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OCR AS Chemistry A H032 for �rst assessment in 2016

Complete Tutor NotesSection:4.1.1 Basic concepts of organic chemistry Naming organic compounds page 123 Chemical formulae page 131 Structural isomerism page 1354.1.2 Alkanes Alkanes page 138 Radical substitution page 140

Naming Organic Compounds Page 123

Naming Organic Compounds

All organic compounds contain carbon. There are millions of organiccompounds and the reason there are so many is that carbon can formsingle, double or triple bonds with other carbon atoms as well as bondingto oxygen, hydrogen, nitrogen, phosphorous and halogens:

O C C

C

C

C C

C

C

Cl

H

H

H HH H

H H

HHH

H

H

H

H

Each carbon makes 4bonds - always checkyour structures to makesure every carbon has4 bonds

Carbon can form single,double or triple bondswith other carbons andbonds readily to otherelements so there arealmost an infinite numberof organic compounds possible

N C

C

C

HH

H HH

H

H H

H

O

O

The simplest organic compounds are the hydrocarbons. These compounds only containhydrogen and carbon.

Hydrocarbons are compounds that contain only hydrogen and carbon.

C

H

H H

H

C

H

H

C

H

H H

H

C

H

H

C

H

H

C

H

H H

H

C

H

H

C

H

H

C

H

H

C

H

H H

H

C

H

H

C

H

H

C

H

H

C

H

H

The simplest of the hydrocarbons are the alkanes. Alkanes have carbons with only single bonds. They are said to be saturated compounds as they are completely saturated with hydrogens - you cannot get any more hydrogens on!

1 x Cmethane

2 x Cethane

3 x Cpropane

4 x Cbutane

5 x Cpentane

6 x C = hexane7 x C = heptane8 x C = octane9 x C = nonane10 x C = decane

You must learn thenames of the firstten alkanes

33


Monkeys Eat Peanut Butter, may help you rememberthe first 4 compounds in the series.

Remember the rest by reference to polygons - e.g. heptagon (7 sides) - heptane (7 carbons) etc.

Unsaturated HydrocarbonsHydrocarbons with one or more carbon-carbon multiple bonds are called unsaturated compounds. Alkenes have at least one carbon-carbon double bond and the names of the alkene series are similar to alkanes except they have the suffix (ending) ene. A difference being you must state the position of the double bond if there is more than one possibility. For example:

C CC

H

H

H

H

HHPropene - double bond can only go in one position sono need to number

C CC

C H

HH

H H

H

H H

1

2 3

4

C CC

C

H

HH

H

HH

H

H1 2

3

4

but-2-ene but-1-ene

2 possibilitiesfor butene

You number the carbons to give the lowest number possible. For example, here you say but-2-ene rather than but-3-ene.

The names of the alkenes are ethene, propene, butene, pentene etc.

Unsaturated compounds are compounds containing one or morecarbon-carbon multiple bonds.

33

Saturated compounds have only single carbon-carbon bonds.


33 Naming Branched Alkanes Using IUPAC Rules

Organic compound names are composed of the:

PREFIX STEM SUFFIX

Find the STEM by finding how many carbons in the longest carbon chain:

C C

H

H

HH

H

HC

C

H

H

H

H

H C H

C

H

H

C

H

H

1 2 3

4

5

123456

The longest carbon chain in thiscompound has 5 carbons. Thiscompound will have the STEMpent

The STEM names for the first ten compounds are meth, eth, prop, but (pronounced bute), pent, hex, hept, oct, non and dec

Once you have identified the longest carbon chain, identify the branches and count the number of carbons in the branches. For example here:

C

H

H

HC

H HC

C

H

H H

C CC

H

H

H HC

H

H

H H

C

H

H

H

H

The longest chain in this branchedalkane is 6 carbons and so theSTEM is hex. There is a reason why this compound is numberedfrom right to left - see later

There are two branches, at position 2 - a one carbon chain, and position 4 - a two carbon chain.


33 The branches form the PREFIX (start) of the name. Branches are named according to how many carbons are present as follows:

Number of carbons

PREFIX methyl1 2 3 4 5 6 7 8 9 10

ethyl propyl butyl pentyl hexyl heptyl octyl nonyl decyl

123456C

H

H

HC

H HC

C

H

H H

C CC

H

H

H HC

H

H

H H

C

H

H

H

H

ethyl branchon position 4

methyl branchon position 2

1 2 3 4 5 6C

H

H

HC

H HC

C

H

H H

C CC

H

H

H HC

H

H

H H

C

H

H

H

H

The full IUPAC name of the compound above is:

4-ethyl-2-methylhexane

hyphens separatenumbers and letters

ethyl comes beforemethyl alphabetically

PREFIX STEMSUFFIX

Numbering of the Carbon ChainAlways number the carbon chain to give the lowest numbers. For example, the compound above could be numbered as follows:

Numbering this way would give thebranches at positions 3 and 5 (adds to8)Numbering from R to L gives branches atpositions 2 and 4 (adds to 6) - use waythat gives lowest total

compound is an alkaneso has suffix ane


33 Some more examples:

1

1

1

2

2

2

3

3

3

4

4

4

5C

H

H

H

H

C

H C

HHH

C

C

H

H H

H

C

CC

C

H

H HC

H

H

H

CH C

H

H

H

H

H

C

H

H

H

H

H

H

carbon chain numbered to givebranch on the lowest number

longest carbon chain has4 carbons so STEM is but

branch has only one carbon so prefix is methyl

compound is an alkaneso has suffix ane

position ofbranch

2-methylbutane

2,2,4-trimethylpentane

three branches onpositions 2, 2 and 4.Note if we numbered from L to R, brancheswould be on 2, 4 and 4-number to give the lowesttotal

commas separatenumbers

all three branches have onecarbon and so prefix will betrimethyl

compound is analkane so has SUFFIX ane

longest carbon chain is 5carbons long so STEM is pent

Naming AlkenesNaming alkenes is similar to alkanes except they have the suffix ene andyou must state the position of the double bond, if there is more thanone possible place it can go.

C C

C H

CC

H

H HH

HH

H

H H

2-methylbut-1-ene

1 carbon branch onposition 2 so PREFIX2-methyl

4 carbons in longest carbonchain so hasSTEM but

The compound isan alkene so hasSUFFIX, ene.Double bond is onposition 1


33 Naming Haloalkanes - compounds that containhalogens F, Cl, Br or I bonded to a carbon. Treat theseas you would branches.

1 2 3

4

CH C

Br H

H H

C H

H

Br

1 2 3CH C

Br H H

H

HH

H

H H

C

C

OC

C

H

H

1,2-dibromopropane

has 2 bromines on positions 1 and 2 so PREFIX is 1,2-dibromo

chain has 3 carbonsso has STEM prop

compound is a haloalkaneso has the SUFFIX ane

2-bromo-3-methylbutane

bromine and methyl groupsin positions 2 and 3 respectivelyso PREFIX is 2-bromo-3-methyl(bromine before methylalphabetically) 4 carbons in

chain so STEMis but

compoundis ahaloalkaneso SUFFIX isane

Functional Groups

The functional group is part of the molecule responsible for its reactions.

These change the suffix. Otherwise, the naming is the same as foralkanes.

C

H HC

C

C

C C

alkene suffixene

O

O

aldehyde suffixanal

C

O

O

carboxylic acid suffixanoic acid

ketone suffixanone

H

alcohol suffixanol


33 Examples of each:

C

5 4 123CH C

H Br H

H

HH

C

H H

C

C C

H

H

H

H

H

CC

H H

HH

H

H

H C

H

O

C

O

O

5 4 123CH C

H H O

HH

H

H H

H

H

C

C C

H

H

C

2-methylpropanal

has a methyl groupon position 2 sohas the PREFIX2-methyl

3 carbons incarbon chainso STEM isprop

compound is an aldehydeso has SUFFIX anal

3-bromo-3-ethylpentanoic acid

bromine and ethyl group onposition 3 so PREFIX is3-bromo-3-ethyl(bromine is placedbefore the ethylalphabetically)

5 carbons inchain so STEMis pent

carboxylic acidso SUFFIX isanoic acid

3-methylpentan-2-onehas a methylgroup on position3 so PREFIX is3-methyl

the chain has5 carbons so STEM is pent

compound isketone so has thesuffix anone. TheC=O is on position2 so has theSUFFIX an-2-one

1 2 3CH C

Br H

H

I

OH

C

H

Cl

2-bromo-1-chloro-3-iodopropan-2-ol

bromine, chlorine and iodine onpositions 2, 1 and 3 respectivelyso has PREFIX 2-bromo-1-chloro-3-iodo (placed in alphabetical order,starting with chloro on position 1- comes alphabetically first)

3 carbons in chainso has STEM prop

compound is an alcoholso has the suffix anol.The OH is on carbon 2so SUFFIX is an-2-ol


Multiples of the Same Group

When you have more than one of the same group attached to a carbonchain, you use the following: di : 2 groups the same tri : 3 groups the same tetra : 4 groups the sameFor example:

5 4 123CCl C

Cl Br

HH

H

H

Cl HH

Br

Br

C

C C

H

Cl

HH C

C

1,1,3-tribromo-4,4,5,5-tetrachloro-2,3-dimethylpentane

33

Chemical Formulae Page 131

34 Empirical Formulae and Molecular Formulae

Empirical formula gives the simplest ratio of each atom of each elementin a compound.

Molecular formula gives the actual number of atoms of each element.

Example:A hydrocarbon has the relative molecular mass of 56.0 and a % composition by mass of 85.7% carbon and 14.3% hydrogen.Calculate the emperical formula.

We can say that 100g of material would have 85.7 g of carbon and14.3 g of hydrogen.

C H

m / g 85.7 14.3

M / g mol-1 12.0 1.0

divide mass, m, by molar mass, M, to get the number of moles

Number of moles 7.14

7.147.14

14.3 divide number of moles by thesmallest value (7.14) to get simplestratio

The simplest wholenumber ratio 1 2

The empirical formula is therefore CH2The compound has a relative molecular mass of 56.0 so we need towork out how many units of CH2 there are in 56

One unit of CH2 has a relative mass of 14.0 so to find out howmany of these are in 56.0:

56 ÷ 14 = 4 units

4 x CH2 = C4H8This is the molecular formula

m

M n

mass of materialin grams

number ofmoles

Molarmass

Chemical Formulae

34 Displayed Formula, Structural Formula andGeneral Formula

The displayed formula is when all the atoms and all the bonds are shown.

For example, 2-methylbutane has the displayed formula:

H

C CH C

H H

H

HO

H

H

H

HH C

C CCH

H

H H H

H

H

C

H

H

C

H

H

H

H

H

C

HH C

C

Its molecular formula would be C5H12. In this case, the empiricalformula (simplest ratio of each atom of each element) would also beC5H12

The structural formula would look like this:

CH3CH(CH3)CH2CH3

Note how the branched methyl group is shown in brackets to show it isa branch at position 2 on the carbon chain.

Here are some more examples:

DisplayedFormula

StructuralFormula

MolecularFormula

propan-2-ol

CH3CH(OH)CH3 C3H8O

H

O

O

CH3CH(CH3)COOH C4H8O2

2-methylpropanoic acid

Chemical Formulae 3

34DisplayedFormula

StructuralFormula

MolecularFormula

C

H H

H

Br

H

CO

C

H

C

H

H

H

H

H

C

H

C

H

H

H

H

H

C

H

H

C

H

C

H

H

H

H

H

C

C

H

H H

H

C

H

H

C

H

H

C

H

CH2(Br)CH2CHO C3H5OBr

3-bromopropanal

Skeletal FormulaThis is where carbon chains are represented by a zig zag line and thecarbon to hydrogen bonds are not shown. For example:

=

=

ethane

propane

2-methylpentane

=

Chemical Formulae

34

C

ClH

Cl

Cl

Cl

H

H

H

C

H

C

H

O

2,2-dichloropropan-1-ol

propene

=

=

=

OH

C CC

C

C C

C C

H HH

HH H

H

H

H H

H

H

HHH H

cyclopentane

For bonds other than the carbon to hydrogen, thesebonds are shown along with the chemical symbolto represent the element

Structural isomerism Page 135

35 Isomerism

There are different types of isomerism. In AS chemistry you need toknow about structural isomerism and Z/E isomerism which is a formof stereoisomerism. You will learn about Z/E isomerism in the Alkenessection.Structural isomers have the same molecular formula but a differentstructural formula.

Examples of structural isomers:

C

H

HH

H H

H

C

H

C

H

H

H

C

C

H

HH

H

H

H

C

H

C

H

OC

H

OH

H

H

H

C

H

C

H

H

C

H

H

H

H

H

H

C

H

C

H

H

HC

butane 2-methylpropaneand

C4H10

C3H8O C3H8O

C4H10the same molecular formula

the same molecular formula

different structural formula

different structural formula

CH3CH2CH2CH3

CH3CH2CH2OH

CH3CH(CH3)CH3

CH3CH(OH)CH3

Another example:

propan-1-ol propan-2-oland

Structural isomerism

35 When trying to find different structural isomers for thesame molecular formula, it is often easiest to useskeletal formulae.

For example, finding the alkane structural isomers of C6H14

start with a straight chain first

1

2

4 5

3

next, move a methyl group around

these two are the samecompound so are notisomers

X

next, move two methyl groups around

If you are unsure whether two structures are different structural isomers,name the two structures using IUPAC rules. If the two structures comeout with the same name, they are the same compound. In the exampleabove, both structures give the name 2-methylpentane.

Structural isomerism Page 137

35

General formula is a mathematical type expression that describes ahomologous series (see later).

Alkanes have the general formula CnH2n+2

It is therefore possible to work out the molecular formula for an alkanewith any number of carbons. For example, an alkane with 24 carbons (n=24), would have the molecular formula would be C24H50

Alkenes have the general formula CnH2n

Homologous SeriesHomologous series have the same functional group but each successivemember of the series differs by CH2

For example, the first 3 compounds in the homologous series ofcarboxylic acids:

CHOH

OCH

OH

OC

H

H

C

H

H

CHOH

OC

H

Hmethanoic acid ethanoic acid propanoic acid

Cyclic CompoundsHere are some cyclic compounds which have the general formula CnH2n

C C C CC

C CC

C

C CC C

H H

H

H

H H

H

H

H

HH

HH

HHH

H

HH

HHH

HH

H

H

cyclopropane cyclobutane cyclohexane

General Formula

Alkanes

36

Alkanes are nonpolar, covalently bonded compounds with a simplemolecular structure. The forces that need to be overcome thereforeto boil them are induced dipole-dipole interactions.

Long chain hydrocarbons have moresurface contact (NOT GREATER SURFACE AREA) thanshort chain hydrocarbons. The induced dipole-dipole interactions for long chain hydrocarbons are greater and so boiling points are higher asmore energy is required to overcomethese forces.

Branched hydrocarbons have less surface contact than theirequivalent straight chain isomers and so increased branching leads toa lowering of boiling points.For example, the structural isomers of C7H16

HeptaneBoiling point = 98°C

2,2,3-trimethylpentaneBoiling point = 80.9°C

2,3-dimethylpentaneBoiling point = 89.5°C

leads to...

leads to...

leads to...

INCR

EASED

BRANCH

ING

LESS SURFA

CECO

NTA

CT

DECR

EASE IN

INDUCED

DIPO

LE-DIPO

LE INTER

ACTIO

NS

DECR

EASE IN

BOILIN

G PO

INTS

Why Do the Alkanes Have Different Boiling Points?

Alkanes Page 139

36 Alkanes as Fuels

You must be able to write equations for the complete and incompletecombustion of these compounds and state the conditions under whichthey occur.

Incomplete CombustionThis occurs when there is limited supply of oxygen. One of the productsis carbon monoxide which is toxic. People that have gas fired boilers areadvised to have carbon monoxide detectors.

Tips for balancing combustion equations:

C3H8(g) + ?O2(g) 3CO(g) + 4H2O(l)

C3H8(g) + 3½O2(g) 3CO(g) + 4H2O(l)

start with just one molecule of the thing you are burning, then work outhow many molecules of carbon monoxide it will make (3 carbons in propane so will make 3 x CO) and how many molecules of water (8 hydrogens in propanewill make 4 x H2O)

You then work out how many molecules of oxygen you need to balance. Wecan see we have 7 atoms of oxygen of the right hand side so we will need3½ molecules of oxygen (it is perfectly acceptable to use half molar quantities)

Complete CombustionWhen the oxygen supply is plentiful, complete combustion occurs andcarbon dioxide is produced rather than carbon monoxide. Whenbalancing the equation, do as before, work out how many molecules ofcarbon dioxide and water will be made and then work out how manymolecules of oxygen will be needed:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Radical Substitution

36 Radical Substitution

Mechanisms show the steps for how we believe a reaction takes place ata molecular level. When writing mechanisms, we draw curly arrows to show the movement of a pair of electrons.

For example, when a covalent bond breaks by heterolytic fission we showthe shaired pair of electrons moving from the bond to one of the atomsto form a negatively charged ion:

A A+ + X-X

A X A. + X.

When bonds break by homolytic fission, the shaired pair of bondingelectrons are split equally resulting in two species with a single, unpairedelectron which we call radicals. When we want to show a single electronmoving we use an arrow with half an arrow head which looks like afish hook:

radicals

Radical Substitution of Alkanes to form HaloalkanesThe overall equation for the radical substitution of an alkane to forma haloalkane is as follows:

C2H6(g) + Cl2(g) C2H5Cl(g) + HCl(g)

We will now go through the mechanism to show the individual stepsfor how we believe this reaction takes place.

1st Stage - Initiation

ClCl Cl. Cl.+

The initiation step involves the halogen, in this case chlorine,undergoing homolytic fission due to the presence of ultraviolet lightto form two chlorine radicals. Radicals are very reactive.

chlorinemolecule

chlorineradicals

u.v. light

u.v. light

Radical Substitution Page 141

36 2nd Stage - Propagation

The propagation stage involves two repeated steps that build up thechain reaction.

This is the stage where the two products are formed as well as furtherradicals.

The first propagation step involves the reaction of the alkane, in thiscase ethane with the chlorine radical. One of the products, hydrogenchloride, is formed along with an alkyl radical:

C

H

H

HH C

H

H

C

H

H

H C

H

H

.Cl

.Cl

.Cl .Cl

Cl.

C

H

H

H C

H

ClCl

ClCl

Cl

H

C

H

H

H C

H

H

.

+

+

H

alkyl radical

The second propagation step involves the alkyl radical formed in theprevious step reacting with a chlorine molecule to produce the otherproduct, in this case chloroethane, and a further chlorine radical.

chloroethane chlorine radical

3rd Stage - Termination

The final stage is the termination stage, which as the name suggests,brings the chain reaction to a close. This is where two radicals cometogether to form a molecule. From the chlorine and alkyl radicals wehave made so far, there are 3 possibilities:

Two chlorine radicals can pair up to form a chlorine molecule.

Radical Substitution

36

C

H

H

H C

H

H

C

H

H

H C

H

H

.

C

H

H

H C

H

H

.

C

H

H

HC

H

H

. C

H

H

H

C

H

H

H C

H

H

C

H

H

Two alkyl radicals can come together to form a longer chain alkane, inthis case butane is produced.

.Cl

.Cl(g) + .Cl(g)

Cl

And chlorine and alkyl radicals can come together to produce a haloalkane.

Radical reactions are difficult to control and will inevitably produce aplethora of products in addition to the desired products.

The mechanism for the reaction above can be written in condensedform as follows. It does not matter which of the ways you learnto describe the mechanism.

1st Stage - Initiation

2nd Stage - propagation

3rd Stage - Termination

Cl2(g)

Cl2(g)

u.v. light

.Cl(g) + C2H6(g)

.Cl(g) + .Cl(g)

.C2H5(g) + HCl(g)

.C2H5(g) + .C2H5(g)

Step 1

Step 2 .C2H5(g) + Cl2(g)

.C2H5(g) + .Cl(g)

C2H5Cl(g) + .Cl(g)

C2H5Cl(g)

C4H10(g)

for ˜rst assessment in 2016 complete tutor notesboomerpublishing.com/asnotes/4.1.1-4.1.2 basic...

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