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Organic Chemistry: Organic Chemistry: IntroductionIntroduction
IB Topic 10IB Topic 10
10.1 Introduction10.1 Introduction
10.1.110.1.1 Describe the features of a homologous series. Describe the features of a homologous series. 10.1.210.1.2 Predict and explain the trends in boiling Predict and explain the trends in boiling
points of members of a homologous series.points of members of a homologous series.10.1.310.1.3 Distinguish between empirical, molecular and Distinguish between empirical, molecular and
structural formulas.structural formulas.10.1.410.1.4 Describe structural isomers as compounds Describe structural isomers as compounds
with the same molecular formula but with different with the same molecular formula but with different arrangement of atoms.arrangement of atoms.
10.1.510.1.5 Deduce structural formulas for the isomers of Deduce structural formulas for the isomers of non-cyclic alkanes up to Cnon-cyclic alkanes up to C66..
10.1.6 Apply IUPAC rules for naming the isomers of 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to Cthe non-cyclic alkanes up to C66..
What is organic chemistry?What is organic chemistry?
Organic ChemistryOrganic Chemistry The study of carbon, the compounds The study of carbon, the compounds
it makes and the reactions it it makes and the reactions it undergoes.undergoes.
Over 16 million carbon-containing Over 16 million carbon-containing compounds are known.compounds are known.
What is organic chemistry?What is organic chemistry?
CarbonCarbon Carbon can form multiple bonds to itself Carbon can form multiple bonds to itself
and with atoms of other elements.and with atoms of other elements. Carbon can only make four bonds since Carbon can only make four bonds since
it has 4 valence electrons and most it has 4 valence electrons and most often bonds to H, O, N and S.often bonds to H, O, N and S.
Because the C-C single bond (348 kJ molBecause the C-C single bond (348 kJ mol--
11) and the C-H bond (412 kJ mol) and the C-H bond (412 kJ mol-1-1) are ) are strong, carbon compounds are stable.strong, carbon compounds are stable.
Carbon can form chains and rings.Carbon can form chains and rings.
What is organic chemistry?What is organic chemistry?
HydrocarbonsHydrocarbons Hydrocarbons are organic compounds Hydrocarbons are organic compounds
that only contain carbon and that only contain carbon and hydrogenhydrogen
Types of hydrocarbons includeTypes of hydrocarbons include AlkAlkaanesnes AlkAlkeenesnes AlkAlkyynesnes AromaticAromatic
10.1.110.1.1 Describe the features of a Describe the features of a homologous series.homologous series.
A homologous series is a series of related compounds that have the same functional group.
Differ from each other by a Differ from each other by a CHCH22 unit unit Can be represented by a general Can be represented by a general
formulaformula examples:examples:
CCnnHH2n+2 2n+2 (alkanes) or C (alkanes) or CnnHH2n 2n (alkenes) or…(alkenes) or…
10.1.110.1.1 Describe the features of a Describe the features of a homologous series.homologous series.
Watch out!!! Homologous compounds DO NOT have the
same empirical formula!
• Have similar Have similar chemical chemical propertiesproperties• Have Have physicalphysical properties that vary properties that vary
in a in a regular manner regular manner as the number as the number of carbon atoms increasesof carbon atoms increases– Example: the alkanesExample: the alkanes
# C# C PrefixPrefix Alkane (ane) Alkane (ane)
CCnnHH2n+22n+2
Alkene (ene)Alkene (ene)
CCnnHH2n2n
11 methmeth CHCH44 methanemethane
22 etheth CC22HH66 ethaneethane CC22HH44 etheneethene
33 propprop
44 butbut
55 pentpent
66 hexhex
10.1.210.1.2 Predict and explain the trends in Predict and explain the trends in boiling points of members of a homologous boiling points of members of a homologous
series.series.
What is the trend?
Why?
AlkaneAlkane FormulFormulaa
Boiling Boiling Pt./Pt./ooCC
methanmethanee
CHCH44 -162.0-162.0
ethaneethane CC22HH66 -88.6-88.6
propanpropanee
CC33HH88 -42.2-42.2
butanebutane CC44HH1010 -0.5-0.5
Trends in boiling points of members of Trends in boiling points of members of a homologous series a homologous series (10.1.2)(10.1.2)
• Melting point and Melting point and boiling point boiling point increase with more increase with more carbon atomscarbon atoms
• Why?Why?– intermolecular intermolecular
forces increaseforces increase– adding a CHadding a CH22 adds adds
more electronsmore electrons• this increases the this increases the
Van der Waal’s Van der Waal’s forcesforces
AlkaneAlkane FormulFormulaa
Boiling Boiling Pt./Pt./ooCC
methanmethanee
CHCH44 -162.0-162.0
ethaneethane CC22HH66 -88.6-88.6
propanpropanee
CC33HH88 -42.2-42.2
butanebutane CC44HH1010 -0.5-0.5
10.1.210.1.2 Predict and explain the trends in Predict and explain the trends in boiling points of members of a homologous boiling points of members of a homologous
series.series.
Intermolecular forces presentIntermolecular forces present Simple alkanes, alkenes, alkynes → van der
Waals’ forces (nonpolar) → lower b.p. Aldehydes, ketones, esters & presence of
halogens (polar) → dipole: dipole forces → slightly higher b.p.
Alcohol, carboxylic acid & amine → hydrogen bonding (w/ O, N, F) → even higher b.p.
Naming song: https://www.youtube.com/watch?v=mAjrnZ-znkY
10.1.210.1.2 Predict and explain the trends in Predict and explain the trends in boiling points of members of a homologous boiling points of members of a homologous
series.series.
10.1.310.1.3 Distinguish between empirical, Distinguish between empirical, molecular and structural formulas.molecular and structural formulas.
Empirical Formula:Empirical Formula:Smallest whole number ratio of
atoms in a molecule
Molecular Formula:Molecular Formula:Formula showing the
actual numbers of atoms
Molecular Molecular FormulaFormula
Empirical Empirical FormulaFormula
CHCH44 CHCH44
CC22HH66 CHCH33
CC66HH1212OO66
CC44HH88
CC88HH1616
10.1.310.1.3 Distinguish between Distinguish between empirical, molecular and structural empirical, molecular and structural
formulas.formulas.Structural FormulaStructural Formula
Bond angles are drawn as though 90o. The true shape around C with 4 single bonds is tetrahedral and the angle is 109.5o.
Show every atom and every bond. Can use condensed structural formulas.
Hexane: CH3CH2CH2CH2CH2CH3 (condensed s.f.) M.F. = C6H14 E.F. = C3H7
Structural formulaStructural formula Structural formulaStructural formula
Can use Can use condensedcondensed structural formulas structural formulas bonds are omitted, bonds are omitted, repeated groups repeated groups
put togetherput together, , side chains put in side chains put in bracketsbrackets CHCH33CHCH22CHCH22CHCH22CHCH22CHCH33
or even CHor even CH33(CH(CH22))44CHCH33 CHCH33CHCH(CH(CH33))CHCH33
10.1.310.1.3 Distinguish between empirical, Distinguish between empirical, molecular and structural formulas.molecular and structural formulas.
Skeletal formulaSkeletal formula– Not accepted in the IB for answers but often Not accepted in the IB for answers but often
used in questions… cuz that’s how they do it used in questions… cuz that’s how they do it – Every “corner” represents a carbonEvery “corner” represents a carbon– Hydrogens are impliedHydrogens are implied
10.1.410.1.4 Describe structural isomers as Describe structural isomers as compounds with the same molecular compounds with the same molecular
formula but with different arrangement of formula but with different arrangement of atoms.atoms.
Structural isomersStructural isomers: : compounds with the same molecular formula, but different arrangement of atoms
The Fuse School: https://www.youtube.com/watch?v=9SX-iWpi98g
10.1.410.1.4 Describe structural isomers as Describe structural isomers as compounds with the same molecular compounds with the same molecular
formula but with different arrangement of formula but with different arrangement of atoms.atoms.
• Different isomers are completely different Different isomers are completely different compoundscompounds
• Have different physical properties such as Have different physical properties such as melting point and boiling pointmelting point and boiling point
Structural Formulas
for C4H10O Isomers
10.1.510.1.5 Deduce structural formulas for Deduce structural formulas for the isomers of non-cyclic alkanes up to Cthe isomers of non-cyclic alkanes up to C66..
You should be able to draw out and write the structural formulas for all isomers that can be formed by: CH4
C2H6 C3H8
C4H10
C5H12
C6H14
Eventually you should be able to name all isomers, as well
10.1.510.1.5 Deduce structural formulas for Deduce structural formulas for the isomers of non-cyclic alkanes up to Cthe isomers of non-cyclic alkanes up to C66..
If there is a branch off of the main chain, put that formula in parentheses
CH3CH(CH3)CH3
CH3CH2CH2CH3
10.1.6 Apply IUPAC rules for naming 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes the isomers of the non-cyclic alkanes
up to Cup to C66..
1. Determine the longest carbon chain2. Use the prefix (next slide) to denote the
number carbons in the chain
11 MMeth-eth-
22 EEth-th-
33 PProp-rop-
44 BBut-ut-
55 Pent-Pent-
66 Hex-Hex-
Monkeys
Eat
Peanut
Butter
3.3. Use the suffix “-Use the suffix “-aneane” to indicate that the ” to indicate that the substance is an alkanesubstance is an alkane
4.4. Number the carbons in the chain Number the carbons in the chain consecutively, starting at the end closest consecutively, starting at the end closest to a substituent (groups attached to the to a substituent (groups attached to the main chain/most busy end)main chain/most busy end)
10.1.6 Apply IUPAC rules for naming 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes the isomers of the non-cyclic alkanes
up to Cup to C66..
MethylpropaneMethylpropane
MethylbutaneMethylbutane DimethylpropaDimethylpropanene
10.1.6 Apply IUPAC rules for naming the 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes up to Cisomers of the non-cyclic alkanes up to C66..
11 Meth-Meth- 66 Hex-Hex-
22 Eth-Eth- 77 Hept-Hept-
33 Prop-Prop- 88 Oct-Oct-
44 But-But- 99 Non-Non-
55 Pent-Pent- 1010 Dec-Dec-
10.1.6 Apply IUPAC rules for naming 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes the isomers of the non-cyclic alkanes
up to Cup to C66..
For chains longer than 4 carbons with side chains:
5. name and number the location of each substituent the name of the substituent will be written before
the main chain and will end with “–yl” (or just memorize the below) CH3 is methyl C2H5 is ethyl C3H7 is propyl
10.1.6 Apply IUPAC rules for naming 10.1.6 Apply IUPAC rules for naming the isomers of the non-cyclic alkanes the isomers of the non-cyclic alkanes
up to Cup to C66..
And with 2 or more side chains: 6. Use prefixes di-, tri-, tetra-, to indicate when
there are multiple side chains of the same type.7. Use commas to separate numbers and hyphens
to separate numbers or letters.8. Name the side chains in alphabetical order.
WHEW!!!!!!!!
How about CHow about C55HH1212? The isomers are:? The isomers are:
Pentane 2-methyl-butane 2,2-dimethyl propanePentane 2-methyl-butane 2,2-dimethyl propane
Nomenclature PracticeNomenclature Practice
CH3 CH3
CH3
CH3
Cl
Name this compound
Step #1: For a branched hydrocarbon, the longest continuous chain of carbon atoms gives the root name for the hydrocarbon
152 43
9
6
87
9 carbons = nonane
Nomenclature PracticeNomenclature PracticeName this compound
CH3 CH3
CH3
CH3
Cl
152 43
9
6
87
9 carbons = nonane
Step #2: When alkane groups appear as substituents, they are named by dropping the -ane and adding -yl.
CH3 = methyl
chlorine = chloro
Nomenclature PracticeNomenclature PracticeName this compound
CH3 CH3
CH3
CH3
Cl
152 43
9
6
87
9 carbons = nonane
CH3 = methyl
chlorine = chloro
Step #3: The positions of substituent groups are specified by numbering the longest chain of carbon atoms sequentially, starting at the end closest to the branching.
1 9 NOT 9 1
Nomenclature PracticeNomenclature PracticeName this compound
CH3 CH3
CH3
CH3
Cl
152 43
9
6
87
9 carbons = nonane
CH3 = methyl
chlorine = chloro
Step #4: The location and name of each substituent are followed by the root alkane name. The substituents are listed in alphabetical order (irrespective of any prefix), and the prefixes di-, tri-, etc. are used to indicate multiple identical substituents.
2-chloro-3,6-dimethylnonane
What about boiling points of isomers???
Pentane 2-methyl-butane 2,2-dimethyl propane
Magnitude of the force depends on…Magnitude of the force depends on…1.1. Number of electrons and size of the Number of electrons and size of the
electron cloudelectron cloud with more electrons, valence electrons with more electrons, valence electrons
are farther away from the nucleus and are farther away from the nucleus and can be polarized more easilycan be polarized more easily
2.2. Shape of moleculesShape of molecules molecules with shapes that have more molecules with shapes that have more
contact area have greater forces contact area have greater forces between them than those don’tbetween them than those don’t
boiling point increases
this flat shape allows it to stick to one another
better
these round shapes do NOT allow them to stick to one
another
10.1 Introduction, cont.10.1.7 Deduce structural formulas for the isomers of the
straight-chain alkenes up to C6. 10.1.8 Apply IUPAC rules for naming the isomers of the
straight-chain alkenes up to C6. 10.1.9 Deduce structural formulas for compounds containing
up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.
10.1.10 Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.
10.1.11 Identify the following functional groups when present in structural formulas: amino (NH2), benzene ring ( ) and esters (RCOOR).
10.1.12 Identify primary, secondary and tertiary carbon atoms in alcohols and halogenoalkanes.
10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in 10.1.9.
10.1.710.1.7 Deduce structural formulas for Deduce structural formulas for the isomers of the straight-chain alkenes the isomers of the straight-chain alkenes
up to Cup to C66.. Remember that structural formulas show the relative
location of atoms around each carbon Hexane: CH3CH2CH2CH2CH2CH3 (condensed s.f.)
M.F. = C6H14
Determine the molecular formulas for the alkenes below. Draw out and write the structural formulas for all isomers that can be formed by each.
C2H4 C3H?
C4H?
C5H?
C6H?
Alkenes have a double bond between two or more of the carbons CnH2n
Draw out and write the structural formulas for all isomers that can be formed by each
– C2H4
– C3H6
– C4H8
– C5H10
– C6H12
45
10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes
up to C6.
1.1. suffix changes to “-ene”suffix changes to “-ene”2.2. when there are 4 or more carbon atoms when there are 4 or more carbon atoms
in a chain, in a chain, the location of the double the location of the double bond is indicated by a numberbond is indicated by a number
3.3. begin begin counting the carbons closest to counting the carbons closest to the end with the C=C bondthe end with the C=C bond
numbering the location of the double numbering the location of the double bond(s) takes precedence over the bond(s) takes precedence over the location of any substituentslocation of any substituents
1-butene 2-butenebut-1-ene but-2-ene
Naming the isomers (IUPAC) of straight chain alkenes up to C6 (10.1.8)
1. Count the number of carbons in a chain
2. Determine the ending of the name, based on # of bonds or functional group
3. Determine any side chains, which will be placed at the front of the name
48
10.1.8 Apply IUPAC rules for naming the isomers of the straight-chain alkenes
up to C6.
Breakin’ it down…
ene
Naming Practice!!!Naming Practice!!!
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
choose the correct ending
ene
determine the longest carbon chain with the double bond
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
assign numbers to each carbon
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
ene
assign numbers to each carbon
CH3 CH2 C2
CH21
CH23
C4
CH25
CH3
CH3
CH36
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
ene
1-hexene ene
attach prefix (according to # of carbons)
CH3 CH2 C2
CH21
CH23
C4
CH25
CH3
CH3
CH36
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
CH3 CH2 C2
CH21
CH23
C4
CH25
CH3
CH3
CH36
determine name for side chains
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
1-hexene 1-hexene
ethyl
methyl
methyl
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
CH3 CH2 C2
CH21
CH23
C4
CH25
CH3
CH3
CH36
2-ethyl-4-methyl-4-methyl-1-hexene
ethyl
methyl
methylattach name of branches alphabetically
group similar branches
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
CH3 CH2 C2
CH21
CH23
C4
CH25
CH3
CH3
CH36
2-ethyl-4-methyl-4-methyl-1-hexene
ethyl
methyl
methyl
group similar branches
CH3 CH2 C
CH2
CH2 C
CH2
CH3
CH3
CH3
CH3 CH2 C2
CH21
CH23
C4
CH25
CH3
CH3
CH36
2-ethyl-4,4-dimethyl-1-hexene
or 2-ethyl-4,4-dimethyl hex-1-ene
ethyl
methyl
methyl
2-butene
propene
CH3 CH CH2
CH3 CH CH CH3
CH3 CH CH C
CH3 CH3
CH3
2,4-dimethyl-2-pentene2,4-dimethyl pent-2-tene
b) same
c) 4,5 dimethyl-2-hexene
a) 3,3-dimethyl-1-pentene
CH2 CH C CH2 CH3
CH3
CH3
CH3 C CH CH2
CH3
CH2 CH3
CH CH CH3
CH3
CC
CH3
CH3
Organic Chemistry Organic Chemistry Introduction: Functional Introduction: Functional
GroupsGroupsTopic 10.1.9 – 10.1.13Topic 10.1.9 – 10.1.13
10.1.9Deduce structural formulas for compounds containing up to six carbon atoms with one of the
following functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.
Functional group: Functional group: a group of atoms that defines the structure of a family and determines its properties
The functional group concept explained: The Chemistry Journey: The Fuse School: https://www.youtube.com/watch?v=nMTQKBn2Iss
10.1.10 Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following
functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.
Functional Group
Formula Structural Formula
Alcohol -OH - O – H
Aldehyde -COH (on the end of a chain)
O- C – H
Ketone -CO- (can’t be on end of chain)
O- C –
Carboxylic Acid
-COOH (on the end of a chain)
O- C – O – H
Halide -Br, -Cl, -F, -I - X
10.1.10 Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following
functional groups: alcohol, aldehyde, ketone, carboxylic acid and halide.
Functional Group
Formula Suffix (or Prefix)
Alcohol -OH -ol
Aldehyde -COH (on the end of a chain)
-al
Ketone -CO- (can’t be on end of chain)
-one
Carboxylic Acid
-COOH (on the end of a chain)
-oic acid
Halide -Br, -Cl, -F, -I bromo-,chloro-, fluoro-,iodo-
Know these 7, only have to recognize the
3 in the
Alcohols: suffix = “ol”Alcohols: suffix = “ol”
propan-1-ol
propan-2-ol
2-methyl propan-2-ol
propanal
Note: an aldeyhde group is always on an end carbon so don’t need a number
butandianal
Aldehydes: suffix = “al”Aldehydes: suffix = “al”
propanone(don’t need C#, must be in between two carbons)
butanone(don’t need C#, must be in between two carbons)
2-pentanone orpentan-2-one
Ketones: suffix = “one”Ketones: suffix = “one”
butandionebutandione
pentan-3-one
butanoic acid
Note: a carboxyl is always on an end carbon
propandioic acid
Carboxylic Acids: Carboxylic Acids: suffix = “oic acid”suffix = “oic acid”
1-bromopropane
2-chlorobutane
1,2-diiodoethane
1,2-difluoroethene
1,2-difluoroethene
1,1,2-trifluorothene
Halides: prefixes = “fluoro, chloro, bromo, iodo”Halides: prefixes = “fluoro, chloro, bromo, iodo”
Functional Group
Formula
Amine - NH2
Ester OR – C – O – R
Benzene
Only identify the following functional groups in structures: (10.1.11)
amino, amino, benzene ring,benzene ring, ester ester
Functional GroupsFunctional Groups
Identify each functional group Identify each functional group by name…by name…
This is a possible idea for This is a possible idea for making flash cards.making flash cards.
10.1.11 Identify the following functional groups when present in structural formulas: amino (NH2), benzene ring ( ) and esters
(RCOOR). Esters are used for fragrances and
flavoring agents since one of their major properties is smell
Benzene is in a family known as the aromatic hydrocarbons… because they smell
10.1.12 Identify primary, secondary and tertiary
carbon atoms in alcohols and halogenoalkanes.
With reference to the carbon that is directly bonded to an alcohol group or a halogen:
Primary = carbon atom is only bonded to one other carbon
Secondary = carbon atom is bonded to two other carbons
Tertiary = carbon atom is bonded to three other carbons
10.1.12 Identify primary, secondary and tertiary
carbon atoms in alcohols and halogenoalkanes.
Draw a… Primary alcohol Secondary halogenoalkane Tertiary alcohol What type are all aldehydes /
carboxylic acids? Why? What type are all ketones? Why?
10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in
10.1.9. Volatility: how easily a substance turns
into a gas The weaker the intermolecular force, the
more volatile it is So, is a nonpolar or polar substance more
volatile? ionic › hydrogen bonding › dipole-dipole › van
der Wall’s (Fig. 10.35 for reference) Therefore, volatility:
vdW › d-d › H alkane › halogenoalkane › aldehyde › ketone ›
amine › alcohol › carboxylic acid
10.1.210.1.2 Predict and explain the trends in Predict and explain the trends in boiling points of members of a homologous boiling points of members of a homologous
series.series.
Intermolecular forces presentIntermolecular forces present Simple alkanes, alkenes, alkynes → van der
Waals’ forces (nonpolar) → lower b.p. Aldehydes, ketones, esters & presence of
halogens (polar) → dipole: dipole forces → slightly higher b.p.
Alcohol, carboxylic acid & amine → hydrogen bonding (w/ O, N, F) → even higher b.p.
Naming song: https://www.youtube.com/watch?v=mAjrnZ-znkY
10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in
10.1.9. Solubility: a solute’s ability to dissolve in a
polar solvent (water) The more polar a substance is, the more
soluble it is Solubility:
If the functional group is soluble (hydrogen bonded), it will be more soluble
Solubility decreases as chain length increases Smaller alcohols, aldehydes, ketones &
carboxylic acids are typically soluble Halogenoalkanes are NOT soluble since they
don’t form hydrogen bonds
10.1.13 Discuss the volatility and solubility in water of compounds containing the functional groups listed in
10.1.9. 1. Which substance is most soluble: ethene,
propene, prop-1-ene or hex-1-ene?2. Rank the following substances in order of
increasing boiling point: C5H12, CH3CH2CH2CH2OH, CH3OCH2CH2CH3
3. Compare the boiling points of C2H6, CH3OH and CH3F
4. Explain, at the molecular level, why ethanol is soluble in water, but cholesterol (C27H45OH) and ethane are not.