always linear
* treat double and triple bonds as if they were singles
UNIT VIII - BONDING
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VII. Lewis Structures – another method for figuring these out
- for example: nitrogen does not always form 3 bonds and have one free pair
- count total number of electrons in the molecule, and “make it work”
- start by single-bonding everything to the central atom, dot the outside atoms
- fix the molecule as needed
A. Simple Molecules.
Give the Lewis Structure, shape, and type of hybrid orbitals for the central atom of the following:
1) CH4 – C 4 ve, 4x H @ 1ve = 8
2) NH3 – 8 ve
H
H
H
H
C
H
N
H
H
-
3
4 pairs of e on C, hybrids sp
4 shared pairs, shape tetrahedron
4 pairs of e- on N, hybrids sp3
3 shared, 1 unshared pyramid
3) H2O
4) HCl
O H
H
H
Cl
4 pairs of e- on O, hybrids sp3
2 shared, 2 unshared bent
4 pairs of e- on Cl, hybrids sp3
1 shared, 3 unshared linear
5) BH3
6) CO2
H
H
B
O
H
3 pairs of e- on B, hybrids sp2
3 shared trigonal planar
C
O
2 pairs of e- on C, hybrids sp (treat double bond as single)
2 shared linear
7) BeH2
H Be
H
2 pairs of e- on Be sp; 2 shared linear
B. Polyatomic Ions - held together by covalent bonds:
• count total # of electrons on molecule (plus extra electrons if – ion, minus if + ion)
• put first atom in center, single bond w/ outside atoms
• move electrons as appropriate to satisfy octet rule
T.
1) NO3-1 5 + 3(6) + 1 = 24 electrons
2) NO2-1 5 + 2(6) + 1 = 18 electrons
-1
N
O
O
O
N
O
O
O
uses 6 electrons
uses other 18
all 24 used up
S.
1. CO3-2 24 electrons
O
N
O
O
N not happy, so move
e- pair into double bond
N
O
2) SO3-2 26 electrons
-2
O
C
O
O
O
-2
O
S
O
O
UNIT VIII - BONDING
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VIII. Polarity - an uneven sharing of electrons in a covalent bond
* electronegativity – a measure of how “badly” an atom wants to gain electrons
* the most electronegative element is fluorine (F); as you move further away, e.n. decreases
* when one atom in a bond has a higher electronegativity than the other atom, the more electronegative
atom gets a “better” share of the electron pair, therefore the molecule is lopsided or polar
Classify the following molecules with one bond as polar or nonpolar:
T.
H
H
Cl
1) HCl
H
2) H2
2 atoms are different polar
2 atoms are same nonpolar
NOTE: If the molecule has more than two atoms in it, look at the center atom.
If all the electron pairs are doing the exact same thing, it is nonpolar, otherwise it is polar.
NOTE: All polyatomic ions are polar
3) CH4
4) CH3Cl
H
H
5) NCl3
H
C
H
H
H
Cl
C
Cl
4 prs e , all 4 doing the same
NONPOLAR
S.
1) CCl4
-
4 prs e , 3 prs w/ H, 1 w/ Cl,
POLAR
2) H2O
Cl
Cl
C
Cl
H
Cl
O H
NONPOLAR
POLAR
3) BH3
4) NH3
H
B
H
H
NONPOLAR
Cl
Cl
H
-
N
H
N
H
POLAR
H
4 prs e-, 3 prs w/ Cl, 1 lone pair
POLAR
UNIT VIII - BONDING
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IX. Molecular Substance - made up of all nonmetals covalently bonded together
* molecule - a group of atoms held together by covalent bonds
* What holds molecules together?
A. polar substances - dipole-dipole interaction causes attraction between molecules
+
+
-
+
single polar molecule has
-
+
+
+
-
+
-
+
-
+
-
+
-
-
+
-
+
+
positive and negative poles
series of polar molecules held
together by dipole-dipole bonds
* as a result:
- resonably high melting points and boiling points
- many polar substances are liquids and solids; some are gases w/ high mp’s
* examples: ethanol (CH3OH) formaldehyde (CH2O), acetone (CH3COCH3)
mp = -114oC
mp = -92oC
mp = -94oC
o
o
bp = 78 C
bp = -19 C
bp = 56oC
B. nonpolar substances - London Dispersion Forces are responsible for attraction between molecules
- VERY WEAK
* temporary, weak dipoles are formed in nonpolar molecules which loosely hold molecules
together in a manner similar to dipole-dipole forces
* as a result:
- most nonpolar substances are gases or volatile liquids with low melting points
* examples: carbon dioxide (CO2), natural gas (CH4), helium (He)
mp = XXX
mp = -182oC
mp = -269.7oC
o
o
bp = -78 C
bp = -161 C
bp = -264oC
X. Network Solid - unique covalent bonding situation in which nonmetal atoms are covalently bonded in a continuous (network) arrangement
* often these substances are called “supermolecules”, because there are no distinct molecules
* only done by carbon and silicon (4 valence electrons)
* as a result:
- atoms are held tightly in place
- substances with this arrangement are all solids with
high melting points
* examples: diamond (Cx), silicon dioxide (SiO2)x
mp = 4440oC mp = 1713oC
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
UNIT VIII - BONDING
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XI. Ionic Crystals - ionic bonds are formed between metals (want to lose electrons) and nonmetals (want to gain
electrons)
* crystal lattice - regular arrangement of ions in a continuous pattern
- ions are held to each other with ionic bonds
+
+
+
+
- +
+ - +
+ - +
+ - +
- +
+ - +
+ - +
+ - +
+
+
+
-
+
+
+
+
+
+
-
+
+
+
+ - +
+
+
-
+
+
+
+ - +
+
+
-
* as a result:
- all substances with this arrangement are solids with
high melting points
*examples: table salt (NaCl), galena (PbS), fluorite (CaF2)
mp = 801oC
mp = 1118oC mp = 1418oC
+
-
XII. Hydrogen Bonding - a very polar bond formed by a “superdipole” between H & O, H & F, or H & N
* bond between H & O is so polar, it is almost ionic
H
H
O
O
H
H
H
H
O
H
H
O
H
H
O
H
O
a network of these H-bonds
H
are formed between molecules
* as a result:
- H in one molecule is attracted to O in another molecule molecules are held
together more tightly than they would with “normal” dipole interaction (see VIII)
- substances with this type of bonding are typically liquids with unusually high
melting points for their mass – (water should boil at -90oC and ammonia at -110oC according to
pattern)
- low vapor pressures, high specific heats, high densities
- adhesion - the ability to “stick” to other substances
- cohesion - the ability to stay bonded to fellow molecules
- surface tension - a “skin” on the surface of the liquid which can handle stress
* examples: water (H2O), ammonia (NH3)
mp = 0oC
mp = -78oC
o
bp = 100 C bp = -33oC
XIII. Relative Strength of Bonds
London Dispersion<< dipole-dipole< H-bond<<covalent<ionic
XIV. Metallic Bonding - since all metals want to lose their electrons, if only other metal atoms are around
there are many “extra electrons” between the atoms - “sea” of electrons
* electrons move freely from atom to atom, “gluing” the substance together
as a result:
- most metals are solids (ex. Hg) with high melting points
- metals are malleable and ductile
- metals are good conductors of electricity and heat
examples: iron (Fe),
gold (Ag),
silver (Au)
mp = 1538oC mp = 962oC mp = 1064oC
NOTE: these are all elements, not compounds - metals only form compounds with nonmetals
UNIT VIII - BONDING
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XV. Exceptions to the Rule of Eight - some heavier nonmetals can have more than 8 electrons around them
* they do this by incorporating d orbitals into their hybridization pattern
orbitals
sp3d
sp3d2
e’s around cent. atom
10 electrons
12 electrons
shape
trigonal bipyramid
octahedron
examples
PF5, SF4
SF6, BrF5
Give the Lewis Structure for the following: for these examples, follow same method from p. 5, but if you have
extra electrons, put them on the central atom (assuming it’s 3rd row or lower)
T
1) PF5 5 + 7(5) = 40 electrons
2) BrF5 7 + 7(5) = 42 electrons
F
F
F
F
F
Br
P
F
S
F
F
F
1) SF6
F
2) SF4
F
F
F
F
S
F
S
F
F
F
F
F
XVI. Multiple Bond Theory - Sigma and Pi Bonds
* there are 2 types of covalent bonds: sigma (σ) bonds, and pi (π) bonds
- sigma bonds result from the overlap of orbitals; lie along axis of the bond
- pi bonds result from the overlap of “p” orbitals; occur outside the bond axis
* make up 2nd bond in double bond and 2nd & 3rd bonds in triple bond
* this is why we treat multiple bonds as single bonds in terms of molecular shape
S,
Molecular Formula Lewis Structure Hybrid Orbital
σ
σ
Hσ C σ C σH
σ
σ
sp3
7
0
sp2
5
1
sp
3
2
H
H
H
H σ
C 2H 4
π
H
H
C 2H 6
σ
π
σ
CσC
σH
H σ
π
C 2H 2
π
H σ C Cσ H
σ
UNIT VIII - BONDING
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XVII. Bond Energies - each bond contains a certain amount of energy
* amount of energy depends on electronegativity of participating atoms
* General Rule:
- When a bond is broken, energy is absorbed ∆H is positive
- When a bond is formed, energy is released ∆H is negative
Table of Bond Energies
Bond
H–H
C–C
C=C
C=C
N–N
N=N
O–O
O=O
F-F
Energy (kJ/mol)
436
347
611
837
159
946
138
498
159
Bond
Cl – Cl
Br – Br
I–I
H–C
H–N
H–O
H–F
H – Cl
H - Br
Energy (kJ/mol)
243
192
151
414
389
464
569
431
368
Bond
H–I
C–O
C – Cl
C – Br
N–F
N – Cl
O–F
O – Cl
F - Cl
Energy (kJ/mol)
297
351
330
276
272
201
184
205
255
Using bond energies, estimate the energy changes (∆H) in the following reactions
__________ 1) H2 + Cl2 2 HCl
T.
H
H
Cl
Cl
H
Cl
H
Cl
(H - H) + (Cl - Cl)
-
2 (H - Cl)
(436)
-
2 (431)
+
(243)
__________ 3) N2 + 3 H2 2 NH3
N
N
H
H
H
O
O
H
H
H
H
H
H
= -183 kJ
(946)
+
H
O
H
H
O
H
N
H
H
N
H
(N = N) + 3 (H - H) -
__________ 2) 2 H2 + O2 2 H2O
H
H
3(436)
H
H
6 (H - N)
-
6 (389)
= -80 kJ
__________ 4) C + 2 H2 CH4
H
C
H
H
H
Note that free elements
H
H
C
H
H
are given values of "0"
2(H - H) + (O = O) 2(436)
+
(498)
4 (H - O)
-
4 (464)
= -486 kJ
(C)
+ 2(H - H)
(0)
+
2(436)
-
4 (H - C)
-
4 (414)
= -784 kJ
UNIT VIII - BONDING
11
__________ 1) 2 C + 3 H2 C2H6
S
C
H
H
H
H
H
H
C
2(C)
2(0)
H
+ 3(H - H)
+
-
3(436)
-
H
H
C
C
H
H
__________ 3) H2 + O2 H2O2
H
6(H - C) - (C - C)
6 (414) - (347)
H
H
= -1523 kJ
H
C
C
H
H
H
Cl
Cl
H
C
H
Cl
(436)
-
2 (464) - (138)
+
(498)
= -132 kJ
__________ 4) O2 + 2 Cl2 2 OCl2
O
C
Cl
Cl
Cl
O
Cl
Cl
Cl
Cl
O
Cl
Cl
H
H
6(H - C) + (C - C) + (Cl - Cl) - 6(H - C) - 2(C - Cl)
(O=O) + 2(Cl - Cl) -
6(414) + (347) + (243)
(498)
- 6 (414) - 2(330)
H
2 (H - O) - (O - O)
H
H
O
-
O
H
O
(H - H) + (O = O)
__________ 2) C2H6 + Cl2 2 CH3Cl
H
H
O
O
= -70 kJ
+ 2 (243)
4 (O - Cl)
-
4 (205)
= +164 kJ
Type of
Crystal
Units at Lattice
Points
Force(s) Holding them
Together
General Properties
Examples
Ionic
Cations and
Anions
Nonmetallic
atoms
Molecules (or
atoms)
Metallic Atoms
Electrostatic Force
Hard, brittle, high mp,
poor conductor
Hard, high mp, poor
conductor
soft, low mp, poor
conductor
soft hard, low high
mp, good conductor
NaCl, LiF, MgO
Covalent
Molecular
Metallic
Covalent Bond
Dispersion Forces, dipoledipole bonds, H bonds
Metallic Bond
C (diamond),
SiO2 (quartz)
Ar, CO2 , I2 , H2O
, C6H12O6
Na, Mg, Fe, Cu,
Au, Ag
UNIT VIII - BONDING
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XVIII. Introduction to Organic Chemistry
* organic chemistry - the study of carbon compounds
- most carbon-based molecules are found in living things
A. Different forms of Carbon - how many bonds can carbon make? _______
* as a result, carbon has at least 4 different allotropes (forms of the same element with different
bonding patterns)
1. DIAMOND - each carbon atom is bonded together in a tetrahedral pattern to 4 others
* these bonds make it the hardest substance on earth
* one type of network solid
* can only be cut in certain planes
2. GRAPHITE - carbon atoms arrange in layers that are bonded together weakly
* used in pencils and as a lubricant
3. AMORPHOUS CARBON - no predictable arrangement of atoms
* forms include charcoal, bone black, coke
4. FULLERENES - recently discovered, globe or “soccer ball” shape
* a.k.a “Buckyballs”
* few uses yet, but one is known to attack an enzyme in the HIV virus
B. The Chemistry of Carbon
1. Because it makes 4 bonds, carbon is able to make large chains and molecules
* proteins, carbohydrates, vitamins, enzymes, etc. are all based on large
carbon chains
2. Because it is a relatively small atom, the bonds it forms hold tightly.
3. There is an infinite variety of carbon-based compounds that could be created
* each year, hundreds of new ones are discovered, many of which are used by
the health industry.
C. Hydrocarbons - compounds containing only carbon and hydrogen
1. all are very nonpolar molecules, poor conductors, low in density, low bp’s and mp’s
2. all are very insoluble in water (oil and water don’t mix)
3. most on earth are found in deposits of oil and natural gas - fossil fuels
D. Hydrocarbon Structures and Formulas
1. Molecular Formula
a. C4H10
b. C4H8
* drawbacks? it is not very descriptive; there are 2 forms (isomers) of C4H10
2. Structural Formula
H
a.
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
C C
C
C
H
H
H
H
H
b.
* drawbacks? it takes up too much space
H
UNIT VIII - BONDING
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3. Condensed Structural Formulas
a.
CH3CH2CH2CH3
b.
CH2CHCH2CH3
E. Saturated Hydrocarbons
1. Alkanes - named based on number of carbons - all single bonds
Number of
Root Word
Number of
Carbons
Carbons
1
meth5
2
eth6
3
prop8
4
but10
* all alkanes’ names end in “-ane”
Root
Word
penthexoctdec-
S. Give the names of the following alkanes
1 Carbon Methane
1. CH4
5 Carbons Pentane
3. C5H12
3 Carbons Propane
2. C3H8
8 Carbons Octane
4. C8H18
Give the molecular formulas of the following alkanes
C6H14
1. hexane
C2H6
2. ethane
C10H22
3. decane
* Note that the number of H’s can be determined by drawing the Lewis Structure, or by
using the pattern illustrated below
* note the patterns:
1) all alkanes follow a CnH2n+2 pattern
2) all the strucures, except methane end with CH3 groups
F. Unsaturated Hydrocarbons - contain at least one double or triple bond between carbon atoms
1. Alkenes - alkanes with one double bond - CnH2n
2. Alkynes - alkanes with one triple bond – CnH2n-2
* example:
CH3CH3 is ethane, CH2CH2 is ethene, CHCH is ethyne
S. Give the structural formula for the following:
1. ethane
2. ethene
H
H
H
C
C
H
H
H
C
H
H
3. ethyne
C
C C
H
H
H
H
Give the structural formula, THEN the name for the following
1. CH2CHCH3
H
H
H
C C
C
H
PROPENE
H
2. CH3CH2CH2CH3
H
H
H
H
H
H
C
C
C
C
H
H
H
H
BUTANE
3. CH3CHCHCH3
H
H
H
H
H
H
C
C
C
C
H
BUTENE
H
H
UNIT VIII - BONDING
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G. Branched Alkanes - have branches in the carbon skeleton
S. Give the three structural isomers of C5H12
1.
H
H
H
3.
C
H
H
H
2.
C1
C2
5
C3 C4 C
H
H
H
C
4
C1 C2 C3 C
H
H
C
H
1
3
C2 C
C
* to name these, you must first find the parent chain (longest chain)
* then number the carbons on the parent chain
* then count how many carbons are on the branch chain
* give the branch chain a name like the alkanes, ending in –yl
* give the branch chain a location number on the parent chain
* if there is more than one of the same type of branch, give it a prefix (i.e. di-, tri-, etc.)
T. Give the IUPAC names for the three isomers of C5H12 above
Pentane (sometimes n-pentane) 1. 5 carbon parent chain
2. 4 carbon parent chain + a 1-carbon branch on carbon #2
2-methyl butane
2,2-dimethyl propane 3. 3 carbon parent chain + 2 1-carbon branches on carbon #2
S. Give the IUPAC names for the following molecules
1.
2.
CH3
1
CH3
2
3
4
CH
CH
5
CH
2
2
6
CH3
CH
2
CH3
1
2
CH3
3
CH2
4
CH
5
CH
2
6
CH
2
CH3
2-methyl hexane
3-methyl hexane
* Note that numbering is set up such that the numbers are as low as possible (i.e. Example S#1 above
would not be numbered R L; therefore, there is no such thing as 5-methyl hexane)
3.
4.
CH3
CH3
1
2
3
CH3
CH
CH
2
CH3
4
5
CH
CH3
2,4-dimethyl pentane
1
2
CH3
C
3
4
CH
CH3
2-methyl, 2-butene
* Note the double bond gets priority
over the branch
H. Cycloalkanes - carbon can be stable in 5 and 6-carbon rings
S. Give the structural and molecular formula for:
1. cyclopentane
2. cyclohexane
C
C
C
C
C
C
* cyclohexane is the basis for all sugar molecules
C
C
C
C
C
UNIT VIII - BONDING
15
I. Benzene
* carbon is very stable in this arrangement, it allows for one group to single bond to each of the
six carbons by replacing a hydrogen
C
C
C
C
C
C
OR
J. Polymers - very large organic compounds made of repeating units
* monomer - the unit that is repeated in a polymer
* starch, cellulose, proteins, DNA are all polymers
* first one made in a lab was nylon in the 1930’s
* example: polyethylene
ethene monomer: CH2=CH2
- bonded repeatedly to make a ..... CH2CH2CH2CH2 .... chain
* all plastics are polymers: Styrofoam, polyester, Teflon, etc.
* because carbon comes from petroleum deposits, it must be recycled
Formula # of Isomers
CH4
1
C 2H 6
1
C 3H 8
1
C4H10
2
C5H12
3
C6H14
5
C7H16
8
C8H18
12
C9H20
17