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10/4/2011 Textbooks CHEM 120 CHEMICAL REACTIVITY Prescribed book ORGANIC CHEMISTRY T.L. Brown, H.E. LeMay Jr, B.E. Bursten, C.J. Murphy, S. Langford and D. Sagatys, Chemistry: The Central Science: A Broad Perspective, 2nd Edition, Pearson, Australia, 2010. Dr. Vincent O. Nyamori Recommended books 19 Lectures 5 Tutorials 2 Quiz 1 Test Organic Chemistry • Surfactant C17H35COO‐ H. Hart, L.E. Craine, D.J. Hart and C.M. Hadad, Organic Chemistry - A Short Course, 12th Edition, Houghton-Mifflin (students not intending to continue with Chemistry). P.Y. Bruice, Organic Chemistry, 6th Edition, Pearson/Prentice Hall (Chemistry major students). Any other relevant Organic textbook - Library worksheets http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx • Carbon 1s22s22p2 The study of carbon‐containing compounds and their properties. The vast majority of organic compounds contain chains or rings of carbon atoms. Glucose C6H12O6 Ascorbic acid HC6H7O6 3 Periodic Table 4 1 10/4/2011 Structure of Carbon Compounds Geometries of carbon compounds sp2 sp3 Three hybridization states Trigonal planar Tetrahedral C C C C sp2 sp3 1.33 Å 1.54 Å each satisfies the octet rule for each carbon! C C sp 1.20 Å C C “Organic Chemistry: A Short Course” 12th Edition, by Hart et al. Chapter 1.14 ‐ 1.18 “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, by LeMay et al. Chapter 21.1 & Chapter 23.1 pg 852 sp Linear 5 Methane: CH4 carbon Hydrogens are in a tetrahedral arrangement around the sp3 Energy hybridized carbon atom. 2s carbon carbon sp2 Ethene: C2H4 sp Hybridization 2p Hybridization Energy 2p Energy Hybridization 1s sp3 Hybridization 1s sp2 Hybridization 2p Hydrogens bond to the carbon sp3 orbitals with 1s orbitals. 2s 6 sp3 2s Hybridization sp 1s 7 Ethyne: C28H2 2 10/4/2011 HYDROCARBONS HYDROCARBONS • compounds composed of only carbon and hydrogen Alkanes • chain of carbon atoms bonded to enough hydrogen atoms to satisfy the octet rule for each carbon • chain is bent because of the 109.5° C–C–C tetrahedral angle e.g. CH2 H H 180° C C H H C C sp3 H 109.5°C H C H 120° H C sp2 C HC H3 C CH3 H How many bonds are • Hydrocarbons with all single carbon‐carbon bonds (no double or triple bonds) • Alkanes are SATURATED • They contain the maximum number of hydrogen atoms Alkenes, alkynes and aromatic compounds CH C sp “Chemistry: The Central Science: A Broad Perspective” ‐ Chapter 21 • UNSATURATED hydrocarbons • contain carbon‐carbon multiple bonds H Π = 3 bonds θ = 16 bonds “Chemistry: The Central Science: A Broad Perspective” ‐ Chapter 23 & 24 ‐ they ARE NOT ALKANES Line notation 9 10 Example Past exam Question 1. Indicate the hybridization for carbons 1 – 10 and their respective geometry. Include bond angles in your answer. Br H H C 1 2 H 3 3 O Cl C3 C C C sp C‐1: _____ C‐3: _____ sp2 sp sp C‐2: _____ C‐4: _____ H C O Solution H 4 F C Hybridization 11 9 7 C 6 8 C 10 N H Geometry Bond angles Carbon 1, 2 and 6: sp3 Tetrahedral Carbon 3, 4 and 9: sp2 Trigonal planar Carbon 7, 8 and 10: sp Linear 109.5° 120° 180° 12 3 10/4/2011 Organic Nomenclature Hydrocarbons • Four basic types: • Three parts to a compound name: ‐ Alkanes CnH2n+2 1 C2H6 Ethane ‐ Alkenes 3 2 1. Prefix CnH2n C2H4 Ethene 2. Base 120° ‐ Alkynes CnHn C2H2 Ethyne 3. Suffix ‐ Aromatic hydrocarbons CnH2n‐2 C6H6 Benzene or H H H H H Chapter 2: “Organic Chemistry”, 5th Edition , Bruice P. Y. H Organic Nomenclature Chapter 21: pg. 807‐9: “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, LeMay et al. Organic Nomenclature Suffix: Base/parent: Tells what type of compound it is. Tells how many carbons are in the longest continuous chain. suffix base What family? 15 suffix How many carbons? What family? 16 4 10/4/2011 Organic Nomenclature ‐ IUPAC Rules Alkanes Prefix: Tells what substituent(s) are attached, if any. prefix What substituent? base suffix How many carbons? What family? Properties 17 To Name a Compound… 3. Number the chain from the end nearest the first substituent encountered. H2C Name?? CH H3C CH2 CH3 3‐Methylheptane 2‐Ethylhexane CH3 CH CH2 CH CH CH3 CH CH3 4. List the substituents as a prefix in alphabetical order along with the number(s) of the carbon(s) to which they are attached. CH2 E1 CH3 2. Find the longest chain in the molecule. ‐ Base CH2 Only van der Waals force: London force. Boiling point increases with length of chain. Combust to give mainly CO2 and H2O Nomenclature suffix “‐ane” Example 1. Determine what type of compound it is. ‐ Suffix H3 C • • • • 19 CH2 CH3 CH3 If there is more than one type of substituent in the molecule, list them alphabetically i.e. name of substituent, not prefix for frequency e.g. di, tri, tetra, etc...are not considered. 3‐Ethyl‐2,4,5‐trimethylheptane 5 10/4/2011 CYCLIC ALKANES Cycloalkanes • Carbon can also form ringed structures. How do we name…. • Five‐ and six‐membered rings are most stable. methyl CH3 – Can take on conformation in which angles are very close to tetrahedral angle. General formula CnH2n 4 1‐Ethyl‐3‐methylcyclohexane cyclohexane cyclopentane H H C H H H H C C H H C C C C C C H 1 Alkenes VSEPR Theory O H C 6 2 22 Carbon can form multiple bonds with itself or other atoms, e.g. N or O. H 5 3 cyclopropane Unsaturated Hydrocarbons ethyl CH2CH3 Ethylmethylcyclohexane – Smaller rings are quite strained. H C H3 C CH 3 C H H 3C C H3 C 120° N S CH3 CH 3 Multiple bonds affect, physical and chemical properties (reactivity of organic molecules). Hydrocarbons that contain one or more double or triple bonds are called unsaturated hydrocarbons. • Contain at least one carbon–carbon double bond • Unsaturated – Have fewer than maximum number of hydrogens – The C atoms on double bond are sp2 hybridized Chapter 23: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition, by LeMay et al. 6 10/4/2011 Structure of Alkenes Structure of Alkenes This creates geometric isomers • Unlike alkanes, alkenes cannot rotate freely about the double bond. – Side‐to‐side overlap makes this impossible without breaking ‐bond. C difference in the spatial arrangement of groups about the double bond C Z‐2‐Pentene Cis‐ isomer “Z”‐isomer Trans‐ isomer “E”‐isomer E‐2‐Pentene 25 26 Example NAMING ALKENES 1. Find the longest unbranched carbon chain containing the double bond. • Name chain according to number of carbon atoms. Name this alkene add ‐ene as a suffix 2. Number the carbon atoms in the main chain. • Start from the end of the chain that is closest to the double bond. location of the double bond is numbered with the CH3 7 1. The longest unbranched chain containing the double bond is seven carbons long, so this is a heptene. 2. The chain numbering starts closest to the double bond. Carbon #2 is the lowest‐numbered carbon in the double bond, so this is a 2‐heptene or hept‐2‐ene lowest‐numbered carbon in the double bond. 27 H 1 2 CH3 CH3 C 3 4 C CH CH CH2 5 6 H CH H2C CH2 28 7 10/4/2011 Properties of Alkenes 3. There are two substituent groups on this alkene: 1 CH3 H2C Example: C4H8 5‐methyl H 2 C 3 4 C CH H CH CH3 5 CH CH2 6 CH2 7 CH3 4‐cyclopropyl Compose the name….. • Add the substituent groups alphabetically to the alkene name 2‐Methyl‐1‐propene bp. ‐7 ⁰C 1‐Butene bp ‐6 ⁰C Cis‐2‐Butene bp +4 ⁰C Trans‐2‐Butene bp +1 ⁰C • Specify the position of each group on the main chain Structure also affects physical properties of alkenes the alkene is: 4‐cyclopropyl‐5‐methyl‐2‐hept ene Can we have more than one double bond? Chapter 22.2: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition 29 Geometric isomers of Alkenes GEOMETRICAL ISOMERS Alkenes exhibit cis‐trans isomerism. Trans‐ isomer “E‐” Cis‐isomer “Z‐” • Identical substituents on opposite sides of the double bond H trans‐ CH3 C CH3 “E‐” • Identical substituents on same side of the double bond cis‐ C CH3 CH3 C H H “Z‐” 1,3‐butadiene C • Cis‐alkenes have similar higher priority elements or group in the chain on the same side of the molecule (or Z‐isomer i.e. have higher priority elements but not necessarily the same on the same side of the molecule) • Trans‐alkenes have similar higher priority elements or group in the chain on opposite sides of the molecule (or E‐isomer i.e. have higher priority elements but not necessarily the same on opposite sides of the molecule). H e.g. Stick diagram Priorities are assigned by the atomic numbers of the atoms bonded to the carbon in the double bond. C5H10 31 1‐Pentene (Z)‐2‐Pentene (E)‐2‐Pentene 32 8 10/4/2011 Examples FUNCTIONAL GROUPS 1. Name the following alkenes and determine whether there are geometric isomers are either Trans‐ (E‐) or Cis‐ (Z‐) isomers if applicable. 1 2 1 a) b) 2 1 1 Each functional group is specified by a suffix or prefix 2 H (E)‐ 3‐Methylhex‐3‐ene (Z)‐ 3‐Bromopent‐2‐ene Equal priority c) No geometric isomers formed. Group / Formula Family Alkane RH Structural Formula Prefix Suffix alkyl- -ane 33 Example Ethane Alkene R2C=CR2 alkenyl- depicted on the nomenclature of the organic molecule Functional groups are given an order of priority to decide on which is the suffix. HOMEWORK! 3‐Fluoro‐2‐methylpent‐2‐ene F -ene Hint: Please refer to your textbooks !! • “Chemistry: The Central Science: A Broad Perspective”, 2nd Edition, by LeMay et al. Chapter 24.1, 25.1, 26.1 & 26.3 • “Organic Chemistry” 5th Edition, by Bruice 34 • Hart et al. “Organic Chemistry: A Short Course” 12th Edition Group / Family Formula Haloalkane Structural Formula Prefix Suffix RX halo- alkyl halide Chloroethane Ethyl chloride Fluoroalkane RF fluoro- alkyl fluoride Fluoromethane Methyl fluoride Chloroalkane RCl chloro- alkyl chloride Chloromethane Methyl chloride Bromoalkane RBr bromo- alkyl bromide Iodoalkane RI iodo- alkyl iodide Ethene Alkyne RC≡CR' alkynyl- -yne Ethyne Benzene derivative RC6H5 RPh phenyl- FUNCTION Acidic, basic, alcohol, etc… The GROUPS are called functional groups. 2 H Br Certain groups of atoms give a molecule a.... -benzene 2-phenylpropane isopropylbenzene Example Bromomethane Methyl bromide Iodomethane Methyl iodide 9 10/4/2011 Group / Family Group Formula Primary amine (1º) RNH2 Secondary amine (2º) R2NH Tertiary amine (3º) R3N Amines Structural Formula Prefix Suffix amino- -amine Example Group / Family Formula Methylamine Alcohol ROH Group / Family Formula Carboxylic acid RCOOH Acyl halide Ester ROR' RCOOR' hydroxy- -ol Example Dimethylamine amino- Ketone RCOR' keto-, oxo- -one Aldehyde RCHO aldo- -al 2-Butanone Methylethyl ketone -amine Trimethylamine CH3 N R4N+X- -ammonium ammonio- Structural Formula Prefix Suffix carboxy- -oic acid H Cl CH3 Trimethylammonium chloride Example haloformyl- Ethanal Acetaldehyde Primary (1°) alcohols and amines alkoxy- Alcohol H 1 R alkyl alkanoate C H OH H -oyl halide alkyl alkyl ether Example General structure Ethanoyl chloride Acetyl chloride Ether Suffix -amine Ethanoic acid Acetic acid RCOX Prefix Methanol amino- H 3C Quaternary ammonium ion Structural Formula 1° alcohols CH3CH2 Propan‐1‐ol 1‐Propanol C H Ethoxy ethane Diethyl ether Ethyl butanoate Ethyl butyrate Amine 1 R N H 1° amine OH H H CH3CH2 N Ethylamine H 10 10/4/2011 Secondary (2°) alcohols and amines H 1 R OH R2 2° Alcohol CH3CH2 1 R Alcohol C OH R 1 R C H CH3CH2 N R2 2° Amine CH3 OH R2 3° alcohols Butan‐2‐ol CH3 2‐Butanol H Amine 3 H C Amine CH3 + R4 N 1 R N 1 R N CH3 Ethyldimethylamine 42 Exercise Example CH3CH2 CH3 CH3CH2 3° Amine 1. Draw the structures of the following alcohols and amines and classify them as either 1°, 2°, 3° or quaternary CH3 R3 C OH CH3 2‐Methylbutan‐2‐ol 2‐Methyl‐2‐butanol R2 Ethylmethylamine R2 Amine CH3CH2 R3 N Quaternary amines General structure Example General structure Example General structure Alcohol Tertiary (3°) alcohols and amines N + a) Pentan‐1‐ol CH3 b) Dimethylamine CH3 Ethyltrimethylammonium ion c) 3‐Ethylhexan‐3‐ol d) Diethylmethylamine e) Butan‐2‐ol f) Triethylmethyl ammonium ion 43 44 11 10/4/2011 Solutions a) Pentan‐1‐ol H HO C H H H C C H H H C H b) Dimethylamine H C H CH2 C CH3 2° amine e) Butan‐2‐ol CH3 CH2CH3 H CH2CH3 C 2° alcohol OH CH2 3° alcohol CH2 CH3 f) Triethylmethyl ammonium ion CH2CH3 H3C CH2CH3 OH 3° amine CH3 N CH3CH2 Name this organic molecule: CH3 OH NO2 CH CH Specify the position of each group on the main chain. Add the substituent groups alphabetically to the name of the alkane (or alkene or alkyne) along with the frequency of each group. CH2CH3 Example….. Quaternary amine A45 Example: Naming Hydrocarbons with Functional Groups Name the other substituent groups, using the prefixes for alkyl groups and the prefixes for any other functional groups CH2 N CH3 H d) Diethylmethylamine N CH3 H 1° alcohol H3C c) 3‐Ethylhexan‐3‐ol 46 Example: Name this organic molecule: OH CH3 CH3 CH NO2 CH CH3 1. This molecule contains a hydroxyl and a nitro group Only the hydroxyl group has priority. So this is an ALCOHOL….. 47 48 12 10/4/2011 Example: Name this organic molecule: OH CH3 CH OH NO2 CH CH3 CH 1 2 CH3 1. This molecule contains a hydroxyl group and a nitro group this is an ALCOHOL. The hydroxyl group is on position 2 4. This molecule has one substituent A nitro group on position 3 Therefore its a “butanol”. 49 3‐nitro‐2‐butanol CH3 1. CH CH2 H2N CH3 CH2 C CH2 CH3 CH2 O CH CH3 C OH CH CH3 CH O CH CH2 C OH CH3 CH3 Identify functional groups carboxyl group Highest priority CH3 CH2 CH 3‐nitrobutan‐2‐ol Name this organic molecule: CH3 CH3 or 50 Example: Name this organic molecule C Butanol CH3 4 3. Number the carbons, starting NEAREST the . functional group An alcohol suffix is ‐OL H2N CH 3 …so this is a 2‐butanol or butan‐2‐ol longest carbon chain containing the hydroxyl group has four carbons. Example: NO2 1. Molecule contains a carboxyl group and an amino group. The carboxyl group has highest priority, hence amino group a carboxylic acid Now determine longest carbon chain 51 52 13 10/4/2011 Example: Name this organic molecule: Example: Name this organic molecule CH3 H2N C CH3 CH2 CH3 CH2 CH3 CH CH C CH2 C CH3 OH CH O CH CH C OH CH2 CH3 CH3 CH3 The longest carbon chain has 8 carbons. The longest carbon CONTAINING the ‐CO2H chain has 7 carbons. BUT………………... so this molecule is based on a heptane. THIS CHAIN DOES NOT CONTAIN THE ‐CO2H group 53 54 Heptanoic acid CH3 H2N C CH2 CH3 CH CH3 CH CH3 H2N 5 C 6 CH2 7 CH3 4 CH2 CH3 CH2 O CH CH2 C CH3 OH CH3 Highest priority functional group is a carboxyl group CH CH3 3 CH 2 CH CH2 1 O C OH CH3 3. Number the carbons starting with the functional group The carboxyl group is on position 1, Suffix ‐OIC ACID Hence… Heptanoic acid Now number chain……. 55 CH3 CH2 O CH CH2 CH3 H2N do not include in the name because the carboxyl group is always a terminal group. 56 14 10/4/2011 Heptanoic acid CH3 5‐methyl 5 H2N 5‐amino C 6 CH2 WHAT IS THE STRUCTURAL FORMULA OF 7 CH3 4 CH2 CH3 3 CH CH 2 CH CH3 3‐isopropyl 1 KETONE GROUP OH CH2 CH3 BUTANONE ? O C CH3 4. This molecule has four substituents i.e. at carbons 2, 3 and two at 5 C H INTERPETING AN IUPAC NAME…... O Br d) Alcohol (2°) OH (Z)‐3‐methylhex‐2‐ene CH2 3,3‐dimethylbutan‐2‐ol H CH3 Ketone CH3 O h) Carboxylic Cl acid 3‐methylhexan‐2‐one O i) Ester OH O 2‐chloropentanoic acid 59 g) 2‐ethylpentanal 6‐bromo‐4‐ethyl‐2,3,7‐trimethyloctane H CH2 Aldehyde f) Alkene c) BUTANAL Give the correct IUPAC name for the following compounds. Bromoalkane Alkylbromide 2,3,7‐trimethyloctane WHY????? Examples Give the correct IUPAC name for the following compounds. b) CH3 58 Examples Alkane CH2 O 5‐amino‐ 2‐ethyl‐ 3‐isopropyl‐ 5‐methyl heptanoic acid a) C CANNOT BE AT END!!!! Compose the name ALPHABETICAL LIST 57 O 4 CARBONS 2‐ethyl O ethylbutanoate 60 15 10/4/2011 Examples Solutions Draw structural formulae for the following compounds: a) 2,3,5‐trimethylhexane a) 2,3,5‐trimethylhexane b) (Z)‐3‐chlorohept‐2‐ene Cl b) (Z)‐3‐chlorohept‐2‐ene c) 3‐ethylnonanol H d) 2,3‐dimethylpentanoic acid d) 2,3‐dimethylpentanoic acid c) 3‐ethylnonanol e) methyl hexanoate O f) 3‐iodohexanal OH g) pentan‐2‐one h) 3‐aminopentane 61 OH 62 Solutions e) methyl hexanoate ISOMERS Two types: (a) Structural isomers (b) Stereoisomers f) 3‐iodohexanal O I O (a) Structural isomers O Molecules with the same chemical formula but different bonds between the atoms Now called... CONSTITUTIONAL ISOMERS H Constitutional isomers have different properties: g) pentan‐2‐one e.g. butane (C4H10) has 2 structural isomers h) 3‐aminopentane These are…..? O n‐Butane: C4H10 H2 N CH3 CH2 CH2 CH3 63 bp = 0 °C mp = ‐138 °C 2‐methylpropane: C4H10 CH3 CH CH3 CH3 bp = ‐12 °C mp = ‐159 °C 64 16 10/4/2011 Example CONSTITUTIONAL (STRUCTURAL) ISOMERS How many constitutional isomers are formed from C5H12? Draw their structures. Solution: THREE CONSTITUTIONAL ISOMERS 1) CH3 CH2 CH2 CH2 CH3 2) CH3 CH CH2 CH3 CH3 2‐methylbutane n‐pentane 3) The general formula for ALKANES is…... n = 1, 2 and 3 1 ISOMER The number of ISOMERS increases with n….. n = 4 2 ISOMERS n = 5 3 ISOMERS n = 6 5 ISOMERS n = 7 9 ISOMERS n = 40 62,491,178,805,831 ISOMERS CH3 C n H 2n 2 n = 8 18 ISOMERS n = 9 35 ISOMERS CH3 C CH3 HOMEWORK: DRAW THE ISOMERS OF C40H82 !!! CH3 2,2‐dimethylpropane n = 10 75 ISOMERS 66 n = 20 366,319 ISOMERS 65 Constitutional isomers for multibonds Constitutional isomers for multibonds Example: Alkene Example: Alkyne C6H10 C6H6 (E,E)‐2,4‐hexadiene H3 C (E)‐1,3‐hexadiene C C C C CH3 2,4‐hexadiyne hexa‐2,4‐diyne (Z,Z)‐2,4‐hexadiene CH3 H (Z)‐1,3‐hexadiene C C C C CH2 1,3‐hexadiyne hexa‐1,3‐diyne (Z,E)‐2,4‐hexadiene 1,5‐hexadiene 67 68 17 10/4/2011 OPTICAL ACTIVITY IN ORGANIC COMPOUNDS OPTICAL ISOMERISM Optical isomerism arises when molecules have a structure such that the mirror image is not superimposable on the original molecule. Stereogenic centre has four different groups attached to a tetrahedral carbon atom W Stereogenic centre occurs whenever there are four different groups bound to the same tetrahedral carbon atom. The carbon involved is called a chiral carbon or stereogenic carbon and the molecule is known as a chiral molecule. Example: 2‐butanol 70 Perspective formula Fischer projection Wedge‐dash notation 2‐Butanol 2‐Butanol Solid wedge represents a bond extending out towards the viewer Simple line for Bonds aligned to the asymmetric center in the plane C* H3C H X Y 69 OH C* Z Some terminology……. Dash shows bond going backwards from the viewer Chiral carbon atom Bond going backwards from the viewer HO Bond extending out towards the viewer CH2CH3 CH3 * Bonds aligned to the asymmetric center in the plane CH2CH3 H Chiral centre Chiral centre 71 72 18 10/4/2011 Question ENANTIOMERISM in ORGANIC CHEMISTRY Identify the chiral carbon (stereogenic centre) and draw the structural formula for each of the following molecules: • 2‐butanol has two optical isomers. • A pair of isomers called enantiomers ‐ (non‐superimposable mirror images of each other) (a) 1‐chloroethanol H OH HO C C H CH3CH2 CH2CH3 (d) 1,3‐dimethlycyclopentane CH3 H E Chapter 22.3: “Chemistry: The Central Science: A Broad Perspective” 2nd (b) 2,3,5‐trimethylhexane * View in 3‐D OH H3C (b) 2,3,5‐trimethylhexane Solution “mirror” View in 3‐D (a) 1‐chloroethanol (c) Methylcyclohexane C* CH3 Cl (c) methylcyclohexane (d) 1,3‐dimethlycyclopentane No chiral carbon * * 74 Edition NAMING OPTICAL ISOMERS CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE Stereogenic centre creates two molecular optical isomers 1. Assign relative priorities to each of the four groups on the stereogenic carbon to describe the configuration. Cl C* H Enantiomers *C OH H3C Two configurations Cl HO H CH3 How do we name these isomers?? Solution: Use R‐S nomenclature system for designating the configuration We assign priorities as in the E, Z system……... 75 The priorities are given by rules: • Higher atomic numbers are given higher priorities. • If necessary, the second atom in each substituent is used to determine the priorities. Chapter 22.2: “Chemistry: The Central Science: A Broad Perspective” 2nd Edition 19 10/4/2011 CAHN‐PRELOG‐INGOLD R,S‐NOMENCLATURE 2‐ butanol OH 2. Draw the molecule with the lowest‐priority group pointing directly into the page…. CH3 CH2 CH3 H OH and the other three groups pointing out of the page in an C Chapter 22; pg. 829‐839. “Chemistry: The Central Science: th Edition. A Broad Perspective” 2 p OH arrangement like a steering wheel or Mercedes‐Benz symbol C Example: 2‐Butanol….. CH3 C CH2 CH3 H Draw molecule as a wedge and dashed line diagram 2‐ butanol H OH CH3 H CH3 THESE ARE THE TWO…. ENANTIOMERS ...or ....OPTICAL ISOMERS 78 2‐Butanol 1 OH 1. Assign priorities to each group: O > CMe = CEt > H CH2CH3 H3CH2C DO NOT FORGET THE OTHER ISOMER………. 1. Assign priorities: C CH2CH3 H3C OH C C H C 4 H C,H,H > H,H,H 3 CEt > CMe CH3 2 CH2CH3 O > CEt > CMe > H 1 OH C H3C 3 2 2. Redraw the molecule with the lowest priority group facing in. CH2CH3 Now what????? 79 80 20 10/4/2011 1 C H3C 1 2‐Butanol OH 2 C 2 CH3 CH3CH2 CH2CH3 3 2‐Butanol OH 3 (R)‐2‐Butanol (S)‐2‐Butanol 3. Look at the direction in which the priorities decrease. If they decrease in a clockwise direction, the stereogenic centre is called “R” or rectus If the priorities decrease in a counter‐clockwise direction the stereogenic centre is called “S” or sinister, which is Latin for “left.” which is Latin for “right.” Or…. 81 82 What are the configurations of the following chiral molecules? Example: NH2 N > CEt > CMe > H C H CH3 CH2CH3 (R)‐2‐butanamine 1 2 H2N CH2CH3 centre in each of the following molecules: NH2 H3C C OH CH2CH3 CH2OH HO C O (A) 2 C H Cl (S)‐2‐butanamine 3 C H CH2CH3 H Cl CH2OH C H3C (B) O 2‐chloro‐3‐hydroxy‐propanoic acid C C A83 Example: Give the configuration of the stereogenic THEY LOOK DIFFERENT BUT ARE THEY??? 3 CH3 1 NH2 84 21 10/4/2011 1. Assign priorities to each group: A) CH2OH 4 3 O,O,O > O,H,H Cl > Ccarb > Calc > H 85 C OH Cl 1 4 C H 1 3 C C OH 1 The priorities decrease anti‐clockwise, so this centre is “S” (S)‐2‐chloro‐3‐hydroxy‐propanoic acid Example: Give the configuration of the stereogenic centre in each of the following molecules: O 3 CH2OH A) Cl Cl B) C H 3. Direction of the priorities decrease? CH2OH 3. Look which way the priorities decrease. O Cl C 2 Cl 2. Redraw the molecule with the lowest priority group facing in. O HO C 2 Cl > Ccarb > Calc > H CH2OH HO C 2 1. Assign priorities to each group 2. Redraw the molecule with the lowest priority group facing in. CH2OH 86 B) Cl > Ccarb > Calc > H O 1 Ccarbox > Calc 1. Assign priorities to each group: 2 Cl C OH O Go to next atoms on tied carbons Cl CH2OH C H Cl > Calc = Ccarb > H C H 3 A) H C OH O C CH2OH HO C O 2‐chloro‐3‐hydroxy‐propanoic acid The priorities decrease anti‐ clockwise, so this centre is “S” THEY LOOK DIFFERENT BUT ARE THEY??? WE JUST FOUND OUT THAT THEY ARE IDENTICAL!!!!! (S)‐2‐chloro‐3‐hydroxy‐propanoic acid 88 WHAT IF????? 22 10/4/2011 If we flip the COOH and H ??????? A) A’) CH2OH C H HO C OH O Cl 3 CH2OH C 3 HO C 2 O “S” configuration 2 2 H3C C Cl O “R” configuration 89 CH3 Cl 1 Draw steering wheel with lowest priority group pointing in……. OH 1 Cl 91 MAKE SURE PRIORITY GOES CLOCKWISE FOR “R” configuration 2 C CH2CH3 3 DRAW MOLECULE….. H3C C 3 CH2 CH3 1 ASSIGN PRIORITIES 90 OH Cl 3 CH2 OH Cl 1 OH 4 1 Assign priorities (R)‐2‐chloro‐2‐butanol 4 DRAW MOLECULAR STRUCTURE Solution: CH2OH C 2 H What is the structural formula of (R)‐2‐chloro‐2‐butanol? Hint: Make use of perspective formula or dash‐wedge notation. Cl O C OH 1 C C CH2OH Cl EXAM QUESTION 2 DRAW MOLECULE TO SEE ALL GROUPS….. C CH2CH3 3 Remember lowest priority group has a “dashed” bond Make sure priority goes clockwise for “R” OH Perspective formula or Dash‐wedge notation C H3C Cl CH2CH3 (R)‐2‐chloro‐2‐butanol 92 23 10/4/2011 * * Diastereomers CH3 CH CH CH3 Br Cl [2R,3R] a Br R b CH3 C H H C Cl R S c CH3 Br H C Br Cl CH3 S Cl C H a & d [2R,3S] [2S,3R] [2S,3S] R S CH3 CH3 d CH3 C H H C Br C H H C Cl S R CH3 CH3 enantiomeric pairs enantiomeric pairs • Enantiomeric pair differ only in optical activity Chapter 22.2, page 838‐9: “Chemistry: The Central Science: A Broad Perspective” Meso compound O 93 O * * a CH3 Br R C H H C Cl R CH3 R H C OH HO C H R CO2H Enantiomeric 170 °C +12° pairs S HO C H H C OH S CO2H 170 °C ‐12° R CO2H S CO2H H C OH HO C H H C OH HO C H S c H C Br Br S Cl S C H CH3 Cl R S CH3 d CH 3 C H H C Br C H H C Cl CH3 S R CH3 Meso compound HOC CH CH COH CO2H b CH3 b & d Diastereomers ‐ are distinct chemical compounds, differing not only in optical activity but also in mp, bp., , solubility etc… molecule with n stereogenic centres may exist in maximum of 2n stereisomeric forms, with maximum of 2n/2 enantiomeric pairs 94 OH OH CO2H b & c a & c CO2H 140 °C 0° R CO2H Meso compounds meso compound ‐ an achiral (optically inactive) diastereomer of compound with stereogenic centres arises because 4 different groups making each of C‐2 & C‐3 stereogenic are same 4 different groups…(!) 95 R CO2H S CO2H R CO2H S CO2H H C OH HO C H H C OH HO C H -------------------- -------------------- -------------------- -------------------HO C H H C OH H C OH HO C H R S CO2H CO2H Enantiomers, Chiral S CO2H R CO2H Identical, achiral, Meso form possess plane of symmetry bisecting central C‐C bond 96 24 10/4/2011 Newman projections Cyclohexane conformations Shows arrangements in space e.g. Ethane C2H6 Staggered H conformation H H H H H H H H H H H H H H H H H “dash‐wedge” Examples H CH3 H H H H H 3 4 H H Cis‐1,2‐Dimethylcyclopentane a 2 e e a a a H H H H H H H H Cyclohexane H H Boat conformation Summary of isomerism 98 Hart et al. “Organic Chemistry: A Short Course” 12th Edition, page 52‐54 Different bond pattern H Structural (constitutional) isomer isomers H H 1 e e H H H e 6 3 4 e e a a 5 Chair conformations H H H H Newman97 H e "flip" e H H H a 5 H “sawhorse” CH3 a 0° H H 2 e H 1,2‐Dimethylcyclopentane CH3 e 6 HH H H e 1 a H Newman H H H H H “dash‐wedge” “sawhorse” Eclipsed conformation 60° a a a Same bond pattern CH3 Trans‐1,2‐Dimethylcyclopentane Stereoisomer Interconvertible by single bond rotation 99 Conformers (rotamers) Not interconvertible by bond rotation Configurational isomers 100 25 10/4/2011 Exercise Solutions 1. Draw the structures of 1. (a) (Z)‐3‐methyl‐2‐pentene (b) (S)‐2‐bromopropanol H a) (Z)‐3‐methyl‐2‐pentene c) (E,Z)‐2,4‐heptadiene b) (S)‐2‐bromopropan‐1‐ol d) (2S,3R)‐3‐bromobutan‐2‐ol 2. Using the Newman projection draw the structure of a staggered conformation of butane. CH3 CH3 H2C (c) (E,Z)‐2,4‐heptadiene H CH3 H3C Solutions C H C HO H Br 102 (4) 1‐bromo‐2‐chlorocyclopropane Staggered conformation of butane CH3 60° H H H Cis‐1‐bromo‐2‐ chlorocyclopropane Newman projection H H Br Cl CH3 H H * COOH H H (S)‐2‐methylpentanoic acid H 3C OH (d) (2S,3R)‐3‐bromobutan‐2‐ol 4. Draw the structure for the cis‐ and trans‐isomers of 101 1‐bromo‐2‐chlorocyclopropane 3. H C Br CH3 3. Using the Fischer projection draw a structure of (S)‐2‐methylpentanoic acid. 2. C H H3C Trans‐1‐bromo‐2‐ chlorocyclopropane H Br H H Cl H Fischer projection C H 2C H 2 C H 3 http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx 103 104 26 10/4/2011 Five important organic reactions Substitution reaction: Arrow Notation Substitution of one atom or group of Uses shows redistribution of valence electrons a molecule atoms by another atom or group of atoms. Addition reaction: defining resonance Addition of one molecule or atom to another chemical reaction direction to give a new molecule . normally on double or triple bond. defining a reaction mechanism. Elimination reaction: Elimination of two atoms or group of atoms from a molecule (reverse of the addition reaction). Oxidation and reduction reactions: Involve the loss or gain of Two common types of electron redistribution: From a bond to an adjacent atom, and From an atom to an adjacent bond. electron density by a carbon respectively. Rearrangement reaction: Conversion of one structure to an isomeric structure. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Terminology Bond breaking Nucleophile (Nu):- electron rich species and/or negatively charged species A B A + B A B A+ + B- Bond forming A + B A Homolytic bond forming B A+ + B- O -OH, Br-, -NH2, Heterolytic bond breaking Homolytic bond breaking examples O C Electrophile (E+):- electron poor species and/or positively charged species A Heterolytic bond forming B examples O H+, M+, H3O+ C C C Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 27 10/4/2011 Reactions of Alkanes PowerPoint to accompany Chapter 21 Alkanes: Scaffolds for Organic Chemistry Because of unreactive. Forcing conditions are required for a reaction structural role, alkanes are relatively They make excellent nonpolar solvents. Combustion: give mainly H2O and CO2 Free-radical reactions: chain reactions In order to be able to predict the product of a reaction, it is important to be able to distinguish between carbon (and hydrogen) types in an alkane. Primary, secondary, tertiary and quaternary carbon atoms. Primary, secondary and tertiary hydrogens. Combustion Classification hydrogen and carbon atoms in alkanes its Combustion of alkane gives CO2 and H2O. Combustion reactions are exothermic. e.g. ethane 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(l) ∆H = -2855 kJ Question Write a balance equation for the combustion of butane Solution Figure 21.25 C4H10(g) + 6½ O2(g) 4 CO2(g) + 5 H2O(l) 2C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(l) Homework Try combustion of pentane C5H12 ?? 28 10/4/2011 Combustion - Napalm The commonly quoted composition is 21% benzene, 33% gasoline, and 46% polystyrene. This mixture is difficult to ignite. Free-Radical Reactions Consider the following halogenation reactions: Reaction 1 Movie – Vietnam. dark CH4 + Cl2 no reaction Reaction 2 CH4 + Cl2 • Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia The free-radical halogenation is also an example of a radical chain reaction. Such reactions are characterised by three steps Initiation Light λ CH3Cl + CH2Cl2 + CHCl3 + CCl4 This type of reaction is called a substitution reaction. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Propagation steps - a series of new reactions which generate other radicals e.g. Propagation Termination radical Initiation produces the first free-radicals by a homolytic bond cleavage. Cl2 Cl Cl λ Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia radical Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 29 10/4/2011 Termination steps: stop the formation of free-radicals neutral molecular species formed e.g. Free-Radical Reactions Given the opportunity, free-radical reactions are regiospecific. More so for bromination than chlorination. radical pair 3°> 2° > 1° most stable least stable 1°> 2° > 3° radical pair most stable least stable R C R = akly group - EDG X C e.g X = halide - EWG Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia PowerPoint to accompany Chapter 23 Electrophilic addition and Substitution reactions The presence of carbon-carbon double or triple bonds in a compound markedly increases its chemical reactivity. Most characteristic reactions of alkene and alkynes are addition reactions: Alkenes and Alkynes Halogenation Hydrohalogenation Hydration The electron-rich nature of a multiple bond leads to enhanced reactivity with electrophiles (electron-loving species). 30 10/4/2011 Mechanism for HX Addition rxn Two-steps mechanism 1. First step is slow, rate-determining step. 2. Second step is fast. Alkenes reactions slow H Br carbocation Fast Figure 23.20 carbocation Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Addition of HX In the first step, -bond breaks and new C—H bond and cation form. Regioselectivity A occurs when a non-symmetric alkene is used. The preference based on intermediate carbocation stability: Primary In the second step, a new bond forms between negative Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia positive carbon. bromide ion and < (Less Stable) Figure 23.22 Secondary < Tertiary (Most Stable) The regioselectivity is stated as Markovnikov’s Rule: “In the addition of HX to an alkene, the hydrogen adds to the carbon atom of the double bond bearing the greater number of hydrogen atoms bonded directly to it.” Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 31 10/4/2011 Addition of HX 1. Which of the following is the most stable carbocation and why? CH3 H 1 C H Exercise H + CH 2 2 C Br Soln: CH3 + CH2 (A ) (D) • It is a tertiary carbocation. (B ) CH3 H 3C + C CH3 • Inductive effect electron rich groups (3˚) from CH3 + (C ) Major product Minor product Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Soln: C C Cl (A) minor product CH3 H H 3C CH3 C H 3C HCl 2-methylbut-2-ene The hydration of alkenes and alkynes requires the addition of a strong acid catalyst e.g. H2SO4 2-chloro-3-methylbutane C H 2. What will be the major product from the reaction by reflux in dark of 2-methylbut-2-ene with hydrochloride in a carbon tetrachloride solution? Name and draw the structure of the product. Addition of H2O CH3 CH3 H (D ) hydronium ion (electrophile) CH 3 CH 3 CCl4 Cl C C CH 3 H H (B) major product The first step is rate determining. Leads to the formation of a carbocation. 2-chloro-2-methylbutane 32 10/4/2011 Addition of H2O (Reversible) Second step is fast Carbocation reaction with excess H2O (nucleophile) leads to the formation of an oxonium ion. The fact that the reaction is always described by equilibria means that the overall reaction is reversible. That is, in the presence of H2SO4, alcohols can be dehydrated to yield alkenes. oxonium ion Loss of H+ from the oxonium ion leads to formation of an alcohol and regenerates the hydronium ion, hence it is catalytic reaction. butan-2-ol 2-butanol hydronium ion regenerated Formation of Alkenes Formation of Alkenes Previously, we have discussed the formation of alcohols from alkenes. The reverse process is also possible. This dehydration reaction is also known as an elimination reaction. I- > H2O > Br- > Cl- >> F- > CH3COO- > OH Requires the elimination of a good leaving group. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 33 10/4/2011 Exercise Halogenation 1. Draw and name the major product in the following reactions OH (a) H+ / H2O OH (b) H One of the easiest and most dramatic reactions is the addition of Br2 to an alkene. pentan-2-ol 2-pentanol H 2SO4 cyclohexene Heat Colour change Useful test for alkenes Figure 23.23 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Reactions of Alkynes Many similar reactions as alkenes. Reactions involve the replacement of -bonds with -bonds. Example H3 C H3C C C Br2 CH 3 CH2Cl2 Br Br C C Br CH3 (E)-2,3-dibromobut-2-ene (1) CH 3 H3C Br (2) CH2Cl2 Br H2C Cl hex-2-yne C H2C C CH3 CH2 Cl2 C F Br C C Br Br 2 Br F 1,1,2,2-tetrabromo1,2-difluoroethane Br C Br CH3 Cl (E)-2,3-dichlorohex-2-ene F CH2Cl2 CH2 C H3C CH2Cl2 Br2 CH3 Br 2,2,3,3-tetrabromobutane C 2. How would you form 1,1,2,2-tetrabromo-1,2-difluoroethane using bromine as one of your reagent? Solutions Br C Problems 1. What will be the product for a reaction between hex-2-yne and one mole of chlorine in dichloromethane under anhydrous condition? Draw the structure of the product and provide it IUPAC name. C Br 2 CH2Cl 2 F (E)-1,2-dibromo-1,2-difluoroethene F C C F 1,2-difluoroethyne 34 10/4/2011 Catalytic Hydrogenation Reaction requires molecular hydrogen (H2) and a metal catalyst Problems Provide the major organic product(s) in the reactions below (1) Br (2) Br 2 CH2Cl 2 HO Br 2 (3) H 2O PowerPoint to accompany Br 1,2-dibromo-3-ethylpentane 3-ethylpent-1-ene Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 3-ethylpentane 3-ethylpent-1-ene Ni, Pd, Pt, Ru typical metals used Because a gas, liquid and solid metal are used, this reaction is known as a heterogeneous catalysis H2 Pd/C 3-ethylpent-1-ene Br 1-bromo-3-ethylpentan-2-ol Conventions in Organic Chemistry Chapter 24 Alcohols, Ethers and Haloalkanes Figure 24.1 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 35 10/4/2011 Alcohol Properties The difference in electronegativity between oxygen and hydrogen allows alcohols to hydrogen bond. Compared to alkanes they: Alcohol Solubility There are differences in solubility: Like dissolves like Table 24.2 Are more soluble in water Have higher boiling and melting points Figure 24.4 Ether Properties Ether Nomenclature Ethers are more polar than alkanes but are unable to hydrogen bond. They are chemically inert. Both these properties mean that ethers are excellent solvents! Figure 24.10 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 36 10/4/2011 Reactions of Alcohols Reactions of Alcohols Alcohols react with sodium metal to form alkoxides and H2(g). Figure 24.12 Alkoxides are excellent nucleophiles and strong bases. Figure 24.13 Homework Fill in the reagents and name type of reaction !! 2 CH3CH2O H + 2 Na 2 CH3CH 2O-Na+ + H2(g) Sodium ethoxide Basicity of Alcohols An alcohol can act as a weak base, generating an oxonium ion under acidic conditions, which then loses water to form carbocation intermediate. CH3CH2 O H + H+ H O+ H CH3CH2 CH3CH2+ + oxonium ion H2O carbocation 2 CH3CH 2O H + 2 Na 2 CH3CH 2O-Na+ + H2(g) The stability of the carbocation is an enormous driving force for the nature of the reactivity of alcohols. Sodium ethoxide 37 10/4/2011 Alcohols to Haloalkanes Haloalkanes can be prepared from alcohols. Haloalkanes are important for introducing alkyl groups in chemical reactions. Two reactions are common Dehydration of Alcohols Dehydration of alcohols is an acid catalysed process. The ratedetermining step is the elimination of water to form the corresponding carbocation. Removal of a small group from a larger molecule is called elimination reaction. Water is classed as a good leaving group because of its stability and relatively unreactive nature. Hydroxide ion, on the other hand, does not eliminate easily and so is classed as a poor leaving group. Depend on the type of alcohol oxonium ion Elimination reaction Nucleophilic Substitution (Haloalkanes) Nucleophilic substitution reactions of haloalkanes proceed by attack of the nucleophile at the carbon bearing the halogen. Because of the difference in electronegativity between carbon and any of the halides, this bond is polarised such that the carbon becomes mildly electrophilic. This is a SN2 or biomolecular (2), nucleophilic (N), substitution (S) reaction. Order of leaving group: I- > H2O > Br- > Cl- >> F- > CH3COO- > OH- > NH2- Example 1. Show the reaction mechanism between sodium hydroxide and 1-bromoethane. Soln H C H Br CH3 + NaOH mechanism SN2 Mechanism Figure 24.15 1-bromoethane. Transition state Transition state ethanol 38 10/4/2011 University of KwaZulu-Natal, Westville Campus, Durban 2. Use reaction mechanisms to explain the why the inversion of the carbon reaction centre is observed when (S)-2-bromobutane is reacted with NaOH. Examinations : November 2010 CHEMICAL REACTIVITY - CHEM120 Duration: 3 hours Total Marks for Examination: 100 Internal Examiners: V.O. Nyamori, H.G. Kruger, M.D. Bala, B.S. Martincigh, P.G. Ndungu and G.D. Dawson Soln HO- CH 2CH 3 C H 3C H Br (S)-2-bromobutan CH2CH3 HO C Br CH3 Transition state _ QUESTION 6 Primary organic halides react with nucleophiles through a SN2 mechanism. In the mechanism inversion of the carbon reaction centre is observed. Provide the mechanism of the SN2 mechanism below. (4) CH2CH3 HO C H CH3 H (R)-butan-2-ol C H2C H3C Br H (a) starting material Inversion:- the (S)-configuration now becomes (R)-configuration. HO SN2 (b) transition state (c) product(s) Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Four Factors affecting the SN2 reaction rate Substrate The nucleophile attacks from the back of the substrate and hence to maximise the rate of the SN2 reaction, the back of the substrate must be as unhindered as possible. Order of reactivity: Primary > secondary > tertiary substrates (Most reactive) (not reactive) Nucleophile Solvent. Solvents may or may not surround a nucleophile, thus hindering or not hindering its approach to the carbon atom. Polar aprotic solvents because polar protic solvents will be solvated by the solvent hydrogen bonding to the nucleophile and thus hindering it from attacking the carbon with the leaving group. Leaving group The more stable the leaving group is, the more likely that it will take the two electrons of its carbon-leaving group bond with it when the nucleophile attacks the carbon. Therefore, the weaker the leaving group is as a conjugate base, and thus the stronger its corresponding acid, the better the leaving group. Examples of good leaving groups are therefore the halides (except fluoride) and tosylate, whereas HO- and H2N- are not Steric hindrance affects the nucleophile's strength e.g. methoxide anion is better nucleophile compared to tert-butoxide. Nucleophile strength is also affected by charge and electronegativity: nucleophilicity increases with increasing negative charge and decreasing electronegativity, e.g. nucleophile OH- > H2O and I- > Br(in polar protic solvents). 39 10/4/2011 University of KwaZulu-Natal, Westville Campus, Durban Examinations : November 2010 CHEM120 - CHEMICAL REACTIVITY Section B Example QUESTION 6 Primary organic halides react with nucleophiles through a SN2 mechanism. In the mechanism inversion of the carbon reaction centre is observed. (4) Provide the mechanism of the SN2 reaction below. H H 3C C H 2C Br H (a) starting material HO SN2 (b) transition state (c) product(s) Soln Nucleophilic Substitution (Haloalkanes) SN1 reaction mechanism R1 Table 24.4 Nu + R3 C R1 R2 C R3 LG This allows two different avenues for the nucleophilic attack, one on either side of the planar molecule. R3 Nu or R2 carbocation R1 C R1 Nu and Nu C R3 If the reaction takes place at a R2 R2 stereocenter yielding a racemic mix of enantiomers. Racemic mixture: Both (R) and (S)-configuration enantiomeric mixtures 40 10/4/2011 Scope of the reaction Example The SN1 mechanism tends to dominate when the central carbon atom is surrounded by bulky groups. SN2 reaction hindered by steric factor. Bulky substituents on the central carbon favour carbocation formation because of the relief of steric strain that occurs. The resultant carbocation is also stabilized by both inductive stabilization and hyperconjugation from attached alkyl groups. The SN1 mechanism dominates in reactions at tertiary alkyl centers and is further observed at secondary alkyl centers in the presence of weak nucleophiles. The carbocation intermediate formed in the reaction's rate limiting step is an sp2 hybridized carbon with trigonal planar molecular geometry. Step 1 Separation of the leaving group Carbocation Step 2 Nucleophilic attack oxonium ion Step 3 Deprotonation Product : 2-methylpropan-2-ol Haloalkanes to Alkenes: -Elimination The process of elimination of a good leaving group. Haloalkanes undergo elimination under basic conditions. This process is called -elimination reaction. It involves the dehydrohalogenation of the haloalkane. For the reaction to proceed, a hydrogen atom on the carbon atom adjacent to the carbon bearing the halide must exist and be removed. The -elimination reaction is a E2 or biomolecular (2) elimination (E) reaction. When there are two sets of -hydrogens present in the haloalkane, there is only one product formed and it is determined by the Zaitsev’s Rule: The major alkene formed by dehydrohalogenation is the more substituted one. major minor 41 10/4/2011 HX C2H5O-Na+ + R R R C C H tetrasubstituted HX C2H5O-Na+ + R Substitution vs Elimination R X R R C C H X Both Substitution reactions (SN1 & SN2) and elimination are important, however, each of these reactions does not work in isolation. H trisubstituted HX disubstituted disubstituted disubstituted Increasing degree of substitution R C2H5O-Na+ HX + C2H5O-Na+ + R monosubstituted HX + H H C C H X C2H5O-Na+ H ethene Figure 24.16 R H H H C C H X E.g. the addition of strong base to an alkyl halide yields an alkene: R H C C H H X H C C H X H R This elimination reaction competes with a substitution reaction which, in this case, forms ethers: H X = Cl, Br, I Substitution vs Elimination Mechanism shows a trigonal planar intermediate formed by concerted bond breaking and bond making processes There are usually stereochemical consequences Elimination vs Substitution Thus, there is a competition between elimination and substitution reactions and the result is dependant on: Size of the alkoxide The accessibility of the carbon bearing the halide The accessibility of the -hydrogen Figure 24.17 42 10/4/2011 Elimination vs Substitution PowerPoint to accompany We looked at the hydration of alkenes Chapter 25 The reverse reaction is also possible! Aldehydes and Ketones The presence of other nucleophiles add another dimension to the reaction. Figure 24.19 Aldehydes Carbonyl Compounds Contain C=O double bond. O Characterised by the functional group CHO. Contains a carbonyl group C Both carbon and oxygen are sp2 hybridised. Include many classes of compounds: aldehydes, ketones, carboxylic acids, esters, amides. O O C C R H Aldehyde R Ketone R' R O O C C O H Carboxylic acid R O O Ester R' R C Amide N R' Figure 25.1 R'' Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 43 10/4/2011 Aldehyde Nomenclature Ketones Name the longest chain containing the carbonyl group and replace ane with al. Name the longest chain containing the carbonyl group and replace ane with one. Must, by definition, occur at one of the ends of a molecule. Remember that aldehyde group takes priority over both methyl and hydroxyl groups. Carbonyl group is bonded to alkyl groups on either side. OH O H H H H H O progesterone Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Ketone Nomenclature Longest chain containing the carbonyl group. Replace ane with one. Constitutional isomers occur for C5 or greater. H O O camphor testosterone Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Preparation of Aldehydes and Ketones Oxidation of alcohols using a range of reagents • Most are chromium(VI) compounds (CrO3). Again, different types of alcohols yield different products • Primary alcohols can lead to aldehydes and carboxylic acids. • Secondary alcohols lead to ketones. • Tertiary alcohols do not oxidise. Figure 25.2 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 44 10/4/2011 Preparation of Aldehydes and Ketones Preparation of Aldehydes and Ketones Using pyridinium chlorochromate (PCC) Preparation of Aldehydes and Ketones ozonolysis Aldehydes and ketones can also be prepared in good yields by the Secondary alcohols can be oxidised by PCC, Jones Reagent (CrO3) or K2Cr2O7 to yield the corresponding ketone. reaction of alkenes with ozone, followed by a mild reduction using dimethyl sulfide. This reaction is called ozonolysis 45 10/4/2011 Reduction Reactions Reactions Aldehydes and Ketones In general Aldehydes and ketones can be reduced to give alcohols Catalytic hydrogenation is one way Reduction reaction Examples Question Reduction Reactions Provide the structure of the reaction products from the following equations. O Hydrides: OH (A) H2 H Pd/C or Pt H NaBH4 OH O LiAlH4 H2 (B) Pd/C or Pt OH O H2 (C) Pd/C or Pt O (D) OH H2 Pd/C or Pt H H Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Figure 25.5 46 10/4/2011 Examples Question Tautomerism in Aldehydes and Ketones Provide the structure of the reaction products from the following equations. OH O (A) LiAlH 4 H H Aldehydes and ketones bearing -hydrogen are able to undergo tautomerism to form enols (i.e. compounds bearing an alkene and an alcohol). Enols are constitutional isomers of their respective keto form (i.e. aldehyde or ketone). Tautomerism is the process by which two isomers are interconverted by a formal movement of an atom or group bearing in mind valency rules. Hence, an enol and its keto form are tautomers of the same structure. OH O LiAlH 4 (B) OH O LiAlH 4 (C) O LiAlH 4 OH (D) H H Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Aldehydes and ketones are in equilibrium with their enol forms and this equilibrium is acid catalysed. The process involves two separate proton transfers: The extent of tautomerisation is dependent on the aldehyde or ketone: Halogenation of Aldehydes and Ketones Aldehydes and ketones undergo a substitution reaction at the αcarbon in the presence of halogen in good yield. This reaction is regiospecific in that only the α-hydrogen is substituted under these conditions. The rate determining step in this reaction is the conversion of the ketone to its enol isomer, e.g. 47 10/4/2011 Carboxylic Acids PowerPoint to accompany Have hydroxyl group bonded to carbonyl group. Carboxylic acids are weak acids. Chapter 26 Carboxylic Acids and their Derivatives Carboxylic Acids Figure 26.3 Carboxylic Acids - Acidity Figure 26.4 Table 26.2 Figure 26.5 48 10/4/2011 Carboxylic Acids - Acidity Question Which of the following organic acids is expected to be the strongest and why? Provide an explanation (A) H H (B) C C O H ethanoic acid acetic acid Soln H F F (C) C C O H3C H C C O pKa = 4.8 H 2,2-dimethylpropanoic acid 2,2,2-trifluoroacetic acid pivalic acid O F C carboxylate anion CH3 CH3 F 2,2,2-trifluoroethanoic acid (B) Stronger acid than alcohols because of resonance stabilisation. O O O F C O- + H+ F Inductive effect from the substituents on the α-carbon. Figure 26.7 F atom is the most electronegative, hence, stabilizes the LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia carboxylate anion Brown, generated Carboxylic Acids - Reactions with Bases Examples Question Provide the structure of the reaction products from the following equations. O O (A) NaOH HO O NaOH + Na-O O (B) OH O-Na+ Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 49 10/4/2011 Esters Esters and Esterification Products of reaction between carboxylic acids and alcohols. O R C Examples O O H + HO R' R C O + R' H O H Table 26.3 Found in many fruits, perfumes and many other everyday products: Figure 26.9 Ester Hydrolysis A Carboxylic Acid Derivatives Saponification = hydrolysis of ester using aqueous base. Figure 26.12 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia 50 10/4/2011 Acid Anhydrides Acid Chlorides Formed by treatment of a carboxylic acid with thionyl chloride. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Formed by the reaction of an acid chloride and carboxylic acid or by dehydration. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Questions (a) Provide the structures for the reaction products from the following equations. O O (i) + HO O OH O (ii) Chapter 28 O F Dilute NaOH O OH H2O + HO F (b) Provide appropriate starting compounds or reagents used to form the products in the following equations. O O CH3 C HO ? PowerPoint to accompany SOCl3 CH3 O CH2 C Cl CH2 ? H3C O C OH O CH3 C H 3C O CH2 NitrogenContaining Compounds of Biological Relevance 51 10/4/2011 Biological Chemistry Biological Chemistry Research at the interface of biology and chemistry. The principles of chemistry are applied to understand the molecular basis of biological processes. We will investigate: Amines and amides Amino acids Peptides, proteins and enzymes Vitamins DNA Figure 28.1 Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Amines General formula: Can be aliphatic, heterocyclic or aromatic: • Next slide shows a number of simple amines. • Classified as 1o, 2o, 3o • Quaternary (4o) ammonium salts are also likely. Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia Figure 28.3 52 10/4/2011 Amine Nomenclature Figure 28.4 Amine Nomenclature Simple amines are named using the same guidelines for alcohols. In many instances, the amino group is named as a substituent. Reactivity of Amines Amines react with acids to form ammonium salts. Amines react with alkyl halides to form ammonium salts. 53 10/4/2011 Formation of Amides Amides Amides are derivatives of carboxylic acids. They are classified as 1o, 2o and 3o H3C O C H C H3C N N CH3 C H3C H H acetamide N-methylacetamide 1° Amides are formed by the reaction of amines with “activated” carboxylic acids. O O N CH3 CH3 O N,N-dimethylacetamide 2° 3° C H3C Cl + O H NCH2CH3 H C H 3C N CH2CH 3 + HCl H Brown, LeMay, Bursten, Murphy, Langford, Sagatys: Chemistry 2e © 2010 Pearson Australia CHEM 120 CHEMICAL REACTIVITY ORGANIC CHEMISTRY Dr. Vincent O. Nyamori 19 Lectures 5 Tutorials 2 Quiz 1 Test worksheets http://cheminnerweb.ukzn.ac.za/Firstyear/chemonetwenty.aspx THANK YOU and All the BEST 216 54