Presentation on theme: "Chemistry 3.5 Describe the structure and reactions of organic compounds containing selected organic groups."— Presentation transcript:
1 Chemistry 3.5Describe the structure and reactions of organic compounds containing selected organic groups
2 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 2.5How well do you know your stuff from last year?The standard then goes on to say….
3 Organic compounds described are limited to those containing no more than 8 carbon atoms Don’t think you get off that easyIt 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)
4 So lets start with Alkanes 1. What’s the general formula for an alkane?CnH2n+2
5 Can you draw the structural formula and molecular formula for the following? MethaneMolecular Formula -CH4Structural Formula
6 Drawing a 3 dimensional Structure How about drawing the 3 dimensional structure for methane?Use the molymods to make a 3 D modelAny ideas on how to draw it?These two lines represent bonds on the same plain as the paperThese lines represent the bond going behind the paperThis wedge represents the bond coming out from the paper
7 What is the bond angle between each of the atoms? Describe the shape of this moleculeTetrahedralIs CH4 polar or non polar?Non polarCan you give a reason why?
8 Condensed Structural Formula Alkane Nomenclature - Give the names, molecular and condensed structural formula for the first ten alkanesNameMolecularCondensed Structural FormulamethaneCH4C2H6(CH3)2ethaneC3H8propaneCH3CH2CH3CH3(CH2)2CH3C4H10butanepentaneC5H12CH3(CH2)3CH3hexaneC6H14CH3(CH2)4CH3heptaneC7H16CH3(CH2)5CH3octaneC8H18CH3(CH2)6CH3nonaneC9H20CH3(CH2)7CH3C10H22decaneCH3(CH2)8CH3
9 Can you remember the alkane order? A simple mnemonic is-ManyElderlyPeopleBuyPentHousesHighOverNorthDakotaMethaneEthanePropaneButanePentaneHexaneHeptaneOctaneNonaneDecane
10 Cyclic AlkanesThese are Alkanes with at least 1 ring of carbons eg cyclohexaneDraw the structure of cyclohexane in your book?CyclohexaneDraw cyclopropaneThe general formula for an alkane with one ring is:CnH2n
11 Do you remember how to name a branched alkane? Steps1. Find the longest chain of continuous carbons (called the parent chain) and name it:ie 5 carbons – name pentane2. Identify any side branches or functional groupsie methyl if there are more than one of the same type use prefixesdi (2) , tri (3) etc. 2 methyls = dimethyl.So the name so far is dimethylpentane3. Number the parent chain from the end that gives the side branch groups the lowest numbermethylPentane (parent chain)methyl
12 Steps continued3. Number the parent chain from the end that gives the side branch groups the lowest number ie from the leftmethyl12345methyl4. Separate numbers by comma’s and separate numbers from words with a dash – now you can name it2,3-dimethylpentane
13 Exercise – Draw the structures for the following 3 – ethylheptane 2,2,4-trimethylpentane
14 Lets look at making some structure Turn to Expt 1 in your bookletIn pairs make and draw the structures for Q 1 to Q 3Answer the questions
16 Structure Classification Name CH3CH2 CH3 C C alkene 2,3-dimethylpent-2-eneH3CCH3CH3alcohol(tertiary)CH3CH2CH2COH2-methylpentan-2-olCH3OH Cesterbutyl methanoateO CH2CH2CH2CH3CH3CH3CH2CHCH2CHCH3alkane2,4-dimethylhexaneCH3
17 Structure Classification Name CH C CH CH3 3-methylbut-1-yne alkyne CH3 CH3CHCH2CHCH3haloalkane2-chloro-4-methylpentaneCH3Clcarboxylic acidCH3CH2CH2CH2CH2CH2COOHheptanoic acidOCH3CH2 Cesterpropyl propanoateO CH2CH2CH3
18 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 ..C8H17CH3CH3HOBr2H 2 with a Pt catalystCH3COOHcholesterol
19 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.C8H17CH3CH3HOBrBr
20 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.C8C17CH3CH3HOHH
21 Ethanoic acid reacts with the hydoxy group to form an ester and water C8H17CH3CH3OH3CCO+ H2O
22 We have some new functional groups to learn this year Title your page Functional groupsUse the photocopied sheet and copy the complete the table neatly into your exercise bookYes the whole table!We must learn these!
23 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)
24 Answers to the left hand column on handout CH3CH2CH2CH2CHCH2CH3ClClass haloalkane Name 3-chloroheptaneOCH3CH2CH2 CClClass acyl chloride Name butanoyl chlorideCH3HOC CH2CHCH3OClass carboxylic acid Name 3-methylbutanoic acid
25 Answers to the left hand column continued CH3CH2CH2CH2NH2Class amine Name 1- aminobutaneOCH3CH2CH2CH2CH2CNH2Class amide Name hexanamideCH3CH2CH2CH2 OC CH2CH2CH2CH3OClass ester Name butyl pentanoate
26 Answers to the Right hand column CH3CH2CH2COCH2CH3Class ketone Name hexan-3-oneCH3CH2CH2CH2CH2OHClass alcohol Name pentan-1-olCH3CH2CH2CH2CH2COHClass aldehyde Name hexanal
27 Answers to the right hand column continued CH3CH2CH2CH2CH2COClClass acyl chloride Name hexanoyl chlorideCH3CH2CHCH3NH2Class amine Name 2-aminobutaneCH3CH2CH2CH2CH2CH2CH2CONH2Class amide Name octanamide27
28 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 C4H10Name: butaneName: 2-methylpropaneBoiling point 36.1o CBoiling point -11.7o C
29 Task – in pairs use the models to make hexane Draw the structural formula for hexaneThen make as many structural isomers of hexane as you canName and draw each oneThere should be 5?Hexane, 2,3 dimethylbutane, 2-methylpentane, 2,2-dimethylbutane , 2,3-dimethylbutane - any others?
30 Structural Isomers of Hexane C6H14HHHHHHHCCCCCCHHHHHHHHCH3HHH2-methylpentaneHCCCCCHHHHHH
31 Structural Isomers of Hexane CH3HH2,2-dimethylbutaneHCCCCHHCH3HHCH3HCH2HH3-methylpentaneHCCCCHHHHH
32 Structural Isomers of Hexane CH3HH2,3-dimethylbutaneHCCCCHHHCH3H
33 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 theC C bondRemember alkanes don’t exhibit geometric isomerism because there is rotation around the single C C bond
34 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-enecis-but-2-enetrans-but-2-eneBpt 3.7o CBpt 0.88 o C* Geometric isomers have similar chemical properties but different physical properties
35 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 attachedBut-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 bondflip it over and it’s the same asTherefore but-1- ene does not have geometric isomers
36 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-eneSame sequence of atomsTwo different geometric isomers exist where atoms are arranged differently in spacecis-but-2-enetrans-but-2-eneBpt 3.7o CBpt 0.88 o C
37 ExerciseDraw the structures of the following alkenes and decide which of them can exist as cis-trans isomers2-methylbut-2-eneb) 3 – methylpent-2-eneforms no cis/trans isomersOccurs as cis trans isomerscis-methylpent-2-enetrans-methylpent-2-ene
38 Identify whether cis trans isomers occur with in the following molecules H3CCH3H3CClHOClCCCCCCClCH3HCH3HCH3NoYesYes
39 H3CHH3CCH3CCCCHCH3HHGeometric 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
40 ClHClClCCCCClHHHHowever cis forms are sometimes polar (as above) and therefore have stronger intermolecular forces between molecules causing higher melting points.
41 Testing for Cis - Trans Isomers Weigh out 2 grams of maleic acid into a 50ml beakerAdd 4mls of water and warm slightly to dissolve the acidPour this into a pear shaped flaskCarefully add 5mls of concentrated HClPlace 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 waterPour the solution into an evaporating dish, pouring off the excess liquid then carefully rinse with water – Then complete task B on page 159
43 Give the names and structural formula for the following substances from their condensed structural formulae.(c) CH3CH2CHCHCH2CH2CH3HHHHHHHHCCCCCCCHhept-3-eneHHHHH(d) (CH3)3COHCH3CH3COH2-methylpropan-2-olCH3
44 Give the names and structural formula for the following substances from their condensed structural formulae.(a) CH3CH2CHClCH2CH3HHClHH3-chloropentaneHCCCCCHHHHHH(b) (CH3)2CHCH2CH2CHCH2CH3HHHH5 -methylhex-1-eneCH3CCCCCHHHH
45 Another form of Stereoisomerism is Optical isomerism
46 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 bondEg 2-chloro-3-methylpent-2-eneCH3CH2 -CH3This isomer has the highest priority groups on opposite sides of double bond therefore is the E isomerThe Z isomer has the highest priority groups on the same side of the double bondCCClCH31st pair Cl highest priority2nd pair CH2CH3 highest priority
47 Optical IsomerismOptical isomers involve an asymmetric carbon – a carbon bonded to four different atoms or groups of atoms such as:H, OH, CH3, C2H5, C3H7 etcA 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
48 3-D diagramof butan-2-olStructural formula ofbutan-2-olC2H5C2H5*CH3*COHCH3COHHHC* = asymmetric carbon
49 3-D optical isomers of butan-2-ol C2H5C2H5**CCH3COHHOCH3HHMirror lineOptical isomers are:mirror images of each otherAre not superimposable on each other
50 Optical activityGlucose 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.
51 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.
52 Molecules which have one or more chiral centres Optical isomers areMolecules which have one or more chiral centresrotate plane-polarised light.Molecules which are mirror images of each otherrotate plane polarised light in opposite directions.
53 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.
54 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.
55 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.
56 Water SucroseAlthough 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.
57 Water SucroseWhen we rotate the top sheet, the film above the sucrose solution is now dark, while light passes through everywhere else.
58 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 lightoptical isomers rotate plane-polarised light by equal angles in opposite directions.
59 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
60 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 mixtureTalk about pthalidamide
62 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).
63 ExerciseDraw the structure of each of the following molecules and thendecide which ones are optically active.Mark the chiral carbon with an asterisk.2-chlorobutane (b) 2-methylpropanoic acid(c) 3-methylpentanal (d) 2-aminobutanoic acidNo optical isomers***
64 Homework Read unit 28 page 111 in pathfinder Complete Q’s 4 and 5 page 114Complete Enantiomers on BestChoice before next Friday please
65 Answer questions on alkanes and alkenes pageGroup work exerciseEach group is to work on problems giving great detailed answers. But all people in the group must have the answers written in their lab booksRandom people from each group will be asked for their groups answer
66 Turn to page 6 in your booklet Properties of alkanes and alkenes
67 Turn to page 145 in the year 13 lab book Demo Alkene additionTurn to page 145 in the year 13 lab bookComparison: Alkanes vs Alkenes
68 Complete Structural isomer starter Complete Worksheet twoQ’s 1 and 2 in organic booklet
69 Alkane ReactionsAlkanes are used as fuels and undergo combustion reactionsIn excess air (oxygen) they form the products H2O and CO2In limited air (oxygen) they form the products H2O and C or CO2Because 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.
70 Properties of Alkanes Insoluble in water Soluble in non polar solvents Don’t conduct – no free electronsFloat on water because H2O is relatively denserBoiling/melting point increases with chain length because as molecular mass increases the intermolecular forces between molecules increases
71 Questions1. Name the type of intramolecular bonding and the type of intermolecular bonding in:Methane b) Liquid pentane2. Explain in terms of bonding why:a) Methane gas can be collected over waterb) Petrol floats on waterc) Oil dissolves in petrol
72 Alkane ReactionsBecause 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)UVlightbutane brominebromobutane +hydrogen bromideWhat’s missing in this reaction?The bromine solution decolourises slowly and the HBr formed is an acidic gas that turns moist blue litmus red
73 These are unsaturated hydrocarbons This means they have at least one double or triple covalent bondThese types of bonds are called functional groups because it’s at these bonds that reactions occurAlkenes (CnH2n) and Alkynes ( Cn H2n - 2 )
74 Alkenes and AlkynesLike alkanes these unsaturated hydrocarbons are non polar and insoluble in water.They undergo combustion the same as alkanes giving the same productsThey have very similar physical and chemical properties to alkanesForm addition reactions because of the reactive double or triple bondAlkenes exhibit a different form of isomerism called geometric isomerisim
75 2-chloro-3-methylbutane Starter ADraw 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-methylbutane2-chloro-2-methylbutane – major productHHHHHCCCCHHClCH3HHCH3HHHCCCCHHClHH
76 Hydrogenation of an Alkene (Addition reaction) Hydrogen can be added across the double C bondThe conditions for this reaction are:platinum catalystO Cpressure of 4 atmospheresThe reaction isC2H4(g) H2(g) CH3CH3(g)ethene ethane
77 The Good OilThis hydrogenation process isused 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.
78 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 sunlightThe reaction isCH2H4(g) Br2(l) CH2 Br CH2 Br (l)ethene ,2-dibromoethanewatch demo of bromine with alkane
79 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
80 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 asHHHH..CCCCHHHH
81 . . . . . . . . . . . . Representation of adddtion polymer reaction H 3 ethene monomers......CCCCCCHHHHHHRepeating monomer unitHeat and a catalyst addedHHHHHHHHHHHHPolyethylenepolymer......CCCCCCCCCCCCHHHHHHHHHHHH
82 Changing the side chain of the monomer in the reaction gives different polymers ie ClHClHClH3 vinyl chloride monomers......CCCCCCHHHHHHRepeating monomer unitHeat and a catalyst addedClClHHClClHHClClHHPolyvinylchloridePolymerAka PVC......CCCCCCCCCCCCHHHHHHHHHHHH
83 This year we will also look at polymer reactions involving condensation reactions
84 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.
85 Haloalkanes (alkyl halides) RX where X = F, Cl, Br, INamed 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
86 The haloalkanes can be classified as: primary - the C atom to which X is attached is only attached to one other C atomegHHHCCBrHHCarbon attached to Br is attached to one carbon
87 Secondary haloalkane - the C atom to which X is attached is attached to two other C atoms egCarbon attached to Br is attached to two other carbonsCH3HCBrCH3
88 Tertiary haloalkane - the C atom to which X is attached is attached to three other C atoms. egCarbon attached to Br is attached to three other carbonsCH3CH3CBrCH3
89 The Lucas TestThe Lucas test is used to distinguish between the primary, secondary and tertiary alcoholsThe Lucas reagent consists of ZnCl2 in concentrated HClThe zinc chloride catalyses a substitution reaction between the alcohol and the concentrated HClChloroalkanes form and appear as a cloudy suspension in the water because they are insoluble
90 The Lucas test *Important The rates at which the different types of alcohol react to form chloroalkanes enable them to be classified as follows:
91 The Lucas Test Type of alcohol Example Observation Product Primary Butan-1-olNo cloudiness, very slow if at all1-chlorobutaneSecondaryButan-2-olCloudiness after 5-15 minutes2-chlorobutaneTertiary2-methylpropan-2-olCloudiness after 1-2 minutes2-methyl-2-chloropaopane
92 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 AlcoholtrengthondncreasesucasHClinc chlorideubstitution
94 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.
95 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/ZnCl2PCl3 or PCl5C2H5OH C2H5ClchloroethaneethanolSOCl2C2H5OH C2H5Cl + HCl +SO2ethanolchloroethaneConc HCl/ZnCl2C2H5OH C2H5Clethanolchloroethane
96 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)HHHCH3H+ HBrCCHCCCBrHHHHH
97 substitution of propane using Br2 substitution of propane using Br2. (forming two products, the bromoalkane and HBr)HHHHHHHCCCH+ Br2HCCCH+ HBrHHHHHBr
99 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 + HXWe can tell whether this reaction has taken place by the presence of the X– ions, which will form precipitates with silver nitrate solution.
100 summary! Tertiary haloalkanes form alcohols in cold water Secondary haloalkanes react when the water is warmPrimary haloalkanes do not react with water, but react to form alcohols with aqueous sodium hydroxide.
101 NucleophilesA 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.egR X(l) + OH- (aq) R OH + X- (aq)
102 Elimination to form an alkene Haloalkanes can also undergo an elimination with hydroxide to form an alkene:R—X + alcOH– → R’=C + HXUse the above reaction as a template to write your own formation of an alkene from a haloalkane
103 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.HClHHHH(Alc)KOHHCCCCHHCCCCH+ HClHHHHHHHH
104 Draw the structural formula for the reaction of the tertiary haloalkane 2-chloro, 2-methylpropane with waterHCH3HHCH3HHCCCH+ H2OHCCCH+ HClHClHHOHH2-methylpropan-2-olTertiary alcohol
105 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.HHHHHCNHHCNCH3HCCNCH3HHHHHHCH3AminomethanePrimary amineN-methylaminomethaneSecondary amineN,N-dimethyaminoethaneTertiary amine
106 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
107 Amines also react with acids to form salts. CH3NH2 + HCl CH3NH3+ Cl aminomethane methyl ammonium chlorideThe 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 onour fish to get rid of the aminesmell
108 Why do you think the ammonia has to be alcoholic? Formation of aminesAnother nucleophilic substitution reaction occurs between haloalkanes and alcoholic ammonia:R—X + NH3(alc) → R—NH2 + HXamineWhy do you think the ammonia has to be alcoholic?It must be alcoholic ammonia: if water is present the alcohol could be formed instead.
109 Write these out give the compound class and name CH3CH2CH2CH2CH2CH2COClClass acyl chloride Name heptanoyl chlorideCH3CH2CHCH2CH3NH2Class amine Name 2-aminopentaneCH3CH2CONH2Class amide Name propanamide
110 Write these out give the compound class and name CH3CH2CH2CH2COCH2CH3Class ketone Name heptan-3-oneCH3CH2CH2CHOHCH2Class secondary alcohol Name pentan-2-olCH3CH2CH2CH2CH2COHClass aldehyde Name hexanal110
111 Draw and name the structural isomers of C4H10O CH3CH2CH2CH2OHbutan-1-olbutan-2-ol2-methylpropan-1-ol2-methylpropan-2-ol
112 Alkene ReactionsAlkenes 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
113 Write these out give the compound class and name CH3CH2CH2CH2CH2NH2Class amine Name 1- aminopentane or pentylamineOCH3CH2CH2CNH2Class amide Name butanamideCH3CH2 OC CH2CH2CH2CH3OClass ester Name ethyl pentanoate113
114 When deciding where to put the double bond in an elimination reaction, apply the rule ‘the poor get poorer’.HClHHHHHCCCCHHCCCCH+ HClHHHHHHHHOne 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
115 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.
116 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.
117 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.
118 The excess NaOH is neutralised with dilute nitric acid. When silver nitrate is added the solution turns cloudy.