Chapter Twenty Two Organic and Biological Molecules 講義

Chapter Twenty Two

Organic and Biological Molecules

講義

Copyright © Houghton Mifflin Company. All rights reserved. 22 | 2

Assignment

4,11,21,25,31,46,47,50,57,65,73,83,95,103,109.


本章投影片主要採用 Atkins書的第 22章：

Copyright © Houghton Mifflin Company. All rights reserved.

Chapter 16 | Slide 3

http://140.117.34.2/faculty/phy/sw_ding/teaching/gchem/gc22.ppt

http://140.117.34.2/faculty/phy/sw_ding/teaching/gchem/gc17.ppt


Figure 22.1 The C-H Bonds

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

Millions of organic compounds (compared to tens of thousands of inorganic compounds), thousands new organic compounds are synthesized every year.



Hydrocarbons and Functional Groups

AlkanesAlkenes and AlkynesAromatic compoundsFunctional Groups:AlcoholsEthersPhenolsAldehydes and KetonesCarboxylic Acids and EstersAmines and Amides


Examples

Structural formula Abbreviated structural formula

Name (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 forcein 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 lesssignificant. 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.

/ ( (>0) is smaller for more symmetric molecules ,

to keep constant, must increase for them)m m m m

m m

S H T S

H T

Obviously, there is no symmetry contributionto boiling point.

Copyright © Houghton Mifflin Company. All rights reserved. 22 | 27Melting (blue) and boiling (pink) points of the first 14 n-alkanes in °C.



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 heat4 2 3CH (g)+ (g) CH (g)+HCl(g)Cl Cl


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.

2 3Cr O3 3 2 2 2C C (g) CH =CH (g)+ (g)H H H


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.

22 2 2 2CH =CH (g)+ ( ) CHCl lHCl C-Cg


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.

22 2 2 2CH =CH (g)+ (gBr B) CH -r BrCH


Aromatic Compounds (Arenes)

蒽

萘


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 2 3 4

5

1

2 3 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 4 3 2

1

5

4 3 2

1

Using the smallest numbers possible.


How to name hydrocarbonsAlkenes 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

CH3CH=CHCH2CHCH3

CH3

1 2 3 4 5 6

5-methyl-2-hexene

6 5 4 3 2 1 2-methyl-5-hexene

The multiple bond has priority over the numbering of substituents


2,3-dimethyl-4-ethylcyclohexene

1

2

34

5

6

1

2

34

5

6

CH3

CH3CH2-CH3

6

1

23

4

5

The multiple bond has priority over the numbering of substituents (Use the smallest numbers to locate the double bond)


2,3-dimethyl-4-ethylcyclohexene

1

2

34

5

6

2

16

5

43

CH3

CH3CH2

-CH3

1,6-dimethyl-5-ethylcyclohexene,5-ethyl-1,6-dimethylcyclohexene,1,2-dimethyl-3-ethylcyclohexene

6

12

3

4

5

CH3

CH3CH2

-CH3

1

2

34

5

6

CH3

CH3CH2-CH3


How to name hydrocarbonsArenes:

-C6H5 arylortho- (o-), meta (m-), para (p-)

CH3

CH3

CH3

CH3

CH3

CH3

1,2-dimethylbenzene(o-Xylene)

1,3-dimethylbenzene(m-Xylene)

1,4-dimethylbenzene(p-Xylene)


Exercise

Name the following hydrocarbons

(a) (CH3)2CHCH2CH(CH2CH3)2

(b)

CH2CH3

CH2CH2CH3

(a) 4-ethyl-2-methylhexane(b) 1-ethyl-3-propylbenzene

(a) (CH3)2CHCH2CH(CH2CH3)2

1 2 3 4 5 6


QuizName 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-hexeneName the following compounds:

CH3

CH3


Functional Groups

AlcoholsEthersPhenolsAldehydes and KetonesCarboxylic Acids and EstersAmines 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 moleculeis 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 Imagingmakes it possible to see insidea 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

R-CH

OR-C

R

O


Smoked meat/fish

Wood smoke contains formaldehyde (formalin) thathas 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.

o

2 2 7 2 4

600 C,Ag3 2 2

Na Cr O (aq),H SO (aq)3 2 3 3 2 3

Aldehydes:CH OH(g)+O 2HCHO(g)+H O(g)

Ketones:CH CH (OH)CH CH CH COCH


Figure 11.13 An aldehyde (left) produces a silver mirror with Tollens reagent, but a ketone (right) does not.

+3 2 3 2

+3 3

Aldehydes:CH CH CHO+Ag (from Tollens Reagent) CH CH COOH+Ag(s)

Ketones:CH COCH +Ag (from Tollens Reagent) no reaction


Carboxylic Acids and Esters

3 2 2 3 2CH CH OH(aq)+O ( ) CH (aq)C +OO H ( )H O lg

+H ,3 2 3 3 2 3 2CH COOH(aq)+H-OCH CH CH CO)O( CH CH +H O

CH3

CH3

COOH

COOH

+ 3O2 (g) +2H2O



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 + waterCH3COOH+NH2CH3CH3CONHCH3+H2O


Amines and Amides

Amines: derivatives from ammonia by replacing hydrogens with organic groups.

Amides: resulted from condensation of amines with carboxylic acids.

CH3-N-H

H

H-N-H

H

CH3-N-CH3

H

CH3-N-CH3

CH3

Methylamine Dimethylamine Trimethylamine

Carboxylic acid + amine amide + waterCH3COOH+NH2CH3CH3CONHCH3+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 hydrocarbonsAlcohols: alkane-ol CH3CH2CHOHCH32-butanolEthers: Name each of hydrocarbon groups attached to the O atom separately and alphabetically. CH3OCH2CH3ethyl 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)CH2CH32-methylbutanal.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. CH3CH2CH2COCH32-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. CH3CH2CH2COOHbutanoic acid. Esters: Change the –ol of the alcohol to –yl and the oic acid of the parent acid to –oate. CH3CH2COOCH3methyl 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)2NCH3diethylmethylamine. Halides: Name the halogen atom as a substituent by changing the –ine part of a=its name to –o. CH3Brbromomethane.


Naming the following compounds

CH3CH(CH3)CHOHCH3

CH3CH2CH2COCH3

(CH3CH2)2NCH2CH2CH3

(a) 3-methyl-2-butanol

(b) 2-pentanone

(c) diethylpropylamine


Exercise

Naming the following compounds:

CH3CH(CH2CH2OH)CH3

CH3CH(CHO)CH2CH3

(C6H5)3N

(a) 3-methyl-1-butanol(b) 2-methylbutanal(c) triphenylamine

CH3CH

CH3

CH2CH2OH

CH3CH

CHO

CH2CH3


Classroom Exercise

Naming the following compounds:CH3CH2CHOHCH2CH3

CH3CH2COCH2CH3

CH3CH2NHCH3

(a) 3-pentanol(b) 3-pentanone(c) ethylmethylamine


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. CH3CH2CH2COOHbutanoic acid. Esters: Change the –ol of the alcohol to –yl and the oic acid of the parent acid to –oate. CH3CH2COOCH3methyl propanoate.


Halides

Name the halogen atom as a substituent by changing the –ine part of a=its name to –o.

CH3Brbromomethane.

CH3CH2Clchloroethane.


Quiz

CH3CH

CH3

CH2CH2OH

CH3CH

CHO

CH2CH3

1. Name the following compounds:

CH3CH2NHCH3

2. What is most important difference between aldelhyde and ketone?

CH3CH2CH2COOHCH3CH2COOCH3

CH3COOH+NH2CH3CH3CONHCH3+H2O

3. Name the following reaction:


Answer

CH3CH

CH3

CH2CH2OH

CH3CH

CHO

CH2CH3

1. Name the following compounds:

CH3CH2NHCH3

2. What is most important difference between an aldelhyde and a ketone?

CH3CH2CH2COOHCH3CH2COOCH3

CH3COOH+NH2CH3CH3CONHCH3+H2O

3. Name the following reaction:

3-methyl-1-butanol2-methylbutanal

butanoic acid ethylmethylamine methyl propanoate

An aldelhyde is a good reducing agent, but a ketone is not.

Condensation reaction


Summary of Nomenclature of Hydrocarbons and Functional Groups

-ane, -ene, -yne-ol, -al, -one, -oic acid, -oate, -amine, -o


Isomers

Structural isomersStereoisomers: 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).


isomer melting point (°C) boiling point (°C)

cis -80 60

trans -50 48

isomer melting point (°C)

boiling point (°C)

cis-but-2-ene -139 4

trans-but-2-ene

-106 1

1,2-dichloroethene

the trans isomer has the higher melting point;the cis isomer has the higher boiling point.

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)


A molecule is chiral if and only if all the (four) attached groups are different


An atom is a chiral center if and only if all the (four) groups attached to it are different


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?

No

CH3

CH3

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 polymersBiopolymers (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 11This 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)

COOH + HO-CH2CH2OHHOOC

HOOC CO

O-CH2CH2OH+ H2O

HOOC CO

O-CH2CH2O C C

O

O--CH2CH2O

O

n

HOOC CO

O-CH2CH2OHHO-CH2CH2OH+

C CO

O-CH2CH2OH

O

HO-CH2CH2O

COOHHOOC+


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--NYLONA 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, 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.


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

CNH3 COO

H

side chain

aminogroup

carboxylgroup


GAF

VLIST

Y

DECM

KR

H

W

NQP


Non-polar Amino Acids

There are 8 non-polar amino acids:

H2N CHC

CH3

OH

OH2N CHC

CH2

OH

O

CHCH3

CH3

H2N CHC

CH

OH

O

CH3

CH3

H2N CHC

CH

OH

O

CH2

CH2

CH2

HN

C OH

O

H2N CHC

CH2

OH

O

CH2

S

CH3

H2N CHC

CH2

OH

O

H2N CH

Proline (P)Isoleucine (I)

C

CH2

OH

O

Alanine (A) Leucine (L)Valine (V)

NH

Methionine (M) Phenylalanine (F) Trptophan (W)

3D structures


Polar, Uncharged Amino Acids

There 7 polar, uncharged amino acids:

H2N CHC

H

OH

O

H2N CHC

CH2

OH

O

OH

H2N CHC

CH

OH

O

OH

CH3

H2N CHC

CH2

OH

O

SH

H2N CHC

CH2

OH

O

C

NH2

O

H2N CHC

CH2

OH

O

CH2

C

NH2

O

H2N CHC

CH2

OH

O

OH

Glycine (G) Serine (S)

Glutamine (Q)

Cysteine (C)Threonine (T)

Asparagine (N)Tyrosine (Y)

3D structures


Polar, Charged Amino Acids

There are 5 polar charged amino acids:

H2N CHC

CH2

OH

O

NNH

H2N CHC

CH2

OH

O

CH2

CH2

CH2

NH2

H2N CHC

CH2

OH

O

CH2

CH2

NH

C

NH2

NH

H2N CHC

CH2

OH

O

CH2

C

OH

O

H2N CHC

CH2

OH

O

C

OH

O

Histidine (H) Lysine (K) Arginine (R)

Glutamic Acid (E) Aspartic Acid (D)

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.

Serotonin, which is derived from tryptophan, function as neurotransmitters and regulators.

H2N CH2

CH2

NNH

Histamine

H2N CH2

CH2

NHHO

Serotonin


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.

HOH

CHO

CH2OH

H

OH

OHC CH2OH

1

2 3

OHH

CHO

CH2OH

H

OH

HOH2C CHO

1

23

L-Glyceraldehyde(S)-Glyceraldehyde

D-Glyceraldehyde(R)-Glyceraldehyde

HNH3

COOH

CH2OH

H

COOH

HOH2C NH31

2

3

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:

Due to the double bond character, the six atoms of the peptide bond group are always planar.

CC

NC

O

H

CC

NC

O

H

CC

NC

OH

H

H2N C C

O

R

H

N C C OH

O

R'

H

HH2N C C

O

R

HNH3 C C OH

O

R'

H

OH + + 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 (psi);Rotation around the Ca and nitrogen bond is called (phi).


Rotation of Amide Planes

If (, are known for all residues, the structure for the entire backbone is known.Some (,) are more likely than others in a folded protein

Positive (,) values correspond to clockwise rotation around bonds when viewed from the CZerois defined when the C=O or N-H bond bisects the R-C-H angle.

(,)=(0,180), two carbonyl oxygens are too close;(,)=(180,0), two amide groups are overlapping;(,)=(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

b) 310 helix

Beta Sheeta) Parallelb) Anti-parallel

Beta-bulgeBeta Turn


The Alpha Helix

The alpha helix is a helical structure. All alpha helices in proteins are right-handed;H-bond patterns of the alpha helix:

a) 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;

b) 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;

c) Proline is not found in -helix except at the beginning of an -helix;d) 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;e) Examples of -helical proteins include -keratin (structural proteins) and

collagen (fibrous protein); f) LinusPauling (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•()=~(-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:

a) Parallel beta-strands (0.325 nm between two residues)b) Anti-parallel beta-strands (0.347 nm between two residues)

N

H

O

R3

N

O

R2

N

H

O

R1

N

O

R0HH

NC

N

H

O

R3

N

O

R2

N

H

O

R1

N

O

R0HH

NC

N

H

O

R3

N

O

R2

N

H

O

R1

N

O

R0HH

N

H O

N

O

N

H O

N

OHH

NC

R0 R2

R1 R3

CN


The Beta SheetsFormed 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 H-bond 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;

N

H

O

R3

N

O

R2

N

H

O

R1

N

O

R0HH

N

H O

N

O

N

H O

N

OHH

NC

R0 R2

R1 R3

CN

Beta bulge Anti-parallel strands

N

H

O

R3

N

O

R2

N

H

O

N

NO

R0H

H

N

H O

N

O

N

HO

N

OHH

N

C

R0 R2

R1R3

C

N

H

O

R1

R-1


Super secondary Structures (I)

Hairpins connect two antiparallel strands;

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 a-helix or loop is on the top of the sheet;

2

1

2

1

1

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) (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 respectto 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

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


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


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)

CHO

OHH

OHH

OHH

OHH

CH2OH

1

2

3

4

56

CHO

OH

OHH

OHH

OHH

CH2OH

1

2

3

4

56


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) c (1,6), dextran (chromatography resins)d (1,4), cellulose (cell walls of all plants)e (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 lens-shaped 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 acidsProteinsAll other moleculesOrganism


Figure 22.2 a-b (a) The Lewis Structure of Ethane (b) The Molecular Structure of Ethane


Figure 22.3 a-b The Structures of (a) Propane and (b) Butane Each Angle

Show in Red is 109.5°


Figure 22.4 a-b (a) Normal Butane and (b) The Branched Isomer of Butane


n-Pentane, Isopentanec and Neopentane


Figure 22.5 a-b (a) The Molecular Structure of Cyclopropane (b) The Overlap of the sp3 Orbitals that Form the C-C Bonds in Cyclopropane


Figure 22.6 a-b The (a) Chair and (b) Boat Forms of Cyclohexane


Figure 22.7 The Bonding in Ethylene


Figure 22.8 The Bonding in Ethane


Figure 22.9 The Two Stereoisomers of 2-Butene: (a) cis-2-Butene and (b) trans-2-Butene


Figure 22.10 The Bonding in Acetylene


Figure 22.11 a-c (a) The Structure of Benzene (b) Two of the Resonance

Structures of Benzene (c.) The Usual Representation of Benzene


Figure 22.12 Selected Substituted Benzenes and their Names


Figure 22.13 Some Common Ketones and Aldehydes


Figure 22.14 Some Carboxylic Acids


Computer-Generated Space-Filling Model of Acetylsaicylic Acid (Aspirin)


Figure 22.15 The General Formulas for Primary, Secondary, and Tertiary Amines


Figure 22.16


Figure 22.17 a-d A Major Use of HDPE is for Blow-Molded Objects such as

Bottles for Consumer Products


Figure 22.18 The 20 Alpha-Amino Acids Found in Most Proteins


A Tripeptide Containing Glycine, Cysteine, and Alanine


Figure 22.20 Hydrogen Bonding within a Protein Chain Causes it to Form a Stable Helical Structure Called the Alpha-Helix


Figure 22.21 Portion of a Protein Chain Showing the Hydrogen-Bonding Interactions


Figure 22.22 When Hydrogen-Bonding Occurs Between Protein Chains, a Stable Structure Called a Pleated Sheet Results


Figure 22.23 a-b (a) Protein Chains Form Superhelix (b) Pleated-Sheet Proteins


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


Figure 22.25 The Permanent Waving of Hair


Figure 22.26 The Thermal Denaturation of a Protein


Figure 22.27 A Tetrahedral Carbon Atom with Four Different Substituents cannot have its Mirror Image Superimposed


Figure 22.28 The Mirror Image Optical Isomers of Glyceraldehyde


Figure 22.29 The Cyclization of D-Fructose


Figure 22.30 The Cyclization of Glucose


Figure 22.31 Sucrose is a Disaccharide Formed from Alpha-D-glucose and Fructose


Figure 22.32 a-b (a) The Polymer Amylose is a Major Component of Starch and is Made Up of Alpha-D-

Glucose Monomers (b) The Polymer Cellulose, which Consists of Beta-D-Glucose Monomers


Figure 22.33 a-b The Structure of the Pentoses (a) Deoxyribose and (b) Ribose


Figure 22.34 The Organic Bases Found in DNA and RNA


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


A Computer Image of the Base Pairs of DNA


Figure 22.36 A Portion of a Typical Nucleic Acid Chain


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


Figure 22.38 DNA Cell Division


Figure 22.39 The Anticodon of the tRNA Must Complement the Codon of

the mRNA


A Butane Lighter Used for Camping


A Worker Using an Oxyacetylene Torch


A Winemaker Drawing Off a Glass of Wine in a Modern Wine Cellar


Ethanol is Being Tested in Selected Areas as a Fuel for Automobiles


Cinnemaldehyde Produces the Characteristic Odor of Cinnamon


Aspirin Tablets


The Soybeans on the Left are Coated with a Red Acrylic Polymer to Delay

Soybean Emergence


A Radio from the 1930s Made of Bakelite


Nylon Netting Magnified 62 Times


A Scanning Electron Microscope Image Showing the Fractured Plane of a Self-Healing Material with a Ruptured Microcapsule in a Thermosetting Matrix


Cross-Linking Gives the Rubber in these Tires Strength and Toughness


Figure 22.16 The Reaction to Form Nylon Occurs Readily


Wallace H. Carothers


PVC Pipe is Widely Used in Industry


The Protein in Muscles Enables Them to Contract


Self-Tanning Products and a Close-Up of a Label Showing the Contents


Photo 22.17 A Bowl of Sugar Cubes


Table 22.1 Selected Properties of the First Ten Normal Alkanes


Table 22.2 The Most Common Alkyl Substituents and Their Names


Table 22.3 More Complex Aromatic Systems


Table 22.4 The Common Functional Groups


Table 22.5 Some Common Alcohols


Table 22.6 Some Common Amines


Table 22.7 Some Common Synthetic Polymers, Their Monomers and Applications


Table 22.8 Some Important Monosaccharides

Chapter Twenty-TwoOrganic 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) 4-Isopropyl-3-methylpentane b) 2-Isopropyl-3-methylpentane c) 1,1,2,3-Tetramethylpentane d) 2,3,4-Trimethylhexane e) 3,4-Dimethylheptane


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.


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


Answer

c) 2-Bromo-1-chloro-4,4-diethyloctane


Choice (a) should be 2,3-dichloropentane. Choice (b) should be pentane.


Question

Which of the following has the lowest boiling point?

a) Butane b) Ethanec) Propaned) Methane


Answer

d) Methane


The smallest of the saturated hydrocarbons will have the lowest boiling point.


Question

How many isomers are there with the formula C2H2Br2? Include both structural and geometric isomers.

a) 2 b) 3c) 4d) 5e) 6


Answer

b) 3

Section 22.2, Alkenes and Alkynes

The formula given could be any of these:cis-1,2-Dibromoethene trans-1,2-Dibromoethene1,1-Dibromoethene


Question

Which of the following is an incorrect name?a) trans-1,2-Dichloroethene b) Propylene c) Ethylene d) cis-1,2-Dichloroethane


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.


Question

How many different “tetramethyl-benzenes” are possible?

a) 2 b) 3c) 4d) 5e) 6


Answer

b) 3

Section 22.3, Aromatic Hydrocarbons

Here are the three possibilities:1,2,3,4-Tetramethylbenzene1,2,3,5-Tetramethylbenzene1,2,4,5-Tetramethylbenzene


Question

H2CCHCH2N(CH3)2 isa) an alkyne and a secondary amine. b) an alkene and a primary amine. c) an alkene and a tertiary amine. d) an alkyne and a tertiary amine.


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.


Question

For which of the following compounds are cis and trans isomers possible?

a) 2,3-Dimethyl-2-butene b) 3-Methyl-2-pentenec) 4,4-Dimethylcyclohexanol d) ortho-Chlorotoluene


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.


Question

Which of the following types of compounds may lack an sp2-hybridized carbon center?

a) Aldehydes b) Ketones c) Alcohols d) Alkenes


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 sp3-hybridized carbon centers.


Question

For which of the following choices do all of the functional groups listed have a C=O bond?

a) Ester, aldehyde, secondary alcohol, ketone b) Alcohol (any), ether, ester c) Secondary alcohol, ketone, aldehyde d) Ester, aldehyde, ketonee) Carboxylic acid, ether, tertiary alcohol


Answer

d) Ester, aldehyde, ketone


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.


Question

Oxidation of secondary alcohols results ina) ketones. b) tertiary alcohols. c) aldehydes. d) esters. e) ethers.


Answer

a) ketones.


Ketones may be prepared from the oxidation of secondary alcohols.


Question

What might be the product of the oxidation of 2-methyl-1-butanol?

a) 2-Methyl-2-butanone b) 2-Methylbutanal c) 2-methylbutanoic acid d) Both b and c e) Both a and c


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 2-methylbutanoic acid.


Question

Which of the following is optically active (i.e., chiral)?

a) HN(CH3)2

b) CH2Cl2 c) 2-Chloropropane d) 2-Chlorobutane e) 3-Chloropentane


Answer

d) 2-Chlorobutane


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.


Question

Oxidation of primary alcohols results ina) ketones. b) tertiary alcohols. c) aldehydes. d) esters. e) ethers.


Answer

c) aldehydes.


An aldehyde results from the oxidation of a primary alcohol.


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.


Answer

b) the hydrogen bonding in methanol and the lack of hydrogen bonding in ethane.


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.


Question

No atoms are lost from the starting material in making which kind of polymer?

a) Condensation polymer b) Polyester polymer c) Addition polymer


Answer

c) Addition polymer

Section 22.5, Polymers

When monomers add to form polymers, no atoms are lost.


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]nWhat is the structure of chief monomer of this wrap?

a) CCl2 CH2

b) Cl2C–CH2

c) Cl2C CH2 CCl2

d) CCl2


Answer

a) CCl2CH2


For the polymer given, the monomeric unit is CCl2=CH2, which adds to form the polymer.


Question

Which of the following steps will increase the rigidity of a polymer?

a) Use shorter polymer chains b) Make chains more branched c) Decrease cross-linking d) Introduce the possibility of hydrogen

bonding between chains


Answer

d) Introduce the possibility of hydrogen bonding between chains


Increasing hydrogen bonding in a polymer can give it greater strength and rigidity.

Documents

Chapter Twenty Two Organic and Biological Molecules 講義