Transcript CH 3

Whenever we cook food, we
are acting as organic
chemists. The proteins, fats,
and carbohydrates that make
up our food are organic
compounds, and cooking the
food causes chemical
reactions to occur that
decompose some of the
compounds into tastier ones.
In this chapter we see a little
of the amazing variety of
organic compounds and
study some of their reactions.
Assignment for Chapter 11
11.32; 11.40; 11.46; 11.57; 11.67;
11.73; 11.77; 11.78; 11.87;11.92; 11.93
Hydrocarbons and Functional Groups
• Alkanes
• Alkenes and Alkynes
• Aromatic compounds
Functional Groups:
• Alcohols
• Ethers
• Phenols
• Aldehydes and Ketones
• Carboxylic Acids and Esters
• Amines and Amides
Examples
Structural formula
Name
Abbreviated structural formula
(ordinary) molecular formula
異戊二烯
The molecule that shocked science
more than once.
Friedrich August von Stradonitz Kekulé
(1829-1896) who proposed the ring
model of benzene in 1861
乙醯柳酸（阿司匹林）
藏茴香酮 （香芹酮）
Alkanes: saturated hydrocarbons
Cycloalkanes
Figure 11.1 The melting and boiling points of the unbranched
alkanes from CH4 to C16H34.
The dominant intermolecular force
in alkanes is the London force
because they are nonpolar.
Melting point of propane (-187 oC) is lower than
that methane (-183 oC) and that of ethane(-172 oC)
because of symmetry of methane is higher than
that of propane. As the number of carbons increases,
the symmetry contribution becomes less and less
significant. This initial “glitch” (the anomalous increase
of melting point of methane and ethane), therefore, is
because of the high symmetry of the two molecules,
which provides extra, entropic contribution to enthalpy
of melting.
Sm  H m / Tm (Sm (>0) is smaller for more symmetric molecules ,
to keep H m constant, Tm must increase for them)
Obviously, there is no symmetry contribution
to boiling point.
Straight Chain
Branched (with side chains)
Figure 11.2 (a) The atoms in neighboring straight-chain
alkanes, represented by the tubelike structures, can lie
close together. (b) Fewer of the atoms of neighboring
branched alkane molecules can get so close together,
so the London forces (represented by double-headed
arrows) are weaker and branched alkanes are more
volatile.
Figure 11.3 The enthalpy changes accompanying
the combustion of methane. Although the bonds in
the reactants are strong, they are even stronger in
the products; and the overall process is exothermic.
Figure 11.4 In an alkane substitution reaction, an incoming
atom or group of atoms (represented by the orange sphere)
replaces a hydrogen atom in the alkane molecule.
UV radiation or heat
CH 4 (g)+Cl2 (g) 
 CH3Cl(g)+HCl(g)
Alkenes and Alkynes
Figure 11.5 The -bond (yellow electron clouds) in an alkene
molecule makes the molecule resistant to twisting around a
double bond, so all six atoms lie in the same plane.
Figure 11.6 The melting point of an alkene is usually lower
than that of the alkane with the same number of carbon atoms.
The values shown are for unbranched alkanes and 1-alkenes
(that is, alkenes in which the double bond is at the end of the
carbon chain).
Figure 11.7
In an elimination reaction, two atoms (the orange and purple
spheres) attached to neighboring carbon atoms are removed
from the molecule, leaving a double bond in their place.
Cr2O3
CH3CH3 (g) 
 CH 2 =CH 2 (g)+H 2 (g)
Figure 11.8 In an addition reaction, the atoms
provided by an incoming molecule are attached to
the carbon atoms originally joined by a multiple
bond. Addition is the reverse of elimination.
CH 2 =CH 2 (g)+Cl2 ( g ) 
 CH 2Cl-CH2Cl
Figure 11.9 When bromine dissolved in a solvent
(the brown liquid) is mixed with an alkene (the
colorless liquid), the bromine atoms add to the
molecule at the double bond, a reaction giving a
colorless product.
CH 2 =CH 2 (g)+Br2 (g) 
 CH 2Br-CH 2 Br
Aromatic Compounds (Arenes)
萘
蒽
Schematic representation of coal
How to name hydrocarbons
• Alkanes:
• (1) identify the longest unbranched chain and give it
the name of the corresponding alkane.
• (2)name the alkyl substituent groups by changing the
suffix –ane into –yl. Use Greek prefix to indicate how
many of each substituents are in the molecule. When
different groups are present, list them in alphabetical
order and attach them to the root name.
• (3) Indicate the locations of the substituents by
numbering the backbone carbone C atoms from
whichever end of the molecule results in the lower
numbers of locations for the substituents. The locations
are then written before each substituent, separated by
commas.
2,2,4-trimethylpentane
1
1
2
5
2
3 3
4
4
5
Except terminals, wherever there is a C or CH, there is substituent(s).
Common mistakes:
2-methyl-2-methyl-4-methylpentane
2-dimethyl-4-methylpentane
4,4,2-trimethylpentane
5
5
1
4
4
33
2
2
1
Using the smallest numbers possible.
How to name hydrocarbons
• Alkenes and Alkynes:
• (1) identify the longest unbranched chain and give it
the name of the corresponding alkene or alkyne.
• (2)name the alkyl substituent groups by changing the
suffix –ane into –yl. Use Greek prefix to indicate how
many of each substituents are in the molecule. When
different groups are present, list them in alphabetical
order and attach them to the root name.
• (3) Indicate the locations of the substituents by
numbering the backbone carbon C atoms from
whichever end of the molecule results in the lower
numbers of locations for the substituents. The locations
are then written before each substituent, separated by
commas.
• (4) Number the C atoms in the backbone in the order
that gives the lower numbers to the two atoms joined
by the multiple bond. The multiple bond has priority
over the numbering of substituents.
Naming an Alkene
6 5 4 3 2 1
1 2 3 4 5 6
CH3CH=CHCH2CHCH3
CH3
2-methyl-5-hexene
5-methyl-2-hexene
The multiple bond has priority over the numbering of substituents
2,3-dimethyl-4-ethylcyclohexene
The multiple bond has priority over the numbering of substituents
(Use the smallest numbers to locate the double bond)
6
6
5
2
4
3
CH2CH3
5
6
4
3
1
5
1
1
2
2
4
3
CH3
CH3
2,3-dimethyl-4-ethylcyclohexene
6
1
5
4
3
2
6
3
5
CH2
CH3
2
4
4
CH2
CH3
1
2
3
CH3
1
5
6
CH3
CH3
1,6-dimethyl-5-ethylcyclohexene,
5-ethyl-1,6-dimethylcyclohexene,
1,2-dimethyl-3-ethylcyclohexene
4
CH2
CH3
3
5
CH3
6
1
CH3
2
CH3
How to name hydrocarbons
• Arenes:
• -C6H5 aryl
• ortho- (o-), meta (m-), para (p-)
1,4-dimethylbenzene
(p-Xylene)
CH3
CH3
CH3
CH3
CH3
1,2-dimethylbenzene
(o-Xylene)
1,3-dimethylbenzene
(m-Xylene)
CH
3
Exercise
• Name the following hydrocarbons
(a) (CH3)2CHCH2CH(CH2CH3)2
1
(b)
CH2CH3
2
3
4
5
6
(a) (CH3)2CHCH2CH(CH2CH3)2
(a) 4-ethyl-2-methylhexane
(b) 1-ethyl-3-propylbenzene
CH2CH2CH3
Quiz
•
•
•
•
Name and give an example of the major reactions of hydrocarbons.
Explain why the melting point and boiling point of a straight-chained
hydrocarbon are higher than that a branched hydrocarbon of equal number of
carbon atoms.
Draw the structures of
4-ethyl-2-methylhexane,
1-ethyl-3-propylbenzene
5-methyl-2-hexene
Name the following compounds:
CH3
CH3
Functional Groups
•
•
•
•
•
•
Alcohols
Ethers
Phenols
Aldehydes and Ketones
Carboxylic Acids and Esters
Amines and Amides
aa
Investigating Matter 11.1 (a) The two orientations of a nuclear spin
have the same energy in the absence of a magnetic field. When a field
is applied, the energy of the  spin falls and that of the  spin
increases. When the separation between the two energy levels is
equal to the energy of a radio-frequency photon, there is a strong
absorption of radiation, giving a peak in the NMR spectrum.
Nuclear Magnetic Resonance
Investigating Matter 11.1 (b) The NMR spectrum of ethanol.
The red letters denote the protons that give rise to the
associated peaks.
The NMR spectrum of a molecule
is like a fingerprint.
Investigating Matter 11.1 (c) An MRI image of a human brain.
The patient must lie within the strong magnetic field
(background) and the detectors can be rotated around the
patient’s head, which allows many different views to be
recorded.
Magnetic Resonance Imaging
makes it possible to see inside
a sample noninvasively.
Alcohols
-OH
Ethers
R-O-R`
Water, CH3CH2-O-H, CH3CH2-O-CH2CH3
Figure 11.12 The boiling points of ethers (given on each
column, in degrees Celsius) are lower than those of isomeric
alcohols, because hydrogen bonding occurs in alcohols but
not in ethers. All the molecules referred to here are
unbranched.
Phenols
百里香酚
(麝香草酚)
丁香油酚
Aldehydes and Ketones
O
R-C
H
O
R-C
R
Smoked meat/fish
Wood smoke contains formaldehyde (formalin) that
has destructive effect on bacteria so smoked food can
be preserved long.
Simplest aldehyde: HCHO
for aroma of cherries and almonds
In oil of cinnamon
in oil of vanilla
Major Properties
• Aldehydes and ketones can be prepared by the
oxidation of alcohols.
• Aldehydes are reducing agents; ketones are not.
Aldehydes:CH 3OH(g)+O 2 
 2HCHO(g)+H 2O(g)
600o C,Ag
Na 2 Cr2 O7 (aq),H 2SO 4 (aq)
Ketones:CH 3CH 2 (OH)CH 3 

 CH 3CH 2 COCH 3
Figure 11.13 An aldehyde (left) produces a silver mirror with
Tollens reagent, but a ketone (right) does not.
Aldehydes:CH3CH 2CHO+Ag + (from Tollens Reagent)  CH3CH 2COOH+Ag(s)
Ketones:CH3COCH3 +Ag + (from Tollens Reagent)  no reaction
Carboxylic Acids and Esters
CH3CH 2OH(aq)+O 2 ( g ) 
 CH3COOH(aq)+H 2 O(l)
CH3
COOH
+ 3O2 (g) 
+2H2O
CH
COOH
3
H+ , 
CH3COOH(aq)+H-OCH2CH3 
 CH3 (CO)OCH2CH3 +H2O
三硬脂精
Figure 11.14 In a condensation reaction, two molecules
are linked as a result of removing two atoms or groups
of atoms (the orange and purple spheres) as a small
molecule (typically, water).
Carboxylic acid + amine amide + water
CH3COOH+NH2CH3CH3CONHCH3+H2O
Amines and Amides
Amines: derivatives from ammonia by replacing hydrogens
with organic groups.
Amides: resulted from condensation of amines with carboxylic
acids.
H
H
H-N-H
CH3-N-H
H
CH3-N-CH3
CH3
CH3-N-CH3
Methylamine Dimethylamine Trimethylamine
Carboxylic acid + amine amide + water
CH3COOH+NH2CH3CH3CONHCH3+H2O
Naming Compounds with Functional Groups
•
•
•
•
•
•
•
•
•
•
Highlight functional groups. Numbering of carbons should results in
lower number for the functional group.
Refer to conventions for hydrocarbons
Alcohols: alkane-ol CH3CH2CHOHCH32-butanol
Ethers: Name each of hydrocarbon groups attached to the O atom
separately and alphabetically. CH3OCH2CH3ethyl methyl ether.
Aldehydes: identify the parent alkane (including C of –CHO in count of
carbon atoms); change the final –e into –al. the –CHO group can occur
only at the end of a carbon chain and is given the number 1 only if other
substituents need to be located. CH3CH(CHO)CH2CH32methylbutanal.
Ketones: Change the –e in parent alkane into –one. the –C=O group is
indicated by selecting a numbering order that gives it the lower number.
CH3CH2CH2COCH32-pentanone.
Carboxylic acids:change the –e of the parent alkane into –oic acid.
Include the C atom of the –COOH in count of carbon atoms.
CH3CH2CH2COOHbutanoic acid.
Esters: Change the –ol of the alcohol to –yl and the oic acid of the parent
acid to –oate. CH3CH2COOCH3methyl propanoate.
Amines:specify the groups attached to the nitrogen atom in alphabetical
order, followed by the suffix –amine. Amines with two amino acids are
called diamines. The –NH2 group is called amino- when it is a substituent.
(CH3CH2)2NCH3diethylmethylamine.
Halides: Name the halogen atom as a substituent by changing the –ine
part of a=its name to –o. CH3Brbromomethane.
Naming the following compounds
(a) CH3CH(CH3)CHOHCH3
(b) CH3CH2CH2COCH3
(c) (CH3CH2)2NCH2CH2CH3
(a) 3-methyl-2-butanol
(b) 2-pentanone
(c) diethylpropylamine
Exercise
Naming the following compounds: CH CH OH
CH CH
(a) CH3CH(CH2CH2OH)CH3
CH
(b)CH3CH(CHO)CH2CH3 CH CH
CH CH
(c) (C6H5)3N
2
3
3
2
3
CHO
(a) 3-methyl-1-butanol
(b) 2-methylbutanal
(c) triphenylamine
3
2
Classroom Exercise
Naming the following compounds:
(a) CH3CH2CHOHCH2CH3
(b) CH3CH2COCH2CH3
(c) CH3CH2NHCH3
(a) 3-pentanol
(b) 3-pentanone
(c) ethylmethylamine
Quiz
1. Name the following compounds:
CH2CH3
CH3CH
CHO
CH3CH2CH2COOH
CH2CH2OH
CH3CH
CH3
CH3CH2NHCH3
CH3CH2COOCH3
2. What is most important difference between aldelhyde and ketone?
3. Name the following reaction:
CH3COOH+NH2CH3CH3CONHCH3+H2O
Answer
1. Name the following compounds:
CH2CH3
CH3CH
CHO
2-methylbutanal
CH3CH2CH2COOH
butanoic acid
CH2CH2OH
CH3CH
CH3
3-methyl-1-butanol
CH3CH2NHCH3
ethylmethylamine
CH3CH2COOCH3
methyl propanoate
2. What is most important difference between an aldelhyde and a ketone?
An aldelhyde is a good reducing agent, but a ketone is not.
3. Name the following reaction:
CH3COOH+NH2CH3CH3CONHCH3+H2O
Condensation reaction
Isomers
• Structural isomers
• Stereoisomers:
Geometrical isomers
Optical isomers
Figure 11.15 A summary of the various types of isomerism
that occur in molecular compounds. Geometrical and
optical isomers are both types of stereoisomers.
Structural Isomers: C4H10
CH3-CH2-CH2-CH3
CH(CH3)3
Structural Isomers: C6H14
Isomer vs Conformation
They are the same isomer but with different conformations:
Exercise: Different Isomers or
Different Conformers
Stereoisomerism I:
Geometric Isomerism
Figure 11.16
A pair of geometrical isomers in which two groups are either
both on the same side of a double bond (cis) or on opposite
sides (trans). Notice that the bonded neighbors of each atom
are the same in both cases, but nevertheless the
arrangements of the atoms in space are different.
cis or trans?
(a) trans (b) cis
Classroom Exercise: cis or trans?
(a) cis (b) trans
Figure 11.17 Compounds with rings can also exhibit
geometrical isomerism. Groups attached to carbon
atoms in a ring can be both on the same face of the
ring (cis) or across the plane of the ring from each
other (trans).
the trans isomer has the higher melting point;
the cis isomer has the higher boiling point.
1,2-dichloroethene
isomer
melting point (°C)
boiling point (°C)
isomer
melting point (°C)
boiling point (°C)
cis
-80
60
cis-but-2-ene
-139
4
trans
-50
48
trans-but-2-ene
-106
1
Why is the boiling point of the cis isomers higher?
There must be stronger intermolecular forces between
the molecules of the cis isomers than between trans isomers.
Why is the melting point of the cis isomers lower?
You might have thought that the same argument would lead to a higher melting point
for cis isomers as well, but there is another important factor operating.
In order for the intermolecular forces to work well, the molecules must be able to pack
together efficiently in the solid.
Trans isomers pack better than cis isomers. The "U" shape of the cis isomer doesn't
pack as well as the straighter shape of the trans isomer.
The poorer packing in the cis isomers means that the intermolecular forces aren't as
effective as they should be and so less energy is needed to melt the molecule - a lower
melting point.
Stereoisomerism II:
Chirality (Enantiomerism)
Figure 11.18 The molecule on the right is the mirror image of
the molecule on the left, as can be seen more clearly by
inspecting the simplified representations in the circles.
Because the two molecules cannot be superimposed, they are
distinct optical isomers.
All enantiomers have a stereogenic center carbon. This makes the molecule chiral
having a non-superimposable mirror image. When we name these enantiomers it
is necessary to distinguish them from one another. As it turns out each enantiomer
in the pair has opposite configuration.
Configuration is the arrangement of the groups attached to a stereogenic center.
In one enantiomer the arrangement is clockwise around the stereogenic carbon
beginning with the highest priority atom or group. This is called the "R"
configuration. The letter "R" comes from the Latin Rectus meaning right. The other
enantiomer of the pair being the non-superimposable mirror image will always have
an arrangement that proceeds counter clockwise around the stereogenic carbon.
This is a different configuration and is called the "S" isomer. The letter "S" comes
from the Latin Sinister meaning left. Now if we were to name the two enantiomers
using the systematic IUPAC nomenclature system, they would have the same name.
We then attach at the beginning of the name the letter "R" or "S" in parenthesis
Exercise:
Structural isomers? Geometric isomers?
Or optical isomers? Conformers
Figure 11.19 Plane-polarized light consists of radiation in
which all the wave motion lies in one plane (as represented by
the orange arrows on the left). When such light passes
through a solution of an optically active substance, the plane
of the polarization is rotated through a characteristic angle
that depends on the concentration of the solution and the
length of the path through it.
Figure 11.20 This polarimeter measures the optical activity of
compounds in solution. Light is plane polarized by passage
through a polarizer and is then sent through a sample. An
analyzer on the right of the sample is rotated until the angle at
which the light is brightest is found. That angle is the angle of
rotation for the sample.
Predicting whether a molecule is chiral
Yes
Classroom Exercise: Chiral?
CH3
CH3
No
CH3
CH3
Example: The significance of isomerism: drug efficiency
Quiz
• Explain the differences in the melting point and
boiling point of trans- and cis- isomers.
• Are the following molecules chiral?
Polymers (macromolecules)
• Synthetic polymers
• Biopolymers (DNA, RNA, Carbohydrates,
Proteins)
Homogenous polymer
Heterogeneous polymer
Tacticity (stereoregularity)
Figure 11.21 The stereoregular polymers produced by using
Ziegler-Natta catalysts may be (a) isotactic (all on one side) or
(b) syndiotactic (alternating). (c) In an atactic polymer, the
substituents lie on random sides of the chain.
Case Study 11
This flexible polyacetylene sheet was peeled from the walls of
the reaction flask in which it was made from acetylene.
Figure 11.22
Collecting latex from a rubber tree in Malaysia, one of its
principal producers.
Figure 11.23 In natural rubber, the isoprene units are
polymerized to be all cis. The harder material, gutta-percha, is
the all-trans polymer.
Condensation Polymerization:
How polymers are synthesized
Example: Synthesis of Dacron (Terylene)
HOOC
COOH + HO-CH2CH2OH
O
+ H2O
HOOC
C
O-CH2CH2OH
O
HO-CH2CH2OH+ HOOC
C
O-CH2CH2OH
O
O
C
C
O-CH2CH2OH
HO-CH2CH2O
+ HOOC
COOH
O
O
HOOC
C
O-CH2CH2O
C
O
C
O--CH2CH2O
n
Figure 11.24 Synthetic fibers are made by extruding liquid
polymer from small holes in an industrial version of the
spider’s spinneret.
Figure 11.25 A scanning electron micrograph of Dacron
polyester and cotton fibers in a blended shirt fabric. The
cotton fibers have been colored green. Compare the smooth
cylinders of the polyester with the irregular surface of cotton.
The smooth polyester fibers resist wrinkles, and the irregular
cotton fibers produce a more comfortable and absorbent
texture.
Figure 11.26 A rather crude nylon fiber can be made by
dissolving the salt of the amine in water and dissolving the
acid in a layer of hexane, which floats on the water. The
polymer forms at the interface of the two layers, and a long
string can be slowly pulled out.
PLAYING AROUND PRODUCES WONDER FIBER--NYLON
A team of organic chemists from Du Pont led by Wallace Hume Carothers had been trying to unravel the composition of natural polymers,
such as cellulose, silk, and rubber. From this knowledge they hoped to develop synthetic materials that mimicked the properties of these
natural polymers. This remarkable group of chemists had developed a group of compounds, polyamides, which had no remarkable or
useful properties.
These compounds were shelved in order to concentrate their work on a more promising series of compounds, polyesters. Polyesters possessed more
desirable properties such as having more soluble products, easier to handle and simpler to work with in the laboratory. Julian Hill, working with
polyester, noticed that if you gathered a small amount of this soft polymer on the end of your stirring rod and drew it out of the beaker, it produced
a silky, fine fiber. One afternoon when their boss, Wallace Carothers, was not in the lab, the chemists decided to see how long a silky thread they
could produce. Hill and his cohorts took a little ball on a stirring rod and ran down the hall and stretched them out into a string. The realization
struck them during this horseplay that by stretching the strand of fiber they were orienting the polymer molecules and increasing the strength
of the product.
The polyesters had very low melting points, too low for textile uses, so they retrieved the polyamides from the shelf and began to experiment with
this need 'cold-drawing process.' They found that the strand of polyamide produced by this cold-drawing technique produced a stron g, excellent
fiber. The patent for the composition of nylon was never applied for by Du Pont, rather they chose to patent the production process -- cold-drawing
-- developed by unsupervised adults playing around in the lab.
In January-February 1939, this consumer product hit the US market. It is without equal in its impact before or since. Nylon stockings were exhibited
at the Golden Gate International Exposition in San Francisco and were sold first to employees of the inventor company Du Pont de Nemours.
On May 15, 1940, nylon stockings went on sale throughout the US, and in New York City alone four million pairs were sold in a matter of hours.
Naming this new polymer too many twists and turns. Initially the name norun was proposed for this new product because it was more resistant to
laddering than silk. But there were problems and the name was then reversed to read nuron. However, it was pointed out that this was too close to
the word neuron which may be construed to be a nerve tonic. Hence, nuron was changed nulon. However this ran into trade mark problems and the
name was again changed to nilon. English speakers differed in their pronunciation of this, so, to remove ambiguity the name finally became nylon.
Two years before the basic patent on nylon had been filed, the discoverer of nylon, Wallace Hume Carothers, suffering from one of his increasingly
frequent attacks of depression, caused by his conviction that he was a scientific failure, drank juice containing potassium cyanide. He would be
pleased to know that half of all the chemists in the US work on the preparation, characterization, or application of polymers.
Figure 11.27 The strength of nylon fibers is yet another sign of
the presence of hydrogen bonds, this time between
neighboring polyamide chains.
Figure 11.28 The two samples of polyethylene in the test tube
were produced by different processes. The floating, lowdensity polymer was produced by high-pressure
polymerization. The high-density polymer at the bottom was
produced with a Ziegler-Natta catalyst. As the insets show,
the higher density results from the greater linearity of the
chains, allowing them to pack together better.
Figure 11.29 Automobile tires are made of vulcanized rubber
and a number of additives, including carbon. The gray
cylinders in the small inset represent polyisoprene molecules,
and the beaded yellow strings represent disulfide (—S—S—)
links that are introduced when the rubber is vulcanized, that is,
heated with sulfur. These cross-links increase the resilience of
the treated rubber and make it more useful than natural rubber.
Figure 11.30 This high-performance race car is made of a
composite material that is stronger than steel and can
withstand great stress.
Biopolymers
• Proteins (polypeptides/polyamino acids)
• Carbohydrates (polysaccharides)
• DNA and RNA (Polynucleotides)
They are all heterogeneous polymers:
DNA/RNA are four-letter sequences
Proteins are 20-letter sequences
Carbohydrates are many-letter sequences
Proteins
• Polymers formed by 20 different residues
of amino acids.
R side chain
C
NH3
amino
group
H
COO
carboxyl
group
G
A
F
V
L
I
S
T
K
R
H
W
Y
D
E
C
M
N
Q
P
Non-polar Amino Acids
• There are 8 non-polar amino acids:
O
H2N CHC OH
CH 3
Alanine (A)
O
H2N CHC OH
CH 2
CH 2
S
CH 3
Methionine (M)
O
H2N CHC OH
CH CH 3
CH 3
Valine (V)
O
H2N CHC OH
CH 2
O
H2N CHC OH
CH 2
CHCH 3
CH 3
Leucine (L)
O
H2N CHC OH
CH CH 2
CH 2
CH 2
Isoleucine (I)
O
C OH
HN
Proline (P)
O
H2N CH C OH
CH 2
3D structures
NH
Phenylalanine (F)
Trptophan (W)
Polar, Uncharged Amino Acids
• There 7 polar, uncharged amino acids:
O
H2N CHC OH
H
Glycine (G)
O
H2N CHC OH
CH 2
CH 2
C O
NH 2
O
H2N CHC OH
CH 2
OH
Serine (S)
O
H2N CHC OH
CH 2
O
H2N CHC OH
CHOH
CH 3
O
H2N CHC OH
CH 2
SH
Threonine (T)
Cysteine (C)
O
H2N CHC OH
CH 2
C O
NH 2
OH
Glutamine (Q)
Tyrosine (Y)
Asparagine (N)
3D structures
Polar, Charged Amino Acids
• There are 5 polar charged amino acids:
O
H2N CHC OH
CH 2
N
NH
Histidine (H)
O
H2N CHC OH
CH 2
CH 2
CH 2
CH 2
NH 2
Lysine (K)
O
H2N CHC OH
CH 2
CH 2
C O
OH
O
H2N CHC OH
CH 2
C O
OH
Glutamic Acid (E)
Aspartic Acid (D)
O
H2N CHC OH
CH 2
CH 2
CH 2
NH
C NH
NH 2
Arginine (R)
3D structures
Amino Acids not found in Proteins
•
Certain amino acids and their derivatives are biochemically important. For example, the visible
symptoms of allergies are caused by the release of histamine in mast cells, a type of cell found in
loose connective tissue. Histamine dilates blood vessels, increases the permeability of capillaries
(allowing antibodies to pass from the capillaries to surrounding tissue), and constricts bronchial
air passages. The molecular mechanism of histamine function is by its specific binding to a
protein called histamine H1 receptor.
H2N CH2
CH2
N
Histamine
HO
NH
•
H2N CH 2
CH 2
Serotonin
NH
Serotonin, which is derived from tryptophan, function as neurotransmitters and regulators.
Optical Activity and Stereochemistry of Amino Acids
•
All amino Acids but glycine are chiral molecules. There are two possible
configurations around C that constitute two non-superimposable mirror
image isomers, or enantiomers. Enantiomers display optical activity in
rotating the plane of polarized light. All natural amino acids are L- isomers.
1
CHO
OH
OH
H
CH 2OH
H
OHC
2
CH 2OH
3
L-Glyceraldehyde
(S)-Glyceraldehyde
1
CHO
H
OH
H
OH
HOH 2C
CH 2OH
3
H
CH 2OH
D-Glyceraldehyde
(R)-Glyceraldehyde
2
COOH
COOH
NH 3
CHO
2
H
HOH 2C
3
NH 3
1
L-Serine
(S)-Serine
Structure of peptide bond
• Two amino acids are joined by the peptide bond,
a reaction catalyzed by the enzyme called
ribosome in all cells:
H
O
H2N C C OH
R
+ NH 3
H O
C C OH
R'
H
O
H2N C C N
R
H
H O
C C OH
R'
• Due to the double bond character, the six atoms of the
peptide bond group are always planar.
H
C
H
N
C
O
C
C
H
N
C
OH
C
C
N
C
O
C
+ H2O
The Level of Protein Structure
• Primary Structrue (1º) refers to the amino acid sequences
of proteins;
• Secondary Structure (2º) refers to segments that constitute
structural conformities, or regular structures in proteins;
• Tertiary Structure (3º) refers to the folding of protein
chains into a more compact three dimensional shape;
• Quaternary Structure (4º) refers to organization of subunits
(one subunit is a single polypeptide chain).
Figure 11.31 A representation of part of an a helix, one of the
secondary structures adopted by polypeptide chains. The
tubes represent the atoms and their bonds, with colors that
correspond to the colors commonly used to represent different
atoms. The narrow lines indicate hydrogen bonds. The methyl
group side chains show that this molecule is polyalanine.
Example:
LSPADKTNVK…
…VKGWAA…
…STVLTSKLYR
Figure 11.32 One of the four polypeptide chains that make up
the human hemoglobin molecule. Each chain consists of
alternating regions of  helix (represented by red ribbons) and
-pleated sheet. The oxygen molecules we inhale attach to the
iron atom (blue sphere) and are carried through the
bloodstream to be released where they are needed.
Restriction by Amide Plane
• Atoms in the peptide bond lie in a plane.
Resonance stabilization energy of this planar
structure is approximately 88 kJ/mol;
• Rotation can only occur around the two bonds
connected to the C atom;
• Rotation around the Ca and carbonyl bond is
called y (psi);
• Rotation around the Ca and nitrogen bond is called
f (phi).
Rotation of Amide Planes
• If (f,y) are known for all residues, the structure for the entire
backbone is known.
• Some (f,y) are more likely than
others in a folded protein
• Positive (f,y) values correspond
to clockwise rotation around bonds
when viewed from the C. Zero
is defined when the C=O or N-H
bond bisects the R-C-H angle.
•
•
•
(f,y)=(0,180), two carbonyl oxygens are too close;
(f,y)=(180,0), two amide groups are overlapping;
(f,y)=(0,0), carbonyl oxygen overlaps with amide group;
Classes of Secondary Structures
• Terms below define all classes of secondary
structures seen in proteins:
• Helix
 -helix
– 310 helix
• Beta Sheet
– Parallel
– Anti-parallel
• Beta-bulge
• Beta Turn
The Alpha Helix
•
•
The alpha helix is a helical structure. All alpha helices in proteins are righthanded;
H-bond patterns of the alpha helix:
– Alpha helix: Carbonyl oxygen of the ith residue forms H-bond with amide proton of
the (i+4)th residue. So there are n-4 H-bonds in a helix of n amino acids;
– 310 helix: carbonyl oxygen of the ith residue forms H-bond with amide proton of the
(i+3)th residue. 3 residues (or 10 atoms) per turn;
– Proline is not found in -helix except at the beginning of an -helix;
– Helix propensity of an amino acid is a measure of the likelyhood for the amino acid
to be in a helix; Glu, Met, Ala, Leu have high propensities;
– Examples of -helical proteins include -keratin (structural proteins) and collagen
(fibrous protein);
– Linus Pauling (Nobel Prize in Chemistry, 1954) figured out the structure of
-keratin helix.
The Alpha Helix
•Residues per turn: 3.6
•Rise per residue: 1.5 Å
•Rise per turn: 5.4 Å
•(f,y)=~(-60º,-45º)
•C=O N-H side chain
•Total dipole moment
Showing dipole moments
The Beta Strands
• Beta strands form beta sheet in proteins;
• H-bond patterns in beta strands:
– Parallel beta-strands (0.325 nm between two residues)
– Anti-parallel beta-strands (0.347 nm between two residues)
C
H
R3
O
H
N
H
R2
O
C
R3
R0
O
H
H
N
R0
O
N
O
H
R2
O
H
R0
H
O
R1
H
O
R3
N
N
O
R1
N
N
N
R2
O
N
N
N
H
R1
N
N
H
O
H
N
O
H
O
N
H
R3
R1
N
N
O
C
N
R0
H
N
N
O
H
R2
C
O
The Beta Sheets
•
•
•
Formed by beta strands. Note that side chains point away from the sheet while main
chains lie on the sheet. Sheets are the most extended form.
Sheets consist of parallel strands are usually larger that those consist of anti-parallel
strands.
A sheet consists of parallel strands distribute hydrophobic residues on both sides of the
sheet while that consist of antiparallel strands distributes hydrophobic residues on one
side.
The Beta Turn (tight turn, or -bend)
• Beta turns connect beta strands and reverse the
direction of beta strands;
• Proline and glycine have high propensity for beta
turns;
• The carbonyl oxygen of the ith residue forms Hbond with the amide proton of the (i+3)th residue;
• Tight turn promotes formation of antiparallel beta
sheets.
The Beta Bulge
•
•
Beta bulge occurs between normal -strands. Comprised of two residues on one strand
and one on the other;
Bulges cause bending of otherwise straight anti-parallel beta strands;
C
C
H
R3
O
H
N
H
N
O
H
R2
O
H
O
R1
H
N
H
R3
N
O
H
R-1
R2
H
H
N
Beta bulge C
R1
O
N
N
O
H
R2
O
H
R0
H
O
R1
H
O
R3
N
R3
O
N
N
N
O
O
N
N
N
R0
R1
O
H
N
N
N
N
R0
N
R0
H
O
N
N
O
H
R2
O
C
Anti-parallel strands
Super secondary Structures (I)
1
• Hairpins connect two antiparallel strands;
2
• Cross-overs connect two parallel beta strands, most common through
an -helix (-- topology). All cross-overs are right-handed. That is,
when placing C-side strand closer and pointing right, the connecting ahelix or loop is on the top of the sheet;
1
1
2
2
Right-handed Cross-over
Left-handed Cross-over
Super Secondary Structures (II)
•
Coiled-coil is a common alpha helix structure found in proteins that participate
in protein folding and protein-protein interactions.
– (a-b-c-d-e-f-g)n, where a and d are
nonpolar that leads to a hydrophobic side
•
Helix bundles refers to three or more helices packing together;
– Knobs into holes packing:
In both kinds of helix packings, slight distortion
of the individual helices and the
inclination of their axes with respect
to each other allows the side chains
of the nonpolar residues to mesh together
Figure 11.33
The sickle-shaped red blood cells that form when a certain
glutamic acid residue in hemoglobin (see Fig. 11.32) is
replaced by valine.
Figure 11.34 The protein made by spiders to produce a web is
a form of silk that can be exceptionally strong.
Figure 11.35 The thread on these spools is synthetic spider
silk, one of the strongest fibers known. It can be used as the
thin, tough thread shown here or wound into cables strong
enough to support suspension bridges.
Carbohydrates
• Carbohydrates are the most abundant organic molecules in
nature
– Photosynthesis energy stored in carbohydrates;
– Carbohydrates are the metabolic precursors of all other
biomolecules;
– Important component of cell structures;
– Important function in cell-cell recognition;
– Carbohydrate chemistry:
• Contains at least one asymmetric carbon center;
• Favorable cyclic structures;
• Able to form polymers
Carbohydrate Nomenclature (I)
• Carbohydrate Classes:
– Monosaccharides (CH2O)n
• Simple sugars, can not be broken down further;
– Oligosaccharides
• Few simple sugars (2-6).
– Polysaccharides
• Polymers of monosaccharides
157
Carbohydrate Nomenclature (II)
• Monosaccharide (carbon numbers 3-7)
– Aldoses
• Contain aldrhyde
• Name: aldo-#-oses (e.g., aldohexoses)
Memorize all aldoses in Figure ?
– Ketoses
• Contain ketones
• Name: keto-#-oses (ketohexoses)
1
CHO
2
H
OH
3
H
OH
4
H
OH
5
6
H
OH
CH2OH
1
CHO
2
H
O
3
H
OH
4
H
OH
5
6
H
OH
CH2OH
Polysacchrides
• Also called glycans;
• Starch and glycogen are storage
molecules;
• Chitin and cellulose are structural
molecules;
• Cell surface polysaccharides are
recognition molecules.
Figure 11.36 The amylose molecule, one component of starch,
is a polysaccharide. A polymer of glucose, it consists of
glucose units linked together to give a structure like this but
with a moderate degree of branching.
Polysacchrides
• Glucose is the monosaccharides of the
following polysacchrides with different linkages
and banches
 (1,4), starch (more branch)
 (1,4), glycogen (less branch)
 (1,6), dextran (chromatography resins)
 (1,4), cellulose (cell walls of all plants)
 (1,4), Chitin similar to cellulose, but C2-OH is
replaced by –NHCOCH3 (found in exoskeletons of
crustaceans, insects, spiders)
Figure 11.37
The amylopectin molecule is another component of starch. It
has a more highly branched structure than amylose.
Figure 11.38 (a) Cellulose is yet another polysaccharide
constructed from glucose units. The linking between the units
in cellulose results in long, flat ribbons that can produce a
fibrous material through hydrogen bonding. (b) These long
tubes of cellulose formed the structural material of an aspen
tree.
DNA and RNA
A
G
C
T
U (in RNA)
Extension of the DNA chain
Figure 11.41 The condensation of nucleotides that leads to the
formation of a nucleic acid—a polynucleotide. The lensshaped object is an attached amine.
Figure 11.42 The bases in the DNA double helix fit together by
virtue of the hydrogen bonds that they can form as shown on
the left. Once formed, the AT and GC pairs are almost identical
in size and shape. As a result, the turns of the helix shown on
the right are regular and consistent.
Figure 11.39 A computer graphics image of a short section of a
DNA molecule, which consists of two entwined helices. In this
illustration, the double helix is also coiled around itself in a
shape called a superhelix.
Figure 11.40 A DNA molecule is very large, even in bacteria. In
this micrograph, a DNA molecule has spilled out through the
damaged cell wall of a bacterium.
The Code of Life
• Three-letter code of DNA  Amino
acidsProteins
• All other molecules
• Organism
Assignment for Chapter 11
11.32; 11.40; 11.46; 11.57; 11.67;
11.73; 11.77; 11.78; 11.87;11.92; 11.93