1435-1436 2014-2015 Alkenes 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. Alkenes tructure Of Alkenes 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 - for non-cyclic alkenes Their general formula is CnH2n-2 - for cyclic alkenes 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) Alkenes Nomencalture Of Alkenes And Cycloalkenes 1. Alkene common names: CH3 H2C CH2 Common: Ethylene CH3-CH CH2 H3C Propylene C CH2 Isobutene Substituent groups containing double bonds are: H2C=CH– Vinyl group H2C=CH–CH2– Allyl group Br Cl Common: Allyl bromide 7 Vinyl chlorride Alkenes 2. IUPAC Nomenclature Of Alkenes Find the longest continuous Carbon chain containing the double bond this determines the root name then add the suffix -ene. Number the C- chain from the end that is nearer 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 H2C 2 3 4 CH CH2CH3 But-1-ene or 1-Butene (not 3-Butene) 1 2 CH3CH 3 4 5 6 CHCH2CH2CH3 Hex-2-ene or 2-Hexene (not 4-Hexene) 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 tri or tetra or penta,--- to indicate their numbers. 8 Alkenes 8 Cl 7 1 4 6 2 5 H3C 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) Alkenes 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. 1 CH3 1 H3C 6 2 5 3 4 CH3 3,5-Dimethyl-cyclohexene (not 4,6-Dimethylcyclohexen) (not 1,5-Dimethyl-2-cyclohexen) 1-Methyl cyclopentene (not 2-Methylcyclopeneten) 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 4 7 6 10 5 3-tert-Butyl-7-isopropyl-cycloheptene (not 3-Isopropyl-7-tert-butylcycloheptene) Alkenes When the longer chain cannot include the C=C, a substituent name is used. 1 1 CH CH2 3 6 5 Vinyl-cyclohexane 2 CH CH2 4 3-Vinyl-cyclohexene Alkenes Geometrical Isomerism In Alkenes G. I. found in some, but not all, alkenes It occurs in alkenes having two different groups / atoms attached to each carbon atom of the = bond A B C A=C or B=D No Cis or transe (G. I. X) G. I. x 12 D A≠C B≠D, A =B or C=D Cis / A =D or C= B transe G. I. G. I. X G. I. G. I. 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 Alkenes Types Of Geometric Isomerism A) Cis / trans isomerism in alkenes Exhibited by alkenes having two H’s and two other similar groups or atoms attached to each carbon atom of the = bond (or generally the alkene have only two types of atom or groups i.e. ABC=CAB) Cis prefix used when hydrogen atoms on both carbon atoms are on the SAME side of C=C bond Trans prefix used when non-hydrogen groups / atoms are on the opposite sides of C=C bond 14 Cis-But-2-ene Trans-But-2-ene Alkenes H H Cl H Cl Cl H Cl cis-1,2-Dichloro-ethene trans-1,2-Dichloro-ethene Cis Groups / atoms are on the Same Side of the double bond Trans Groups / atoms are on Opposite Sides across the double bond = H Cis-Oct-4-ene 15 H Trans-Oct-4-ene Alkenes B) Z/ E isomerism in alkenes If the groups attached to the C=C are different, we distinguish the two isomers by adding the prefix Z (from German word Zusammen) if the higher-priority groups are together in the same side or E (from German word Entgegen) if the higherpriority groups are opposite sides depending on the atomic number of the atoms attached to each end of the C=C. Atoms with higher atomic numbers receive higher priority I> Br > Cl > F > O > N > C > H 16 Alkenes I Cl Br CH3 I CH3 Br Cl Z-2-bromo-1-chloro-1-iodopropene E-2-bromo-1-chloro-1-iodopropene O CH3 HC CH3 I CH2 Cl H HO Z Z 17 Alkenes Exercise Q1-Which of the following compounds can exhibit cis / trans isomerism a) 2-Methylpropene b) 1-Butene c) 2-Methyl-2-pentene d) 2-Butene e) 3-Methyl-2-hexene Q2- Name the following compounds according to IUPAC system a) 18 b) c) Alkenes 1 3 4 2 3 5 2 Br 4 1 3-Bromo propene 2-Ethyl-4-methyl pentene CH3 5 4 Cl CH3 6 1 2 3 5 4 3 CH3 4-Chloro-3,6-dimethylcyclohexene Br C H 3-Chloro-2,5-dimethylcyclohexene 1 H C 3 Br C 1 2 Cl CH3 6 C H 3 H C CC 2 3 H H H Cis-1,2-Dibromoethene Geometrical isomerism 2 C H C H 2 3 45 E-3-Methyl-2-pentene 4 Cl C Cl 1,1-Dichloroethene not geometrical isomerism 6 1 3 5 Trans-trans-2,4-heptadiene Cis-cis-2,5-heptadiene 7 Trans-1,3,5-heptatriene Home work 19 Alkenes Physical Properties of Alkenes 20 Alkenes are nonpolar compounds thus: Insoluble in water Soluble in nonpolar solvents ( hexane, benzene,…) The boiling point of alkenes increase as the number of carbons increase. Alkenes Preparation Of Aalkenes 1- Dehydration of alcohols ( removal of OH group and a proton from two adjacent carbon atoms ) using mineral acids such as H2SO4 or H3PO4 + CH 3CH 2OH H / heat H2C Ethanol CH2 Ethene OH + H / heat + H2O H cyclohexanol 21 + H2O cyclohexene Alkenes 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 22 Alkenes 2- Dehydrohalogenation of alkyl halides using a base or NaOH 23 Alkenes 3. Dehalogenation of vicinal dihalides Zn/AcOH Br Br For example: Dehalogenation of 1,2-Dibromobutane leads to the formation of 1Butene. In the presence of catalyst. 24 Alkenes Reactions Of Alkenes Oxidation Reactions KMnO4 Reactions of Alkenes Ozonolysis Addition(Electrophilic) reaction: - Hydrogenation - Halogenation - Hydrohalogenation - Halohydrin formation -Hydration 25 Alkenes An electrophile, an electron-poor species,(from the Greek words meaning electron loving). It is a species (any molecule, ion or atom) that accept a pair of electrons to form a new covalent bond. , H , B r , C l , I , e tc ., A lC l 3 , B F 3 , F e C l 3 , F e B r 3 , e tc . C A nucleophile, an electron-rich species, ,(from the Greek words meaning nucleus loving). It is a species (any molecule, ion or atom) that donate an electron pair to form a new covalent bond. C , OH , Br , Cl , I , etc., H2N, HS, etc., H2O , CH3OH , RNH2 , R2NH , R3N , , etc. 26 Alkenes Electrophilic Addition Reaction 1- Additions To The Carbon-Carbon Double Bond 1.1 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 27 Alkenes 1.2.Addition of Halogens( Halogenation) A A A + A H3C X2 A A CH3 A A X + Cl 2 CCl 4 (X= Cl or Br) X Cl H3C CH3 Cl Br + Br2 CCl 4 Br CH3 Br CH3 + CH3 28 Br2 CCl 4 Br CH3 Trans-1,2- Dibromo-1,2-Dimethyl cyclohexane Alkenes 1.3. Addition of Hydrogen Halides Only one product is possible from the addition of these strong acids to symmetrical alkenes 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 1methylcyclohexene, the reagent may add in two different ways to give two isomeric products. This is shown for 1-propene in the following equation. 29 Alkenes Br CH3CHCH3 HBr CH3CHCH3 2o Carbocation maijor Br CH3CH=CH2 Br CH3CH3CH3 CH3CH2CH2Br minor 1o Carbocation Stability of carbocation H C 3 C H C 3H 3 C C H C H 3 C H 3 o 3 30 o 2 C H C H 2 2 C H 3 o 1 Alkenes However when the addition reactions to such unsymmetrical alkenes are carried out, it was found that 2-bromopropane is nearly the exclusive product. Thus it said the reaction proceeded according to Markovnikov’s rule Markovnikov’s rule stats that : 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 31 Alkenes 1.4. Addition of HOX halogen in aqueous solution ( -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 32 Alkenes 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 OH 33 Alkenes 1.5. 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 H2 O H A A A A H OH OH + H2 O H CH3 H3C H However, addition reactions to unsymmetrical alkenes will result in the formation of Markovonikov’s product preferentially. Unsymmetrical akenes H H 34 CH3 + H2O OH CH3 Alkenes Oxidation Reaction: 1- Ozonolysis (Oxidative cleavage): 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 O O ( A= H or R) i) O3 O + ii) Zn /H2O H i) O3 ii) Zn /H2O i) O3 35 Zn /H2O - H2O2 O + O ii) Zn /H2O O O O A A O + A O A Alkenes 2- Oxidation with KMnO4 (Oxidative addition): OH KMnO4 / OH OH Cis- doil 36 Alkenes Thank You for your kind attention ! Questions? Comments 37