Sunday 13 April 2014

Topic 20.6: Stereoisomerism

20.6.1 Describe stereoisomers as compounds with the same structural formula but with different arrangements of atoms in space

Most molecules are not a rigid species. They usually have free rotation about the single bonds, although in certain cases, this might not be the case. There is a strict rotation about double bonds, effectively fixing the atoms into fixed positions.

Two molecules with identical formulas may have their arrangements such that the relationship between the atoms is in some way different. These are called stereoisomers.

There are two types of stereoisomerism, geometric and optical.



20.6.2 Describe and explain geometrical isomerism in non-cyclic alkenes

Geometric isomerism is caused by the lack of rotation between the carbon carbon double bonds

If the alkene has two different substituents on each carbon atom then it is possible to produce two structures that have the same name and molecular formulas.

1,2 - dichloroethene is an example of geometrical isomers.

When both chlorine atoms are on the same side, it is represented with (cis-) and when chlorine atoms are on opposite sides, it is represented with (trans-)



20.6.3 Describe and explain geometrical isomerism in C3 and C4 cycloakanes

Cyclic molecules such as cyclopropane or cyclobutane also have restricted rotation about the carbon - carbon double bonds, as rotation would break the ring structure. This gives the possibility of geometric isomerism, with a difference between sites above the plane of the heterocycle and those below.

1,2 - dichlorocyclobutane

When both chlorine atoms are on the same side, it is represented with (cis-) and when chlorine atoms are on opposite sides, it is represented with (trans-)



20.6.4 Explain the difference in the physical and chemical properties of geometrical isomers

Geometric isomers have different physical and chemical properties

Physical properties

Cis - 1,2-dichloroethene has both of the chlorine atoms on the same side as each carbon - chlorine bond is polar the overall molecule itself is polar. However, the trans, the individual dipoles in the bonds then cancel out and the molecule is non-polar overall.

Cis - 1,2 - dichloroethane  = 60.2

Trans - 1,2 - dichloroethane = 48.5

The two geometric isomers of butenedioic acid (maleic and fumaric acid) also demonstrate different physical properties. The melting points are very different as the close proximity of the two -COOH groups allows for the formation of intramolecular hydrogen bonds. This decreases the possibility of intermolecular hydrogen bonding and reduces the melting points.

Cis-butenedioic acid + 135

Trans-butenedioic acid = 287


Chemical properties

Action of heat allows cis to bond with each other and form a cyclo.




20.6.5 Describe and explain optical isomerism in simple organic molecules


Chiral carbons must be joined to four different atoms or groups, and are also as a description of the molecule itself.

Enantiomers are optical isomers. Racemic mixture = an equimolar mixture of two optical isomers whose opposite rotation effect on plane polarised light cancels out.


20.6.6 Outline the use of a polarimeter in distinguishing between optical isomers

Plane polarised light is light with which all light waves are vibrating in the same orientation. This can be achieved by passing normal light through a lens which contains thousands of vertical lines called polarised filters.

The lines that emerges from the polarised filter has only vertical orientated light.

When the light passes through the solution of an optical isomer, the angle rotated is dependent on three factors.

The type of enantiomers, path length and concentration of enantiomers

The light emerging from the sample solution can be analysed with another filter. The two enantiomers rotate the polarised light the same angle, only different direction.



20.6.7 Compare the physical and chemical properties of enatiomers

The physical properties of enantiomers are identical except for different angles on plane polarised light.

The chemical properties are identical as well except with enzymes. React differently with other optical isomers. This is because enzymes are used to optically active enantiomers.


Topic 20.5: Reaction pathways

20.5.1 Deduce reaction pathways given the starting materials and the product

The IB requires you to know all the reaction in the diagram below


Halogenoalkane


--> Ethene

Hot Hydroxide in ethanoic acid

Hydrogen halide is released


--> Nitrile

SN2 reaction

Substitution reaction

Bromine is released


--> Amine

SN2 reaction

Substitution reaction

Bromine is released



Nitrile


--> Amine

+Hydrogen and nickel catalyst

Hydrogenation reaction



Carboxylic acid + alcohol

--> Ester

Warm sulphuric acid

redox reaction


Carboxylic acid + amine

--> Amide

Condensation polymerisation reaction



Topic 20.4: Condensation reactions

20.4.1 Describe, using equations, the reactions of alcohols with carboxylic acids to form esters, and state the uses of esters

Ester is often used for solvents. Responsible for smells of fruits



20.4.2 Describe, using equations, the reactions of amines with carboxylic acids

Water is produced from the condensation reaction. Small molecule is produced, does not necessary have to be water.



20.4.3 Deduce the structures of the polymers formed in the reactions of alcohols with carboxylic acids

Polyester

Put the brackets across the repeating unit. This is the example used in the syllabus.



20.4.4 Deduce the structures of the polymers formed in the reactions of amines with carboxylic acids

Polyamide

This is nylon made from condensation reaction. This example was used in the syllabus.


Note that they have double ended functional group.


20.4.5 Outline the economic importance of condensation reactions

The development of condensation polymers gave another push to the blossoming plastic and artificial fibre industries. Condensation polymers were used to artificial silk and cotton.

Kelvar is a polyamide that is used as a bullet-proof vest. PVC is used for insulating wires as they are durable.



Topic 20.3: Elimination reactions

20.3.1 Describe, using equations, the elimination of HBr from bromoalkanes

Bromoalkane give alkene by dehydrohalogenation using heat and concentrated sulphuric acid




20.3.2 Describe and explain the mechanism for the elimination of HBr from bromoalkanes

The mechanism first involves removal of a photon (abstraction) by the base. This happens to the hydrogen atom which is on the carbon adjacent to the carbon holding the bromine atom. This hydrogen atom is said to be an 'alpha-hydrogen'


This produces a negatively charged species (a carbocation), which then rearranges by moving the negatively charge lone pair into a pi orbital between the two carbon atoms with the loss of the bromine atom as bromide ion


Topic 20.2: Nucleophilic substitution reactions

20.2.1 Explain why the hydroxide ion is a better nucleophile than water

Nucleophile are compounds which are attracted to positive compounds. They are a chemcial species that are able to donate an outer electron pair to an electron deficient species

Water has no charge hence it does not bond with compounds. Hydroxide ions are negative and using the knowledge of opposite attract, this shows hydroxide ions are better nucleophile than water.



20.2.2 Describe and explain how the rate of nucleophilic substitution in halogenoalkanes by the hydroxide ion depends on the identity of the halogen

Bond breaking is an endothermic process

The stronger the carbon - halogen bond, the more energy is required to break this bond, the less energetically favourable the reaction

The more mass the halogen have, the more electron shielding thus less attraction. Iodine bonds are weakest and easiest to break while fluorine has the strongest bonds.



20.2.3 Describe and explain how the rate of nucleophilic substitution in hydrogenoalkanes by the hydroxide ion depends on whether the hydrogenoalkane is primary, secondary or tertiary

The reaction involving the intermediate carbocations (SN1) reacts faster than the formation of the activated complex (SN2). This is because the formation of an activated complex requires a larger activation energy (collision energy)

Tertiary halogenoalkanes react faster than secondary and primary react the slowest



20.2.4 Describe, using equations, the substitution reactions of hydrogenoalkanes with ammonia and potassium cyanide

The ammonia molecule has a lone pair, but it is neutral and must also lose a hydrogen ion before arriving at the final product.


The problem with this reaction is that the primary amine that is formed can further react with other molecule of halogenoalkane, as it still has the lone pair on the nitrogen. To prevent this, they use excess nitrogen to prevent further substitution reaction.

The cyanide ion is a good nucleophile as there is a formal negative charge and a lone pair on the carbon atom.


The reaction is very useful as it adds a carbon atom to the chain.


20.2.5 Explain the reactions with primary hydrogenoalkanes with ammonia and potassium cyanide in terms of the SN2 mechanism

Primary halogenoalkanes are attacked by the lone pair of the nitrogen atom

This makes an unstable transition state with a five membered carbon atom

The halogen atom bond can then break in a concerted process in which the nitrogen atom always loses a hydrogen ion to restore the neutrality of the product molecule


The cyanide ion is provided by potassium cyanide by ensuring that it is in an alkaline medium (pH>7)

This then attacks the primary halogenoalkane in the same way as the hydroxide ion:



20.2.6 Describe, using equations, the reactions of nitriles using hydrogen and a nickel catalyst

Nitrile can be transformed into an anime by heating with hydrogen at 150 degree celcius in the presence of a nickel catalyst.



Topic 20.1: Introduction

20.1.1 Deduce structural formulas for compounds containing up to six carbon atoms with one of the following functional groups: amine, amide, ester and nitrile

Amine (-amine)



This is propan-1-amine



There are different degrees of amines, it is dependent on how many carbons are connected to the nitrogen.

The naming of the compound depends on the longest carbon chain it is linked to

Propan-1-amine

N - methyl propan-1-amine

N,N - dimethyl propan-1-amine


Amide (-amide)


This is ethanamide


There are different degrees of amines, it is dependent on how many carbons are connected to the nitrogen.

The naming of the compound depends on where the double bond oxygen is.

Propanamide

N - methyl propanamide

N,N - dimethyl proapanamide


Ester (-anoate)


To name an ester, name R' first then R

These are a few examples



Nitrile (-nitrile)


Always remember to count the carbon in the nitrile



20.1.2 Apply IUPAC rules for naming compounds containing up to six carbon atoms with one of the following functional groups: amine, amide, ester and nitrile

Amine (-amine)

Amide (-amide)

Ester (-anoate)

Nitrile (-nitrile)

Topic 20: Organic chemistry

Topic 20 of the IB HL Chemistry syllabus is the Organic chemistry. IBO recommends to spend 10 hours on this topic.

This topic has 6 sub-chapters: "Introduction", "Nucleophilic substitution reactions", "Elimination reactions", "Condensation reactions", "Reaction pathways" and "Stereoisomerism". Each are separated with numerical values in order of mentioned.

These are advanced HL syllabus statements, it is recommended to bring a Casio Graphical Calculator instead of Texas. Casio Calculators have the periodic table installed already.

Topic 10.6: Reaction pathways

10.6.1 Deduce reaction pathways given the starting materials and the product

Alkene --> Ketone

Alkane --> Alcohol

Halogenalkane --> Aldehyde

Every arrow on the picture represents a reaction. IB requires you to remember all reaction represented


Alkene

--> Polyalkene

+multiple alkene

Polymerisation reaction


--> 1,2 - Dihalogenoalkane

+ halogen

Halogenation reaction


--> Halogenoalkane

+ Hydrogen halide



--> Alkanes

+ Hydrogen (Nickel catalyst, heat)

Hydrogenation reaction


--> Alcohols

+ Water (Concentrated sulphuric acid catalyst)

Hydration reaction



Alkane

--> Halogenoalkane

+ Halogens (UV light catalyst)

Free radical mechanism required


Halogenoalkane

--> Dihalogenoalkane

+ Halogen (UV light catalyst)



--> Alcohol

+ Aqueous hydroxide (warm)

Substitution reactions



1st degree Alcohol

--> Aldehyde

+ Heat and potassium dichromate catalyst

Oxidation reaction


2nd degree Alcohol

--> Ketone

+ Heat and potassium dichromate catalyst

Oxidation reaction


Aldehyde

--> Carboxylic acid

+ Heat and potassium dichromate catalyst

Oxidation reaction

Topic 10.5: Halogenoalkanes

10.5.1 Describe, using equations, the substitution reactions of halogenoalkanes with sodium hydroxide

Halogenoalkanes can be classified according to their structure as primary, secondary or tertiary.

This class of structure plays an important role in the reactions of the halogenoalkanes. Halogenoalkaenes contain polar carbon - halogen bonds that make them easy to convert to other products

Halogenoalkanes will undergo nucleophilic substitution (Sn) reactions. This involves the attack of a negatively charged / neutral electron rich species (called a nucleophile) on the slightly positive carbon atom bonded to the halogen

This causes the polar carbon - halogen bond to break so that the halogen is substituted by the nucleophile

The are two mechanisms for nucleophilic substitution reactions, called SN1 and SN2. This stands for substitution nucleophilic first or second order.



10.5.2 Explain the substitution reactions of halogenoalkanes with sodium hydroxide in terms of SN1 and SN2 mechanisms

Tertiary halogenoalkanes react in a SN1 mechanism

The rate of this reaction is dictated by one factor, the concentration of the halogenoalkane

Only the halogenoalkane is involved in the rate expression and therefore involved in the rate determining step

This is therefore a first order reaction that is unimolecular process with a molecularity of one

The first stage of the mechanism involves the breaking of the carbon - halogen bond to give an intermediate carbocation

This is bond breaking is called heterolytic fission. This is a slow process and so is the rate-determining step

The second stage involves the nucleophile bonding with the carbocation intermediate. This happens very quickly. It therefore does not appear in the rate expression


Note: To gain all the marks off this questions, this is directly pulled off the answers of past papers

SN1:
Curly arrow showing Cl leaving;
Representation of tertiary carbocation;
curly arrow going from lone pair / negative charge on O in OH- to C+;
Do not allow arrow orginating on H in OH-;
Formation of organic product
(CH3)3COH and CL-


SN2 mechanism has primary halogenoalkanes reacting this way.

The rate of reaction is dictated by two factors. By both the concentration of the halogenoalkane and the concentration of the nucleophile

Both the halogenoalkane and the nucleophile are involved in the rate expression and therefore both are involved in the rate determining step

This is a second order reaction, bimolecular process with a molecularity of two.

This mechanism involves the simultaneous attack of the nucleophile and loss of the halogen

An activated complex is formed during the process


Note: To gain all the marks off this questions, this is directly pulled off the answers of past papers

SN2:
Curly arrow going from lone pair / negative charge on O in OH- to C;
Do not allow curly arrow originating on H in OH-:
Curly arrow showing Br leaving;
Accept curly arrow either going from bond between C and Br to Br in bromoethane or in the transition state;
Representation of transition state showing negative charge, square bracket and partial bonds;
Do not penalize if HO and BR are not a t 180 to each other
Do not award M3 if OH---C bond is represented
Formation of organic product CH3CH2OH and Br-

Topic 10.4: Alcohols

10.4.1 Describe, using equations, the complete combustion of alcohols

The complete combustion of alcohols produces water and carbon dioxide. It is extremely important to remember the alcohols already have one oxygen atom in them already.


The equation should look something like this.



10.4.2 Describe, using equations, the oxidation reactions of alcohols

Oxidation reactions of alcohol require heating with an acidic solution of potassium dichromate (IV)

There are three types of alcohols depending on their degree of carbon, they will undergo different reactions.


Primary alcohols are oxidised to aldehydes, with the alcohol group becoming a carbonyl group

Repeating the same process for aldehydes would create carboxylic acids.



Secondary alcohols are oxidised to ketones. The ketones are unable to oxidise further.


Tertiary alcohols are unable to be oxidised.



10.4.3 Determine the products formed by the oxidation of primary and secondary alcohols

The products formed by the oxidation of primary alcohols are aldehydes then carboxylic acids.

The products formed by the oxidation of secondary alchols are ketones.