Download 9.1-10.5 Organic Chemistry

Unit A: Organic Chemistry
Innisfail High School
Chemistry 30
Ms. Littke
Spring 2014
Assessment

Formative

Students must participate in lessons


Students are expected to self assess and seek help if
necessary



Complete daily activities and review new concepts regularly
F/S - Quizzes will be given periodically throughout
the unit


- complete daily notes and examples.
You should use these to your advantage as they test
smaller sections of the curriculum and help prepare you
for the Unit Exam
Labs – 2 labs per unit
Summative Unit Exam – test format will be given to
you later
Organic Chemistry
Today’s Objectives:
1) Define organic compounds as compounds containing carbon, recognizing inorganic
exceptions such as carbonates, cyanides and carbides
2) Identify and describe significant organic compounds in daily life, demonstrating
generalized knowledge of their origins and applications
3) STS: Demonstrate an understanding that science and technology are developed to
meet societal needs and expand human capability
Unit A: Chapters 9 and 10
Organic
Introduction
Today’s Agenda:
1) Introduce organic chemistry and review the origins and
applications of some major organic compounds
2) “Carbon – The element of life” video
3) Are You Ready pg. 354 #1-6 – due tomorrow
Section 9.1 (pg. 354-361)
What is Organic Chemistry?


The early definition related to compounds obtained only from
living things.
Today, it is a major branch of chemistry that deals with
compounds of carbon, called ORGANIC compounds*.
 *Carbon compounds that are exceptions and considered
INORGANIC are compounds like:
Oxides carbon monoxide (CO(g) ) and carbon dioxide (CO2(g) ), and
 Ionic compounds of carbon-based ions, such as carbonate CO32-,
cyanide CN-, and carbide ions, SiC (silicon carbide)


The major source of carbon compounds is still living or
previously living things, such as plants, animals and all types
of fossil fuels.
Organic or Inorganic??
Formula
CaCO3(s)
C25H52(s)
Organic or Inorganic?
Inorganic (carbonate ion)
Organic
Ca2C(s)
CCl4(l)
CH3COOH(l)
CO2(g)
KCN(s)
C12H22O11(s)
Inorganic (carbide ion)
Organic
Organic
Inorganic (oxide)
Inorganic (cyanide)
Organic
Why is carbon special?

There are millions or organic compounds and only a thousand
inorganic compounds. WHY?
 Carbon has a bonding capacity of 4
 Remember Lewis Dot Diagrams from Chem 20??
 This means carbon can bond extensively and can bond
together to form chains effectively = called Polymerism
 Carbon covalently bonds by sharing 4 pairs of electrons.
These bonds may be single, double or triple, all producing
stable compounds
 Compounds can form with same number of each type of atom
but different structures = Isomerism
Determining Lewis Formulas

So why do we care about bonding capacity?
we know how many bonding e-’s an atom has, we can
predict what structure a molecular compound will have
 If
Atom
H
Number of
Number of
Bonding capacity
valence electrons bonding electrons
carbon
4
4
4
nitrogen
5
3
3
oxygen
6
2
2
halogens
7
1
1
hydrogen
1
1
1
I.e. Carbon can form 4 single bonds, 2 double bonds, 1 triple and 1 single, or 1 double and 2 singles
Polymers

Examples of repeating carbon chains:
Isomers


Compounds with the same number of each type of
atom but different structures (C4H10)
We will talk about this in more detail later
Importance of Organic Chemistry





Building units of all living matter: carbohydrates, proteins, fats
All foods are organic compounds
Photosynthesis is a reaction that makes carbon a part of our
food. Carbon is passed along through food chains and sugar
from photosynthesis is modified and combined with other
materials.
Dead organisms are food for other organisms, or are buried in
the earth and converted to fossil fuels like peat, coal and
petroleum
Petroleum is the source of fuel and starting material for plastics,
fabrics and industrial chemicals


The carbon cycle is
an illustration of the
interrelationship of
all living things with
the environment and
with technologies
that refine and use
fossil fuels
We will continually
outline the
importance of
organic compounds
in our daily lives
Carbon: The Element of Life
Complete the worksheet provided as you watch
the following video (20 min)
Today’s homework



Ensure video worksheet is complete and
in your notes to review later
Work on Are you Ready pg. 354-355
# 1- 6 – due tomorrow
What is coming up tomorrow?
 Naming
alkanes, branched alkanes and
cycloalkanes
Naming Organic
Compounds
Today’s Objectives:
1) Name and draw structural, condensed structural, and line diagrams and
formulas for saturated and unsaturated aliphatic (including cyclic)
• Containing up to 10 carbon atoms in the parent chain/cyclic structure
• Containing only one type of a functional group or multiple bond
• Using the IUPAC nomenclature guidelines
2) Identify types of compounds from the functional groups, given the structural
formula
3) Define structural isomerism and relate to variations in properties
Section 9.2 (pg. 366-374)
Four Types of Formulas
1.
Molecular Formulas
Not very useful for organic
compounds because so many
isomers can exist
2.
Structural Formulas
1.
Condensed Structural Formulas
2.
C5H10(g)
Line Diagrams
– end of line segment represents carbon
– it is assumed to satisfy each carbon’s octet
Naming Organic Compounds

Aliphatic Hydrocarbons – contains only hydrogen and carbon atoms


Straight line chains of carbon atoms
Alicyclic hydrocarbons have carbon atoms forming a closed ring. Still
considered aliphatic
Alkanes
Alkenes
Alkynes
Only single C-C
bonds
Double C-C Bond
present
Triple C-C bond
present
General formula
CnH2n+2
General formula:
CnH2n
General formula:
CnH2n-2
Saturated
Unsaturated
Unsaturated
Naming Organic Compounds

In organic chemistry, names
have a root and a suffix.
 The
root describes the number
of carbons present in the chain
or ring.
 The suffix describes the type of
compound it is.
Naming Organic Compounds
Naming ALKANES
1.
Find the parent chain (the longest continuous chain
of carbon atoms). Use the appropriate root and
the suffix-ane.
Naming ALKANES
1.
2.
3.
4.
Find the parent chain. Use the appropriate root and suffix.
Number the carbon atoms, starting from the end closest to the
branch(es) so that the numbers are the lowest possible
Identify any branches and their location number on the parent chain
(us the suffix –yl for branches)
Write the complete IUPAC name, following the format: (number of
location, if necessary) – (branch name) (parent chain)
2-methylheptane
Naming ALKANES


If more than one of the same branch exist, use a multiplier to
show this (di, tri). Remember to include all numbers
Draw 2,4,6-trimethylheptane
Naming ALKANES

If different branches exist, name them in alphabetical order
ethyl before methyl (e before m in the alphabet)
Naming ALKANES



If there is more than one
branch of the same type, a
locating number is given to
each branch and a prefix
indicating the number of that
type of branch is attached to
the name.
This numbering prefix does
not affect the alphabetical
order of the branches
Draw the structural formula
for 3,4-dimethylhexane
Summary of Naming Alkanes
1.
2.
3.
4.
5.
6.
Find the parent chain. Use the appropriate root and suffix.
Number the parent chain carbon atoms, starting from the end
closest to the branch(es) so that the numbers are the lowest
possible
Identify any branches and their location number on the
parent chain (us the suffix –yl for branches)
If more than one of the same branch exist, use a multiplier
(di, tri) to show this. Remember to include all numbers
If different branches exist, name them in alphabetical order
Separate numbers from numbers using commas, and
numbers from words using dashes (no extra spaces)
Don’t forget


Questions will specifically
ask about structural,
condensed structural or
line structural formulas.
You must be comfortable
drawing any of the three
Practice

Write the IUPAC name for the following
2,5-dimethyl-4-propyloctane
Correct the following names:

4-ethyl-2-methylpentane

ACTUALLY 2,4-dimethylhexane
 ALWAYS
LOOK FOR LONGEST CHAIN!!
Correct the following name:

4,5-dimethylhexane

Actually 2,3-dimethylhexane
 **Want
branch numbers to be as low as possible
CYCLOALKANES
Based on evidence, chemists believe that organic carbon compounds
sometimes take the form of cyclic hydrocarbons:

Cycloalkanes: Alkanes that form a closed ring

General Formula CnH2n




Two less hydrogens are present than in straight chain alkanes because
the two ends of the molecule are joined
Are these considered saturated?? Yes, because they have only single
bonds and the max amount of hydrogen's bonded to the carbons
Cyclo-compounds will have a higher boiling point than their straight
chain partners (because there is an additional bond present)
Naming CYCLOALKANES
Cycloalkanes are named by placing the prefix cyclo in front
of the alkane name, as in cyclopropane and cyclobutane
If branches are present, treat the cycloalkane as the parent
chain and identify the branches.



Since there is no end at which to start the numbering, use the
lowest numbers possible
Name the following:
1.
1,2-dimethylcyclopentane
2.
ethylcyclohexane
**Why don’t we need a number?
Today’s homework

Pg. 370 – 71 #7 – 11
Pg. 372 #5-6

What is coming up tomorrow?

 Naming
Alkenes, Alkynes, Cycloalkenes and
Cycloalkynes
 Comparing properties of Isomers
Naming Organic
Compounds:
Alkenes and Alkynes
Today’s Objectives:
1) Name and draw structural, condensed structural, and line diagrams and
formulas for saturated and unsaturated aliphatic (including cyclic
• Containing up to 10 carbon atoms in the parent chain/cyclic structure
• Containing only one type of a functional group or multiple bond
• Using the IUPAC nomenclature guidelines
2) Identify types of compounds from the functional groups, given the structural
formula
3) Define structural isomerism and relate to variations in properties
Section 9.3 (pg. 374-380)
Review:

Find and name all of the isomers of pentane (C5H12(l))
Structural Isomerism

Compound with the same molecular formula but
different structures
 They
will have different chemical and physical
properties – based on their different structures
Alkenes and Alkynes

Alkenes – hydrocarbons containing a double C-C bond
 General


-
(like cycloalkanes)
Alkynes – hydrocarbons containing a triple C-C bond
 General

formula (CnH2n)
formula (CnH2n-2) – (like cycloalkenes)
Alkenes and Alkynes are considered unsaturated
compounds. They do not have the maximum number of
hydrogen atoms surrounding each carbon.
Reactivity: Alkynes (highest), Alkenes, Alkanes (lowest)
Naming Alkenes and Alkynes
1.
2.
3.
Find the parent chain. It MUST contain the multiple bond.

If the bond is a double, the suffix for the parent chain will be -ene

If the bond is a triple, the suffix for the parent chain will be –yne
Count carbon atoms so that the multiple bond will be on the lowest
possible number. Indicate the number that the multiple bond falls on
directly before the suffix
Name branches as before
Naming Alkenes and Alkynes

Draw the following as condensed structural formulas:
4-methylpent-2-yne
methylpropene
(why don’t we need a number?)
Naming Alkenes and Alkynes

Name the following:
3-methylbut-1-ene
5-methylhex-2-ene
Naming Alkenes and Alkynes
4.
It is possible for a molecule to have more than one
double bond. These are called alkadienes and have
the same general formula as alkynes (CnH2n-2)

If this is the case, indicate both numbers where the
double bond is formed, and change the suffix to –diene.
a) Draw buta-1,3-diene:
b) What is the IUPAC name for the following:
buta-1,2-diene
Cycloalkenes and -ynes

The rules for naming cycloalkenes and cycloalkynes
are the same as naming cycloalkanes
 The
numbering for the carbon atoms begins with the
double bond; the carbons of the double bond are
carbons 1 and 2; lowest numbers possible
 Draw
3-methylcyclohexene
as a condensed structural
formula
Today’s homework

Pg. 377 #1-5
Pg. 380 #6, 7, 11

What is coming up tomorrow?

 Hydrocarbon
Quiz #1
 Naming Aromatics
 Boiling point and Chemical Properties Analysis
Naming Organic
Compounds:
Aromatics
Today’s Objectives:
1) Name and draw structural, condensed structural, and line diagrams and
formulas for aromatic carbon compounds
• Using the IUPAC nomenclature guidelines
2) Identify types of compounds from the functional groups, given the structural
formula
3) Define structural isomerism and relate to variations in properties
Section 9.4 (pg. 381-385)
Hydrocarbon Quiz #1
Aromatics


Originally, organic compounds with an aroma or
odour were called aromatic compounds
Now, aromatics refer to compounds containing a
benzene-ring structure
 Benzene’s
formula is C6H6, which would suggest a highly
unsaturated and reactive compound
 Benzene is actually quite unreactive and is considered
more stable than alkenes and alkynes
 Did You Know?? Benzene is a carcinogen and is found
naturally in petroleum – why would this be a problem?

What do we know about benzene?
 Formula
is C6H6 (3D link)
 Unreactive – so no true double or triple bonds
 Carbon-carbon bonds are the same length and strength
 Each carbon is bonded to a hydrogen
 So what does benzene look like??
The three double bonds resonate resulting in an overall
bond length somewhere in between a single and a
double bond, explaining benzene’s stability
We will use this line
structural formula to
represent benzene in
compounds
Common Aromatic Compounds

Include Aspirin and Vanillin (one of the flavour molecules in vanilla)
 You will notice many aromatic molecules are often depicted
using a condensed structural formula except for the benzene
ring, which is shown as a line structural formula.
 This combination is commonly used by chemists, and we will use
this method when drawing aromatics.
Naming Aromatics
If an alkyl branch is attached to a benzene ring, the
compound is named as an alkylbenzene.
1.

Alternatively, the benzene ring may be considered as a
branch of a large molecule: in this case, the benzene ring
is called a phenyl branch. Which has a phenyl branch?
An alkylbenzene
Contains a phenyl branch
Naming Aromatics
If more than one alkyl branch is attached to a benzene
ring, the branches are numbered using the lowest
numbers possible, starting with one of the branches.
2.

Given the choice between two sets of lowest numbers, choose the
set that is in both numerical and alphabetical order
1-ethyl-2,4-dimethylbenzene
3-phenyl-4-propyloctane
Practice Naming Aromatics
Draw 1,2-dimethylbenzene


Are there any isomers of this compound?
1,2-dimethylbenzene
o-dimethylbenzene

1,3-dimethylbenzene
m-dimethylbenzene
1,4-dimethylbenzene
p-dimethylbenzene
There is also classical way of naming these isomers. The
arrangements are denoted by the prefixe:s ortho (o), meta (m) and
para (p). These names are still used in industry so you may
encounter them in other references.
Practice Naming Aromatics

Draw the line structural formula for 1-ethyl-3-methylbenzene

Draw the line structural formula for 2-phenylpentane
Practice Naming Aromatics

Draw 3-phenylpent-2-ene
Are the hydrogen’s
wrong??

Name the following
propylbenzene
Why is no number needed?
Summary

We have now learned about both aliphatic and
aromatic hydrocarbons. You will need to be
comfortable naming all of the following:
Teacher Note:


End if not enough time for two dry labs
Can use during the Station Lab Periods
Review of Intermolecular Forces

London Forces – temporary dipoles resulting from an
uneven distribution of e- in all molecules
Temporary (-) end will repel e- in neighbouring molecules
and so on
 Depends on size of molecule (number of e-’s)
 Weakest of the intermolecular forces


Dipole-Dipole – only exists in polar molecules


Attraction between + and – ends of molecule
Hydrogen Bonding – super strong force

Only exists when H-N, H-F, H-O bonds are present
Applications

Read pg. 384 Lab Exercise 9.A – complete the parts in red
 Purpose: To test the generalization that aromatic hydrocarbons
react like saturated rather than unsaturated hydrocarbons
 Design: cyclohexane, cyclohexene and benzene are all mixed
with potassium permanganate (purple). Evidence for a reaction
is a change in the initial purple colour of the solution.
 Prediction: Based on your current knowledge, predict the order
in which the compounds will react, from least reactive to most
reactive. Explain your reasoning.
 Analysis: On the basis of the evidence, determine the order of
the reaction rate for the three compounds.
 Evaluation: Determine if your prediction was verified or
falsified. Was the generalization about aromatic hydrocarbons
acceptable based on the evidence? Was the purpose of the
investigation accomplished?
Sample lab report for 9.A
Applications

Read pg. 384 Lab Exercise 9.B – complete the parts in red
 Purpose: To test the ability of the concept of London forces to predict
the relative boiling points of aliphatic and aromatic compounds





(Remember: the more electrons in a compound = the stronger the
intermolecular forces = the higher the boiling point needed to pull the
molecules apart)
Problem: What is the relative order of the boiling points of hexane,
hex-1-ene, cyclohexane, cyclohexene, and benzene?
Prediction: Determine the number of electrons in each molecule and
use these numbers to determine the order of boiling points
Analysis: On the basis of the evidence given, determine the order of
the boiling points. (from lowest to highest)
Evaluation: Determine if your prediction was verified or falsified.
Was the predictive power of the concept of London forces judged to
be acceptable based on the evidence? Was the purpose of the
investigation accomplished?
Extra Practice for Lab Reports

Complete pg. 355 #6
 Complete
the prediction, analysis and evaluation
 Remember lab report guidelines in textbook starting on
pg. 790
Today’s homework

Pg. 385 #3-5 – due tomorrow

What is coming up tomorrow?
 Review
Aromatic and Aliphatic Compounds
Organic Chemistry:
Organic Halides and Alcohols
Today’s Objectives:
1) Name and draw structural, condensed structural and line diagrams and
formulas for organic halides and alcohols
2) Identify types of compounds from their functional groups, given the structural
formula and name of the functional groups
Section 10.2 and 10.3 (pg. 417-435)
Today’s Agenda: Review


You will have another Hydrocarbon Quiz next
class covering all aliphatic and aromatic
compounds
Today’s agenda:




Review Quiz – go over common mistakes
Homework Book page 3 and 4
Pg. 380 #6 (c-e), 7 (c-e)
Pg. 385 #6-8
**7d) 4-methyl (not 3-methyl)

Make the change in your book if not done
already
Hydrocarbon Quiz #2
Aliphatic and Aromatic Compounds
Review of Intermolecular Forces

London Forces – temporary dipoles resulting from an
uneven distribution of e- in all molecules
Temporary (-) end will repel e- in neighbouring molecules
and so on
 Depends on size of molecule (number of e-’s)
 Weakest of the intermolecular forces


Dipole-Dipole – only exists in polar molecules


Attraction between + and – ends of molecule
Hydrogen Bonding – super strong force

Only exists when H-N, H-F, H-O bonds are present
Organic Halides

Organic compounds where one or more hydrogen
has been replaced with halogens (F, Cl, Br, I)
 Common

example: CFC (chlorofluorocarbons)
Nomenclature is similar to naming branch chains of
hydrocarbons, but the branch name used is based
on the halogen used
 chloro-,
fluoro-, bromo-, iodo-
Organic Halides

What do you need to know about organic halides?
 May
by polar or nonpolar molecules or may have a
relatively nonpolar (hydrocarbon) end and a polar
(halide) end (**Remember Electronegativity differences)
 Have higher boiling points than similar hydrocarbons
 Have very low solubility in water but higher solubility
than similar hydrocarbons
 Are typically good solvents for organic materials such
as fats, oils, waxes, gums, resins or rubber
 Usually toxic or ecologically damaging (DDTs and PCBs)
Practice Naming Organic Halides

Draw 1,2-dichloroethane

Draw 2,2,5-tribromo-5-methylhexane
Practice Naming Organic Halides

Name the following:
CH2Cl2
dichloromethane
1,2-dibromoethene
Bonus: Try 1,2-dibromo-1,2-dichloroethene
chlorobenzene
Did You Know?


There is always a
new concept to
learn that extends
what you have
already learned.
In addition to the
structural isomers
that you know
about, there are cis
and trans isomers
Stop and Practice

Homework Book pg. 5 and 6
Alcohols


An alcohol is an organic compound that contains the –OH
functional group (hydroxyl)
 General formula is R-OH (R = rest of molecule)
Alcohols are classified as primary, secondary or tertiary
depending on the number of carbons bonded to the carbon
that contains the hydroxyl group
Common Alcohols



Methanol (also called wood alcohol) is extremely toxic, causing
death and blindness
Ethanol (also known as grain alcohol) is the alcohol found in
alcoholic beverages and is used in the production of vinegar
Gas line antifreeze,
windshield de-icer,
windshield washer
fluid – all contain
methanol
Naming Alcohols
1.
2.
3.
Locate the longest chain that contains an –OH group attached to
one of the carbon atoms. Name the parent alkane
Replace the –e at the end of the name of the parent alkane with
–ol (i.e. butane becomes butanol)
Add a position number before the suffix –ol to indicate the
location of the –OH group

REMEMBER to number the main chain of the hydrocarbon so that
the hydroxyl group has the lowest possible position number
propan-1-ol
Naming Alcohols
If there is more than one –OH group (called polyalcohols), leave
the –e in the name of the parent alkane and put the appropriate
prefix before the suffix –ol (i.e. diol, triol, tetraol)
4.
Name and number any branches on the main chain. Add the
names of these branches to the prefix.
4.

Draw 2,3-dimethylbutan-2-ol
Practice Naming Alcohols

Draw line structural formulas for:
These are the
only two cyclic or
aromatic alcohols
you will need to
know as they get
very complicated
1.
cyclohexanol
2.
phenol
1.
The three isomers of C5H11OH that are pentanols
Alcohols

What do you need to know about alcohols?
Question: Why is the propane used in a barbecue a gas at
room temperature, but propan-2-ol (also known as rubbing
alcohol) a liquid at room temperature?
Answer: Propane is a non-polar hydrocarbon with weak
intermolecular forces, thus it has a low boiling point and is a
gas at room temperature.
Propan-2-ol is an alcohol, with a polar hydroxyl group and
strong intermolecular forces, thus it has a higher boiling point
than propane and is a liquid at room temperature
Alcohols

What do you need to know about alcohols?
Question: Glycerol (propane-1,2,3-triol) is more viscous than
water, and can lower the freezing point of water. When
added to biological samples, it helps to keep the tissues from
freezing, thereby reducing damage. From your knowledge of
the molecular structure of glycerol, suggest reasons to account
for these properties of glycerol.
Answer: Each molecule of glycerol contains three hydroxyl
groups which can hydrogen-bond with water, interfering with
the attractions between water molecules and thus interfering
with the freezing of water. When water in tissues does not
freeze, there is less damage to the tissues.
Application Question


Predict the order of increasing boiling points for the
following compounds, and give reasons for your answer.
butan-1-ol
pentane
1-chlorobutane
Answer: (Lowest b.p.): pentane, 1-chlorobutane, butan-1-ol
Why? All molecules have a similar number of electrons.
Pentane has the lowest boiling point, because it is non-polar so
will only have London forces between the molecules. 1chlorobutane is polar so will have dipole-dipole forces as well
as London forces. Butan-1-ol has the highest boiling point
because its molecules will have all three intermolecular forces,
most importantly, hydrogen bonding
Today’s homework




Homework Book pg. 5 and 6
Pg. 419 #3
Pg. 430 #5,6,7a,9a
 Due tomorrow
What is coming up tomorrow?
 Carboxylic
Acids and Esters
Organic Chemistry:
Carboxylic Acids and Esters
Today’s Objectives:
1) Name and draw structural, condensed structural and line diagrams and
formulas for carboxylic acids and esters
2) Identify types of compounds from their functional groups, given the structural
formula and name of the functional groups
Section 10.4 (pg. 436-443)
Carboxylic Acids

A carboxyl group is composed of a carbon atom double
bonded to an oxygen atom and bonded to a hydroxyl
group (-COOH)
 Note: Because the carboxyl group involves three of the
carbon atom’s four bonds, the carboxyl is always at the end
of a carbon chain or branch
Examples:
methanoic acid
ethanoic acid
Carboxylic acids are
weak organic acids
Naming Carboxylic Acids
1.
2.
3.
Name the parent alkane
Replace the –e at the end of the name of than parent
alkane with –oic acid
The carbon atoms of the carboxyl group is always given
position number 1. Name and number the branches that
are attached to the compound.
Draw 3-methylbutanoic acid
Remember COOH or HOOC
can also represent the
carboxyl group
HOOC
Naming Carboxylic Acids
1.
Draw trichloroethanoic acid (key ingredient in chemical peels)
CCl3COOH
2.
Draw 3-propyloctanoic acid
Do you need the CH3 here?
No – but sometimes you will
see it written this way. Don’t
be confused because it doesn’t
change the meaning
Esters

The reaction between a carboxylic acid and an alcohol produces
an ester molecule and a molecule of water

This reaction is known as a condensation or esterification reaction

The ester functional group –COO– is similar to that of a carboxylic
acid, except that the H atom of the carboxyl group has been
replaced by a hydrocarbon branch.
Esters are responsible for natural and artificial fragrance and
flavourings in plants and fruits.

Esters

In naming an ester you have to recognize that an ester
has 2 distinct parts. The main part contains the C=O
group which comes from the parent acid. The second
part is the alkyl group.
Naming Esters
1.
2.
3.
4.
Identify the main part of the ester, which contains the C=O group.
This part comes from the parent acid.
Begin by naming the parent acid but replace the –oic acid ending
of the name with –oate. (propanoic acid becomes propanoate)
The second part is the alkyl group that is attached to the single
oxygen atom. Name this as you would any other alkyl group (in
this case = methyl)
Put the names together. Note that esters are named as two words.
Naming Esters

Name the following ester and the acid and alcohol from
which it can be prepared.
butanoic acid
ethanol
A strong acid catalyst,
such as H2SO4(aq) is used
along with some heating
to increase the rate of the
organic reaction
ethyl butanoate
water
Tip: The branch attached to the oxygen (of the –COO) comes first in the
name, the chain attached to the carbon (of the –COO) comes second
Naming Esters

Name the following ester and the acid and alcohol from
which it can be prepared.
benzoic acid
ethanol
ethyl benzoate
What is missing from this esterification reaction?
water
Today’s homework

Pg. 438 #1,2
Pg. 441 #3-5, 6(a-c only)
 Due tomorrow

What is coming up tomorrow?

 Physical
Properties of Organic Compounds
 Crude Oil Refining
 Full Naming Quiz (in two days)
Today’s Objectives:
Properties of Organic
Compounds & Crude Oil
Refining
1) Compare boiling points and solubility of organic compounds
2) Describe fractional distillation and solvent extraction
3) Describe major reactions for producing energy and economically important
compounds from fossil fuels
Section 9.5 and 9.6 (pg. 386-400)
Physical Properties of Simple Hydrocarbons
Alkanes
Non-polar molecules
Only intermolecular forces are London Force
Boiling point and melting point increase with number of carbons
All insoluble in water (like dissolves like) – nonpolar and polar don’t mix
1-4Cs = gas, 5-16Cs = liquid 17 and up = solid at SATP
Alkenes
Non-polar molecules, therefore insoluble in water
Boiling points slightly lower than alkanes with the same number of carbons
due to less electrons (unsaturated), resulting in lower London Forces
Alkynes
Non-polar molecules, therefore insoluble in water
Higher boiling points than alkanes and alkenes with similar C #s
Accepted explanation: Linear structure around triple bond allows electrons to
come closer together than in alkanes/enes, resulting in greater London Force
Branching
The more branching, the less significant the London Force (~lower b.p.)
- more surface area in straight chain hydrocarbons allows more
separation of charge, resulting in greater London Force
- see Table #3 pg. 378 (i.e. pentane (with 5Cs) has a b.p. of 36oC
which is much higher than dimethylpropane (5Cs) -12oC) = because
branching decreased the strength of the London force
Physical Properties of Hydrocarbon Derivatives
Alcohols
Much higher boiling points than hydrocarbons (1-12Cs are liquids at SATP)
due to hydrogen bonding between hydroxyl groups of adjacent molecules
Small alcohols are totally miscible in water, but the larger the hydrocarbon
part of the alcohol (nonpolar part), the more nonpolar the alcohol is
Carboxylic
Acids
Like alcohols they have hydrogen bonding, but is more significant due to the
C=O. This means greater bps and solubility than alcohols with same number
of Cs.
Carboxylic acids with 1-4Cs are
completely miscible in water
Esters
Compound
Boiling Point (oC)
butane
-0.5
butan-1-ol
117.2
butanoic acid
165.5
Fruity odour in some cases
Polar but they lack the –OH bond therefore do not have hydrogen bonding, so
lower bps than both alcohols and carboxylic acids
Esters with few carbons are polar enough to be soluble in water
Summary Table - Organic
In your homework book (pg. 7), there is a summary table
of all of the organic compounds we have studied. This
will be a good page to reference when studying
Sample Question

Predict the relative order of boiling points of the following
compounds (lowest to highest). Explain your reasoning.
butanol
but-1-ene
cyclobutane
butanoic acid
butane
Lowest -------------------------------------------------------------> Highest
but-1-ene
butane
cyclobutane
butanol
butanoic acid
Reasoning: but-1-ene has lower LF’s for than butane because it is
unsaturated, cyclobutane has an additional bond because cyclic,
butanol has H-bonding, butanoic acid has stronger H-bonding)

Which would be soluble in water?

Butanol and butanoic acid – because they are the only polar molecules
and like dissolves like!
Crude Oil Refining

Crude oil is a complex mixture of hundreds of thousands
of compounds, all of which have different boiling points
We can take advantage of these different b.p.’s and
physically separate the different components using heat
 This process is called fractional distillation or fractionation

Electronic Visual




A fractional distillation tower
contains trays positioned at
various levels.
Heated crude oil enters near
the bottom of the tower.
The bottom is kept hot, and the
temperature gradually
decreases toward the top of
the tower.
As compounds cool to their
boiling point, they condense in
the cooler trays. The streams of
liquid (called fractions) are
withdrawn from the tower at
various heights along the tower.

A more detailed look…





The vaporized components of the
crude oil rise and gradually cool.
To get from one level to the next, the
vapours are forced to bubble through
the liquid condensed in each tray.
The figure shows the bubble caps used
to allow this to happen.
If a gas cools to its boiling point, it will
condense and be piped out through
the draining tube
Crude oil is heated in the
fractionation tower without air being
present to reduce the risk of mixtures
starting to burn or explode
Q: How does the number of carbon atoms in a hydrocarbon chain
affect its boiling point?
 Smaller molecules have fewer electrons, so weaker London forces
compared with larger molecules. The fractions with higher boiling points
are found to contain much larger molecules (see Table 1 pg. 387)
Cracking

Cracking: large hydrocarbons are broken into smaller fragments

Historically, thermal cracking used extremely high temperatures but
created large quantities of solid coke.

Now, catalytic cracking uses a catalyst to speed up the reaction and
produce less residual products like tar, asphalt and coke

Example: C17H36(l)  C9H20(l) + C8H16(l) + C(s)
larger molecules  smaller molecules + carbon

In 1960, hydrocracking was developed, which combines catalytic
cracking and hydrogenation and produces no coke.

Example: C17H36(l) + H2(g)  C9H20(l) + C8H16(l)
larger molecule + hydrogen  smaller molecules
Oil Refining

The refining of crude oil can be divided into two main categories:
Physical Processes
1.
 Fractional
Distillation: see previous slides
 Solvent Extraction: solvent is added to selectively dissolve and
remove an impurity or to separate a useful product from a
mixture
Chemical Processes
2.



Cracking – larger molecules are broken down into smaller ones
Reforming – large molecules are formed from smaller ones
These chemical processes are needed because fractional distillation
does not produce enough of the hydrocarbons that are in demand
(i.e. gasoline) and produces too much of the heavier fractions
Catalytic Reforming and Alkylation


Catalytic Reforming: improves the quality of the gasoline
aliphatic molecule  aromatic molecule + hydrogen
Alkylation: increases the branching; improves the quality of the fuel
aliphatic molecule  more branched molecule

(AKA: isomerization because it converts molecules into a branched isomer)
FYI pg. 392 on
Octane Numbers
Combustion Reactions

Burning of hydrocarbons in the presence of oxygen

Complete Combustion: abundant supply of oxygen;
products are carbon dioxide, water vapour and heat
 Ex. C3H8(l) + 5O2(g)  3CO2(g) + 4H2O(g)

Incomplete Combustion: limited supply of oxygen; products
are carbon monoxide, soot (pure carbon) or any
combination of carbon dioxide, carbon monoxide and soot
in addition to water vapour and heat
 Ex. 2C8H18(l) + 17O2(g)  16CO(g) + 18H2O(g)
 OR
2C8H18(l) + 9O2(g)  16C(s) + 18H2O(g)
** Assume complete combustion unless specified otherwise
Balancing FYI
Ex. 2C8H18(l) + 17O2(g)  16CO(g) + 18H2O(g)
Can also be balanced using a fraction (you need to be
comfortable using this method) – divide each number by 2
C8H18(l) + 17/2 O2(g)  8CO(g) + 9H2O(g)
Ex.
2C8H18(l) + 9O2(g)  16C(s) + 18H2O(g)
can also be balanced as …
C8H18(l) + 9/2 O2(g)  8C(s) + 9H2O(g)
Today’s homework

Read pages 410-416
Pg. 388 #2
Pg. 391 #11
Pg. 397 #4 b, c, d only
Pg. 430 #11

What is coming up tomorrow?




 Hydrocarbon
 Use
Quiz #3 (all organic compounds)
Homework Book pg. 8 as practice quiz
 Hydrocarbon
 Addition,
Reactions
Substitution and Elimination
Hydrocarbon Reactions:
Addition, Substitution and
Elimination
Today’s Objectives:
1) Define, illustrate and provide examples of simple addition, substituion,
elimination and esterification (condensation)
2) Predict products and write and interpret balanced equations for the above
reations
Section 10.2,10.3 and 10.4 (pg. 419-444)
Hydrocarbon Quiz #3
All Organic Compounds
Hydrocarbon Reactions
1.
Addition
alkenes and alkynes + H2(g)  alkanes (hydrogenation)
alkenes and alkynes + HX (or X2)  organic halides
2.
Substitution
alkanes and aromatics + X2  organic halides
3.
Elimination
alcohols  alkenes + water (dehydration)
organic halides + OH-  alkenes + halide ion + water
4.
Esterification (already covered)
carboxylic acid + alcohol  ester + water
Addition Reactions: reaction of alkenes and alkynes with hydrogen
gas, a halogen compound, or a hydrogen halide compound.
1.

a)
Addition reactions usually occur in the presence of a catalyst
Addition with H2(g) (also called hydrogenation)
Addition Reactions: reaction of alkenes and alkynes with hydrogen
gas, a halogen compound, or a hydrogen halide compound
1.
b)
Addition of a halogen
1,2-dichloroethane is a useful
solvent and is used to
produce chloroethene, the
monomer used to make PVC.
See Pg. 417 equation
Since addition reactions
involving multiple bonds are
very rapid, the alkene
product 1,2-dibromoethene
can easily undergo a second
addition step to produce
1,1,2,2-tetrabromomethane.
Excess bromine promotes
this second step.
Addition Reactions: reaction of alkenes and alkynes with hydrogen
gas, a halogen compound, or a hydrogen halide compound
1.
Addition of an HX (hydrogen halide) molecule
c)

Show both possible isomers when predicting the products
The addition of hydrogen halides (HF, HCl, HBr or HI) to unsaturated
compounds can produce structural isomers, since the hydrogen halide
molecules can add in different orientations. However, the type of isomers
produced are not always equal.
For example only a tiny amount of 1-chloropropane is produced
compared to 2-chloropropane in the above reaction.
Practice Addition Reactions

For each of the following questions, draw condensed structural
diagrams and name all products
+ bromine

1.
ethene
2.
hydrogen chloride + ethene 
3.
3-methylbut-1-yne + excess hydrogen 
__
1,2-dibromoethane
chloroethane

2-methylbutane
Substitution Reactions – breaking of a C-H bond in an alkane
or an aromatic ring and replacing it with another atom or
group of atoms
2.



Usually occur slowly at room temperature, so light may be
necessary as a catalyst
Often substitutes a halogen for a hydrogen
No change in saturation
Propane contains hydrogen atoms bonded to end carbons and the middle carbon
atom, so two different products (isomers) are formed, in unequal proportions
2.
Substitution Reactions – breaking of a C-H bond in an alkane
or an aromatic ring and replacing it with another atom or
group of atoms
Benzene rings are stable, like alkanes, so they react slowly with halogens,
even in the presence of light.
Would all three isomers be created equally?
NO
As with alkanes, further substitution can occur in benzene rings, until all hydrogen
atoms are replaced by halogen atoms (in the presence of excess halogens)
Practice Substitution Reactions

1.
For each of the following questions, draw condensed structural
diagrams and name all products
propane + fluorine

F-F
2.
ethane
+
chlorine
1-fluoropropane + 2-fluoropropane +
hydrogen fluoride
H-F
 chloroethane + hydrogen chloride
3.
Elimination Reactions – involves eliminating atoms or groups of atoms
from adjacent carbon atoms; decreases the level of saturation
a)
Alkane cracked into an alkene (uses high temperatures)
b)
Alcohol is reacted with a catalyst to produce an alkene and water
(dehydration – removes a water molecule from the alcohol)
a)
Alkyl halide reacts with a hydroxide ion (OH-) to produce an alkene
(dehydrohalogenation – removes a hydrogen and halogen atom)
Practice Elimination Reactions


Write a structural formula equation for the preparation of but2-ene from chlorobutane, in the presence of a strong base
Write a structural formula equation for the preparation of but2-ene from butan-2-ol
Reaction
type
Complete
Combustion
Addition
Reactants
Hydrocarbon* ene or yne +
HOH, X2, HX,
+ O2
or H2
Products
CO2(g) +
H2O(g) or (l)
Elimination
Substitution
Esterification
alcohol or
alkyl halide
alkane or
alcohol +
aromatic + X2 carboxylic
acid
alcohol, alkyl alkene + HOH alkyl halide
halide*,
or HX
alkane
ester +
water
Other
triple or
double to
single
single to
double
acid
catalyst
needed
Other
more bonds*
fewer bonds*
slow; needs
uv light
Today’s homework

Homework Book Pg. 9
Pg. 422 #7,8 (Addition and Substitution)
Pg. 433 #18-19 (Elimination)

What is coming up tomorrow?


 Polymerization
 STS
Connections
Organic Chemistry:
Polymerization Reactions
Today’s Objectives:
1) Define, illustrate, and give examples of monomers, polymers, and
polymerization in living and non-living systems
STS: 2) Illustrate how science and technology are developed to meet societal
needs and expand human capabilities
STS: 3) Illustrate how science and technology have both intended and
unintended consequences
Section 10.5 (pg. 445-459)
Polymerization

Polymers are large molecules made of
chains of monomers, small molecules that
link together.
 Polymerization is the formation of
polymers from these small units

Polymers can occur naturally (proteins,
carbohydrates) and can be
synthesized (nylon, Teflon, polyethylene)

They play an integral part in the
function of life systems and have
revolutionized the way society functions
Addition Polymerization


Many plastics (synthetic polymers) are made by this process
The polymerization process is initiated with a free radical (a species
with an unpaired electron). The free radical attacks and breaks the
double bond forming a new free radical that attacks another monomer



Addition Polymerization always results in one product, the polymer
Requires unsaturated hydrocarbon monomers and bond saturation
occurs when the polymer is made
Common polymers produced by addition polymerization:
Things to know about addition polymers…


The polymer names end in –ene (i.e. polystyrene, polypropene). Does this
mean they have double bonds?
 No, the double bonds are saturated by adjacent monomers, as the
polymer is made. The name refers to the starting monomer (i.e.
polyethene is started by the monomer ethene)
What properties make Teflon a good product for non-stick materials?
 Teflon is made up of C-F bonds which are very strong (not C-H bonds).
These very strong bonds make the Teflon highly unreactive (non-sticking),
it has a high melting point and it has a slippery surface
Condensation Polymerization


Monomers combine to form a polymer and a bi-product. Each
time a bond forms between monomers, small molecules, such as
water, ammonia, or HCl are “condensed” out.
The polymerization of nylon:
• For condensation polymerization
to occur, monomers must be
bifunctional, meaning they have
at least two functional groups.
• If they only had one functional
group, then only one bond would
form.
Condensation Polymerization


Condensation polymerization also produces natural polymers,
called proteins.
Amino acids (monomers) polymerize to make peptides (short
chains of amino acids) or proteins (long chains of amino acids)
Comparison of Addition and
Condensation Polymerization
Addition


Needs a double or
triple bond in the
monomer
Produces only one
product, the polymer
Condensation


Needs bifunctional
monomers (have two
functional groups)
Produces two products:
the polymer and the
biproduct (water,
ammonia or HCl)
Polyester


When a carboxylic acid reacts with an alcohol in an esterification
reaction, a water molecule is eliminated and a single ester molecule
is formed.
This esterification reaction can be repeated so many esters are
joined in a long chain… a polyester


This is created using a dicarboxylic acid (an acid with a carboxyl group
at each end) and a diol (an alcohol with a hydroxyl group at each end)
The ester linkages are formed end to end between alternating acid and
alcohol molecules
Polyester: Dacron

Another example of a polyester:

Note the two carboxyl groups in the dicarboxylic acid and
the two hydroxyl groups in the polyalcohol that start the
chain reaction
Natural Polymer Examples

Starch, wood, silk, DNA
Today’s homework



Pg. 448 #2-4
Pg. 452 #13
Pg. 455 #16, 17


Due tomorrow
What is coming up tomorrow?
 Stations
Lab – will take two days
Learning Tip Pg. 441
Unit Exam
Chemistry 30 Organic Review