Topic 4: Organic and Biological
Chemistry
4.1
Systematic Nomenclature
Key Ideas
Intended Student Learning
The presence or absence of functional groups in
an organic compound determines its physical
and chemical properties.
Identify the functional groups in the structural
formulae of alcohols, aldehydes, ketones,
carboxylic acids, amines, esters, and amides.
Organic compounds are named systematically to
provide unambiguous identification.
State, given its structural formula, the systematic
name of an organic compound containing:
up to eight carbon atoms arranged as either a
straight chain or a branched chain
one or more of the same functional groups (with
these limited to hydroxyl, aldehyde, ketone,
carboxyl, or primary amino groups).
The structural formula of an organic compound
can be deduced from its systematic name.
Given its systematic name, draw the structural
formula of an organic compound containing:
up to eight carbon atoms arranged as either a
straight chain or a branched chain
one or more of the same functional groups (with
these limited to hydroxyl, aldehyde, ketone,
carboxyl, or primary amino groups).
Esters are named as derivatives of a carboxylic
acid.
State the systematic names of methyl and ethyl
esters of straight-chain acids containing up to
eight carbon atoms.
The structural formula of an ester can be
deduced from its systematic name.
Given its systematic name, draw the structural
formula of an organic methyl or ethyl ester of a
straight-chain acid containing up to eight carbon
atoms.
Identify the functional groups in the structural formulae
of alcohols, aldehydes, ketones, carboxylic acids,
amines, esters, and amides.
"A group of atoms or small groups of atoms which show a characteristic reactivity
when treated with certain reagents and impart some characteristic chemical and
physical properties to the main molecule is termed as function group."
Group
Formula
Class name Example
IUPAC name
Common name
Halide
CH3-I
Iodomethane
Methyl iodide
Alcohol
CH3-CH2OH
Ethanol
Ethyl alcohol
Ether
CH3-CH2-ODiethyl ether
CH2-CH3
Ether
Amine
CH3-NH2
Aminomethane
Methyl amine
Aldehyde
CH3-CHO
Ethane
Acetaldehyde
Ketone
CH3-CO-CH3 Propanone
Acetone
Carboxylic
acid
CH3-COOH
Ethanoic acid
Acetic acid
Ester
CH3-CO2CH2-CH3
Ethyl ethanoate
Ethyl acetate
Amide
CH3-CON(CH3)2
N,NDimethylethanamide
N,NDimethylacetamide
•State, given its structural formula, the systematic name of an organic compound
containing:
•up to eight carbon atoms arranged as either a straight chain or a branched chain
•one or more of the same functional groups (with these limited to hydroxyl, aldehyde,
ketone, carboxyl, or primary amino groups).
Key Ideas
Intended Student Learning
The melting points and boiling points of
organic compounds that contain the same
functional group increase with the length of
carbon chain.
Predict and explain the melting points and
boiling points of an organic compound in
comparison with those of other compounds
that contain the same functional group.
The boiling points of organic compounds that
display hydrogen bonding between
molecules are higher than those of
compounds of similar molar mass that do not
display hydrogen bonding.
Predict and explain the boiling points of
alcohols in comparison with those of
aldehydes and ketones of similar molar
mass.
The boiling points of esters are lower than
those of isomeric acids because of the
absence of hydrogen bonding between
molecules of the ester.
Predict and explain the boiling points of
esters in comparison with those of isomeric
acids.
Organic compounds are generally insoluble
in water.
Explain the insolubility in water of most
organic compounds.
Hydrogen bonding between functional groups
and water can explain the solubility in water
of some smaller organic compounds.
Predict and explain the solubility in water of
the smaller amino acids, carboxylic acids,
alcohols, aldehydes, and ketones.
The solubility in water of an organic
compound depends on its molar mass and
the functional groups present.
Predict and explain the relative solubilities in
water of two organic compounds, given their
structural formulae.
4.2
Physical Properties
The melting points and boiling points of
organic compounds that contain the same
functional group increase with the length of
carbon chain.
Predict and explain the melting points and
boiling points of an organic compound in
comparison with those of other compounds
that contain the same functional group.
As the chain increases the molar mass increases. Therefore dispersion force increases
• Complete Q4.8 page 248
4-2 Physical Properties
bp and mp affected by secondary bonding –
especially hydrogen bonds
Need to assess relative polarity
Remember H-Bonds between H and N, O or F
Mark on polarity
Prentice-Hall © 2002
General Chemistry: Chapter 27
Slide 12 of 76
The polarity of functional groups and
boiling points
• Aldehydes, alcohols and ketones
Its important to work out the polarity of
functional group and then explain how it
affects the boiling point
•
•
•
•
For example:
Compound
Propan-1-ol
Propanal
Propanone
molar mass
60
58
58
Boiling point
97.2
47.9
56.1
Alcohol has the highest- why??
Functionol group present is polar hydroxyl group 0-H.
This polar groups form strong hydrogen bonds with
each other.
On the other hand
Aldehyde and ketone both have polar carbonyl group.
Dipole- dipole interactions between these polar
groups
4-2 Physical Properties
In organic the H- bond will increase bp and mp.
Bishop ch 14 p555-556
Prentice-Hall © 2002
General Chemistry: Chapter 27
Slide 15 of 76
4.2
Physical
Properties
Assessing Polarity
Glucose
Least polar
Most polar
All six carbons, but different
number of oxygens and hydroxy
groups.
4,5,6 Trihydroxy 2,3 dioxy hexanoic acid
Prentice-Hall © 2002
General Chemistry: Chapter 27
Slide 16 of 76
• Complete Q4.9 pg 249
Organic compounds are generally insoluble
in water.
Explain the insolubility in water of most
organic compounds.
Hydrogen bonding between functional groups
and water can explain the solubility in water
of some smaller organic compounds.
Predict and explain the solubility in water of
the smaller amino acids, carboxylic acids,
alcohols, aldehydes, and ketones.
The solubility in water of an organic
compound depends on its molar mass and
the functional groups present.
Predict and explain the relative solubilities in
water of two organic compounds, given their
structural formulae.
Esters and carboxylic acids
• When comparing boiling points- clear pattern
is evident
• A carboxylic acid that has the same molecular
formula as a particular group of esters is called
an isomeric acid of those esters. In other
words the esters and carboxylic acid form a
group of structural isomers
For example:
• Propanoic acid- CH3CH2COOH
• Methyl ethanoate- CH3COOCH3
• Ethyl methanoate- HCOOCH2CH3
All have molecular formula of C3H6O2
Therefore propanoic acid is isomeric acid of the two esters.
Same molecular formula:same molar mass: same dispersion
forces.
HOWEVER – the different functional groups will have an effect
on boiling points!
Continued….
• Ester- lower boiling point- functional group in ester
have only one polar group as part of their structure
O-C=O. therefore dipole dipole interaction occurs
between the groups
• On the other hand
• Carboxylic acid – higher boiling point- Carboxyl
groups in carboxylic acid have both polar C=O and OH groups as part of their molecule. Hydrogen
bonding operates between both of these groups
• Fig 4.2 important
The solubility of organic compounds in water
• Non polar organic compounds, such as hydrocarbons are not
soluble in water because they cannot form hydrogen bonds
with the polar water molecules.
• Molecules with polar functional groups but long chains( 6
Carbon atoms or more) are classified as non polar because
non polar C chain is the dominant feature
• The influence of functional groups is increased when more
than one functional group is present such as diols,
dicarboxylic acids and diamines.
• Read page 251-252
• Complete question on pg 253
An organic compound which has the
general formula CnH2n+1OH, they
consist of hydrocarbon chains
terminated by hydroxyl groups, O-H.
Smaller members are water soluble,
flammable and are useful as organic
solvents and fuels. As with
hydrocarbons, each member differs
from the previous by an additional CH2- group.
Polar molecule +/ or able to make H-bonds with water
Give the expected trend (lowest to highest) in boiling points for
the following series of compounds :
4.3. Alcohols
• Alcohols are organic compounds which
contain a hydroxyl (—OH) group covalently
bonded to a carbon atom.
The boiling points of organic compounds that
display hydrogen bonding between
molecules are higher than those of
compounds of similar molar mass that do not
display hydrogen bonding.
Predict and explain the boiling points of
alcohols in comparison with those of
aldehydes and ketones of similar molar
mass.
Predict and explain the boiling points
Key Ideas
Intended Student Learning
Ethanol is produced by the fermentation
of glucose, which can be derived by the
hydrolysis of complex carbohydrates.
Describe the conditions, and write
equations, for the hydrolysis of
polysaccharides and disaccharides, and
the production of ethanol by the
fermentation of glucose.
Alcohols are classified as primary,
secondary, or tertiary.
Identify a hydroxyl group in an alcohol as
primary, secondary, or tertiary, given the
structural formula.
Primary and secondary alcohols can be
distinguished from tertiary alcohols by
their reaction with acidified dichromate
solution.
Describe how primary and secondary
alcohols can be distinguished from
tertiary alcohols by their reaction with
acidified dichromate solution.
The type of product obtained by oxidising
an alcohol depends on whether the
alcohol is primary or secondary.
Predict the structural formula(e) of the
product(s) of dichromate oxidation of a
primary or secondary alcohol, given its
structural formula.
Ethanol is produced by the fermentation of
glucose, which can be derived by the hydrolysis
of complex carbohydrates.
Describe the conditions, and write equations, for
the hydrolysis of polysaccharides and
disaccharides, and the production of ethanol by
the fermentation of glucose.
4-0 Hydration Reaction
One way to form ethanol. Requires high temperature,
high pressure and catalyst (phosphoric acid, sulfuric
acid or aluminium oxide). Ethanol produced in this
way is called synthetic or industrial ethanol
Prentice-Hall © 2002
General Chemistry: Chapter 27
Slide 35 of 76
2nd method- fermentation of glucose
• Glucose is a monosaccharide which is
obtained directly from fruits.
• If there is disaccharides or polysaccharidesthese cannot undergo fermentation.
Therefore undergoes hydrolysis first to form
monosaccharide.
Equations:
• Hydrolysis of polysaccharides to monosaccharides
(C6H10O5)n + nH2O
nC6H12O6
• Hydrolysis of disaccharides to monosaccharides
C12H22O11 + H2O
2C6H12O6
• Fermentation of glucose to ethanol and carbon dioxide
Special condition required: read page 254 (
important)
• Optimum temperature for fermentation- btw 20 and 30.
• Since fermentation is an exothermic process, it is necessary to
cool the fermentation vessel to avoid excessive high
temperature.
• Acidic condition required ( changing of pH of the solution
affects the structure of the enzymes which in turn produces
different fermentation products.
• Fermentation is anaerobic process and oxygen must be totally
excluded. If oxygen enters, ethanol undergoes oxidation to
produce ethanal and ethanoic acid
• Yeasts- micro organisms which produces enzymes to speed up
fermentation. Once the concentration of alcohol exceeds 14%
yeasts die.
• Complete Q4.11 page 255
Alcohols are classified as primary,
secondary, or tertiary.
Identify a hydroxyl group in an alcohol
as primary, secondary, or tertiary,
given the structural formula.
Alcohols are classified as primary, secondary and tertiary
depending on the position of the hydroxyl group in the
molecular structure
• Complete Q4.12 page 256
Primary and secondary alcohols
can be distinguished from
tertiary alcohols by their
reaction with acidified
dichromate solution.
Describe how primary and
secondary alcohols can be
distinguished from tertiary
alcohols by their reaction with
acidified dichromate solution.
Samples of alcohols are added to acidic dichromate solutions, which are orange.
The rate at which the orange color converts to blue/green chromium (III) is
compared for the three types of alcohols
Oxidation of Alcohols- acidified potassium dichromate
Primary Alcohol
Aldehyde
Cr2O72-/H+/
CH3CH2OH →
Carboxylic acid
CH3COH → CH3COOH
Ketone
Secondary Alcohol
CH3CHOHCH3 → CH3COCH3
Tertiary Alcohol
CH3COHCH3CH3 → no reaction
This lack of reaction with acidified potassium dichromate could
be used to distinguish primary and secondary alcohols from
tertiary alcohols
Prentice-Hall © 2002
General Chemistry: Chapter 27
Slide 44 of 76
Note:
• Primary & secondary alcohols and their
oxidation products all have the same number
of carbon atoms.
• Primary and secondary alcohols- reduce
orange dichromate ions to green chromium
(III) ions while tertiary alcohols do not react
(no colour change). Oxidation num reduces
from +6 to +3
• Note: alcohols and their oxidation products
are colourless ; it is the reduction of
dichromate to chromium (III) ions that is
responsible for the colour change.
• Half equation for the reduction of dichromate:
Given the structural formula of the aldehyde or ketone, draw the structural
formula of the alcohol from which it could be produced by oxidation, and
describe the necessary reaction conditions.
• Complete Question 4.13 and 4.14 page 257
4.4
Aldehydes and Ketones
Key Ideas
Intended Student Learning
Aldehydes and ketones are produced by
the oxidation of the corresponding
primary and secondary alcohols
respectively. Aldehydes are readily
oxidised and so must be distilled off from
the reaction mixture as they are formed.
Given the structural formula of the
aldehyde or ketone, draw the structural
formula of the alcohol from which it
could be produced by oxidation, and
describe the necessary reaction
conditions.
Aldehydes can be oxidised to form
carboxylic acids or, in alkaline solutions,
carboxylate ions.
Draw the structural formula of the
oxidation product of a given aldehyde in
either acidic or alkaline conditions.
Ketones cannot readily be oxidised. This
difference in properties between
aldehydes from ketones can be used to
distinguish one from the other.
Describe how acidified dichromate
solution and Tollens’ reagent
(ammoniacal silver nitrate solution) can
be used to distinguish between aldehydes
and ketones.
Aldehydes and ketones are produced by
the oxidation of the corresponding
primary and secondary alcohols
respectively. Aldehydes are readily
oxidised and so must be distilled off from
the reaction mixture as they are formed.
Given the structural formula of the
aldehyde or ketone, draw the structural
formula of the alcohol from which it
could be produced by oxidation, and
describe the necessary reaction
conditions.
•Aldehydes and ketones are organic compounds which incorporate a carbonyl
functional group, C=O. The carbon atom of this group has two remaining bonds
that may be occupied by hydrogen or alkyl or aryl substituents. If at least one of
these substituents is hydrogen, the compound is an aldehyde. If neither is
hydrogen, the compound is a ketone.
Aldehydes and ketones
• One thing in common??
• Difference?
Preparation of aldehydes and ketones fig 4.7
Oxidation of aldehydes by two ways
Oxidation of aldehydes by two ways
1. Aldehydes heated with acidified potassium
dichromate, undergo oxidation to carboxylic
acid ( orange dichromate ions reduced to
green chromate ions)
2nd way
2. Known as silver mirror test or tollen’s reagent
Process
• aldehydes undergo oxidation to carboxylate ions
when heated with ammoniacal silver nitrate solution
(Tollen’s reagent). Oxidising agent is silver diammine
ion which is reduced to metallic silver and this forms
a deposit on the inside of the reaction vessel. The
formation of silver mirror is a positive test for the
preence of an aldehyde functional group in an
organic compound
Oxidation
The presence of that
hydrogen atom makes
aldehydes very easy to
oxidise. Or, put another way,
they are strong reducing
agents.
The electron-half-equation for the reduction of of the
diamminesilver(I) ions to silver is:
Combining that with the half-equation for the oxidation of
an aldehyde under alkaline conditions:
. . . gives the overall equation:
Using Tollens' reagent (the silver mirror test) to distinguish between an
aldehyde and a ketone
To carry out the test, you add a few drops of the aldehyde or ketone to
the freshly prepared reagent, and warm gently in a hot water bath for a
few minutes.
Ketone- No change in the colourless solution.
In aldehydethe colourless solution produces a grey precipitate of silver,
or a silver mirror on the test tube. Aldehydes reduce the
diamminesilver(I) ion to metallic silver. Because the solution is alkaline,
the aldehyde itself is oxidised to a salt of the corresponding carboxylic
acid.
• Complete Q 4.15 – 4.17 page 260
Using acidified potassium dichromate(VI) solution
A small amount of potassium dichromate(VI) solution is acidified with dilute
sulphuric acid and a few drops of the aldehyde or ketone are added. If
nothing happens in the cold, the mixture is warmed gently for a couple of
minutes - for example, in a beaker of hot water.
ketoneNo change in the orange solution.aldehydeOrange solution turns
green.The orange dichromate(VI) ions have been reduced to green
chromium(III) ions by the aldehyde. In turn the aldehyde is oxidised to the
corresponding carboxylic acid.
The electron-half-equation for the reduction of dichromate(VI) ions is:
Combining that with the half-equation for the oxidation of an aldehyde
under acidic conditions:
. . . gives the overall equation:
• Step One: Grain Receiving, Storing and Milling
• We receive corn by rail and by truck, at which point we
inspect, weigh and unload the corn in a receiving building and
then transfer it to storage bins. On the grain receiving system,
a dust collection system limits particulate emissions. Truck
scales and a rail car scale weigh delivered corn. Corn
unloading and storage systems include independent
unloading legs and concrete and steel storage bins. From its
storage location, corn is conveyed to scalpers to remove
debris before it is transferred to hammermills or grinders
where it is ground into a flour, or "meal."
• Step Two: Conversion and Liquefaction, Fermentation and
Evaporation Systems
• The meal is conveyed into slurry tanks for enzymatic processing.
The meal is mixed with water and enzymes and heated to break the
ground grain into a fine slurry. The slurry is routed through pressure
vessels and steam flashed in a flash vessel. This liquefied meal,
now called "mash," reaches a temperature of approximately 200°F,
which reduces bacterial build-up. The sterilized mash is then
pumped to a liquefaction tank where additional enzymes are added.
This cooked mash continues through liquefaction tanks and is
pumped into one of the fermenters, where propagated yeast is
added, to begin a batch fermentation process.
• The fermentation process converts the cooked mash into carbon
dioxide and beer, which contains ethanol as well as all the solids
from the original feedstock. We employ the batch fermentation
process in which the mash is kept in one fermentation tank for
approximately two days. Circulation through external plate and
frame heat exchangers, designed for high solids content and easy
cleaning, keeps the mash at the proper temperature.
•
•
Step Three: Distillation and Molecular Sieve
After batch fermentation is complete, beer is pumped to the beer well and
then to the distillation column to vaporize and separate the alcohol from the
mash. The distillation results in a 96%, or 190-proof, alcohol. This alcohol is
then transported through a rectifier column, a side stripper and a molecular
sieve system where it is dehydrated to produce 200-proof anhydrous
ethanol. The 200-proof alcohol and up to 5% denaturant constitute ethanol
ready for sale.
•
•
Step Four: Liquid—Solid Separation System
The residue corn mash from the distillation stripper, called stillage, is
pumped into one of several decanter type centrifuges for dewatering. The
water, or thin stillage, is then pumped from the centrifuges back to mashing
as backset or to an evaporator where it is dried into a thick syrup. The solids
that exit the centrifuges, known as "wet cake," are conveyed to the wet cake
storage pad or the gas-fired rotary dryer for removal of residual water. Syrup
is added to the wet cake as it enters the dryer, where moisture is removed.
The end result of the process is the production of dried distillers grains with
solubles, or DDGS.
• Step Five: Product Storage
• Final storage tanks hold the denatured ethanol product prior
to being transferred to loading facilities for truck and rail car
transportation. Each of our plants has one 190-proof storage
tank and one 200-proof storage tank. Our Iowa Falls plant has
two 750,000 gallon final storage tanks with another two
750,000 gallon final storage tanks to be be added in the fourth
quarter of 2006. Our Fairbank plant has two 1,500,000 gallon
final storage tanks. These final storage tanks will
accommodate ten days of storage per plant.
Given the structural formula of the aldehyde or ketone, draw the structural
formula of the alcohol from which it could be produced by oxidation, and
describe the necessary reaction conditions.
Draw the structural formula of the oxidation product of a given aldehyde
in either acidic or alkaline conditions.
Describe how acidified dichromate solution and Tollens’ reagent
(ammoniacal silver nitrate solution) can be used to distinguish
between aldehydes and ketones.
Predict and explain the melting points and boiling points of an organic compound in
comparison with those of other compounds that contain the same functional group.