Presentation on theme: "Chemistry 3.5 Describe the structure and reactions of organic compounds containing selected organic groups."— Presentation transcript:
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 Dont 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. Whats the general formula for an alkane? C n H 2n+2
Can you draw the structural formula and molecular formula for the following? Methane Molecular Formula - Structural Formula CH 4
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? This wedge represents the bond coming out from the paper These lines represent the bond going behind the paper These two lines represent bonds on the same plain as the paper Drawing a 3 dimensional Structure
Describe the shape of this molecule Tetrahedral Is CH 4 polar or non polar? Non polar Can you give a reason why? What is the bond angle between each of the atoms? 109 o
Alkane Nomenclature - Give the names, molecular and condensed structural formula for the first ten alkanesNameMolecular Condensed Structural Formula Condensed Structural Formula methane CH 4 C 7 H 16 C 6 H 14 C 4 H 10 C3H8C3H8C3H8C3H8 C2H6C2H6C2H6C2H6 C 5 H 12 C 8 H 18 C 9 H 20 C 10 H 22 ethane propane butane pentane hexane heptane octane nonane decane (CH 3 ) 2 CH 3 (CH 2 ) 2 CH 3 CH 3 (CH 2 ) 5 CH 3 CH 3 (CH 2 ) 6 CH 3 CH 3 (CH 2 ) 7 CH 3 CH 3 (CH 2 ) 8 CH 3 CH 3 (CH 2 ) 3 CH 3 CH 3 (CH 2 ) 4 CH 3 CH 3 CH 2 CH 3
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: C n H 2n
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 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 12 345 4. Separate numbers by commas and separate numbers from words with a dash – now you can name it 2,3-dimethylpentane
Exercise – Draw the structures for the following a) 3 – ethylheptane b) 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
StructureClassification Name alkynepent-2-yne CH 3 CH 2 CHCH 2 CH 2 CH 2 CH 2 CH 3 OH alcohol (secondary) octan-3-ol CH 3 CH 2 CH C O CH 3 OH carboxylic acid 2-methylbutanoic acid C C H H3CH3C CH 2 CH 3 H alkene cis-pent-2-ene CH 3 CHCH 2 CH 3 Br haloalkane2-bromobutane CH 3 CH 2 CH 2 CH 2 CHCH 2 CH 3 CH 3 alkane3-methylheptane CH 3 CH 2 C CCH 3
StructureClassification Name alkene2,3-dimethylpent-2-ene CH 3 CH 2 CH 2 C CH 3 OH alcohol (tertiary) 2-methylpentan-2-ol C C H3CH3C CH 3 esterbutyl methanoate CH 3 CH 2 CHCH 2 CHCH 3 CH 3 alkane 2,4-dimethylhexane CH 3 CH 2 CH 3 H C O O CH 2 CH 2 CH 2 CH 3 CH 3
HO CH 3 C 8 H 17 Cholesterol is a major component of gallstones. From the following structure of the compound predict its reaction with.. cholesterol (a)Br 2 (b) H 2 with a Pt catalyst (c) CH 3 COOH
HO CH 3 C 8 H 17 The reaction with Br 2 results in addition of bromine to the double bonded carbons forming a single carbon bond. The solution would change colour from orange to colourless. Br
With H 2 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. HO CH 3 C 8 C 17 H H
CH 3 C 8 H 17 Ethanoic acid reacts with the hydoxy group to form an ester and water OH3CH3CC O + H 2 O
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)
CH 3 CH 2 CH 2 CH 2 CHCH 2 CH 3 Cl Class haloalkane Name 3-chloroheptane CH 3 CH 2 CH 2 C O Cl Class acyl chloride Name butanoyl chloride C CH 2 CHCH 3 HO CH 3 O Class carboxylic acid Name 3-methylbutanoic acid Answers to the left hand column on handout
CH 3 CH 2 CH 2 CH 2 NH 2 Class amine Name 1- aminobutane CH 3 CH 2 CH 2 CH 2 CH 2 C O NH 2 Class amide Name hexanamide Class ester Name butyl pentanoate Answers to the left hand column continued CH 3 CH 2 CH 2 CH 2 O O C CH 2 CH 2 CH 2 CH 3
CH 3 CH 2 CH 2 COCH 2 CH 3 Class ketone Name hexan-3-one CH 3 CH 2 CH 2 CH 2 CH 2 OH Class alcohol Name pentan-1-ol Class aldehyde Name hexanal Answers to the Right hand column CH 3 CH 2 CH 2 CH 2 CH 2 C O H
CH 3 CH 2 CH 2 CH 2 CH 2 COCl Class acyl chloride Name hexanoyl chloride CH 3 CH 2 CHCH 3 NH 2 Class amine Name 2-aminobutane Class amide Name octanamide Answers to the right hand column continued CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CONH 2
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 C 4 H 10 Name: 2-methylpropane Name: butane Boiling point 36.1 o C Boiling point -11.7 o 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?
C 6 H 14 H CH H H CH H H C H H C H H C H H C H Hexane CH 3 2-methylpentane H CH H H CH H C H H C H H C H Structural Isomers of Hexane
H CH H C H C H H C H CH 3 H H CH H C H C H H C H CH 2 H H CH 3 Structural Isomers of Hexane 2,2-dimethylbutane 3-methylpentane
H CH H C H C H C H CH 3 H H Structural Isomers of Hexane 2,3-dimethylbutane
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 dont exhibit geometric isomerism because there is rotation around the single C C bond Geometric (cis and trans) Isomers
e.g. The Geometric Isomers of but-2-ene cis-but-2-ene trans-but-2-ene Bpt 3.7 o C Bpt 0.88 o C * Geometric isomers have similar chemical properties but different physical properties To exist as geometrical isomers the C atoms at both ends of the double bond must each have two different groups (or atoms attached).
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 Therefore but-1- ene does not have geometric isomers flip it over and its the same as
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 cis-but-2-enetrans-but-2-ene Same sequence of atoms Two different geometric isomers exist where atoms are arranged differently in space Bpt 3.7 o CBpt 0.88 o C
Exercise Draw the structures of the following alkenes and decide which of them can exist as cis-trans isomers a) 2-methylbut-2-ene b) 3 – methylpent-2-ene forms no cis/trans isomers Occurs as cis trans isomers cis-methylpent-2-enetrans-methylpent-2-ene
Identify whether cis trans isomers occur with in the following molecules HO C H Cl C CH 3 C H Cl C CH 3 H3CH3C C Cl C CH 3 H3CH3C NoYes
C H C 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 H3CH3C CH 3 C H C H H3CH3C
C H Cl C H C H C 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
C H H C CC O O H O O H C H H C COOH Maleic acid (cis isomer)
Give the names and structural formula for the following substances from their condensed structural formulae. (c) CH 3 CH 2 CHCHCH 2 CH 2 CH 3 (d) (CH 3 ) 3 COH H H C H H CH H C H H C H C CH 3 COH CH 3 hept-3-ene 2-methylpropan-2-ol H CH H H C H CH 3
Give the names and structural formula for the following substances from their condensed structural formulae. (a) CH 3 CH 2 CHClCH 2 CH 3 (b) (CH 3 ) 2 CHCH 2 CH 2 CHCH 2 H H CH H H CH H C H Cl C H H C H CH 3 H H C H C H C H H C H H C 3-chloropentane 5 -methylhex-1-ene
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 C Cl CH 3 C CH 2 -CH 3 CH 3 1 st pair Cl highest priority 2 nd pair CH 2 CH 3 highest priority 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
Optical Isomerism Optical isomers involve an asymmetric carbon – a carbon bonded to four different atoms or groups of atoms such as: H, OH, CH 3, C 2 H 5, C 3 H 7 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
C OH H H3CH3C C2H5C2H5 C H CH 3 C2H5C2H5 Structural formula of butan-2-ol 3-D diagram of butan-2-ol * C* = asymmetric carbon *
C OH H H3CH3C C2H5C2H5 C HO H CH 3 C2H5C2H5 ** 3-D optical isomers of butan-2-ol Mirror line Optical isomers are: mirror images of each other Are not superimposable on each other
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. Optical activity
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 rotate plane-polarised light. Molecules which are mirror images of each other rotate plane polarised light in opposite directions. Optical isomers are
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.
When we rotate the top sheet, the film above the sucrose solution is now dark, while light passes through everywhere else. Water Sucrose
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.
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
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). 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).
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. (a)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 Qs 4 and 5 page 114 Complete Enantiomers on BestChoice before next Friday please
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 Answer questions on alkanes and alkenes page 172 -174
Turn to page 6 in your booklet Properties of alkanes and alkenes
Demo Alkene addition Turn to page 145 in the year 13 lab book Comparison: Alkanes vs Alkenes
Complete Structural isomer starter Complete Worksheet two Qs 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 H 2 O and CO 2 In limited air (oxygen) they form the products H 2 O and C or CO 2 Because the carbon atoms in alkane molecules are saturated with hydrogens (ie they dont have any double or triple bonds) they are called saturated hydrocarbons.
Properties of Alkanes Insoluble in water Soluble in non polar solvents Dont conduct – no free electrons Float on water because H 2 O 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: a) 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) butane + brominebromobutane +hydrogen bromide Whats missing in this reaction? The bromine solution decolourises slowly and the HBr formed is an acidic gas that turns moist blue litmus red UV light
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 its at these bonds that reactions occur Alkenes (C n H 2n ) and Alkynes ( C n H 2n - 2 )
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 Alkenes and Alkynes
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 CH H C H C H C H CH 3 Cl H H H CH H C H C H C H CH 3 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 150 - 200 O C pressure of 4 atmospheres The reaction is C 2 H 4 (g) + H 2 (g) CH 3 CH 3 (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 CH 2 H 4 (g) + Br 2 (l) CH 2 Br CH 2 Br (l) ethene 1,2-dibromoethane
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. H C H H C H Can be drawn as H C H H C H..
H C H H C H.. H C H H C H.. H C H H C H.. H C H H C H.. H C H H C H.. H C H H C H.. Heat and a catalyst added 3 ethene monomers Repeating monomer unit Polyethylene polymer Representation of adddtion polymer reaction H C H H C H H C H H C H H C H H C H
Cl C H H C H.. C H H C H.. C H H C H.. C H H C H.. C H H C H.. C H H C H.. Heat and a catalyst added 3 vinyl chloride monomers Repeating monomer unit Polyvinylchloride Polymer Aka PVC Changing the side chain of the monomer in the reaction gives different polymers ie C H H C H Cl C H H C H C H H C 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 CH H H CBr 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 H CH 3 CBr CH 3 Carbon attached to Br is attached to two other carbons
Tertiary haloalkane - the C atom to which X is attached is attached to three other C atoms. eg CH 3 CBr CH 3 Carbon attached to Br is attached to three other carbons CH 3
The Lucas Test The Lucas test is used to distinguish between the primary, secondary and tertiary alcohols The Lucas reagent consists of ZnCl 2 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 ExampleObservationProduct Primary PrimaryButan-1-ol No cloudiness, very slow if at all 1-chlorobutane SecondaryButan-2-ol Cloudiness after 5-15 minutes 2-chlorobutane Tertiary2-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 trengthond ncreases ucas HClinc chloride ubstitution
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. PCl 3, PCl 5,SOCl 2 or conc HCl/ZnCl 2 C 2 H 5 OH C 2 H 5 Cl PCl 3 or PCl 5 ethanol chloroethane C 2 H 5 OH C 2 H 5 Cl + HCl +SO 2 SOCl 2 ethanol chloroethane C 2 H 5 OH C 2 H 5 Cl Conc HCl/ZnCl 2 ethanol chloroethane
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: a) addition of HBr to propene (forming only one product) CH 3 C H H C H + HBr H CCH H H CBr H H H
b) substitution of propane using Br 2. (forming two products, the bromoalkane and HBr) H CCH H CH H H + Br 2 HH H CCH H CH H HBrH + HBr
Like tertiary alcohols, tertiary haloalkanes are easily substituted. A tertiary haloalkane will react with cold water to form an alcohol: RX + H 2 O ROH + 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. Substitution of haloalkanes to form alcohols
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. summary!
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-, H 2 O and NH 3 (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: RX + 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 C H H Cl C H H C H H C H H H C H H C H C H H C H H + HCl (Alc)KOH
Draw the structural formula for the reaction of the tertiary haloalkane 2-chloro, 2-methylpropane with water + H 2 O H CCH H CH CH 3 ClHH + HCl H CCH H CH CH 3 OHHH 2-methylpropan-2-ol Tertiary alcohol
Amines (aminoalkanes) Amines are named as substituents eg aminomethane, CH 3 NH 2. 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 RNH 2, secondary R 2 NH, tertiary R 3 N. H CNH H HH H CNH CH 3 HH H CNH H H C H 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 C 5, 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 2 + H 2 O RNH 3 + + OH
Amines also react with acids to form salts. CH 3 NH 2 + HCl CH 3 NH 3 + 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
Formation of amines Another nucleophilic substitution reaction occurs between haloalkanes and alcoholic ammonia: RX + NH 3 (alc) RNH 2 + 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.
CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 COCl Class acyl chloride Name heptanoyl chloride CH 3 CH 2 CHCH 2 CH 3 Class amine Name 2-aminopentane Class amide Name propanamide CH 3 CH 2 CONH 2 NH 2 Write these out give the compound class and name
CH 3 CH 2 CH 2 CH 2 COCH 2 CH 3 Class ketone Name heptan-3-one CH 3 CH 2 CH 2 CHOHCH 2 Class secondary alcohol Name pentan-2-ol Class aldehyde Name hexanal CH 3 CH 2 CH 2 CH 2 CH 2 C O H Write these out give the compound class and name
Draw and name the structural isomers of C 4 H 10 O CH 3 CH 2 CH 2 CH 2 OH 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
CH 3 CH 2 CH 2 CH 2 CH 2 NH 2 Class amine Name 1- aminopentane or pentylamine CH 3 CH 2 CH 2 C O NH 2 Class amide Name butanamide Class ester Name ethyl pentanoate Write these out give the compound class and name CH 3 CH 2 O O C CH 2 CH 2 CH 2 CH 3
When deciding where to put the double bond in an elimination reaction, apply the rule the poor get poorer. H C H H Cl C H H C H H C H H H C H H C H C H H C H H + HCl 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.
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.