Organic Chemistry: Introduction IB Topic 10. 10.1 Introduction 10.1.1Describe the features of a homologous series. 10.1.2Predict and explain the trends

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.


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.


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


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


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)


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-


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


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


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


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


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


10.1.9. 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


10.1.9. 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.

Documents

Organic Chemistry: Introduction IB Topic 10. 10.1 Introduction 10.1.1Describe the features of a homologous series. 10.1.2Predict and explain the trends