Introduction
The
presentation of many concepts in organic and biochemistry is greatly
facilitated by the use of graphic materials.
In some cases, visuals clarify difficult ideas. In other cases, reaction mechanisms are best
understood through illustrations. In a
few situations, graphics are almost essential.
It is difficult to imagine a really effective explanation of topics such
as geometric isomerism without some kind of visual.
Many
texts are now rich with visuals of all kinds.
But instructors are often still at a disadvantage in their lecture
presentations. The construction of some
graphics is complex and time consuming.
Sometimes accuracy of completeness must be sacrificed just to produce a
usable blackboard sketch in the time available.
This
set of visual masters is designed to ameliorate that problem. We have assembled seventy-one visual masters
on some of the most common concepts in organic and biological chemistry. The masters can be used directly in an
opaque projector, or they can be
duplicated as overhead transparencies by the Thermofax process. It is also possible to make spirit masters
from these originals so that every student in class will have his or her own
copy of any desired graphic. Permission
is hereby granted for the use of these masters in any of the above ways.
The
commentary following this introduction is intended to give some general
information on each of the concepts illustrated. For more detailed information, the instructor should refer to the
book that this set accompanies, The Chemistry of Carbon Compounds: Introducing Organic and Biochemistry, or
to any standard text in the field.
We
strongly encourage instructors to think about the possibility of using
three-dimensional model kits in conjunction with these visuals. Just as two-dimensional sketches and
drawings are helpful in understanding some ideas, three-dimensional models are
even more useful, and in some cases, indispensable. A number of kits are now available commercially and, although
very expensive, are worth having in at least small quantities.
Sets
of questions are provided at the ends of certain groups of visuals. They provide a simple method of reviewing
basic information contained in the visuals in each groups.
Dr.
David E. Newton
Professor
of Chemistry
Salem
State College
Salem,
MA 01970
Background for Teachers
A note
after the paragraph heading refers to the location of a transparency
master. For example after #1. below the
notation PIM IV, p. 21 refers the reader to that PIM and page for a
transparency master.
1.
Differences Between
Organic and Inorganic Compounds - PIM III, page 21.
Most
students come to organic chemistry with some background in inorganic. A useful point of departure, therefore, may
be to illustrate the ways in which organic and inorganic chemistry differ from
each other. The instructor should be
able to provide concrete examples from both fields to illustrate the
generalizations contained in the table.
It is essential to point out that the distinctions made here are not
absolute. There are exceptions to every
generalization made. But as an
introduction to some of the ways in which lectures and laboratories will
proceed in organic (as opposed to inorganic) chemistry, these may be a useful beginning
to the course.
2.
Organic Families and
Functional Groups - PIM III, page 22.
One of the
first points about organic chemistry that is likely to capture students'
attention is the sheer magnitude of the subject. One is staggered by the challenge of learning organic chemistry,
with over three million compounds to study.
This provides a logical rationale for the notion of organic families and
functional groups. A functional group
can be defined as a distinctive arrangement of atoms by which one can recognize
a set of similar organic compounds. The
members of those sets constitute the organic families whose names are given in
Visual Master 2. The concept of organic
families and functional groups also gives the instructor an opportunity to
review the familiar connection between structure and function, since the
structure of organic families (their functional groups) is directly related to
their functions (their behaviors).
3. and
4. Tetrahedral Carbon Atom: 1,11 - PIM III, pages 23,24.
The first
theoretical point with which most instructors deal is the tetrahedral character
of the carbon atom. The four valence
electrons of carbon (2s2 2pxl 2pyl
) are hybridized to yield and four identical sp3 orbitals. In the formation of compounds, these are employed
to produce four equivalent bonds that are oriented in space as far from each
other as possible, i.e., directed towards the vertices of a tetrahedron. The task of distributing four mutually
repulsive regions of like charges is really a problem in solid geometry. It may be treated as such with students who
have a suitable background in mathematics.
The methane molecule is treated here as the simplest possible example of
the tetrahedral shape of carbon bonding.
The propane molecule illustrates a more complex example of the
tetrahedral model.
5. Isomers of Pentane - PIM IV, page 3.
The
extraordinary number of organic compounds can also serve as an entrée to
another fundamental feature of organic chemistry: the existence of
isomers. The ability of even a
relatively small number of atoms to arrange themselves in a variety of ways
(isomers) is one factor that accounts for the amazing number of organic
compounds. This master can be used profitably with a model kit that will show
the same isomers three-dimensionally. The point that needs to be made is that
isomers do not result from the twisting and bending of a chain. They can only
be produced by tearing bonds apart and physically moving atoms from one place
to another on the chain.
6. Polar and Nonpolar Bonds - PIM IV, page 2.
Covalent
bonds are formed when two atoms share a pair of electrons between them. The
bond exists because the nuclei of both atoms exert a pull on the shared
electrons. The relative ability of an atom to pull on the electrons in a
covalent bond is defined as the electronegativity of that atom. It is fairly
rare that both atoms pull equally strongly on a shared pair of electrons. The electrons are, therefore, likely to be
some-what closer to one atom than to the other. In most covalent bonds, then, one end of the bond is slightly
more negative ( 15 -) , than the other ( 15 +). Such a bond is said to be
polar. In fact, about the only truly nonpolar covalent bonds are those between
two identical atoms, as in a diatomic molecule. All bonds between carbon and
hydrogen, oxygen, sulfur, nitrogen, or phosphorus, or between oxygen and
hydrogen, and so on, are polar bonds.
7.
Polar and Nonpolar Molecules - PIM IV,
page 4.
A molecule
made entirely of polar bonds mayor may not be polar overall. The methane molecule is a good example. All four carbon-hydrogen bonds are polar
because of carbon's greater electronegativity.
The molecule as a whole is nonpolar, however, since there is no
separation of charge on the exterior of the molecule. A “trip around the outside of the molecule” would provide an
observer with an identical view from any point. (This is not a very elegant test for polarity, but it is one that
has worked with students before.) This
master shows two common molecules, one of which is polar, one nonpolar. Carbon dioxide, at the top, contains two
polar bonds, but the molecule overall is nonpolar. Its linear arrangement means that opposite ends of the molecule
are identical, and, as a result, carbon dioxide has no dipole moment. On the other hand, water is polar. Although it contains the same number of
atoms as carbon dioxide, the hydrogen-oxygen bonds are not arranged linearly
(at 180°), but at an angle of about 109°.
This makes the oxygen portion of the molecule visibly more negative than
the hydrogen ends. The symbols + and –
indicate the visibly positive charge in this case.
8. The
Periodic Table - PIM IV, page 5.
The
periodic table is provided here as a general reference for use at any point in
the course.
9. and
10. Projection of Ball-and-Stick Model, Space-Filling Model - PIM III, pages 23
and 24.
A variety of methods have been developed to represent the
shape of organic molecules. Almost
certainly the best of these are the "Tinker-Toy" disk or
ball-and-stick kits that are now commercially available. Visual Masters 9 and
10 describe the three most familiar two-dimensional methods of representation. In the ball-and-stick method, various atoms
are represented by spheres of different color and/or size. The covalent bonds between atoms are
represented by short sticks. The use of
the ball-and-stick method shown here on an extensive scale would obviously be a
time-consuming inconvenience for textbook writers and instructors. The representations we normally see for
organic compounds (structural formulas) are actually two-dimensional
projections of three-dimensional ball-and-stick models. They are clearly much simpler to draw and use
than any form of the ball-and-stick model.
Given the
very great significance of molecular geometry, however, it is worth reminding
students of the differences between structural formulas and the reality of molecular
architecture. The soon-to-be encountered case of optical isomers may be cited
as a case in which two-dimensional representations fail completely. Space-filling models attempt to represent
more faithfully the three-dimensional space actually occupied by electrons in
various atoms. The virtue of accuracy provided by space-filling models is
somewhat moderated by their relative lack of clarity for some students.
QUESTION
SET No.1
·
1. List six important ways in
which organic compounds and organic reactions tend to differ from their
inorganic counterparts.
·
2. What is meant by the term
tetrahedral carbon atom? How can you explain the tetrahedral shape of the
carbon bonding?
·
3. Define an isomer. How do
we know if two compounds are isomers of each other?
·
4. What determines whether a
bond will be polar or nonpolar? Whether a molecule will be polar or nonpolar?
·
5. Describe the three common
methods for representing an organic molecule. What advantages and disadvantages
do you see for each?
11.
Naming Alkanes (IUPAC System) - PIM V, page 17.
The IUPAC
system of nomenclature is a brilliantly simple way of naming any organic
compound. This master outlines the
basic steps in that system. Begin by
finding the longest continuous chain of carbon atoms. That provides the stem name for the compound. In this case, the longest chain contains
five carbon atoms, so the stem name is pentane. Each substituted group is indicated by a
characteristic prefix. The -CH3
group is called methyl, and a single chlorine, chloro. The presence of two chlorines is evidenced
by the use of the prefix di. Finally,
the position of each substituent is indicated by a number. That makes the correct name for this alkane 1,5-dichloro-2-methylpentane.
12.
Nomenclature Exercise - PIM V, page 18.
This
nomenclature exercise is provided as a class review of instruction on naming
provided by the text and/or in lecture.
Problems using both expanded and condensed structural formulas are
given. The answers for these problems
are:
1.
3-iodo-2,4-dimethylhexane
2.
1-bromo-5-chloro-3,4-dimethylheptane
3. 3-methyl-3-ethylpentane
4. 3-bromo-2,5-dimethylhexane
5. 2-methyl-3,3-diethylpentane
6. 2-chloro-l,3-dimethylcyclohexane
13. A
Typical Substitution Reaction
One of the
few chemical reactions in which alkanes take part is substitution. The example with chlorine is used on this
master. The reaction goes only very slowly in the dark and at room temperature. Heat or ultraviolet light (uv) initiates the
reaction by inducing the dissociation of a chlorine molecule into two chlorine
atoms. One of these bonds to and then
removes a hydrogen atom from the alkane molecule. Hydrogen chloride is thus
produced as one of the reaction products.
The highly reactive alkyl radical that remains then attacks a second
chlorine molecule, forming the second reaction product, an alkyl halide, and a
free chlorine atom. The chlorine atom
is then free to begin a second turn of this cycle. Certain characteristics of the substitution reaction that
distinguish it from addition reactions should be noted :
a. A catalyst is
required to initiate the reaction;
b. Only one chlorine
atom in the original molecule substitutes in the alkane in the reaction.
c. Hydrogen halide is always formed as one of the products.
14.
Drilling for Petroleum and Natural Gas
Coal,
petroleum, and natural gas are the decomposition products of organisms that
lived long ago. These organisms died
and then were buried in swamps, marshes, and other watery environments before
they could decay in the usual fashion.
The hydrocarbons of which these fossil fuels consist are, therefore,
decomposition products formed under rather special circumstances. In most cases, petroleum and natural gas are
found in geological formations that trap these fossil fuels between Impermeable
barriers (rock and/or water) above and below.
Collecting the petroleum and natural gas requires tapping into the
saturated layers of sandstone or into actual pools of oil and gas.
15. Refining
Petroleum
Petroleum
as it comes from the ground is virtually useless. As pitch, it has had some historical value. But its significance to modern societies was
recognized only when methods were devised for separating the raw material into
useful fractions. The process of
fractional distillation involves heating crude oil to a high temperature at the
bottom of a tall still. All but the
highest-boiling-point components vaporize and rise in the tower. As the vapors rise, they are cooled and
eventually condense. At a half-dozen or
so points in the tower, platforms collect all compounds whose boiling point is
lower than some temperature. The
fractions obtained at each level have characteristic names: kerosene, gasoline,
light oils, and so on. Each fraction consists of a complex mixture of saturated
and unsaturated, aliphatic and aromic hydrocarbons. Further separation of anyone fraction is possible. Gaseous hydrocarbons escape from the top of
the tower and a semisolid fraction (asphalt, pitch, tar) is drawn from the
bottom.
16.
Products of the Refining of Crude Oil
The table
shown here represents one way of classifying the fractions obtained in the
refining of petroleum. The names and temperature ranges vary in different
classification systems, but the criteria tend to be the same: boiling-point
range and size of hydrocarbon molecule. A list of specific materials obtained
by the fractional distillation of petroleum would contain thousands of names.
17. and
18. Rotational Forms of 1,2-Dichloroethane , Conformational Isomers of Ethane
The single
covalent bond that joins two carbon atoms (a sigma bond) does not constrain the
relative positions of those two atoms in any way. They are free to assume any configuration with respect to each
other. There is said to be “free
rotation” around the single bond. In fact, since molecules are constantly in
motion (translational, rotational, and vibrational), one would expect the two
to progress through an unlimited number of possible spatial conformities. Two clearly identifiable arrangements (shown
in Visual Master 18) are those in which (1) substituents are exactly lined up
with each other and (2) substituents of one carbon are aligned in the gaps
between those of the other. These are
known as (1) eclipsed and (2) skewed forms of the compound. Conformational isomers are a subject of
rather recent interest and significance in organic chemistry, but tend to be of
relatively little interest for beginning students in an introductory organic
course.
19.
Geometric Isomerism
This
master introduces a second type of isomerism students will encounter regularly:
geometric isomerism. Disubstituted
cyclic compounds all have the potential for existing in two forms: one in which
both substituents appear on the same side of (above or below) the ring; the
other in which they occur on opposite sides of the ring. The former form is known as the cis form,
and the latter as the trans form. Three-dimensional models should be used to
demonstrate that one form is not convertible into the other by any method other
than destruction of the molecule. Some
alkenes also occur as geometric isomers.
No comparable master exists for this, so the instructor may choose to
come back to this master to review the concept when the alkenes are studied. In
the case of alkenes, three-dimensional models are even more important in
illustrating the role of the pi bond in preventing free rotation around the
double bond.
20.
Conformational Isomers
The
tetrahedral character of the carbon atom means that some ring compounds may
exist in more than one conformation.
Again, the use of three-dimensional models to illustrate this point is
highly recommended. The two forms shown
on this master are commonly referred to as the "chair" and
"boat'. conformations. The former
is recognizable by the fact that adjacent hydrogens are staggered with respect
to each other , while those in the boat form are eclipsed with respect to each
other. Conformational isomers of cyclic
hydrocarbons are generally not emphasized in introductory courses for general
students.
21. A
Typical Addition Reaction - PIM IV,
page 6.
The
movement of electrons and atoms during an addition reaction is illustrated in
this master. The pi bond in an alkene
is a region of negative electrical charge that will attract positively charged
species like the hydrogen ion (a proton).
Addition of the proton to one of the carbons in the carbon-carbon double
bond produces a localized positive charge on the other carbon. Anything with a negative charge-the chlorine
end of an HCI molecule in this case-is attracted to this positive region. Fissure of the hydrogen chlorine bond
results in the major organic product, an alkyl halide, and a free proton with which
to initiate a repetition of the cycle.
Three important characteristics that distinguish this reaction from a
substitution reaction are:
a. All of
the inorganic molecule is added to the organic molecule in a single cycle of
the reaction;
b. Energy
is not required to initiate the reaction ;
c. Only a
single product is formed in the reaction .
22.
Markownikoff's Rule - PIM IV, page 7.
If either
reactant in an addition reaction is symmetrical, only one addition product is
possible. If both are asymmetrical, two
products are possible. The addition of
hydrogen halide to propylene illustrates that point in this master. In 1872 the Russian chemist Vladimir
Markownikoff found that when two products are formed, one is preferred over the
other. His empirical observation was
that the more positive part of the addend (usually a proton) adds to the carbon
in the carbon-carbon double bond which is more negative. This is generally the one with the greater
number of hydrogens already attached to it.
The rule is sometimes referred to as the "them that has, “gits”
rule, since the carbon with the greater number of hydrogens to begin with
.”gits” the hydrogen from the addend.
23.
Natural Rubber - PIM IV, page 8.
Natural
rubber is a polymer of 2-methyl-1,3-butadiene (isoprene). In its natural form, rubber has relatively
few uses. The major disadvantage is its
tendency to become soft and gooey when warmed.
In 1839 Charles Goodyear accidentally discovered a method for overcoming
this handicap. The addition of sulfur
to a batch of molten rubber results in a product that is tough, pliable, and
thermosetting. “Vulcanized”. rubber
maintains its rigidity when warmed. The
process is explained chemically as an addition reaction in which element sulfur
adds to the carbon-carbon double bond on adjacent polyisoprene molecules. These intermolecular bonds act as “pegs”
that prevent rubber molecules from sliding over each other when they are
heated. The process has its natural counterpart in the aging of rubber . Oxygen-chemically similar to sulfur-also
adds intermolecularly in what might be described as “natural
vulcanization.”. As more and more
double bonds are broken and oxygen atoms added, the rubber becomes less and
less pliable. It eventually “ages”
(oxidizes) to a point at which it is brittle and useless for most purposes.
24.
Terpenes
Terpenes
are natural products formed by the condensation of two or more isoprene
units. This master shows the structural
formula for isoprene, a common abbreviation for this structure, and some
typical terpenes. Those which contain two isoprene un its are known as
monoterpenes ; those with three isoprene units, sesquiterpenes; and those with
four isoprene units, diterpenes. The
terpenes illustrated here are only a small sample of a large class of naturally
occurring substances.
25.
Benzene - PIM V, page 20.
Determining
the structure of benzene was one of the great chemical challenges of all
time. Over a century passed between the
discovery of this compound and an acceptable explanation of its chemical
structure. The molecular formula and
chemical properties for the compound were in dramatic contrast to all
predictions made that were based on existing chemical theory. This master allows discussion of some
important elements of this story. At one time, it was thought that the
structure of benzene was a compromise between two related forms, differing from
each other in the location of carbon-carbon double bonds in the ring. A more satisfactory explanation made use of
the concept of resonance, the idea that there were neither double nor single
bonds between specific carbons, but that all bonds were of a form intermediary
between these. Chemists have explored
various ways of representing this phenomenon, the hexagon with alternating
double and single bonds having been one of the most popular. Today there seems to be increasing use of
the last of these representations, a hexagon with a circle drawn
internally. This suggests a cyclic
hydrocarbon with six carbon-carbon sigma bonds and an additional six electrons
shared equally by all carbons in the ring.
26.
Substituted Benzene Derivatives
Given the
molecular formula for benzene, C6 H6, one would predict
addition as the typical chemical property for the compound. It is not.
Benzene substitutes in a typically “alkane” method. In fact, a very wide range of useful
substitution products can be obtained from benzene. Some common examples are given on this master. It can be used to show the IUPAC system for
naming monosubstituted products and to illustrate the common names associated
with some of these.
27.
Directing Influence of Substituted Groups
Benzene
can be multisubstituted, like the alkanes, up to the number of hydrogens which
can be replaced (six in this case). The
location of a second substituent does not occur at random, however. It is determined by the substituent already
present in the molecule. The methyl
group is a familiar ortho and para directing substituent. The amino group, hydroxyl group, ethyl
group, and halogens are also ortho and para directors. The nitro group is a meta directing group, along
with cyanide, carboxyl, and carbonyl groups.
28.
Some Halogenated Hydrocarbons
The
halogenated hydrocarbons (alkyl halides) are an exceedingly important group of
organic compounds. They are virtually
unknown in nature. The ones that
students encounter are likely to have been synthesized by chemists for very
specific purposes. Their effectiveness
as pesticides and herbicides is perhaps their most dramatic claim to fame in
modern society. In many cases, however,
their introduction as the solution to one set of problems has resulted in the
evolution of a whole new set of social and environmental issues. DDT and Freon propellants are classic cases
of the mixed blessings that result from modern chemical technology.
29.
Structure of the Benzene Molecule - PIM
V, page 19.
This
master gives an idea of the electron distribution in the benzene molecule. The
doughnut-shaped cloud represents the space occupied by the six electrons shared
among the six carbon atoms. The carbon atoms are, of course, imbedded within
this cloud with the hydrogen atoms projecting out of it.
QUESTION
SET No. 2
1. State the IUPAC rules by which an alkane is named. How are
those rules different for naming an alkene?
2. Write a chemical equation that shows how a substitution
reaction occurs. Do the same for an
addition reaction. List three important
ways in which these reactions differ from each other .
3. How are petroleum and natural gas formed? Under what conditions are they found in the
earth? Describe the process by which petroleum is refined.
4. Invent an example that illustrates that you understand
Markownikoff's rule.
5. Write a pair of chemical equations that show what happens
in a rubber molecule during the process of (1) vulcanization and (2) aging.
6. List six halogenated hydrocarbons and give an important use
of each.
30.
Hydrogen Bonding
Water
molecules are polar. The hydrogen
portions of the molecule tend to be somewhat more positive than the oxygen
portion. This master shows how polar
water molecules form weak bonds with each other. These hydrogen bonds are exceedingly important in determining a
number of special properties of water.
Alcohols can be thought of as derivatives of water in which one hydrogen
has been replaced by an a alkyl group. As such, they are also polar molecules, but in a way somewhat
different from water .
The
carbon-oxygen bond in alcohols is much less polar than the hydrogen-oxygen bond
in water . The hydrogen bonds that form
in alcohols are, therefore, weaker than those in water because only one end of
the alcohol molecule is capable of forming such a bond. Hydrogen bonding in alcohols is,
nonetheless, of great significance in biological systems. It provides a means of bonding species
together in a bond which, although weak, is significant. Students should recognize that hydrogen
bonding will be a major consideration in later discussions of biochemical
reactions.
31.
Primary, Secondary, and Tertiary Alcohols
- PIM V, page 21.
The carbon
to which the hydroxyl group in an alcohol is attached is known as the carbinol
carbon. One method of classifying alcohols is by specifying the number of
carbons attached to the carbinol carbon. The examples given here illustrate a
primary alcohol, in which only one carbon is attached to the carbinol carbon; a
secondary alcohol, in which there are two carbons; and a tertiary alcohol, in
which there are three. The three
alcohols differ from each other not only in their structures, but also in some
chemical behaviors.
32.
Lucas Test
In the
reaction between an alcohol and a hydrogen halide (with a catalyst of zinc
chloride), a halogen from the latter replaces the hydroxyl group of the
former. The alkyl halide produced in
this reaction is insoluble in this reaction mixture. The rates at which the reaction (ROH + HX) occurs differ for
primary, secondary, and tertiary alcohols.
It takes place instantaneously with tertiary alcohols, within thirty
minutes with a secondary alcohol (or in about five minutes with moderate heat),
and only after an hour or more with primary alcohols. Performed in this way, the reaction is known as the Lucas test
and is an effective and simple way of distinguishing among the three classes of
alcohols.
33.
Addition on the Carbonyl Group
Addition
on the carbon-oxygen double bond is roughly comparable to the corresponding
reaction on the carbon-carbon double bond in alkenes. A variety of substances
undergo the reaction, given the proper conditions. Hydrogen, water, alcohols,
hydrogen halides, and sodium bisulfite are typical addends. In such additions,
the more positive part of the addend (usually hydrogen) adds to the oxygen of
the carboxyl group, while the remainder adds to the carbon.
34.
Acetal Formation
By far the
most important application of this reaction in biochemistry is the formation of
hemiacetals and acetals, hemiketals and ketals. The reaction between an aldehyde and an alcohol is illustrated
here. In the first step, the hydroxyl
hydrogen adds to the carboxyl oxygen, while the remainder of the alcohol adds
to the carboxyl carbon. The product-a
hemiacetaJ-is nearly always unstable, with reversion to the reactants most
likely. The single most common
exception, involving monosaccharides, is an important one. The stable intramolecular hemiacetal formed
in a monosaccharide is the form in which the members of this family are usually
found in water solutions. Students
should be alerted to this important exception as one that surfaces later in the
course (Visual Master 43) .
A second
reaction occurs with excess alcohol.
This time water is eliminated between the hemi-acetal and the second
alcohol molecule. The resulting acetal
is usually stable. Comparable reactions
between ketones and alcohols yield hemiketals and ketals.
35.
Ester Formation
Esterification
is an important typical reaction of carboxylic acids. A proton from an alcohol and a hydroxyl group from the acid are
removed in the production of the ester.
Students should be made aware of the equilibrium nature of this
reaction, and of the role of the hydrogen ion as catalyst in either
direction. Saponification and the
influence of the hydroxide ion on the reverse reaction may also be introduced
at this point. Mention can also be made
of the numerous applications of the esterification reaction, ranging from the
production of artificial flavors and odors to the biochemical synthesis of
lipids.
36.
Production of Nylon - PIM V, page 22.
A reaction
of considerable commercial interest is illustrated in this master. Nylon was one of the first and most
important synthetic fibers invented by chemists. The term nylon refers to a variety of polyamides, all
formed in the reaction between a dicarboxylic acid and a diamine. The distinguishing numbers that follow the
word nylon tell the number of carbon atoms contained in each of the
reactants. The polymer is rather easily
made into fibers that are exceptionally strong and abrasion resistant.
37.
Some Important Heterocyclic Compounds
This
master contains formulas for six heterocyclic compounds to which reference may
be made at various times during a course.
Besides introducing the concept of heterocycles, the master can be used
to illustrate the use of monosaccharide terminology (furan/furanose; pyran/pyranose)
and to discuss nitrogen bases (pyrimidine and purine) and the derivatives that
occur in nucleic acids and certain energy-transfer compounds (uracil and
adenine).
QUESTION
SET No.3
·
1. Make a sketch that shows
how and why hydrogen bonding in alcohols is different from hydrogen bonding in
water. Would you expect ethers to show
hydrogen bonding? Why or why not?
·
2. Describe the Lucas test.
Explain how it is performed. what observations are possible, and what
conclusions can be drawn from each observation.
·
3. Write an equation that
shows how a hemiacetal is formed. Write
another equation showing how an acetal is formed. Write comparable equations for the reactions of a ketone with
excess alcohol.
&n