Chemistry 3.5 Describe the structure and reactions of organic compounds containing selected organic groups.

Chemistry 3.5 Describe the structure and reactions of organic compounds containing selected organic groups

You will notice the first thing the 3.5 standard says is :
You will be expected to know the principles of Organic Chemistry you covered in AS 2.5 How well do you know your stuff from last year? The standard then goes on to say….

Organic compounds described are limited to those containing no more than 8 carbon atoms
Don’t think you get off that easy It then goes on to say ……… Larger organic molecules may be used in questions involving the application of organic principles (e.g. the identification of functional groups)

So lets start with Alkanes
1. What’s the general formula for an alkane? CnH2n+2

Can you draw the structural formula and molecular formula for the following?
Methane Molecular Formula - CH4 Structural Formula

Drawing a 3 dimensional Structure
How about drawing the 3 dimensional structure for methane? Use the molymods to make a 3 D model Any ideas on how to draw it? These two lines represent bonds on the same plain as the paper These lines represent the bond going behind the paper This wedge represents the bond coming out from the paper

What is the bond angle between each of the atoms?
Describe the shape of this molecule Tetrahedral Is CH4 polar or non polar? Non polar Can you give a reason why?

Condensed Structural Formula
Alkane Nomenclature - Give the names, molecular and condensed structural formula for the first ten alkanes Name Molecular Condensed Structural Formula methane CH4 C2H6 (CH3)2 ethane C3H8 propane CH3CH2CH3 CH3(CH2)2CH3 C4H10 butane pentane C5H12 CH3(CH2)3CH3 hexane C6H14 CH3(CH2)4CH3 heptane C7H16 CH3(CH2)5CH3 octane C8H18 CH3(CH2)6CH3 nonane C9H20 CH3(CH2)7CH3 C10H22 decane CH3(CH2)8CH3

Can you remember the alkane order?
A simple mnemonic is- Many Elderly People Buy Pent Houses High Over North Dakota Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane

Cyclic Alkanes These are Alkanes with at least 1 ring of carbons eg cyclohexane Draw the structure of cyclohexane in your book? Cyclohexane Draw cyclopropane The general formula for an alkane with one ring is: CnH2n

Do you remember how to name a branched alkane?
Steps 1. Find the longest chain of continuous carbons (called the parent chain) and name it: ie 5 carbons – name pentane 2. Identify any side branches or functional groups ie methyl if there are more than one of the same type use prefixes di (2) , tri (3) etc. 2 methyls = dimethyl. So the name so far is dimethylpentane 3. Number the parent chain from the end that gives the side branch groups the lowest number methyl Pentane (parent chain) methyl

Steps continued 3. Number the parent chain from the end that gives the side branch groups the lowest number ie from the left methyl 1 2 3 4 5 methyl 4. Separate numbers by comma’s and separate numbers from words with a dash – now you can name it 2,3-dimethylpentane

Exercise – Draw the structures for the following 3 – ethylheptane
2,2,4-trimethylpentane

Lets look at making some structure
Turn to Expt 1 in your booklet In pairs make and draw the structures for Q 1 to Q 3 Answer the questions

Structure Classification Name CH3CH2C CCH3 alkyne pent-2-yne alcohol
(secondary) CH3CH2CHCH2CH2CH2CH2CH3 OH octan-3-ol O carboxylic acid 2-methylbutanoic acid CH3CH2CH C CH3 OH H H C C cis-pent-2-ene alkene H3C CH2CH3 CH3CHCH2CH3 Br haloalkane 2-bromobutane CH3CH2CH2CH2CHCH2CH3 CH3 alkane 3-methylheptane

Structure Classification Name CH3CH2 CH3 C C alkene
2,3-dimethylpent-2-ene H3C CH3 CH3 alcohol (tertiary) CH3CH2CH2C OH 2-methylpentan-2-ol CH3 O H C ester butyl methanoate O CH2CH2CH2CH3 CH3 CH3CH2CHCH2CHCH3 alkane 2,4-dimethylhexane CH3

Structure Classification Name CH C CH CH3 3-methylbut-1-yne alkyne CH3
CH3CHCH2CHCH3 haloalkane 2-chloro-4-methylpentane CH3 Cl carboxylic acid CH3CH2CH2CH2CH2CH2COOH heptanoic acid O CH3CH2 C ester propyl propanoate O CH2CH2CH3

Cholesterol is a major component of gallstones
Cholesterol is a major component of gallstones. From the following structure of the compound predict its reaction with .. C8H17 CH3 CH3 HO Br2 H 2 with a Pt catalyst CH3COOH cholesterol

The reaction with Br2 results in addition of bromine to the double bonded carbons forming a single carbon bond. The solution would change colour from orange to colourless. C8H17 CH3 CH3 HO Br Br

With H2 and a Pt catalyst a hydrogenation reaction would occur and H atoms would be added across the double bond forming a single C – C bond. C8C17 CH3 CH3 HO H H

Ethanoic acid reacts with the hydoxy group to form an ester and water
C8H17 CH3 CH3 O H3C C O + H2O

We have some new functional groups to learn this year
Title your page Functional groups Use the photocopied sheet and copy the complete the table neatly into your exercise book Yes the whole table! We must learn these!

Complete the task on the handout , glue into your lab book
Use your chart to help you classify and name the listed compounds (complete in pencil)

Answers to the left hand column on handout
CH3CH2CH2CH2CHCH2CH3 Cl Class haloalkane Name 3-chloroheptane O CH3CH2CH2 C Cl Class acyl chloride Name butanoyl chloride CH3 HO C CH2CHCH3 O Class carboxylic acid Name 3-methylbutanoic acid

Answers to the left hand column continued
CH3CH2CH2CH2NH2 Class amine Name 1- aminobutane O CH3CH2CH2CH2CH2C NH2 Class amide Name hexanamide CH3CH2CH2CH2 O C CH2CH2CH2CH3 O Class ester Name butyl pentanoate

Answers to the Right hand column
CH3CH2CH2COCH2CH3 Class ketone Name hexan-3-one CH3CH2CH2CH2CH2OH Class alcohol Name pentan-1-ol CH3CH2CH2CH2CH2C O H Class aldehyde Name hexanal

Answers to the right hand column continued
CH3CH2CH2CH2CH2COCl Class acyl chloride Name hexanoyl chloride CH3CH2CHCH3 NH2 Class amine Name 2-aminobutane CH3CH2CH2CH2CH2CH2CH2CONH2 Class amide Name octanamide 27

Structural Isomers (are also called constitutional isomers)
These are molecules with the same molecular formula but different structural formula. The isomers of a particular molecule will have different physical properties e.g. melting and boiling points. They may also have different chemical properties. Draw and name the structural isomers of C4H10 Name: butane Name: 2-methylpropane Boiling point 36.1o C Boiling point -11.7o C

Task – in pairs use the models to make hexane
Draw the structural formula for hexane Then make as many structural isomers of hexane as you can Name and draw each one There should be 5? Hexane, 2,3 dimethylbutane, 2-methylpentane, 2,2-dimethylbutane , 2,3-dimethylbutane - any others?

Structural Isomers of Hexane
C6H14 H H H H H H H C C C C C C H H H H H H H H CH3 H H H 2-methylpentane H C C C C C H H H H H H

CH3 H H 2,2-dimethylbutane H C C C C H H CH3 H H CH3 H CH2 H H 3-methylpentane H C C C C H H H H H

CH3 H H 2,3-dimethylbutane H C C C C H H H CH3 H

Geometric (cis and trans) Isomers
Geometric Isomers will only occur if…. The compound has a double or triple bond where there can be no rotation around the C C bond Remember alkanes don’t exhibit geometric isomerism because there is rotation around the single C C bond

e.g. The Geometric Isomers of but-2-ene
To exist as geometrical isomers the C atoms at both ends of the double bond must each have two different groups (or atoms attached). e.g. The Geometric Isomers of but-2-ene cis-but-2-ene trans-but-2-ene Bpt 3.7o C Bpt 0.88 o C * Geometric isomers have similar chemical properties but different physical properties

Geometric isomerisim does not occur if one of the carbon atoms in the fixed (ie the double bond) has two identical atoms or groups of atoms attached But-1-ene does not have geometric isomers, because it has two groups, H atoms, attached to the C atoms on either side of the double bond flip it over and it’s the same as Therefore but-1- ene does not have geometric isomers

Geometric isomers are a form of stereoisomerisim
Stereoisomerism – are where the atoms are bonded in the same sequence but are arranged differently in space in a molecule. e.g. but-2-ene Same sequence of atoms Two different geometric isomers exist where atoms are arranged differently in space cis-but-2-ene trans-but-2-ene Bpt 3.7o C Bpt 0.88 o C

Exercise Draw the structures of the following alkenes and decide which of them can exist as cis-trans isomers 2-methylbut-2-ene b) 3 – methylpent-2-ene forms no cis/trans isomers Occurs as cis trans isomers cis-methylpent-2-ene trans-methylpent-2-ene

Identify whether cis trans isomers occur with in the following molecules
H3C CH3 H3C Cl HO Cl C C C C C C Cl CH3 H CH3 H CH3 No Yes Yes

H3C H H3C CH3 C C C C H CH3 H H Geometric isomers have the same chemical properties, but different physical properties. Why? Because cis isomers have bulky side groups and cannot be packed closely together, this causes weaker intermolecular forces between molecules and therefore lower melting points than the trans version

Cl H Cl Cl C C C C Cl H H H However cis forms are sometimes polar (as above) and therefore have stronger intermolecular forces between molecules causing higher melting points.

Testing for Cis - Trans Isomers
Weigh out 2 grams of maleic acid into a 50ml beaker Add 4mls of water and warm slightly to dissolve the acid Pour this into a pear shaped flask Carefully add 5mls of concentrated HCl Place a condenser on top of the flask and secure it in a retort stand with a water bath. Then warm the solution until a solid forms. Cool the solution to room temperature by placing the flask in a cold water bath ie a 250ml beaker of cold water Pour the solution into an evaporating dish, pouring off the excess liquid then carefully rinse with water – Then complete task B on page 159

Maleic acid (cis isomer)
COOH COOH C C H H Maleic acid (cis isomer) H O O H O C C O C C H H

Give the names and structural formula for the following substances from their condensed structural formulae. (c) CH3CH2CHCHCH2CH2CH3 H H H H H H H H C C C C C C C H hept-3-ene H H H H H (d) (CH3)3COH CH3 CH3 C OH 2-methylpropan-2-ol CH3

Give the names and structural formula for the following substances from their condensed structural formulae. (a) CH3CH2CHClCH2CH3 H H Cl H H 3-chloropentane H C C C C C H H H H H H (b) (CH3)2CHCH2CH2CHCH2 CH3 H H H H 5 -methylhex-1-ene CH3 C C C C C H H H H

Another form of Stereoisomerism is Optical isomerism

E,Z Nomenclature of bond geometry
In cis and trans – nomenclature the ‘like” groups are identified and used to specify the type of isomer. With E,Z rules the pair of substituent's at each end of the double bond are given a priority. Highest priority = atom of highest atomic no attached directly to the double bond Eg 2-chloro-3-methylpent-2-ene CH3 CH2 -CH3 This isomer has the highest priority groups on opposite sides of double bond therefore is the E isomer The Z isomer has the highest priority groups on the same side of the double bond C C Cl CH3 1st pair Cl highest priority 2nd pair CH2CH3 highest priority

Optical Isomerism Optical isomers involve an asymmetric carbon – a carbon bonded to four different atoms or groups of atoms such as: H, OH, CH3, C2H5, C3H7 etc A molecule with an asymmetric carbon is known as a chiral molecule. The two forms of the chiral molecule are known as enantiomers or optical isomers

3-D diagram of butan-2-ol Structural formula of butan-2-ol C2H5 C2H5 * CH3 * C OH C H3C OH H H C* = asymmetric carbon

3-D optical isomers of butan-2-ol
C2H5 C2H5 * * C C H3C OH HO CH3 H H Mirror line Optical isomers are: mirror images of each other Are not superimposable on each other

Optical activity Glucose is an optically active compound. On the straight-chain form of glucose shown here, all four of the carbons in the middle of the chain are chiral centres – they each have four different groups attached to them.

Four different groups can be arranged around a central atom in two different ways. These are optical isomers and are mirror images of each other and cannot be superimposed — just as your left and right hands are mirror images and cannot be superimposed. The structure for glucose shows four chiral centres, which means a total of 16 different forms are possible. Glucose is just one of those forms.

Molecules which have one or more chiral centres
Optical isomers are Molecules which have one or more chiral centres rotate plane-polarised light. Molecules which are mirror images of each other rotate plane polarised light in opposite directions.

If we add a second grill at right angles to the first, it will block the red wave.
Two polarising filters placed at right angles will block all light.

Two sheets of polarising film are placed on an overhead projector stage. The arrows on the sheets show their orientation. On the left we see that when the two sheets are orientated in the same direction light passes through them, but when the sheets are at right angles (right) the light is blocked.

Beakers of water and sucrose solution (which is cheaper than glucose and also optically active) are placed on a sheet of polarising film sitting on the overhead stage. A second sheet of polarising film is on top of the beakers, at right angles to the first.

Water Sucrose Although the film above the water beaker is dark, light shines through the film above the sucrose solution. The sucrose has rotated the light waves sufficiently so that they are able to pass through the second film.

Water Sucrose When we rotate the top sheet, the film above the sucrose solution is now dark, while light passes through everywhere else.

Different wavelengths (colours) of light are rotated by different amounts, so that as the polarising film is rotated by different angles, we see these different colours. Remember: Optically-active solutions rotate plane-polarised light optical isomers rotate plane-polarised light by equal angles in opposite directions.

Optical Isomer Properties
Many organic compounds have optical isomers including hormones and enzymes involved in biochemical reactions. The shape of a molecule is very important in these reactions and this means that the mirror image (ie the optical isomer) of an enzyme will not work properly in the body. Talk about pthalidamide

Optical Isomer Properties
Optical isomers have identical chemical and physical properties except that they rotate the plane of polarised light in opposite directions. Optical isomers react differently with other optically active compounds. an equal mixture of both enantiomers is called a racemic mixture Talk about pthalidamide

The drug thalidomide, prescribed as a treatment for morning sickness in early pregnancy in the 1960s, tragically caused the development of serious birth defects (badly deformed limbs, or none at all). In fact, only one of the two optical isomers of thalidomide appear to cause these birth defects, although it could be that once ingested each isomer readily changes into the other form. Today thalidomide is being used successfully as a treatment for leprosy (although not for pregnant women).

Exercise Draw the structure of each of the following molecules and then decide which ones are optically active. Mark the chiral carbon with an asterisk. 2-chlorobutane (b) 2-methylpropanoic acid (c) 3-methylpentanal (d) 2-aminobutanoic acid No optical isomers * * *

Homework Read unit 28 page 111 in pathfinder
Complete Q’s 4 and 5 page 114 Complete Enantiomers on BestChoice before next Friday please

Answer questions on alkanes and alkenes
page Group work exercise Each group is to work on problems giving great detailed answers. But all people in the group must have the answers written in their lab books Random people from each group will be asked for their groups answer

Turn to page 6 in your booklet
Properties of alkanes and alkenes

Turn to page 145 in the year 13 lab book
Demo Alkene addition Turn to page 145 in the year 13 lab book Comparison: Alkanes vs Alkenes

Complete Structural isomer starter
Complete Worksheet two Q’s 1 and 2 in organic booklet

Alkane Reactions Alkanes are used as fuels and undergo combustion reactions In excess air (oxygen) they form the products H2O and CO2 In limited air (oxygen) they form the products H2O and C or CO2 Because the carbon atoms in alkane molecules are saturated with hydrogens (ie they don’t have any double or triple bonds) they are called saturated hydrocarbons.

Properties of Alkanes Insoluble in water Soluble in non polar solvents
Don’t conduct – no free electrons Float on water because H2O is relatively denser Boiling/melting point increases with chain length because as molecular mass increases the intermolecular forces between molecules increases

Questions 1. Name the type of intramolecular bonding and the type of intermolecular bonding in: Methane b) Liquid pentane 2. Explain in terms of bonding why: a) Methane gas can be collected over water b) Petrol floats on water c) Oil dissolves in petrol

Alkane Reactions Because alkanes have no double bonds they react slowly with halogens in the presence of UV light. This reaction is called a substitution reaction in which an H atom is replaced by a halogen atom (eg Cl or Br) UV light butane bromine bromobutane + hydrogen bromide What’s missing in this reaction? The bromine solution decolourises slowly and the HBr formed is an acidic gas that turns moist blue litmus red

These are unsaturated hydrocarbons
This means they have at least one double or triple covalent bond These types of bonds are called functional groups because it’s at these bonds that reactions occur Alkenes (CnH2n) and Alkynes ( Cn H2n - 2 )

Alkenes and Alkynes Like alkanes these unsaturated hydrocarbons are non polar and insoluble in water. They undergo combustion the same as alkanes giving the same products They have very similar physical and chemical properties to alkanes Form addition reactions because of the reactive double or triple bond Alkenes exhibit a different form of isomerism called geometric isomerisim

2-chloro-3-methylbutane
Starter A Draw and name the 2 possible products formed when HCl is added to 2-methylbut-2-ene.Name the products and identify which is the major product. 2-chloro-3-methylbutane 2-chloro-2-methylbutane – major product H H H H H C C C C H H Cl CH3 H H CH3 H H H C C C C H H Cl H H

Hydrogenation of an Alkene (Addition reaction)
Hydrogen can be added across the double C bond The conditions for this reaction are: platinum catalyst O C pressure of 4 atmospheres The reaction is C2H4(g) H2(g) CH3CH3(g) ethene ethane

The Good Oil This hydrogenation process is used to harden plant oils to commercially produce margarine. Natural plant oils contain many double bonds per molecule, and because several double bonds exist after hydrogenation, the margarine is said to be polyunsaturated.

Bromination of an Alkene (Addition reaction)
Alkenes and alkynes undergo addition reactions with halogens to form a dihaloalkane (or tetrahaloalkane). The common test for an unsaturated hydrocarbon (ie a hydrocarbon with a C C or C C bond) is therefore the rapid decolourisation of an orange bromine solution in the absence of sunlight The reaction is CH2H4(g) Br2(l) CH2 Br CH2 Br (l) ethene ,2-dibromoethane watch demo of bromine with alkane

Alkene molecules can create polymers (plastics) by addition reactions where many alkene monomer units are joined together in the presence of heat and a catalyst

The process involves the breaking of one of the bonds in the double bond in each alkene molecule.
Each of the two electrons from the bond go to each end of the molecule to create a bond with another molecule that has undergone the same process. This creates long chains of joined monomers to create a polymer. Can be drawn as H H H H . . C C C C H H H H

. . . . . . . . . . . . Representation of adddtion polymer reaction H
3 ethene monomers . . . . . . C C C C C C H H H H H H Repeating monomer unit Heat and a catalyst added H H H H H H H H H H H H Polyethylene polymer . . . . . . C C C C C C C C C C C C H H H H H H H H H H H H

Changing the side chain of the monomer in the reaction gives different polymers ie
Cl H Cl H Cl H 3 vinyl chloride monomers . . . . . . C C C C C C H H H H H H Repeating monomer unit Heat and a catalyst added Cl Cl H H Cl Cl H H Cl Cl H H Polyvinylchloride Polymer Aka PVC . . . . . . C C C C C C C C C C C C H H H H H H H H H H H H

This year we will also look at polymer reactions involving condensation reactions

Addition polymers are formed when alkene monomers undergo addition to form a polymer eg. polythene from ethene, P.V.C. from vinyl chloride (chloroethene), polypropene from propene.

Haloalkanes (alkyl halides) RX
where X = F, Cl, Br, I Named as a chloroalkane or bromoalkane etc, with the position of the halogen given by the appropriate number of the carbon that it is attached to in the chain. Exist as primary, secondary, tertiary

The haloalkanes can be classified as:
primary - the C atom to which X is attached is only attached to one other C atom eg H H H C C Br H H Carbon attached to Br is attached to one carbon

Secondary haloalkane - the C atom to which X is attached is attached to two other C atoms
eg Carbon attached to Br is attached to two other carbons CH3 H C Br CH3

Tertiary haloalkane - the C atom to which X is attached is attached to three other C atoms.
eg Carbon attached to Br is attached to three other carbons CH3 CH3 C Br CH3

The Lucas Test The Lucas test is used to distinguish between the primary, secondary and tertiary alcohols The Lucas reagent consists of ZnCl2 in concentrated HCl The zinc chloride catalyses a substitution reaction between the alcohol and the concentrated HCl Chloroalkanes form and appear as a cloudy suspension in the water because they are insoluble

The Lucas test *Important
The rates at which the different types of alcohol react to form chloroalkanes enable them to be classified as follows:

The Lucas Test Type of alcohol Example Observation Product Primary
Butan-1-ol No cloudiness, very slow if at all 1-chlorobutane Secondary Butan-2-ol Cloudiness after 5-15 minutes 2-chlorobutane Tertiary 2-methylpropan-2-ol Cloudiness after 1-2 minutes 2-methyl-2-chloropaopane

Starter: Write out and fill in the missing words
The s_____ of the C-OH b____ i_______ from tertiary to secondary to primary alcohols (which have the strongest C-OH bond.) The test for the C-OH strength is called the l____ test which uses concentrated ____ and Z_________ as the catalyst. The speed of this s_________ reaction of the OH for a Cl indicates the type of Alcohol trength ond ncreases ucas HCl inc chloride ubstitution

Properties of haloalkanes

Haloalkanes do not form hydrogen bonds, so they have lower boiling points than alcohols and are not miscible in water. However, they are polar compounds, so have higher boiling points than their parent alkanes. The lowest mass haloalkanes are gases at room temperature, but the rest are volatile liquids.

To make a haloalkane we can substitute the OH on an alcohol using eg
To make a haloalkane we can substitute the OH on an alcohol using eg. PCl3, PCl5,SOCl2 or conc HCl/ZnCl2 PCl3 or PCl5 C2H5OH C2H5Cl chloroethane ethanol SOCl2 C2H5OH C2H5Cl + HCl +SO2 ethanol chloroethane Conc HCl/ZnCl2 C2H5OH C2H5Cl ethanol chloroethane

A monohaloalkane eg. 2-bromopropane can be formed by:
Haloalkanes are relatively nonpolar overall (despite the polarity of the C-X bond) and are insoluble in water. A monohaloalkane eg. 2-bromopropane can be formed by: addition of HBr to propene (forming only one product) H H H CH3 H + HBr C C H C C C Br H H H H H

substitution of propane using Br2
substitution of propane using Br2. (forming two products, the bromoalkane and HBr) H H H H H H H C C C H + Br2 H C C C H + HBr H H H H H Br

Lab book Page 201 Preparation of a Haloalkane

Substitution of haloalkanes to form alcohols
Like tertiary alcohols, tertiary haloalkanes are easily substituted. A tertiary haloalkane will react with cold water to form an alcohol: R—X + H2O → R—OH + HX We can tell whether this reaction has taken place by the presence of the X– ions, which will form precipitates with silver nitrate solution.

summary! Tertiary haloalkanes form alcohols in cold water
Secondary haloalkanes react when the water is warm Primary haloalkanes do not react with water, but react to form alcohols with aqueous sodium hydroxide.

Nucleophiles A nucleophile is any species which loves nuclei, that is anything attracted to a positive charge. Nucleophiles are therefore species that carry a negative charge or a lone pair of electrons, eg OH- , H2O and NH3 (in alcohol) The C -X bond is polar and the slighty positive C atom is prone to attack from a negative nucleophile. eg R X(l) + OH- (aq) R OH + X- (aq)

Elimination to form an alkene
Haloalkanes can also undergo an elimination with hydroxide to form an alkene: R—X + alcOH– → R’=C + HX Use the above reaction as a template to write your own formation of an alkene from a haloalkane

When deciding where to put the double bond in an elimination reaction, apply the rule ‘the poor get poorer’. One carbon of the double bond will be the carbon that lost the halogen. To decide whether the bond goes to the left or right of that carbon, look at the number of hydrogen atoms on each of those carbons. The carbon to lose the hydrogen atom (and thus become the other half of the double bond) is that carbon which has the fewer hydrogen atoms already bonded to it. H Cl H H H H (Alc)KOH H C C C C H H C C C C H + HCl H H H H H H H H

Draw the structural formula for the reaction of the tertiary haloalkane 2-chloro, 2-methylpropane with water H CH3 H H CH3 H H C C C H + H2O H C C C H + HCl H Cl H H OH H 2-methylpropan-2-ol Tertiary alcohol

Amines (aminoalkanes)
Amines are named as substituents eg aminomethane, CH3NH2. These may be classed as primary, secondary or tertiary, but unlike the haloalkanes the classification depends on the number of C atoms attached to the N atom. Primary RNH2, secondary R2NH, tertiary R3N. H H H H H C N H H C N CH3 H C C N CH3 H H H H H H CH3 Aminomethane Primary amine N-methylaminomethane Secondary amine N,N-dimethyaminoethane Tertiary amine

Amines (aminoalkanes)
Amines have an unpleasant “fishy” smell. The smaller amines, up to C5, are soluble in water but larger amines are insoluble, as the size of the non-polar hydrocarbon chain cancels out the effect of the polar amino functional group. Like ammonia itself, water soluble amines form alkaline solutions. They react with water by proton transfer to form OH- ions. This means aqueous solutions of amines turn litmus blue. RNH H2O  RNH OH

Amines also react with acids to form salts. CH3NH2 + HCl  CH3NH3+ Cl
aminomethane methyl ammonium chloride The formation of an ionic salt increases the solubility of the amine in acidic solutions (compared to their solubility in water). This is why we put lemon juice on our fish to get rid of the amine smell

Why do you think the ammonia has to be alcoholic?
Formation of amines Another nucleophilic substitution reaction occurs between haloalkanes and alcoholic ammonia: R—X + NH3(alc) → R—NH2 + HX amine Why do you think the ammonia has to be alcoholic? It must be alcoholic ammonia: if water is present the alcohol could be formed instead.

Write these out give the compound class and name
CH3CH2CH2CH2CH2CH2COCl Class acyl chloride Name heptanoyl chloride CH3CH2CHCH2CH3 NH2 Class amine Name 2-aminopentane CH3CH2CONH2 Class amide Name propanamide

CH3CH2CH2CH2COCH2CH3 Class ketone Name heptan-3-one CH3CH2CH2CHOHCH2 Class secondary alcohol Name pentan-2-ol CH3CH2CH2CH2CH2C O H Class aldehyde Name hexanal 110

Draw and name the structural isomers of C4H10O
CH3CH2CH2CH2OH butan-1-ol butan-2-ol 2-methylpropan-1-ol 2-methylpropan-2-ol

Alkene Reactions Alkenes react readily by adding small molecules across the double C C bond. These reactions are known as addition reactions because molecules add “across” the double bond

CH3CH2CH2CH2CH2NH2 Class amine Name 1- aminopentane or pentylamine O CH3CH2CH2C NH2 Class amide Name butanamide CH3CH2 O C CH2CH2CH2CH3 O Class ester Name ethyl pentanoate 113

When deciding where to put the double bond in an elimination reaction, apply the rule ‘the poor get poorer’. H Cl H H H H H C C C C H H C C C C H + HCl H H H H H H H H One carbon of the double bond will be the carbon that lost the halogen. To decide whether the bond goes to the left or right of that carbon, look at the number of hydrogen atoms on each of those carbons. The carbon to lose the hydrogen atom (and thus become the other half of the double bond) is that carbon which has the fewer hydrogen atoms already bonded to it. 114

Haloalkanes are molecular substances, so they do not contain free X– ions.
When silver nitrate solution is added to 1-bromo-butane no cream precipitate of silver bromide forms.

When silver nitrate solution is added to 2-chloro, 2-methyl propane, the water in the solution reacts with the haloalkane, forming an alcohol and releasing chloride ions which then react with the silver nitrate to form a white precipitate.

Primary haloalkanes can also be converted into alcohols, but a stronger nucleophile is needed: OH–.
Dilute sodium hydroxide solution is added to 1-bromo butane and shaken.

The excess NaOH is neutralised with dilute nitric acid.
When silver nitrate is added the solution turns cloudy.

Chemistry 3.5 Describe the structure and reactions of organic compounds containing selected organic groups.

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Chemistry 3.5 Describe the structure and reactions of organic compounds containing selected organic groups.

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