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Dr. Shatha I Alaqeel 108 Chem Learning Objectives Chapter two discusses the following topics and the student by the end of this chapter will: Know the structure, hybridization and bonding of alkenes Know the common and IUPAC naming of alkenes Know the geometry of the double bond i.e. cis/trans isomerization Know the physical properties of alkenes Know the different methods used for preparation of alkenes (elimination reactions ; dehydrogenation, dehydration and alkenes stability (Zaitsev’s rule) play an important role in understanding these reactions Know the addition reactions of alkenes and the effect of Markovnikov’s rule in determining the regioselectivity of this reaction. 2 Structure Of Alkenes (Olefines) They are unsaturated hydrocarbons – made up of C and H atoms and contain one or more C=C double bond somewhere in their structures. Their general formula is CnH2n cyclic alkenes - for non- Their general formula is CnH2n-2 alkenes - for cyclic 3 Alkenes sp2 Hybridization Of Orbitals In Alkenes The electronic configuration of a carbon atom is 1s22s22p2 Thus 2s2 2s1 2p2 promotion 2p3 3 x sp2 hybridization Trigonal planar 2p Alkenes Orbital Overlap In Ethene In ethylene (ethene), each carbon atom use an sp2 orbital to form a single C-C bond. Because of the two sp2 orbitals overlap by endto- end the resulting bond is called σ bond. The pi (π) bond between the two carbon atoms is formed by side- by-side overlap of the two unhybridized p- orbitals (2p–2p ) for maximum overlap and hence the strongest bond, the 2p orbitals are in line and perpendicular to the molecular plane. This gives rise to the planar arrangement around C=C bonds. Also s orbitals of hydrogen atoms overlap with the sp2 orbitals in carbon atoms to form two C-H bonds with each carbon atom. The resulting shape of ethene molecule is planar with bond angles of 120º and C=C bond length is 1.34 Å 5 Alkenes Orbital Overlap In Ethene sp2 hybridized carbon atoms two 2p orbitals overlap to form a pi bond between the two carbon atoms 6 two sp2 orbitals overlap to form a sigma bond between the two carbon atoms s orbitals in hydrogen atoms overlap with the sp2 orbitals in carbon atoms to form C-H bonds the resulting shape is planar with bond angles of 120º and C=C (1.34 Å) Nomenclature Of Alkenes and Cycloalkenes 1. Alkene common names: CH3 H2C CH2 CH3-CH Common: Ethylene CH2 Propylene Substituent groups containing double bonds are: H2C=CH– Vinyl group H2C=CH–CH2– Allyl group H3C C CH2 Isobutene Br Common: Allyl bromide Cl Vinyl chlorride The substituent is named in a similar way to the parent alkene. It is named based on the number of carbon atoms in the branch plus the suffix -yl. i.e. alkenyl. Vinyl group (=Ethenyl group) 7 IUPAC RULES Select the longest continuous Carbon chain containing the double bond , the ending of the name is changed from alkane to alkene . The C- chain is numbered starting from the end closet to the double bond. Indicate the location of the double bond by using the number of the first atom of the double bond just before the suffix ene or as a prefix. 1 2 3 4 1 2 3 4 5 6 CH3CH CHCH2CH2CH3 H2C CH CH2CH3 Hex-2-ene or 2-Hexene But-1-ene or 1-Butene (not 4-Hexene) (not 3-Butene) Indicate the positions of the substituents using numbers of carbon atoms to which they are bonded and write their names in alphabetical order (N.B. discard the suffixes tert-, di, tri,--when alphabetize the substituents) and if more than one substituent of the same type are present use the prefixes di- or 8 tri or tetra or penta,--- to indicate their numbers. 8 Cl 7 1 4 6 2 5 H3 C 1 6 4 C 2 CH 3 Br 1 3 CH3-CH2-CH2-CH=CH-C-CH3 6 7 5 4 3 2 1 Br OCH3 = (CH3CHCHCH2OCH3) 2 CH3 1-Methoxy-but-2-ene (not 4-Methoxy-but-2-ene) 2,2-Dibromo-3-heptene (not 6,6-Dibromo-4-heptene) 1 2 1 3 5 4 CH3 4 2-Methyl-but-2-ene or 2-Methyl-2-butene (not 3-Methyl-2-butene) 1CH 3 3-Chloro-2-hexene (not 2-Chloro-1-methyl-1-pentene) 6-Methyl-2-octene 4 5 3 2 3 CH3 6 7 9 8 2,3,7-Trimethyl-non-3-ene (not 2-Isopropyl-6-methyl--2-octene) 3 2 5-Methylcyclopenta-1,3-diene An ''a'' is added due to inclusion of di put two consonants consecutive 3 2 1 5 4 CN 4-Cyano-2-ethyl-1-pentene (not 2-Ethyl-4-cyano-1-pentene) 9 In cycloalkenes, the double bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule. CH3 1 1 2 H3C 1-Methyl cyclopentene (not 2-Methylcyclopeneten) 6 2 5 3 4 CH3 3,5-Dimethyl-cyclohexene (not 4,6-Dimethylcyclohexen) (not 1,5-Dimethyl-2-cyclohexen) CH3 5 4 5 4 Cl 3 2 3 1 1 2 CH3 4-Chloro-3,6-dimethylcyclohexene NOT CH3 1 6 6 3-Chloro-2,5-dimethylcyclohexene 3 2 5-Methylcyclopenta-1,3-diene An ''a'' is added due to inclusion of di put two consonants consecutive 10 If the substituents on both sides of the = bond are at the same distance, the numbering should start from the side that gives the substituents with lower alphabet the lower number. 2 1 3 7 4 6 5 3-tert-Butyl-7-isopropyl-cycloheptene (not 3-Isopropyl-7-tert-butylcycloheptene) When the longer chain cannot include the C=C bond, a substituent name is used 1 2 1 CH CH2 3 6 5 Vinyl-cyclohexane CH CH2 4 3-Vinyl-cyclohexene 11 Geometrical Isomerism (G.I) In Alkenes G. I. found in some, but not all, alkenes It occurs in alkenes having two different groups / atoms attached to each carbon A B atom of the = bond C G. I. x G. I. X A = C or B = D No Cis or trans (G. I. X) D G. I. G. I. A≠C B≠D A = B or C = D Cis or Z A =D or C = B trans or E G. I. 12 Alkenes Geometrical Isomerism In Alkenes It occurs due to the Restricted Rotation of C=C bonds so the groups on either end of the bond are ‘fixed’ in one position in space; to flip between the two groups a bond must be broken. X Geometrical isomers can not convert to each at room temperature. 13 Geometric (cis-trans) isomerism * The restricted rotation about C=C and the planer geometry give rise to a type of isomerism. * The prefix cis- is used when the two similar atoms or groups are on the same side of the double bond. * The prefix trans- is used when they are on opposite sides of the double bond. Cis-But-2-ene Trans-But-2-ene = Cis-Oct-4-ene 14 H H Trans-Oct-4-ene E-Z NOTATION FOR GEOMETRIC ISOMERISM If the groups attached to the C=C are different, we distinguish the two isomers *If the two first-priority atoms are together on the same side of the double bond, you have Z isomer. *If the two first-priority atoms are on opposite sides of the double bond, you have E isomer. *The priorities of the substituents are determined by the atomic number with atoms of higher atomic number having higher priority. I> Br > Cl > F > O > N > C > H 15 1 H3C CH3 C 2 H C 3 CH2CH3 4 5 E-3-Methyl-2-pentene 16 Alkenes * Exercise Q1-Which of the following compounds can exhibit cis / trans isomerism a) b) c) d) e) 2-Methylpropene 1-Butene 2-Methyl-2-pentene 2-Butene 3-Methyl-2-hexene Q2- Name the following compounds according to IUPAC system a) 17 b) c) * Physical Properties of Alkenes physical states C1-C4 are gases C5-C17 are liquids more than 18 carbon atoms are solids. * Solubility Alkenes are non polar compounds. Insoluble in water. Soluble in non polar organic solvents ( hexane, benzene,…). They are less dense than water. * Boiling point The alkenes has a boiling point which is a small number of degrees lower than the corresponding alkanes. The boiling point of alkenes increase as the number of carbons increase. 18 Preparation Of Alkenes Elimination 1- Dehydration of alcohols: ( removal of OH group and a proton from two adjacent carbon atoms ) using mineral acids such as H2SO4 or H3PO4 (H+). + CH 3CH 2OH H / heat H2C Ethanol CH2 + H2O Ethene OH + H / heat + H2O H cyclohexanol cyclohexene 19 Zaitsev’sRule If there are different protons can be eliminated with the hydroxyl group or with halogen atom, in this case more than one alkene can be formed, the major product will be the alkene with the most alkyl substituents attached to the double bonded carbon. H2C CH3 + H2O H3C CH3 H / Heat 1- Butene Minor OH H3C CH3 + H2O 2- Butene Major Zaitsev rule: an elimination occurs to give the most stable, more highly substituted alkene 20 2- Dehydrohalogenation of alkyl halides using a base: (removal halogen atom with hydrogen atom ) Heating an alkyl halide with a solution of KOH, in alcohol, yields an alkenes. 21 3- Dehalogenation of vicinal dihalides alcohol Example: 22 Reactions Of Alkenes Substitution reaction Oxidation Reactions KMnO4 Reactions of Alkenes Ozonolysis Epoxidation Addition(Electrophilic) reaction: - Hydrogenation - Halogenation - Hydrohalogenation - Halohydrin formation -Hydration 23 1-Substitution reactions Halogenation at High Temperature Involve the saturated alkyl chain at High temperature or UV light Addition reaction: low temperature Absence of light CH2=CH-CH3 Substitution reaction: High temperature or Ultraviolet light 24 Electrophilic Addition Reaction 2- Additions To The Carbon-Carbon Double Bond 2.1 Addition Of Hydrogen: Hydrogenation A A + A H2 Pt or Ni or Pd A A A A H H An alkane An alkene H2C H3C CH2 A + Pt H2 CH2 + H2 Pt CH3 H3C CH3 CH3 CH3 + CH3 H3C H2 Pt CH3 Cis-1,2-Dimethyl cyclohexane 25 2.2.Addition of Halogens( Halogenation) A A + A X2 CCl4 A A A A A X (X= Cl or Br) X Cl H3C CH3 + Cl 2 CCl 4 H3C CH3 Cl Br + Br2 CCl 4 Br 26 2.3. Addition of Hydrogen Halides (Hydrohalogenation) when HX is added to symmetrical alkenes, there is only one possible product from this addition by strong acids such as ethene, 2-butene and cyclohexene. A A A + A A A A HX (x= Cl or Br or I) A H X Cl + HCl CH3 H3C H + H HI I However, if the double bond carbon atoms are not structurally equivalent, i.e. unsymmetrical alkenes as in molecules of 1- propene, 1-butene, 2-methyl-2-butene and 1-methylcyclohexene, the reagent may add in two different ways to give two isomeric products. This is shown for 1-propene in the following equation. 27 CH3CHCH3 HBr Br CH3CHCH3 2o Carbocation Br CH3CH=CH2 CH3CH2CH3 maijor Br CH3CH2CH2Br minor 1o Carbocation Stability of carbocation H3C CH3 C CH CH3 CH3 3o 28 H3C 2o CH2CH2 1o CH3 Markovnikov’s rule In addition of unsymmetrical reagent to unsymmetrical alkenes the positive ion adds to the carbon of the alkene that bears the greater number of hydrogen atoms and the negative ion adds to the other carbon of the alkene. CH3 CH3 CH3 + H3C Cl HCl CH3 H3C 29 Anti Markovnikov’s Rule In the addition of an acid to an alkene the hydrogen will go to the vinyl carbon that already has the lowest number of hydrogens. 30 2.4- Addition of Sulfuric acid heat (H-OSO3H) Addition of Sulfuric acid to alkenes also follows Markovnikov’s rule, as the example H2O heat Propene Isopropyl hydrogen sulfate 2-Propanol 31 2.5. Addition of HOX ( -OH, X+): Halohydrin formation Only one product is possible from the addition of HOX acids (formed from mixture of H2O and X2) to symmetrical alkenes such as ethene and cyclohexene. Symmetrical akenes A A A + A H2 O / X2 A A A A (x= Cl or Br ) OH X Cl + H2O / Cl2 CH3 H3C OH However, addition reactions to unsymmetrical alkenes will result in the formation of Markovonikov’s product preferentially. Unsymmetrical akenes + H2O / Cl2 OH Cl + CH2Br H2O / Br2 32 OH 2.6. Addition of H2O: Hydration Only one product is possible from the addition of H2O in presence of acids as catalysts to symmetrical alkenes such as ethene and cyclohexene. Symmetrical akenes A A A + A H H2 O A A A A H OH OH + H H2 O CH3 H3C H However, addition reactions to unsymmetrical alkenes will result in the formation of Markovonikov’s product preferentially. Unsymmetrical akenes H H CH3 + H2O 33 OH CH3 3-Oxidation of alkenes 3.1- Ozonolysis ( Oxidation of alkenes by ozone O3 ) This reaction involves rupture of the C=C to give aldehydes or ketones according to the structure of the original alkene. A A A + A A A O3 A A O Zn /H2O - H2O2 O A A O + O A A O ( A= H or R) i) O3 O + ii) Zn /H2O O H i) O3 ii) Zn /H2O i) O3 O + O O ii) Zn /H2O O 34 3.2- Oxidation of alkenes with Permanganate (Baeyer test) OH KMnO4 / OH OH 3.3- Oxidation of alkenes with peroxy acid RCO 3H O Epoxide H3O + OH anti hydroxylationOH 35 Polymerization (Free radical) A) B) 36 Thank You for your kind attention ! Questions? 37