Biochemistry
PRMSU, 1st Semester ST 2020-2021
College of Education
Module 1: Review of Organic Chemistry
and Organic Chemical Reactions
This module covers topics of Organic Chemistry that is deemed necessary in understanding the
behavior and chemical reactions that is happening inside a living system/organism. Specifically,
it covers the fpllowing topics:
A. Overview of the Nature, Classification and Structure of Organic Compounds
B. Experimental Determination of Organic Compounds
C. Reaction Mechanisms
Types of Organic Reactions
Oxidation Reactions of C=C
Electrophilic Addition Reactions
Electrophilic Aromatics Substitution Reactions
Reactions of Alcohols
Reactions of Aldehydes and Ketones
Interconversion of Carboxylic acids and their Derivatives
At the end of this module, the students are expected to:
Understand why Carbon is fundamental to biochemistry;
Explain the experimental procedures that is performed to determine the
composition, formula and structures of oragnic and biomolecular substances;
Find out the characteristics and reactivity of different functional groups;
Calculate the percentage composition;
Determine the empirical and molecular formulas from experimental data;
Derive possible structural formula from a given molecular formula;
Be able to recognize what distinguishes each functional group from the
others;
Be able to recognize the functional group within the context of a larger
molecule;
Be able to name the functional group; and
Classify organic reactions as addition, elimination, substitution,
INTRODUCTION:
rearrangement, oxidation, or reduction;
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Biochemistry
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College of Education
The science of organic chemistry deals with the relationship between molecular structure
and properties. Since biochemistry deals with organic substances that are seen in living
organisms, it is therefore important for students understand the nature and reactions of
substances which play an important role in the metabolic processes seen in a living cell. To begin
your study of biomolecules, it is fitting to begin your study on the structural features of organic
compounds.
In this module, you will also take a brief look at the various classes of organic
compounds that are to be studied in this course. Youy will also be taught of the processes and
procedures to deternine the elemental composition, molecular and structural formula
determination, as well as important chemical reactions that are basis in understanding various
metabolic pathways.
DISCUSSION AND EXAMPLES:
Kekuld (Gerrnany) and Couper (Scotland) were the first to recoguize that carbon is
tctrnvnlcrrt - it cau fortn four bonds, not only with other atoms (nrainly I{, O, N) but also with
other carbon atoms, The most distinctive feature of carbon atoms is thcir capacity to form very
strong covalent bonds with each other. No othcr element, includir-rg silicou, can clo this.
Siliconsilicon boncls are easily cleaved. Silicon-oxygen-silicou boncls are quite strorig but do
not result in thc incrccliblc clivcrsity for-rnd in carbon compoutrds.
Carbon atoms can link with each other to form linear, brauched or cyclic backbones for a
wide variety of organic molecules. Take for exatnple only four carborr atonrs and see the number
of ways by which they can be linked to one another
You can also imagine how many more diverse molecular framework can result if there
are more carbon atoms that are linked to one another. Moreover, carbon atoms may participate in
single, double or triple bonds.
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Biochemistry
PRMSU, 1st Semester ST 2020-2021
College of Education
The ability of carbons to form single as well as multiple bonds, not only with another
carbon atom, but also with atoms like nitrogen, sulfur, phosphorous and oxygen, results in
greater diversity of molecular frameworks whuich constitute the skeletons of small and simple
molecules or big and complex biochemical found in living organisms.
Organic molecules exhibit isomerism, a term derived from the Latin iso (equal) and mer
(part). There are two general types of isomerism: structural isomerism and stereoisonrerism.
Structural isomers have the same molecular fonnula br.rt different structural fonnulas, i.e.. the
same atons can be put together in dilferent ways resulting in different cornpounds with entirely
differerrt properties. For examplc, two structural fbrmulas can be derived from the molecular
formula C2H6O, and each one corresponds to a different compound.
As the number of carbon atoms increases, the number of possible structures
corresponding to a molecular formula also increases..If you take a look into Chemical Abstracts,
where known organic compouncls are catalogued, you will see that there are dozens and
sometimes hundieds of different cJmpounds that have the same molecular formula.
The four bonds to carbon are directed to the four corners of a regular tetrahedron. As a
direct consequence of this, many different three-dimensional structures can arise in organic
molecules. This brings us to stereoisomerism (derived from the Greek stereos - solid) - a more
subtle kind of isomerism which is particularly important in compounds found in living
organisms. Stereoisomers have the same molecular and structural formulas but diifer from each
other in their three-dimensional structures.
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Biochemistry
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College of Education
EXERCISES:
A. Give four reasons for the occurrence of a large number of organic compounds and great
diversity in their structures.
1.
______________________________________________________________________________
___________________________________________________________________________.
2.
______________________________________________________________________________
____________________________________________________________________________
3.
______________________________________________________________________________
_____________________________________________________________________________
4.
______________________________________________________________________________
___________________________________________________________________________
B. From the structures given below, choose a pair that would illustrate each of the following (use
the letter labels):
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_____ and _____ 1. ways by which three carbon atoms can be link to one another by single
bonds
_____ and _____ 2. a pair of structural isomers
_____ and _____ 3. a pair of stereoisomers
_____ and _____ 4. molecular frameworks that result from the introduction of a C = C into a
linear chain of four carbon atoms
Qualitative organic analysis, the identification of organic compounds based on their
physical and chemical properties, is analogous in some ways to the identification of plants and
animals according to their taxonomy—their structural features and presumed natural
relationships. To classify an organic compound into a given family requires first detecting a
specific functional group (characteristic set of atoms) in the molecules of organic compounds.
Because functional groups influence the physical, chemical, and spectral properties of an
organic compound, a chemist can identify a compound's functional groups by measuring certain
physical properties, observing its chemical behavior with different classification reagents, and
studying other spectral data.
In this experiment we will subject a series of organic compounds to specific chemical
reactions in order to identify which class of functional group the substance belongs to.
Some common organic functional groups
Functional group name General formula*
R C C R'
Alkene
Alkyl halide R Cl or R Br
Alcohol R O H
O
Aldehyde R C H (R–CHO)
O
Amide R C N R'
H
R N H R N H R N R''
Amine
H R' R'
O
Carboxylic acid
R C O H
O
Ester
R C O R'
Ether R O R'
O
Ketone
R C R'
1. Alkane
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The “alkane” is the most basic unit in organic chemistry, made up of carbon/carbon single bonds
and carbon/hydrogen single bonds. In fact, we often don’t consider the alkane to be a functional
group at all, but rather the molecular scaffold on which functional groups are placed. Think of
the alkane as the canvas on which a picture is painted. We can use the term “alkane” to refer to a
whole molecule that contains nothing but carbon and hydrogen single bonds, or we can refer to a
portion of a molecule as the alkane portion of the molecule. Alkanes are a type of hydrocarbon,
a molecule that contains only carbon and hydrogen.
Some other important common terms associated with alkanes include straight-chain alkane,
branched alkane, and cyclic alkane.
Examples of alkanes:
cyclic
straight-chain branched
The molecule below is cholesterol. We might say that the molecule has an “alkane portion”
which is circled. An older term that is still common is to say that cholesterol contains an
“aliphatic portion” because it is all carbon and hydrogen single bonds.
HO
2. Alkenes
The alkene is also a hydrocarbon because it contains only carbon and hydrogen, but it also
contains a carbon-carbon double bond. We say that the alkene is unsaturated because it is not
“saturated with hydrogen atoms—it does not contain as many C-H bonds as possible. The base
unit of the alkene is the C=C, with hydrogen atoms or “R” groups attached. The “R” group is the
generic variable of any alkane group. We refer to this as an “alkyl group.
R or H R or H
C C base unit of an alkene
R or H R or H
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We might use the term alkene in two ways: we might refer to the C=C bond itself as an alkene,
or we might refer to a whole molecule containing the carbon-carbon double bond as an alkene.
For example, we might say that there are three alkenes shown below, or we might say that the
three molecules below all contain the alkene functional group. Because of their structural
similarities, all alkenes have some similarities in their properties and reactivity.
a cyclic alkene
3. Alkyne
The alkyne is similar to the alkene except that alkynes contain carbon-carbon triple bonds. They
are also unsaturated hydrocarbons. A terminal alkene has at least one hydrogen directly attached
to the triple bonded carbon. An internal alkene has two R groups. The prime symbol in R’
designates that this R group may be different than the other R group.
R C C H R C C R'
terminal alkyne internal alkyne
4. Aromatic
In this course, we will learn quite a bit about aromatic compounds. For now, we will discuss the
most common aromatic structure: benzene. Benzene is a six member carbon ring with three
double bonds.
benzene aromatic compounds
5. Alcohol
An alcohol is a molecule that contains the hydroxyl group, -OH. Therefore, the generic
representation of an alcohol is ROH. The base unit is carbon single bonded to oxygen, which is
single bonded to hydrogen.
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R or H
R' or H C O H
OH OH
R' or H
base unit of moleucle A molecule B
alcohol
The two example molecules above both contain the base unit C-O-H. All of the following
statements would be proper to make:
Molecules A and B are alcohols.
Molecule A contains a hydroxyl group.
Molecule B is an alkene.
Molecule B is a cyclic molecule containing a hydroxyl group.
6. Phenol
Alcohols that are directly connected to benzene have a special name: phenol. It is important to
recognize that the phenol involves the hydroxyl directly connected to the benzene, not simply in
the same molecule.
OH
OH
a phenol
not a phenol--an alcohol
5. Ether
The ether is a functional group similar to the alcohol, but there is one crucial distinction: the
oxygen atom is bonded to two carbon atoms. Therefore the base unit of the ether is C-O-C.
O O
a symmetrical ether
a cyclic ether O
an unsaturated ether
6. Amine
The distinctive of the amine group is that it contains a nitrogen atom bonded to hydrogen and/or
alkyl groups. Amines can look quite different, as in the examples below.
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H
N
H2N
N
H
7. Nitrile
A nitrile contains the cyano functional group, which is a carbon-nitrogen triple bond. Due to the
nature of nitrogen bonding, the cyano group is always a terminal group in a compound. It is
often written in its condensed formula, “CN”. Therefore, the general representation of a nitrile is
RCN.
CN
N
an aromatic nitrile an aliphatic nitrile
7. Thiol
Thiols are the equivalent of alcohols that have sulfur instead of oxygen. Therefore, the base unit
of a thiol is C-S-H and the general designation for a thiol is RSH.
8. Alkyl halide
The alkyl halide is a alkane skeleton with a halogen bonded. The halogens (fluorine, chlorine,
bromine, and iodine) can be represented generically as “X”, and therefore, the general alkyl
halide can be designated RX.
Cl
Br I
9. Ketone
The ketone is the first of many compounds that contains the carbonyl group, which is a carbon-
oxygen double bond. A compound is a ketone if the carbonyl carbon is directly attached to two
carbon atoms, typically alkanes or alkenes.
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O O
O
C C C
base unit of
ketone cyclic keton an unsaturated ketone
10. Aldehyde
An aldehyde is very similar to a ketone, but in an aldehyde, the carbonyl is attached to carbon on
one side and hydrogen on the other. Therefore, the aldehyde is always at the end of a molecule.
In the condensed formula, the aldehyde is designated with “CHO” which does not mean that
hydrogen is single bonded to carbon and oxygen, but is meant to distinguish the group from a
hydroxyl group. The general designation of an aldehyde is RCHO.
O
O
H
C C H
CH3CH2CHO
base unit of a condensed formula
aldehyde of an aldehyde
an aromatic aldehyde
11. Carboxylic acid
A carboxylic acid contains a carbonyl attached to an R group on one side and a hydroxyl group
on the other side. NOTICE: The carboxylic acid is NOT a ketone and alcohol, but rather, a
whole new group. Therefore it is proper to say that all carboxylic acids contain a hydroxyl
group, but not all hydroxyl groups are carboxylic acids. Condensed formula for a carboxylic
acid is COOH, and these acids are generally written RCOOH.
Two carboxylic acids in the Citric Acid Cycle
O succinate fumarate
C C OH
COOH COOH
base unit of HOOC HOOC
aldehyde
an unsaturated carboxylic acid
12. Ester
The ester contains a carbonyl attached to an R group on one side and an OR group on the other
side.
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O O O
C C OR
O O
base unit of
ester
13. Amide
An amide contains a carbonyl with an R group on one side and a nitrogen atom on the other.
The nitrogen may be attached to other R groups or hydrogens. NOTICE: The amide is NOT a
ketone attached to an amine, but a whole new group called an amide.
O
O O
H2N
C C N HN
N
base unit of H
O
ester
Functional group identification tests
Chemists have developed a number of simple chemical tests that are positive only for
compounds having certain kinds of functional groups. These are the four tests we will be
performing today:
Compounds that are easily oxidized react with a solution of chromium(VI) oxide in
sulfuric acid, commonly referred to as "chromic acid." Primary and secondary alcohols
react within 2-3 seconds to form an opaque blue-green suspension. Aldehydes give the
same result but usually take 10 seconds or more to react.
Aldehydes and ketones both react with 2,4-dinitrophenylhydrazine (DNPH) reagent to
yield yellow or orange precipitates within a few minutes.
Alkenes react readily with dilute aqueous potassium permanganate to form a brown
precipitate as the purple color of the permanganate disappears.
Alkyl halides give a green flame in the Beilstein test, which involves heating a copper
wire moistened with the unknown in a burner flame.
Procedure
1. First, you will be performing the functional group identification tests on known
compounds (indicated in each of the test procedures) for which each of the tests
should produce a positive result. Note your observations carefully in your
laboratory notebook after performing each of these tests, as they will be helpful in
identifying an unknown compound later.
2. Steps 3-6 can be performed in any order. One of the following tests will be set up
in one of four fume hoods in the lab. When you are finished with a test, move to
another hood to perform another test.
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3. 2,4-Dinitrophenylhydrazine (2,4-DNPH) test: Add 1 drop of the compound to be
tested (in this case, reagent-grade acetone will give a positive test) to ~1 mL of
the DNPH reagent in a small disposable test tube. (The disposable test tubes hold
roughly 1.2 mL of liquid, so simply estimate the volume) Stopper the tube with a
small rubber septum (the instructor has these) and shake the test tube and let the
mixture stand for 15 minutes. Carefully record your observations. The septa are
expensive and are not disposable so please rinse them and return them to your
instructor when finished.
4. Chromic acid test: Dissolve one drop of the compound to be tested (in this case,
ethanol will give a positive test) in ~1 mL of reagent-grade acetone in a small test
tube. Add 1 drop of the chromic acid reagent, shake the mixture, and observe it
for at least one minute, noting the time required for any positive test. Carefully
record your observations.
5. Potassium permanganate test: Dissolve 1 drop of the compound to be tested (in
this case, limonene will give a positive test) in ~1 mL of 95% ethanol, and slowly
add 5-10 drops of 0.1 M potassium permanganate with shaking. Note what
happens after each drop is added.
6. Beilstein's test: Make a small loop in the end of a 10-cm length of copper wire.
Heat the loop to redness in a burner flame. (Keep the unknown and any other
flammable liquids at a safe distance.) Dip the loop into a little bit of your
compound to be tested (in this case, bromobenzene will give a positive test) and
hold it in the lower outside part of the flame. Note the color of the flame
produced.
7. Obtain a numbered unknown from your instructor and record its number in your
laboratory notebook. Important: cap your unknown vial tightly after removing
the required amount for the tests, as these compounds may have strong smells
and will evaporate very quickly! Perform each of the four functional group
classification tests as in steps 3-6, but use your unknown as the test compound as
opposed to the control compounds in the instructions above. Carefully observe
and record the results of your tests, and decide whether they are positive or
negative. (Positive tests should look similar to the tests with the control reagents
in steps 3-6). Note: some tests may sometimes provide ambiguous results. The
purity of the test reagent and the unknown compound can have adverse effects on
these tests, so it is especially important to prevent cross-contamination today. The
2,4-DNPH reagent has been known to crystallize a small amount on its own, so
you should look for heavy crystallization as a sign of a positive test. Consult with
the instructor and/or run a test again if the result seems ambiguous.
8. Using the organic modeling kits, make models of each of each of the four
standard chemicals used in the positive tests (acetone, ethanol, limonene, and
bromobenzene), as well as 1-bromobutane, 2-butanone, cyclohexene, 1-butanol.
(a web search will help you identify the structures of acetone and limonene).
Show these models to your instructor.
9. Use your reasoning skills to deduce which functional group(s) may be present in
your unknown compound. Based on the functional group tests, identify your
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unknown compound by choosing it from the following possible unknowns:
1-bromobutane, 2-butanone, cyclohexene, 1-butanol.
Organic Nomenclature Report.pdf
Formula of Organic Molecules - The Chemistry Journey - The Fuse School.mp4
The functional group concept explained - The Chemistry Journey - The Fuse School.mp4
Elemental Analysis:
1.1 Qualitative Elemental Analysis
To determine rvhat elements are present in an organic compound, it is necessary to
decompose the compound so that the covalently bonded elements are convefted to r,vater-soluble
ions which can then be detected by qualitative inorganic analysis. 'fhe sirnplest and most useful
rnethod for the decomposition of organic compottnds is sodium fusion.
1.2 Quantitative Elemental Analysis
The amounts of carbon and hydrogen in an organic compound can be measured by
combustion of a weighed sample in a stream of oxygen in an apparatus called a combustion train.
All of the carbons in the organic compound combine with oxygen to form carbon
dioxide, while all of the hydrogens combine with oxygen to form water.
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Figure 1. Schematic representation of a combustion train for the quantitative determination of C
and H.
The resulting water and carbon dioxide are absorbed in tubes that contain magnesium
perchlorate and soda lime, respectively. The tubes are then weighed in order to determine the
amounts of water and carbon dioxide produced from combustion of the weighed sample. From
these measurements, the masses and percentages of carbon and hydrogen in the sample can be
calculated as follows:
The quantitative determination of other elements is accomplished by using methods
which make use of the same principle as above, i.e., to transform an element to a compound
whose amount can easily be measured.
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The halogens (X = Cl, Br, I) are quantitatively determined by the Carius Method (using
hot fuming HNO3) or the Parr Method (using sodium peroxide).
The amount and percentage of halogen are calculated as follows:
Sulfur is also determined by the Carius or Parr Method.
The amount and percentage of sulfur are calculated as follows:
Nitrogen is determined by the Dumas method. In this procedure, the nitrogen present in
the organic sample is transformed to nitrogen gas and its volume is measured. To calculate the
amount of nitrogen the Ideal Gas Law is used:
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The amount and percentage of nityrogen in the sample arte calculated as follows:
There is no direct way of determining the % of oxygen. The amount of oxygen is
determined inderectly by difference. If the sum of the percentages of the other elements present
is not 100% the difference is due to oxygen.
EMPIRICAL AND MOLECULAR FORMULA DETERMINATION
The empirical formula expresses the relative number (not the actual number) of the
different atoms present in a molecule. It can easily be calculated from theoh composition. Some
examples of empirical formulas are the following:
The molecular formula, which expresses the actual number of each kind of atom present,
can be derived from the empirical formula. The molecular formula is a whole number multiple of
the empirical formula. In some cases the molecular formula is the same as the empirical formula.
The compounds with molecular formulas C2Ha, CrHe, C+Hs all have the empirical formula
CHz.
In order to derive a molecular formula from an empirical formula, the molecular weight
(molecular mass) of the compound being analyzed,must first be known.
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There are several ways of determining molecular weight. For gaseous organic compounds
and liquid substances of moderate volatility, molecular weights may be determined by vapor
density experiments. The molecular weight of high-boiling liquids and nonvolatile solids can be
determined by measurements of freezing-point depression or boiling-point elevation. Highly
accurate molecular weights are determined by mass spectrometry.
The following example illustrates how calculations for o/o composition, empirical and
molecular formulas can be carried out from elemental analysis data.
Example:
An 8.00 mg sample of an organic compound was subjected to quantitative combustion to give
15.33 mg of carbon dioxide and 9.36 mg of water. Th molecular weight was found to be 46.
calculate the yo composition, empirical and molecular formulas of the organic compound.
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EXERCISES:
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DERIVATION OF STRUCTURAL FORMULAS FROM A MOLECULAR FORMULA
From the molecular formula, the next step is to derive possible structural formulas. A
structural formula tells us how the atoms in a molecule are bonded to each other. oftentimes for a
given molecular formula there are several ways of bonding the atoms to one another.
Example:
Propose possible structural formulas for each of the following molecular formulas:
a. C2H6O
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b. C3H6O
For a molecular formula, we should calculate for the value of ρ so that we will know the
total number of rings and/or double and triple bonds (degree of unsaturation) that may be present
in the structural formulas.
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EXERCISES:
Propose possible structural formulas for each of the following molecular formulas:
a. C3H4
b. C2H7N
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c. C3H5Br
Using the molecular formula C3H6)2, draw a structural formula that contains the designated
functional groups:
a. aldehyde and ether
b. ester
c. carboxylic acid
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d. ketone and alcohol
REACTION MECHANISMS
In addition reactions, there is an increase in the number of atoms or gloups attached to a
carbon(s) in the substrate. A saturated product is obtained from an unsaturated substrate. (A
saturated compound.contains single bonds only while an unsaturated compound contains double
or triple bonds.)
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Each carbon atom in the unsaturated substrate gets one more atom resulting in a saturated
product.
If you compare the structure of the substrate with that of the product, you can see quite
clearly that each carbon atom in the unsaturated substrate'jets one more atom (one carbon atom
gets an H and the other a cl atbm) resuliing in a saturated product.
Elimination reactions are the exact opposite of addition reactions. In elimination
reactions, an atom or u gro.tf of atoms is removed from the substrate, thus there is a decrease in
the number of atoms or groups attached to carbon. The substrate is transformed to a more
unsaturated product. Consider the reaction below:
An.unsaturated organic p'oduct is obtained through the removal of an H atom and OH
group from the saturated substrate.
Substitution reactions involve the replacement of an atom or a group of atoms in the
substrate by another atorn or group of atoms. An eximpie is given below:
The organic product results from the replacement of the chlorine atom in the substrate by
the hydroxyl group ( -OH ).
Many rearuangement reactions involve migration of an atom or group of atoms from one
atom to another. To illustrate:
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In this reaction, bromine migrates from an outer oarbon atom to the imer carbon atom.
Oxidation and reduction reactions in organic chemistry are usually defined according to
the increase or decrease in oxygen or hydrogen content because the transfer of electrons cannot
be easily monitored as in inorganic reactions.
An oxidation reaction occurs when there is an increase in the oxygen content and/or a
decrease in the hydrogen content of the substrate. For example:
Comparing the structure of the organic product with the substrate, notice that there is an
increase in the oxygen content (by one oxygen) as well as a decrease in the hydrogen content (by
two hydrogens).
In a reduction reaction, there is an increase in the hydrogen content and/or a decrease in
the oxygen content of the substrate. Consider the following reaction:
EXERCISES:
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HYDROXYLATION REACTION
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OXIDATIVE CLEAVAGE OF ALKENE AND ALKYNES
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EXERCISES:
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REFERENCES AND RESOURCES:
Ball, P. (1994). Designing the Molecular World. New Jersey: Princeton University Press.
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Brock, w. H.. (1993). The Norton History of chemistry. New york: w. w. Norton and Co.
Brown, W. H. (1982). Introduction to Organic Chemistry (3rd ed.). Boston: Wilier Grant Press.
carey,
F. A. (1992). organic Chemistry (2nd ed.). New york: McGraw-Hill,Inc.
McMurry, J. (1992). Organic Chemistry (3rd ed.). Califomia: Brooks/Cole Publishing Co.
Morrison, R.T. & Boyd. R. N. (1989). organic Chemistry (5th ed.). Boston: Allyn and Bacon.
Solomons, T.w.G. (1994). Fundamentals of organic Chemistry (4thed.). New york: John Wiley
and Sons, Inc.48.
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