Chapter Twenty
Two
Organic and
Biological
Molecules
講義
Assignment
4,11,21,25,31,46,47,50,57,65,
73,83,95,103,109.
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Chapter
16
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Figure 22.1 The C-H Bonds
Millions of organic compounds
(compared to tens of thousands of
inorganic compounds), thousands
new organic compounds are synthesized
every year.
Carbon is unique:
(i) It has four valence electrons:
0+4=4=8-4
(ii) It has only two shells of electrons—
the only inner shell is 1s electrons
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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
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Examples
Structural formula
Name
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Abbreviated structural formula
(ordinary) molecular formula
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異戊二烯
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The molecule that shocked science more
than once.
Friedrich August von Stradonitz Kekulé
(1829-1896) who proposed the ring
model of benzene in 1861
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乙醯柳酸（阿司匹林）
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藏茴香酮 （香芹酮）
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Alkanes: saturated hydrocarbons
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Cycloalkanes
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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.
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Melting
(blue) and boiling (pink) points of the first 14 n-alkanes in °C.
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Straight Chain
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Branched (with side chains)
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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.
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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.
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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)
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Alkenes and Alkynes
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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.
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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).
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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)
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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
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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
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Aromatic Compounds (Arenes)
萘
蒽
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Schematic representation of coal
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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.
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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
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4,4,2-trimethylpentane
5
5
1
4
4
33
2
2
1
Using the smallest numbers possible.
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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.
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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
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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
1
5
2
4
3
6
4
2
4
5
3
1
5
CH2CH3
3
CH3
CH3
1
2
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2,3-dimethyl-4-ethylcyclohexene
6
1
5
4
3
2
3
6
5
CH2
CH3
2
4
CH2
CH3
1
4
2
3
CH3
1
5
6
CH3
5
CH3
1,6-dimethyl-5-ethylcyclohexene,
5-ethyl-1,6-dimethylcyclohexene,
1,2-dimethyl-3-ethylcyclohexene
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CH3
4
CH2 3
CH3
6
1 CH3
2
CH3
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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)
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CH
3
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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
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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
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Functional Groups
Alcohols
Ethers
Phenols
Aldehydes and Ketones
Carboxylic Acids and Esters
Amines and Amides
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aa
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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
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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.
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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.
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Alcohols
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-OH
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Ethers
R-O-R`
Water, CH3CH2-O-H, CH3CH2-O-CH2CH3
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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.
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Phenols
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百里香酚
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(麝香草酚)
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丁香油酚
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Aldehydes and Ketones
O
R-C
H
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O
R-C
R
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Smoked meat/fish
Wood smoke contains formaldehyde (formalin) that
has destructive effect on bacteria so smoked food can
be preserved long.
Simplest aldehyde: HCHO
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for aroma of cherries and almonds
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In oil of cinnamon
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in oil of vanilla
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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
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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:CH
COCH
+Ag
(from
Tollens Reagent)  no reaction
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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
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三硬脂精
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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
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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
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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 aminowhen it is a substituent. (CH3CH2)2NCH3diethylmethylamine.
Halides: Name the halogen atom as a substituent by changing the –
ineCopyright
part ©of
a=its
name
toAll –o.
CH3Brbromomethane.
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Naming the following compounds
CH3CH(CH3)CHOHCH3
CH3CH2CH2COCH3
(CH3CH2)2NCH2CH2CH3
(a) 3-methyl-2-butanol
(b) 2-pentanone
(c) diethylpropylamine
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Exercise
CH CH OH
Naming the following compounds:
CH CH
CH3CH(CH2CH2OH)CH3
CH
CH3CH(CHO)CH2CH3
CH CH
(C6H5)3N
CH CH
2
2
3
3
2
3
3
CHO
(a) 3-methyl-1-butanol
(b) 2-methylbutanal
(c) triphenylamine
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Classroom Exercise
Naming the following compounds:
CH3CH2CHOHCH2CH3
CH3CH2COCH2CH3
CH3CH2NHCH3
(a) 3-pentanol
(b) 3-pentanone
(c) ethylmethylamine
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Carboxylic acids and Esters
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.
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Halides
Name the halogen atom as a substituent
by changing the –ine part of a=its name to –
o.
CH3Brbromomethane.
CH3CH2Clchloroethane.
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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
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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
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Summary of Nomenclature of
Hydrocarbons and Functional Groups
-ane, -ene, -yne
-ol, -al, -one, -oic acid, -oate, -amine, -o
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Isomers
Structural isomers
Stereoisomers:
Geometrical isomers
Optical isomers
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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.
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Structural Isomers: C4H10
CH3-CH2-CH2-CH3
CH(CH3)3
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Structural Isomers: C6H14
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Isomer vs Conformation
They are the same isomer but with different conformations:
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Exercise: Different Isomers or Different
Conformers
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Stereoisomerism I:
Geometric Isomerism
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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.
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cis or trans?
(a) trans (b) cis
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Classroom Exercise: cis or trans?
(a) cis (b) trans
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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).
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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-2ene
-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.
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Stereoisomerism II:
Chirality (Enantiomerism)
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A molecule is chiral if and only if all the (four) attached
groups are different
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An atom is a chiral center if and only if all the (four) groups attached to it
are different
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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.
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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
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Exercise:
Structural isomers? Geometric isomers? Or optical
isomers? Conformers
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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.
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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.
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Predicting whether a molecule is chiral
Yes
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22 | 114
Classroom Exercise: Chiral?
CH3
CH3
CH3
CH3
No
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22 | 115
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Example: The significance of isomerism: drug efficiency
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Quiz
Explain the differences in the melting point
and boiling point of trans- and cis- isomers.
Are the following molecules chiral?
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Polymers (macromolecules)
Synthetic polymers
Biopolymers (DNA, RNA, Carbohydrates,
Proteins)
Homogenous polymer
Heterogeneous polymer
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Tacticity (stereoregularity)
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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.
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Case Study 11
This flexible polyacetylene sheet was peeled from the walls of the reaction
flask in which it was made from acetylene.
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Figure 11.22
Collecting latex from a rubber tree in Malaysia, one of its principal producers.
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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.
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Condensation Polymerization:
How polymers are synthesized
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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
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C
O
C
O--CH2CH2O
n
22 | 128
Figure 11.24 Synthetic fibers are made by extruding liquid polymer from small
holes in an industrial version of the spider’s spinneret.
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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.
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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.
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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.
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Figure 11.27 The strength of nylon fibers is yet another sign of the presence
of hydrogen bonds, this time between neighboring polyamide chains.
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Figure 11.28 The two samples of polyethylene in the test tube were produced
by different processes. The floating, low-density 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.
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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.
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Figure 11.30 This high-performance race car is made of a composite material
that is stronger than steel and can withstand great stress.
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22 | 138
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
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Proteins
Polymers formed by 20 different residues
of amino acids.
R side chain
C
NH3
amino
group
H
COO
carboxyl
group
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22 | 140
G
A
F
V
L
I
S
T
K
R
H
W
Y
D
E
C
M
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N
Q
P
22 | 141
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
CHCH 3
CH 3
Leucine (L)
O
H2N CHC OH
CH 2
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)
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Trptophan (W)
22 | 142
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
3D structures
OH
Glutamine (Q)
Tyrosine (Y)
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Asparagine (N)
22 | 143
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)
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O
H2N CHC OH
CH 2
CH 2
CH 2
NH
C NH
NH 2
Arginine (R)
3D structures
22 | 144
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
Histamine
H2N CH 2
CH 2
Serotonin
N
HO
NH
NH
Serotonin, which is derived from tryptophan, function as neurotransmitters and
regulators.
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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 Lisomers.
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
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NH 3
1
L-Serine
(S)-Serine
22 | 146
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
C C OH
R'
H
O
H2N C C OH
R
H
O
H2N C C N
R
H
H O
C C OH
R'
+ NH
Due to the double
bond
character, the six atoms of the
3
peptide bond group are always planar.
H
C
H
N
C
C
C
O
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H
N
C
OH
+ H2O
C
C
N
C
C
O
22 | 147
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).
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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
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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.
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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).
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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;
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Classes of Secondary Structures
Terms below define all classes of secondary
structures seen in proteins:
Helix
) -helix
b) 310 helix
Beta Sheet
a) Parallel
b) Anti-parallel
Beta-bulge
Beta Turn
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The Alpha Helix
The alpha helix is a helical structure. All alpha helices in proteins are righthanded;
H-bond patterns of the alpha helix:
a)
b)
c)
d)
e)
f)
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.
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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
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The Beta Strands
Beta strands form beta sheet in proteins;
H-bond patterns in beta strands:
a)
b)
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
R0
N
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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
O
C
22 | 156
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.
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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.
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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
O
H
R2
O
H
R0
H
O
R1
H
O
R3
R0
H
O
Beta bulge C
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O
N
N
N
N
R1
N
N
R3
O
N
O
O
N
N
N
R0
R1
O
H
N
N
N
N
R0
N
N
N
O
H
R2
O
C
Anti-parallel strands
22 | 159
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 righthanded. That is, when placing C-side strand closer and pointing
right, the connecting a-helix or loop is on the top of the sheet;
1
1
2
2
Right-handed Cross-over
Left-handed Cross-over
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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)
(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;
a)
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
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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.
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22 | 162
Figure 11.34 The protein made by spiders to produce a web is a form of silk
that can be exceptionally strong.
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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.
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22 | 164
Carbohydrates
Carbohydrates are the most abundant organic molecules
in nature
a) Photosynthesis energy stored in carbohydrates;
b) Carbohydrates are the metabolic precursors of all other
biomolecules;
c) Important component of cell structures;
d) Important function in cell-cell recognition;
e) Carbohydrate chemistry:
•
•
•
Contains at least one asymmetric carbon center;
Favorable cyclic structures;
Able to form polymers
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22 | 165
Carbohydrate Nomenclature (I)
Carbohydrate Classes:
a) Monosaccharides (CH2O)n
•
Simple sugars, can not be broken down further;
b) Oligosaccharides
•
Few simple sugars (2-6).
c) Polysaccharides
•
Polymers of monosaccharides
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Carbohydrate Nomenclature (II)
Monosaccharide (carbon numbers 3-7)
a) Aldoses
• Contain aldrhyde
• Name: aldo-#-oses (e.g., aldohexoses)
Memorize all aldoses in Figure ?
b) Ketoses
•
•
Contain ketones
Name: keto-#-oses (ketohexoses)
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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
22 | 167
Polysacchrides
Also called glycans;
Starch and glycogen are storage
molecules;
Chitin and cellulose are structural
molecules;
Cell surface polysaccharides are
recognition molecules.
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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.
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22 | 169
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Polysacchrides
Glucose is the monosaccharides of the
following polysacchrides with different
linkages and banches
)
)
c)
d)
e)
(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)
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22 | 172
Figure 11.37
The amylopectin molecule is another component of starch. It has a more
highly branched structure than amylose.
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22 | 173
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.
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22 | 174
DNA and RNA
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22 | 176
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22 | 177
A
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22 | 178
G
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C
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22 | 180
T
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U (in RNA)
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Extension of the DNA chain
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Figure 11.41 The condensation of nucleotides that leads to the formation of a
nucleic acid—a polynucleotide. The lens-shaped object is an attached amine.
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22 | 186
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.
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22 | 187
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.
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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.
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22 | 189
The Code of Life
Three-letter code of DNA  Amino
acidsProteins
All other molecules
Organism
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22 | 190
Figure 22.2
a-b (a) The
Lewis
Structure of
Ethane (b)
The
Molecular
Structure of
Ethane
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22 | 191
Figure 22.3 a-b The Structures of (a)
Propane and (b) Butane Each Angle
Show in Red is 109.5°
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22 | 192
Figure 22.4 a-b (a) Normal Butane and
(b) The Branched Isomer of Butane
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n-Pentane, Isopentanec and
Neopentane
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Figure 22.5 ab (a) The
Molecular
Structure of
Cyclopropane
(b) The
Overlap of
the sp3
Orbitals that
Form the C-C
Bonds in
Cyclopropane
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Figure 22.6 a-b The (a) Chair and (b)
Boat Forms of Cyclohexane
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22 | 196
Figure 22.7 The Bonding in Ethylene
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Figure 22.8 The Bonding in Ethane
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Figure 22.9
The Two
Stereoisomers
of 2-Butene: (a)
cis-2-Butene
and (b) trans2-Butene
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Figure 22.10 The Bonding in Acetylene
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Figure 22.11 a-c (a) The Structure of
Benzene (b) Two of the Resonance
Structures of Benzene (c.) The Usual
Representation of Benzene
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Figure 22.12 Selected Substituted
Benzenes and their Names
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Figure 22.13 Some Common Ketones
and Aldehydes
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Figure 22.14
Some
Carboxylic
Acids
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22 | 204
Computer-Generated Space-Filling
Model of Acetylsaicylic Acid (Aspirin)
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22 | 205
Figure 22.15 The General Formulas for
Primary, Secondary, and Tertiary Amines
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22 | 206
Figure 22.16
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22 | 207
Figure 22.17 a-d A Major Use of HDPE
is for Blow-Molded Objects such as
Bottles for Consumer Products
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22 | 208
Figure
22.18 The
20 AlphaAmino
Acids
Found in
Most
Proteins
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22 | 209
A Tripeptide Containing Glycine,
Cysteine, and Alanine
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22 | 210
Figure 22.20
Hydrogen
Bonding
within a
Protein Chain
Causes it to
Form a
Stable Helical
Structure
Called the
Alpha-Helix
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Figure 22.21
Portion of a
Protein Chain
Showing the
HydrogenBonding
Interactions
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Figure 22.22 When Hydrogen-Bonding
Occurs Between Protein Chains, a Stable
Structure Called a Pleated Sheet Results
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Figure 22.23
a-b (a)
Protein
Chains Form
Superhelix (b)
PleatedSheet
Proteins
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22 | 214
Figure 22.24 a-e Summary of the Various
Types of Interactions that Stabilize the
Tertiary Structure of a Protein: (a) Ionic, (b)
Hydrogen Bonding, (c) Covalent, (d) London
Dispersion, and (e) Dipole-Dipole
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Figure
22.25 The
Permanent
Waving of
Hair
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22 | 216
Figure 22.26
The Thermal
Denaturation
of a Protein
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22 | 217
Figure 22.27
A
Tetrahedral
Carbon Atom
with Four
Different
Substituents
cannot have
its Mirror
Image
Superimpose
d
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22 | 218
Figure 22.28 The Mirror Image Optical
Isomers of Glyceraldehyde
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22 | 219
Figure
22.29 The
Cyclization
of DFructose
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Figure 22.30 The Cyclization of Glucose
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Figure 22.31 Sucrose is a Disaccharide
Formed from Alpha-D-glucose and Fructose
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Figure 22.32 a-b (a) The Polymer Amylose is a Major
Component of Starch and is Made Up of Alpha-DGlucose Monomers (b) The Polymer Cellulose, which
Consists of Beta-D-Glucose Monomers
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Figure 22.33 a-b
The Structure of
the Pentoses (a)
Deoxyribose
and (b) Ribose
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Figure 22.34 The Organic Bases
Found in DNA and RNA
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Figure 22.35 a-b The Base and Sugar
Combine to Form a Unit that in Turn Reacts
with Phosphoric Acid to Create the
Nucleotide, which is an Ester
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22 | 226
A Computer
Image of the
Base Pairs of
DNA
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22 | 227
Figure 22.36
A Portion of
a Typical
Nucleic Acid
Chain
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22 | 228
Figure 22.37 a-c (a) The DNA Double Helix
Contains Two Sugar-Phosphate Backbones, with
the Bases from the Two Strands Hydrogen Bonded
to each other; The Complementarity of the (b)
Thymine-Adenine and (c) Cytosine-Guanine Pairs
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Figure 22.38
DNA Cell
Division
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22 | 230
Figure 22.39 The Anticodon of the
tRNA Must Complement the Codon of
the mRNA
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22 | 231
A Butane Lighter Used for Camping
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22 | 232
A Worker Using an Oxyacetylene Torch
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22 | 233
A Winemaker
Drawing Off a
Glass of Wine
in a Modern
Wine Cellar
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22 | 234
Ethanol is
Being Tested
in Selected
Areas as a
Fuel for
Automobiles
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22 | 235
Cinnemaldehyde Produces the
Characteristic Odor of Cinnamon
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22 | 236
Aspirin Tablets
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22 | 237
The Soybeans on the Left are Coated
with a Red Acrylic Polymer to Delay
Soybean Emergence
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22 | 238
A Radio from the 1930s Made of
Bakelite
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22 | 239
Nylon Netting Magnified 62 Times
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22 | 240
A Scanning Electron Microscope Image Showing
the Fractured Plane of a Self-Healing Material with
a Ruptured Microcapsule in a Thermosetting Matrix
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22 | 241
Cross-Linking Gives the Rubber in
these Tires Strength and Toughness
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Figure 22.16
The Reaction
to Form
Nylon Occurs
Readily
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Wallace H. Carothers
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22 | 244
PVC Pipe is Widely Used in Industry
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22 | 245
The Protein
in Muscles
Enables
Them to
Contract
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22 | 246
Self-Tanning Products
and a Close-Up of a
Label Showing the
Contents
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Photo 22.17 A Bowl of Sugar Cubes
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Table 22.1 Selected Properties of the
First Ten Normal Alkanes
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22 | 249
Table 22.2
The Most
Common
Alkyl
Substituents
and Their
Names
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22 | 250
Table 22.3 More Complex Aromatic
Systems
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22 | 251
Table 22.4
The Common
Functional
Groups
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22 | 252
Table 22.5 Some Common Alcohols
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22 | 253
Table 22.6 Some Common Amines
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22 | 254
Table 22.7 Some Common Synthetic Polymers,
Their Monomers and Applications
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22 | 255
Table 22.8 Some Important
Monosaccharides
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22 | 256
Chapter
Twenty-Two
Organic and
Biological
Molecules
問題/答案
Question
A student gave a molecule the following name:
3-methyl-4-isopropylpentane
The teacher pointed out that, although the
molecule could be correctly drawn from this
name, the name violates the IUPAC rules. What
is the correct (IUPAC) name of the molecule?
a)
b)
c)
d)
e)
4-Isopropyl-3-methylpentane
2-Isopropyl-3-methylpentane
1,1,2,3-Tetramethylpentane
2,3,4-Trimethylhexane
3,4-Dimethylheptane
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22 | 258
Answer
d) 2,3,4-Trimethylhexane
Section 22.1, Alkanes: Saturated Hydrocarbons
The molecule would have six carbons in the
longest chain and three methyl groups. The correct
name is 2,3,4-trimethylhexane.
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22 | 259
Question
Which of the following names is a correct
one?
a) 3,4-Dichloropentane
b) cis-1,3-Dimethylpropane
c) 2-Bromo-1-chloro-4,4-diethyloctane
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22 | 260
Answer
c) 2-Bromo-1-chloro-4,4-diethyloctane
Section 22.1, Alkanes: Saturated
Hydrocarbons
Choice (a) should be 2,3-dichloropentane.
Choice (b) should be pentane.
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22 | 261
Question
Which of the following has the lowest boiling
point?
a)
b)
c)
d)
Butane
Ethane
Propane
Methane
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22 | 262
Answer
d) Methane
Section 22.1, Alkanes: Saturated
Hydrocarbons
The smallest of the saturated hydrocarbons
will have the lowest boiling point.
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22 | 263
Question
How many isomers are there with the
formula C2H2Br2? Include both structural and
geometric isomers.
a)
b)
c)
d)
e)
2
3
4
5
6
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22 | 264
Answer
b) 3
Section 22.2, Alkenes and Alkynes
The formula given could be any of these:
cis-1,2-Dibromoethene
trans-1,2-Dibromoethene
1,1-Dibromoethene
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22 | 265
Question
Which of the following is an incorrect name?
a)
b)
c)
d)
trans-1,2-Dichloroethene
Propylene
Ethylene
cis-1,2-Dichloroethane
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22 | 266
Answer
d) cis-1,2-Dichloroethane
Section 22.1, Alkanes: Saturated Hydrocarbons;
Section 22.2, Alkenes and Alkynes
Given that 1,2-dichloroethane is a saturated
hydrocarbon, no cis or trans designation is
necessary.
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22 | 267
Question
How many different “tetramethyl-benzenes”
are possible?
a)
b)
c)
d)
e)
2
3
4
5
6
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22 | 268
Answer
b) 3
Section 22.3, Aromatic Hydrocarbons
Here are the three possibilities:
1,2,3,4-Tetramethylbenzene
1,2,3,5-Tetramethylbenzene
1,2,4,5-Tetramethylbenzene
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22 | 269
Question
H2CCHCH2N(CH3)2 is
a)
b)
c)
d)
an alkyne and a secondary amine.
an alkene and a primary amine.
an alkene and a tertiary amine.
an alkyne and a tertiary amine.
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22 | 270
Answer
c) an alkene and a tertiary amine.
Section 22.2, Alkenes and Alkynes; Section
22.4, Hydrocarbon Derivatives
This species is a tertiary amine because
three carbons are bonded to the nitrogen
and the molecule contains a C=C bond.
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22 | 271
Question
For which of the following compounds are
cis and trans isomers possible?
a)
b)
c)
d)
2,3-Dimethyl-2-butene
3-Methyl-2-pentene
4,4-Dimethylcyclohexanol
ortho-Chlorotoluene
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22 | 272
Answer
b) 3-Methyl-2-pentene
Section 22.2, Alkenes and Alkynes; Section 22.3, Aromatic
Hydrocarbons; Section 22.4, Hydrocarbon Derivatives
Choices (c) and (d) do not contain different groups across a
double bond. For choice (a), the second carbon contains
two methyl groups, so the cis-trans designation is not
necessary.
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22 | 273
Question
Which of the following types of compounds
may lack an sp2-hybridized carbon center?
a)
b)
c)
d)
Aldehydes
Ketones
Alcohols
Alkenes
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22 | 274
Answer
c) Alcohols
Section 22.4, Hydrocarbon Derivatives
Since aldehydes and ketones have C=O bonds
and alkenes and benzene have C=C bonds, these
derivatives have sp2-hybridized carbon centers.
Alcohols have C–O–H bonds and can have sp3hybridized carbon centers.
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22 | 275
Question
For which of the following choices do all of the
functional groups listed have a C=O bond?
a)
b)
c)
d)
e)
Ester, aldehyde, secondary alcohol, ketone
Alcohol (any), ether, ester
Secondary alcohol, ketone, aldehyde
Ester, aldehyde, ketone
Carboxylic acid, ether, tertiary alcohol
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22 | 276
Answer
d) Ester, aldehyde, ketone
Section 22.4, Hydrocarbon Derivatives
Table 22.4 lists the functional groups. Carboxylic
acids, esters, aldehydes, and ketones all contain
C=O bonds. Alcohols contain the –OH group.
Ethers contain the –O– group.
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22 | 277
Question
Oxidation of secondary alcohols results in
a)
b)
c)
d)
e)
ketones.
tertiary alcohols.
aldehydes.
esters.
ethers.
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22 | 278
Answer
a) ketones.
Section 22.4, Hydrocarbon Derivatives
Ketones may be prepared from the oxidation
of secondary alcohols.
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22 | 279
Question
What might be the product of the oxidation
of 2-methyl-1-butanol?
a)
b)
c)
d)
e)
2-Methyl-2-butanone
2-Methylbutanal
2-methylbutanoic acid
Both b and c
Both a and c
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22 | 280
Answer
d) Both b and c
Section 22.4. Hydrocarbon Derivatives
The primary alcohol could be oxidized to an
aldehyde or a carboxylic acid, so the
answers are 2-methyl-1-butanal and 2methylbutanoic acid.
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22 | 281
Question
Which of the following is optically active (i.e.,
chiral)?
a)
b)
c)
d)
e)
HN(CH3)2
CH2Cl2
2-Chloropropane
2-Chlorobutane
3-Chloropentane
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22 | 282
Answer
d) 2-Chlorobutane
Section 22.4, Hydrocarbon Derivatives
For a molecule to be optically active, it must have
a carbon atom bonded to four different species.
For 2-chlorobutane, the carbon at the 2 position is
bonded to a methyl group, a chlorine, a hydrogen,
and an ethyl group.
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22 | 283
Question
Oxidation of primary alcohols results in
a)
b)
c)
d)
e)
ketones.
tertiary alcohols.
aldehydes.
esters.
ethers.
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22 | 284
Answer
c)
aldehydes.
Section 22.4, Hydrocarbon Derivatives
An aldehyde results from the oxidation of a
primary alcohol.
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22 | 285
Question
The boiling point of methanol is much higher than
that of ethane. This is primarily due to
a) the significant difference in the molar masses of
methanol and ethane.
b) the hydrogen bonding in methanol and the lack of
hydrogen bonding in ethane.
c) the significant difference in the molecular sizes of
methanol and ethane.
d) the carbon–oxygen bond in the methanol.
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22 | 286
Answer
b)
the hydrogen bonding in methanol and the
lack of hydrogen bonding in ethane.
Section 22.4, Hydrocarbon Derivatives
Methanol is a polar molecule with hydrogen
bonding; ethane is a nonpolar molecule with
London dispersion forces. The boiling point will be
much higher for polar molecules with hydrogen
bonding.
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22 | 287
Question
No atoms are lost from the starting material
in making which kind of polymer?
a) Condensation polymer
b) Polyester polymer
c) Addition polymer
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22 | 288
Answer
c) Addition polymer
Section 22.5, Polymers
When monomers add to form polymers, no
atoms are lost.
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22 | 289
Question
The structure of the polymer used in a freezer
wrap can mainly be described as follows:
[CCl2 – CH2 – CCl2 – CH2 – CCl2 – CH2 – CCl2 – CH2]n
What is the structure of chief monomer of this wrap?
a) CCl2 CH2
b) Cl2C–CH2
c) Cl2C CH2 CCl2
d) CCl2
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22 | 290
Answer
a)
CCl2CH2
Section 22.5, Polymers
For the polymer given, the monomeric unit is
CCl2=CH2, which adds to form the polymer.
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22 | 291
Question
Which of the following steps will increase the
rigidity of a polymer?
a)
b)
c)
d)
Use shorter polymer chains
Make chains more branched
Decrease cross-linking
Introduce the possibility of hydrogen
bonding between chains
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22 | 292
Answer
d) Introduce the possibility of hydrogen
bonding between chains
Section 22.5, Polymers
Increasing hydrogen bonding in a polymer
can give it greater strength and rigidity.
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22 | 293