Chapter 10 Organic Chemistry Notes

8/9/2019 Chapter 10 Organic Chemistry Notes

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Chapter 10 Organic Chemistry

1. Unique properties of carbon

- Able to form multiple bonds (up to 4) with other atoms

-

Able to undergo catenation i.e. form long chains and ring structures- Able to form stronger bonds with other atoms (in contrast to Silicon, the next

member of group IV)

- Unable to expand beyond its octet i.e. forms more kinetically stable compounds

(in contrast to compounds of Silicon)

2. Homologous Series

- Members have the same general formula (i.e. same elements in the same ratio)

-

Members have the same functional group; a functional group is a group of atomsin a compound which gives its members their characteristic chemical properties

- Members differ by -(CH2)- groups

- have similar chemical properties

- show a gradual change (gradation) in physical properties

3. Formulas representing organic compounds

type of formula description example

empirical formula Shows the simplest whole number ratio of all atoms.CH2

molecular formula Shows the actual number of the different atoms andhow many of each; no information on how the atoms

are arranged.C6H14

structural formula Shows how atoms are arranged together in the

molecule; a full structural formula (sometimes called a

graphic formula or displayed formula) shows every

atom and bond.

condensed structural

formula

Shows order in which atoms are arranged but which omits

bonds.

CH3CH2CH2CH2CH2CH3

or CH3(CH2)4CH3

4. Nomenclature

-

When naming an organic compound we want to give a lot of information in its

name which is why the name of an organic compound consists of at least two

parts: one to indicate the number of carbon atoms in the chain and the other the

functional group. Other parts will indicate length and number of branches or if the

compound is cyclic.


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Example: prop ane

this part tells us how many carbons

atoms there are in the molecule;

this part could be:

meth- = means it has 1 carbon atom

eth- = 2 carbon atoms

prop- = 3 carbon atoms

but- = 4 carbon atoms

pent = 5 carbon atoms

hex = 6 carbon atoms benz- = benzene ring

this part tells us the functional group it

has or which homologous series it

belongs to;

the ending could be:

-ane which means it belongs to the

alkanes (=homologous series)

-ene = alkene

-anol = alcohol

-anal = aldehyde

-anone = ketone

-anoic acid = carboxylic acid.

Naming of halogenoalkanes

-

In the case of the halogenoalkanes, the name begins with the name of the halogen and

not the number of carbon atoms.

-

The name should also indicate the position and the number of halides if there is more

than two.

-

For example: chloro, bromo, fluoro, iodo, .. in names such as 2-bromopropane, 1,2-dichloroethane.

5. Alkanes

- Alkanes are a homologous series in which all non-cyclic compounds have the

following general formula: CnH2n + 2

- Alkanes only contain carboncarbon single bonds and carbonhydrogen single

bonds, hence they are also known as hydrocarbons.


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Structure of alkanes

structure 1

straight-chain: all carbon atoms can be

joined by a continuous line.

structure 2

branched-chains: molecules have side groups and these are

referred to as alkyl groups.

alkyl group formula

methyl CH3-

ethyl CH3CH2-

propyl CH3CH2CH2-

butyl CH3CH2CH2CH3-

C4H10- butane

2-methylpropane

Naming branched alkanes

A branch is a carbon atom or group of carbon atoms bonded onto a larger carbon chain.

The name of that branch (or side group) should indicate the number of carbon atoms in it and

should end with yl to indicate it is a branch e.g. methyl, ethyl, propyl. You need to identify the

longest chain in the branched compound and any carbon atoms not in it must be part of a branch.The name of the branched compound should include:

the name of the branch

the number of branches if there is two or more

and, using a number, the position of the branch(es) on the straight chain; the position

should be decided from the end of the longest chain which gives the lowest number(s).

the position of the functional group has priority over alkyl.

Examples: 2-methylpentane, 2,2-dimethylpentane, methylpropan-2-ol, 2,2,4-trimethylpentane


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Physical properties of alkanes: melting/boiling points, viscosity, density

Melting points, boiling points, viscosity (resistance to flow) and density increase with

increasing number of carbon atoms

Explanation:The larger the molecules (the greater the molar mass)

the more sites/surface area there are for intermolecular attraction,

or the greater the number of electrons, the greater the polarisation within the molecule,

or the greater their mass/inertia

and therefore the greater the attraction betweenthe molecules, the more energy is needed to

overcome the intermolecular forces of attraction and cause the alkane to melt or boil; the

greater the viscosity and the greater the density.

Structural isomerism

Structural isomers are compounds with the same molecular formula but the sequence or

arrangement in which the atoms are bonded is different which is why they have a different

structural formula.

Branching has a different effect on the melting point than on the boiling point.

Straight chained alkanes have higher boiling points than branched alkanes of similar

molecular mass as there is more contact/larger surface area between the straight-chained

molecules and therefore more sites for induced polarity and stronger Van der Waals forces;

the higher the number of branches, the lower the boiling points. As branching decreases

surface area, it increases volatility and decreases density.

However, straight chained alkanes generally have lower melting points than branched alkanes

(in particular with symmetrical structures). Branching increases m.p. as branched molecules

can fit more closely together and more energy is needed to separate them; the lattice of a

solid straight chained alkane is like wet spaghetti: molecules can easily slide over each other.


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name

Structural formulamolecular

formula

boiling point

(oC)

pentane C5H12 36

2-methylbutane C5H12

28

2,2-

dimethylpropane

C5H12 10


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Chemical properties of alkanes

Alkanes are unreactive because:

Large bond enthalpies: the covalent bonds between carbon and hydrogen and carbon and

carbon have large bond enthalpies; this is because both hydrogen and carbon are smallatoms so bonding pairs are attracted strongly by both nuclei.

Low or no difference in electronegativity: due to the very small difference inelectronegativity between carbon and hydrogen the covalent between the two atoms very

low polarity and does therefore not attract other reagents.

The only reactions they easily undergo are combustion and substitution reactions.

(a)

Combustion of alkanes

Complete combustion: (in plentiful of oxygen) products are water and carbon dioxide

CxHy + (x + y/4) O2 xCO2 + (y/2)H2O

Examples: CH4 + 2O2 CO2 + 2H2O

C3H8+ 5O2 3CO2 + 4H2O

Incomplete combustion:products are water and carbon monoxide or carbon (=black

smoke) depending on the extent of the lack of oxygen (incomplete oxidation); water is

always produced.

Examples:

CH4+ 1 O2 CO2 + 2H2O

CH4+ O2 C + 2H2O

C3H8+ 4O2 CO2 + 2CO + 4H2O


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(b)

Free Radical Substitution

Bond fission

When a covalent bond is broken the bonding pair electrons are redistributed between the twoatoms; there are two ways of redistributing these two electrons:

Homolytic f ission

* occurs in non-polar bonds or bonds with a very low polarity when bonding

electrons are fairly equally shared in the bond

* each atom gets one of the bonding electrons; each atom has now an unpaired

electron and is therefore unstable and reactive; such a particle with an

unpaired electron is called a free radical

* free radicals have a strong tendency to react and usually have a short

existence; tend to be intermediates in reactionsA B A + B

Heterolytic fission

* Both bonding electrons go to one of the atoms forming a negative and positive

ion for example a carbocation and carbanion

* occurs in polar bonds

* ions are unstable and highly reactive sites

A B A+ + B-

During a substitution reaction, a hydrogen atom on the carbon chain is replaced by a halogen

atom. The reaction needs sunlight/UV as UV/sunlight has the corresponding amount of

energy to break the halogen bonds; no reaction will occur in the dark.

Examples of substitution reactions:

UV

CH4 (g) + Cl2(g) CH3Cl (g) + HCl (g)

methane + chlorine chloromethane + hydrogen chloride

(substitution with chlorine: tetrachloromethane is formed if proportion of chlorine is high

compared to the proportion of methane).


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UV

C2H6 (g) + Br2(l) C2H5Br (g) + HBr (g)

ethane + bromine 1-bromoethane + hydrogen bromide

Homolytic free radical substitution reaction

The sequence of steps/collisions by which substitution reaction takes place is called the

reaction mechanism. The reaction is an example of a free radical substitution reaction.

step 1:initiation

The UV light causes homolytic fission of the halogen molecule; each atom

takes one of the electrons in the covalent bond. The two species formed are

not atoms but are called free radicals and each has one electron from the bond.

A free radicalis the name given to a species containing an unpaired electron.

They are very reactive, because they have an unpaired electron, and so have a

strong tendency to pair up with an electron from another molecule. A dot is

used to represent the unpaired electron.

Using the example of the reaction between methane and chlorine:

Homolytic fission reaction Cl - Cl Cl + Cl

to form free radicals free radicals

step 2:

propagation

Chain reaction during which the free radicals react with molecules forming

more free

radicals and other molecules:

free radical + molecule new free radical + new molecule

Free radical knocks an atom off the molecule which itself becomes a free

radical while the radical becomes a molecule.


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In the case of the chlorine and methane reaction:

One of the chlorine free radicals reacts with a methane molecule by

substituting itself for a hydrogen atom. A new free radical is formed (orpropagated).

CH4 + Cl CH3 + HCl

The CH3 reacts with the Cl2to form CH3Cl and the free radical Cl:

CH3 + Cl2 CH3Cl + Cl

step 3:

termination

Occurs when two free radicals react to form a stable molecule.

In the case of the chlorine and methane reaction:

CH3 + Cl CH3Cl

Cl + Cl Cl2

CH3 + CH3 C2H6

If excess Cl2 is used further substitutions may take place until all the hydrogen atoms are

substituted for:

CH3Cl + Cl2 CH2Cl2 + HCl

Followed by

CH2Cl2 + Cl2 CHCl3 + HCl

And then

CHCl3 + Cl2 CCl4 + HCl


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6.

Alkenes

Structure of alkenes

There are 2 types of structures you need to know:

straight chain

branched chains

Physical properties of alkenes

Same trends and explanations as in alkanes; however, boiling points/melting points are

generally a little lower than the boiling points/melting points of the alkanes as the alkenes

have a lower molecular mass.

Chemical properties of alkenes

As they are unsaturatedalkenes are reactive. The second bond of the double bond is weaker

than a single carbon-carbon bond as it has a lower enthalpy changeless energy needed to

break it.

Alkenes undergo addition reactions; atoms are added to the carbon chain using the double

bond

As they are unsaturatedalkenes are reactive. The second bond of the double bond is weaker

than a single carbon-carbon bond and is broken much easier.

It is because of this greater reactivity that alkenes, especially ethene, are important starting

materials in organic synthesis of useful chemicals.

It is important to note that alkenes also easily combust and undergo both complete and

incomplete combustion.

Alkanes undergo addition reaction that means that atoms are added to the molecule at either

side of the double bond so any addition reaction increases the number of atoms in the

molecule. During the addition reaction the double bond is converted to a single bond. Theweaker second bond of the double bond is replaced by a stronger single bond; this increases

the stability of the molecule.


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Reactions

The reagents (hydrogen, halogen, water and hydrogen halide molecules) are attracted to the

double bond; the double bond breaks open and is replaced by two single bonds to 2 new

atoms or groups of atoms. One part of the reagent bonds to one carbon atom while the other

part of the reagent bonds onto the second carbon atom of the double bond.

Reaction with bromine (bromination)

Occurs under room conditions (even in the dark); the product is a colourless di-

halogenoalkane.

molecular C2H4 (g) + Br2(l) C2H4Br2 (g)

equation ethene + bromine 1,2-dibromoethane

Test for unsaturation

The bromination reaction is useful as it can be used to distinguish between an alkane (no

decolourization of bromine water occurs as it remains yellow/brown) and an alkene (bromine

water which is yellow or brown becomes colourless as 1,2-dibromoethane is a colourless

compound).

Reaction with hydrogen (hydrogenation)

Requires a nickel catalyst and temperature of 180C.

molecular C2H4 (g) + H2(g) C2H6(g)

equation ethene + hydrogen ethane

Reaction with water (=hydration)

(a)Industrial hydration of ethene into ethanol

Reaction in which H and OH are introduced in the molecule: needs a strong concentrated

acid (e.g. H3PO4) as catalyst, 300 C and 70 atm. The reaction is used to make industrial

ethanol.


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Example:

molecular C2H4 (g) + H2O (l) C2H5OH (l)

word ethene + water ethanol

(b)Lab preparation of alkenes into alcohols

Reaction in which H and OH are introduced in the molecule: requires concentrated sulfuric

acid and water at room temperature

Example: equations

molecular C2H4 (g) + H2SO4(aq) C2H5OSO3H (l)

C2H5OSO3H (l) + H2O (l) C2H5OH (l) + H2SO4(aq)

word ethene + sulfuric acid ethyl hydrogensulfate

ethyl hydrogensulfate + water ethanol + sulfuric acid

Reaction with hydrogen halides

Concentrated aqueous solutions of hydrogen halide at room temperature; product is a

halogenoalkane.

molecular C2H4 (g) + HCl (l) C2H5OH (l)

word ethene + water ethanol

Economic importance of reactions of alkenes

-

Hydrogenation of vegetable oils in the manufacture of margarine.-

Hydration of ethene in the manufacture of industrial ethanol which is used as a

solvent, antiseptic or fuel.

- Polymerization in the manufacture of plastics.

- Synthesis of drugs.

- Synthesis of ethane-1,2-diol which is used as antifreeze.


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7.

Alcohols

- Most important alcohol is ethanol which is used as a fuel, solvent, antiseptic, to

make esters.

Primary, secondary and tertiary alcohols

Based on the number of alkyl groups or carbons bonded onto the carbon that carries theOH

group.

primary alcohol secondary alcohol tertiary alcohol

Has one alkyl group or 1

carbon atom onto the carbon

which carries theOH group

Has two alkyl groups or 2

carbon atoms onto the carbon


Has three alkyl groups or 3



This distinction is important as the product of the same type of reaction e.g. an oxidation will

be different for each type of alcohol. As such the three different types of alcohols can be

considered different homologous series.

Complete combustion of alcohols

Alcohols can be combusted completely (very exothermic reaction) producing carbon dioxide

and water as shown by the symbol equation of the combustion of ethanol.

C2H5OH + 3O2 2CO2 + 3H2O


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Oxidation of alcohols using an oxidising agent

Primary alcohols, such as ethanol, can be oxidised by heating them with an oxidising agent

e.g. potassium dichromate(VI) which needs to be acidified (by adding H2SO4) and which

goes from orange (oxidation state +6) to green (oxidation state +3). The reaction involves the

hydrogen atoms bonded onto the carbon that carries the hydroxyl group.

The oxidation of primary alcohols can yield two different products depending on the

conditions under which it is carried out.

1. Partial oxidation to form an aldehyde, e.g. ethanal from ethanol:

Heating excess alcohol with oxidizing agent; however, to obtain a high yield of the aldehyde

it needs to be distilled (it has a lower boiling point than ethanoic acid) as soon as it is formed

as otherwise it will oxidize further to a carboxylic acid (full oxidation).

Equation for the partial oxidation of ethanol:

full equation: 3C2H5OH + Cr2O72- + 8H+ 3CH3CHO + 2Cr

3+ + 7H2O

simpler version: C2H5OH + [O] CH3CHO + H2O

2.

Full oxidation: if an excess of the oxidizing agent is used and the mixture is heated

under reflux, the alcohol oxidises to a carboxylic acid.

Equation for the further oxidation from ethanol to ethanoic acid:

full equation: 3C2H5OH + 2Cr2O72- + 16H+ 3CH3COOH + 4Cr

3+ + 11H2O

simpler version: C2H5OH + 2[O] CH3COOH + H2O

Secondary alcohols can only be oxidized to ketones by heating under reflux with acidified

potassium dichromate(VI). This is because the ketone does not have any hydrogen atom left

on the carbon of the carbonyl group.

Tertiary alcohols cannot be oxidized at all as they have no hydrogen atoms onto the carbon

that carries the hydroxyl group.


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8.

Halogenoalkanes

Primary, secondary and tertiary halogenoalkanes

- As with alcohols, the distinction is again based on the number of alkyl groups or

carbons bonded onto the carbon which carries the halogen.

primary secondary tertiary

Has one alkyl group or 1

carbon atom onto the carbon

which carries the halogen

Has two alkyl groups or 2


which carries the halogen

Has three alkyl groups or

3 carbon atoms onto the

carbon which carries the

halogen

SN1 and SN2 mechanisms: SN= nucleophilic subsitution

Halogenoalkanes react with hot dilute sodium hydroxide solutionin nucleophilic

substitution reactions of which there are two types. The type of nucleophilic substitution

reaction which is favoured or predominant depends on the structure of the halogenoalkane i.e.

primary, secondary or tertiary.

During a nucleophilic substitution reaction the halogen is substituted by the hydroxide group

to form an alcohol.

In these reactions the hydroxide ion in sodium hydroxide, which is a negative ion, is called

the nucleophile as it seeks a positive nucleus i.e. the carbon atoms that carried the halogenatom as the bond C-Hal is polar.


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There are two different ways in which a nucleophilic substitution reaction can occur: SN1

and the SN2.

SN1

Example: 2-bromo-2-methylpropane with sodium hydroxide to form 2-methyl propan-2-ol

and sodium bromide.

Equation: (CH3)3 CBr + NaOH (CH3)3 COH + NaBr

Mechanism (multistep):

unimolecular(hence the 1 in SN1 ): rate of reaction depends on 1 molecule only i.e. the

tertiary halogenoalkane;

multistep as it involves 2 steps - involves the formation of an intermediate i.e. carbocation

first order with respect to the halogenoalkane

molecularity of first step is 1

rate = k [halogenoalkane]

steps:

step 1

(slow

step)

Heterolytic fission results in dissociation or spontaneous ionisation ofhalogenoalkane resulting in the formation of a carbocation- which is the

intermediate and can be isolated - and a halide ion; this step is the slowest

and therefore the rate determining step.

step 2

(fast)

Nucleophilic attack of the carbocation formed in step 1.


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The rate of this multistep mechanism depends on the slowest reaction which is the first step

and therefore the rate of the reaction depends on the concentration of the halogenoalkane.

SN2

Example: 1-bromoethane with sodium hydroxide to form ethanol and sodium bromide.

Equation: CH3CH2Br + NaOH CH3CH2OH + NaBr

Mechanism (one step) (overall second order):

bimolecular(hence the 2 in SN2 ): as both reactants (halogenoalkane and nucleophile) are

involved in the same single step which is also the rate determining step; overall reaction

is a one step process and involves a transition state/activated complex which produces the

product instead of an intermediate.

rate = k [Halogenoalkane][Incoming nucleophile]

In the single step, a new bond is made at the same time as an old bond is broken:

Bond

broken

as the carbon carrying the halogen is positively charged because of the polar

C-X bond, the carbon is attacked by the OH-on the side of the carbon atom

opposite to the halogen (the halogen atom affects approach by nucleophile);

during this process the carbon-nucleophile bond is formed; the nucleophile

donates its electron pair to the carbon to form the bond.

Bond

made

at the same time, the halogen breaks away from the molecule producing a

negative halogen ion; the energy needed to break the bond comes from the

bond made between the OH-and the carbon atom; the halogen needs to break

away from the molecule which is a very unstable arrangement as there are 5

groups bonded onto the carbon atom; the electron pair of the CX bond is

donated to the halogen.


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In summary, the nucleophile donates an electron pair to the carbon that carries the halogen

atom which causes the bonding pair of the halogen to move away; no intermediate is formed

as the transition compound cannot be isolated; a transition state is a dynamic process during

which bonds are broken and made.

Factors which affect the rate of nucleophillic reactions in halogenoalkanes

Factor 1: Identity of the nucleophile

The hydroxide ion is a stronger nucleophile than water because it has a negative charge and

will therefore be attracted more strongly towards the partial positively charged carbon atom

bonded onto the halogen. As a result, the rate of a substitution with hydroxide (or hydrolysis)

is always faster than with water.

You are already familiar with both SN1 and SN2 reaction mechanisms or pathways. In this

section we will focus mainly on factors that affect the rate of both pathways in reactions withhydroxide ions. This will help us to decide which pathway will be favoured with a certain

combination of reactants.

Factor 2: Nature of the halogenoalkane

The first elementary step in the SN1 pathway involves the formation of a carbocation

intermediate as the halogen leaves the molecule as an anion.

The greater the stability of the carbocation, the lower the activation energy of its formation

and the faster it is formed; the more its formation is favoured, the more SN1 is favoured.

The stability of carbocations depends on the number of alkyl groups on the carbon that

carries the positive charge as the electron-releasing alkyl groups make the carbon atom less

positive (delocalises the positive charge more). As a result, the greater the number of alkyl

groups, the more stable the carbocation.

Therefore, in decreasing stability of the carbocation (most stable first):

tertiary carbocation secondary carbocation primary carbocation

Tertiary carbocations, which are formed from tertiary halogenoalkanes, are more stable/less

reactive as the electron-releasing alkyl groups make the carbon atom less positive (delocalises

the positive charge more).

With primary halogenoalkanes, the formation of a reactive (unstable) primary carbocation has

an activation energy which is higher than the activation energy of the elementary step in

which the halogenoalkane and the nucleophile collide with each other which is why SN2 is

favoured with primary halogenoalkanes.

Another way of explaining favoured pathways is by considering how likely the transition

state in SN2 can be formed. In the transition state of SN2 carbon needs to accommodate five

groups; this is easier in a primary halogenoalkane as the carbon atom has 2 hydrogens on it which


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leaves space for a 5thatom or ion. The alkyl groups on the carbon atom in a tertiary

halogenoalkane allow less space for a fifth particle to bond onto the carbon.

Factor 3: mechanism pathway taken by the halogenoalkane

primary halogenoalkanes

SN2 - slowest

secondary

halogenoalkanes

SN1 or SN2

tertiary halogenoalkanes

SN1 - fastest

As SN1 reactions generally occur faster(they are unimolecular), nucleophilic substitutions

involving tertiary halogenoalkanes are usually faster than those involving secondary or

primary halogenoalkanes.

Factor 4: Identity of halogen

As the iodinecarbon bond is the weakest bond iodoalkanes tend to react at greater rates

than other haologenoalkanes in both SN1 and SN2.

Relative rates of reaction: fastest/highest rate: I Br Cl F.

9.

Other functional groups

functional

group

condensed structural formula of

functional group

amine

(weak bases as

they are

derivatives of

NH3)

-NH2

amide -CONH2

ester COOC

nitrile - CN


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Rules for naming compounds with the above functional groups

rules examples

amines

Take the name the longest alkane chain bonded

onto the NH2

Drop thee

Add amine

If carbon chain is benzene ring: have phenyl as

prefix

Secondary amine: name both alkyl groups but inalphabetical order

Can also be: name carbon chain and have prefix -

amino

methanamine,

ethanamine,

butan-1-amine,

dimethylamine,

methylethanmine,

ethylmethylamine,

2-aminobutane

amides

Take the name the longest alkane chain onto theCONH2including the C of CONH2

Drop thee and add amide

Any other shorter chain becomes an alkyl group

and is named first

methanamide,

ethanamide,

butanamide,

methylpropanamide,

methylethanamide

esters

The carbon chain from the alcohol is named first

and ends withyl; it is the chain bonded onto the

single bonded oxygen

The chain from the acid i.e. the chain bonded witha double bond onto the oxygen is theanoate part.

Leave space between both parts.

If phenyl is attached to C in COO than benzoate

methyl methanoate,

ethyl ethanoate, propyl

methanoate,

nitriles

Take the name the longest alkane chain onto the

CN including the C of CN

Add -nitrile

Other shorter chain becomes an alkyl group and is

named first

methanenitrile,

butanenitrile

Reactions involving halogenoalkanes and other nucleophiles

1. Nucleophilic substitution of primary halogenoalkanes with ammonia to form a

primary amine and a hydrogen halide acid.

- Ammonia is a stronger nucleophile than water because nitrogen is less

electronegative than oxygen and donates an electron pair more readily.

- Conditions: warm with excess NH3.


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The SN2 mechanism for this reaction is shown below:

transition state

The nucleophile now is a molecule and not a negative ion such as the hydroxide ion. As a

result, after the halogen has left as an ion, the remaining particle has a positive charge and

therefore needs to lose a hydrogen proton to become neutral product.

In the SN2 mechanism this happens in the transition state as shown by the curly arrow

showing the loss of the hydrogen ion; the bonding pair between the nitrogen and the

hydrogen becomes the lone pair on the nitrogen.

In the SN1mechanism this de-protonation occurs as a step 3 after the nucleophile substitution

has been completed.

2.

Nucleophilic substitution of primary halogenoalkanes with potassium cyanide to form

a nitrile and a potassium halide salt.

- Cyanide ions are very strong nucleophiles because of the negative charge. In this

reaction the non-bonding pair donated is from the carbon. The carbon atom also

carries the negative charge of the molecular ion.

-

As a result of this substitution reaction the chain length of the halogenoalkane hasbeen increased by 1 carbon atom.

- For example, using 1-bromopropane as a typical primary halogenoalkane:

CH3CH2CH2Br + CN- CH3CH2CH2CN + Br-

- The full equation: CH3CH2CH2Br + KCN CH3CH2CH2CN + KBr

- The bromine (or other halogen) in the halogenoalkane is simply replaced by a -CN

group - hence a substitution reaction. In this example, 1-butanenitrile is formed

from 1-bromopropane.

- Conditions:reflux with KCN dissolved in ethanol.


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Reactions of nitriles

1. Acidic hydrolysis

- Nitriles can undergo hydrolysis with a dilute acid (such as H2SO4) to form

carboxylic acids and the ammonium ion. (It forms ammonia from the hydrolysis

reaction but the ammonia undergoes an acid-base reaction with the acid to form

the ammonium ion)

2. Reduction

- A reduction reaction is the addition of hydrogen. The reduction of nitriles results

in the formation of a primary amine as shown by the equation below showing how

1-butanenitrile is converted into 1-butanamine

- Example: CH3CH2CH2CN + 2H2 CH3CH2CH2CH2NH2

10.Elimination reactions

Elimination of HBr from bromoalkanes

-

Bromoalkanes in the presence of nucleophiles can undergo either:

A nucleophilic substitution reaction to form, for instance, an alcohol

Elimination reaction to form an alkene

Whenever a bromoalkane is in the presence of a nucleophile, e.g. OH -, competition between

these two types of reactions occurs. Which type of reaction actually takes place depends on

the conditions like temperature, solvent, structure of the halogenoalkane and the identity ofthe nucleophile e.g. if the nucleophile is a strong base or not.

Example:

A hot alcoholic hydroxide ion (e.g. NaOH/KOH dissolved in ethanol) can either substitute

the halogen atom or effect an elimination (remove H+) to produce an alkene.

However, in these conditions, hot alcoholic (condition = heat and alcoholic OH-), the

elimination is favoured. In fact, the OH- reacts with the ethanol to form water and the

ethoxide ion, C2H5O

-

, which then acts as a base in the reaction with the bromoalkane(stronger base than OH-).

Mechanism for Elimination Reactions

In each mechanism a base (e.g. C2H5O-, ethoxide ion, from a hot alkanoic solution of NaOH)

abstracts a hydrogen atom from a carbon which is the carbon next to the carbon which

carries the halogen:


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Overall equation: NaOH + CH3CH2Br CH2CH2 + H2O + NaBr

or C2H5ONa + CH3CH2Br CH2CH2 + NaBr + C2H5OH

E2:

1 step (similar to SN2); bimolecular; transition state; this involves:

reagent acts as a strong base and abstracts hydrogen (it accepts a hydrogen ion) from

carbon adjacent to the carbon which carries the halogen (instead of attacking

positive carbon); it does this by donating an electron pair to the carbon (Lewis base);

at the same time: the carbon-hydrogen bond breaks and is used to make a double

bond between the two carbons

at the same time: the halogen leaves from the other carbon taking the bonding pair

Rate = k [RHal] [CN-]

Overall second order

Favoured by primary halogenoalkanes

E1:

2 steps; first step unimolecular; this involves:

1ststep: formation of cation/intermediate as halogen leaves (ionisation) and takes bond (at

this point, either a E1 or a SN1 can occur as there is competition between the two

pathways)

2ndstep: reagent acts like a strong Bronsted base as it abstracts hydrogen from a carbon

attached to the carbon to which the halogen was attached; electrons from this carbon-

hydrogen bond are used to make the double bond.

rate = k [RHal]

overall first order

favoured by tertiary halogenoalkanes: order of reactivity: R3CHal R2CHHal RCH2Hal


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11.Condensation Reactions

- Condensation reactions are reactions between two different reactants that react

together to form a larger molecule with the elimination of a small molecule such as

water.

Reactions between alcohols and carboxylic acids esterification

- Conditions for esterification: heatand concentrated sulphuric acid.

-

For example when ethanol and ethanoic acid are heated in the presence ofconcentrated sulphuric acid an ester called ethyl ethanoate is formed.

- CH3CH2OH + CH3COOH CH3CH2OOCCH3 + H2O

In the above reaction, the hydrogen on the alcohol is eliminated whilst the hydroxide is

eliminated on the acid molecule and both join to form water. It is the carbon-oxygen bond in

the acid molecule that breaks.

Esters have sweet and fruity smells and are used in: perfumes, flavourings in food, solvents,

pain killers and in the production of fibres e.g. polyester

Reactions of amines with carboxylic acids

-

Amines and carboxylic acids react to form amides.

-

The OH group on the acid is eliminated and forms water with the hydrogen

eliminated from the amine group.


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Condensation polymerization

Condensation polymerization requires 2 different monomers with each monomer having two

functional groups so that a new bond can be made between the two monomers. The

monomers can have the same functional groups on them as those used in the examples below

or they can have two different functional groups like in amino acids.

Polymerization involving alcohols and carboxylic acids to form polyesters

Example:ethane-1,2-diol and benzene-1,4-dicarboxylic acid.


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Polymerization involving carboxylic acids and amines to form polyamides

Example: polymerization involving 1, 6-diaminohexane and 1, 6-hexanedioic acid.

12.Stereoisomerism

Stereoisomers are compounds with the same molecular formula andsame structural formulabut with different arrangements of atoms in space i.e.the order in which atoms are bonded

along the chain is the same but the spatial arrangement around certain (carbon) atoms is

different.

There are two types of stereoisomers: geometric isomerism and optical isomerism. Within

each type of stereoisomers there are only two isomers.

Geometric isomerism (cis-trans isomerism):

This type of isomerism occurs because free rotation of a carbon-carbon bond is prevented

either because it is a double bond or because the carbon-carbon bond is part of a ring

structure.

As a result geometric isomerism happens in:

Non-cyclic alkenes in which there are 2 different atoms (or groups of atoms) on each

carbon atom on either side of the double bond. The bond prevents the C=C from

rotating and changing the position of the atoms on either side. These 2 different atoms on

the one carbon atom can be the same different atoms on the other atom or they can be

different. If both sets of different atoms are all different than we cannot use the terms cis

and trans anymore.


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In cycloalkanes when the same atom (other than hydrogen) is found on two carbons inthe ring; no rotation possible without breaking the ring.

Non-cyclic alkenes

Refers to a different arrangement in terms of orientation of atoms or groups attached to a

double bond (free rotation is impossible as there are two regions of overlap within the

bond; above and below the axis between both nuclei; rotation would cause the overlapping

orbitals to separate and therefore break the bond) (a single sigma can rotate as the area of

overlap is between both nucleii).

The two isomeric forms cannot be interconverted without breaking a carbon-carbon bond;

this means that both are chemically different compounds!

Two possible isomers:

cis: two identical groups are on the same side of the double bond;

trans: two identical groups on opposite side of the double bond.

Example 1: cis-but-2-ene trans-but-2-ene

melting point

(C)

- 139 - 106

boiling point

(C)

4 1

Example 2:

cis-1,2-dichloroethene

trans-1,2-dichloroethene


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melting point

(C)

- 80 - 50

boiling point

(C)

60 48

Geometrical isomers have different physical properties and sometimes also, chemical

properties.

Examples:

Physical:

The trans isomers may have lower boiling points than cis-isomers - this is the case

because the trans-isomer is a non-polar molecule whilst the cis-isomer has a dipole

moment so therefore stronger intermolecular forces.

However, trans-isomers tend to have higher melting points as their molecules can fit

more tightly together (more like a mosaic of bricks) than the molecules in the cis-isomers.

As intermolecular forces can only act over a short distance, packing molecules more

closely makes the intermolecular forces more effective needing more energy to melt.

More examples: physical properties of cis/trans isomers of but-2-ene-1,4-dioic acid

(1) The trans form: density = 1.64 g cm-3, solubility in water 0.7g/100 cm3at 25 o C,

melting point 287 o C,

(2) The cis form: density = 1.59 g cm-3, solubility in water 78.8g/100 cm3at 25 o C,

melting point 130 o C,

In this example, the trans form has a much higher melting point as a result of

intermolecularhydrogen bonding. However, the cis form, because both COOH groups

are near each other there is intramolecularhydrogen bonding (i.e. between both COOH

groups within the same molecule) and therefore less intermolecular hydrogen bonding.

Intramolecular bonding does not determine any melting and boiling point

Chemical: when cis-but-2-ene-1,4-dioic acid is heated gently, water is eliminated and the

two carbons at either end of the chain are linked to form a cyclic anhydride (contains

O=C-O-C=O) while the trans isomer only sublimes at that temperature.

Explanation of reaction of cis-isomer: because bothCOOH are on the same side (close

together), they react, eliminating a water molecule and forming a cyclic anhydride. In the

trans isomer, both COOH groups are too far apart for such a reaction to happen.


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+ H2O

Cyclo-alkanes

Geometric isomerism also occurs in cyclic alkanes.

Free rotation around a single carbon-carbon bond is also restricted in cyclic alkanes due to

the inflexibility of the ring. Cyclic alkanes have a planar structure which allows cyclo-

alkanes to have other atoms, e.g. halogens, connected on the same side of the plane or one

halogen on one side and one halogen on the other side of the ring.

Geometric isomerism can be recognised by looking for:

a double bond with two different atoms/groups joined to the atoms at either end of the

double bond;

a ring structure with two different atoms or groups of atoms joined to any two carbon

atoms in the ring.

Optical isomerism

Optical isomers (or enantiomers) are two molecules with the same molecular and structural

formulas that are mirror images from each other and that cannot be superimposed. A

substance which has two mirror-image molecules which cannot be superimposed is calledchiral.


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You can determine if a substance is chiral by looking for one or more chiral centres in its

molecule.

A chiral centre is an asymmetric carbon atom which is a carbon atom that has four different

atoms of groups of atoms bonded onto it.

These four different groups can be arranged differently in two different ways; each way is

called an enantiomer.

However, some molecules with chiral centres have mirror-images which are super-imposable

and are therefore not chiral or optically active.

Similarities between enantiomers

enantiomers have identical melting and boiling points as the different spatial arrangement

does not affect the strength of intermolecular forces; also because of this solubility is the

same;

have identical chemical energetic stability and identical reaction paths (same activation

energies) which means that when a chiral molecule is formed during a reaction both

enantiomers are formed in equal amounts (=racemic mixture or racemate);

Differences

chemical: they react differently with other chiral molecules; which is why enzymes

(which are chiral) only react with a specific protein (chiral molecules) to catalyse a

reaction;

physical: the chiral centre in each enantiomer rotate the plane of polarised light in theopposite direction (non-chiral centres do not rotate the light at all!); plane-polarised light is

light that emerges from a polariser and is characterised by the fact that the electric field of

the light oscillates in one plane only e.g. perpendicular to the direction in which the light

travels. Using a polarimeter, which measures the angle of rotation, as shown below, the


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degree of rotation can be measured. Both enantiomers will rotate the light over the same

angle but in the opposite direction

Chapter 10 Organic Chemistry Notes

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