Organic Chemistry. Alkenes (2)

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For updated version, please click on http://ocw.ump.edu.my Organic Chemistry Alkenes (2) by Dr. Seema Zareen & Dr. Izan Izwan Misnon Faculty Industrial Science & Technology seema@ump.edu.my & iezwan@ump.edu.my Alkenes By Seema Zareen http://ocw.ump.edu.my/course/view.php?id=152

Expected Outcomes In the completion of this chapter, student will have the ability to: - Describe characteristics and physical properties of alkene - Draw and name alkene using IUPAC nomenclature - Write alkene reaction mechanism

Contents Alkenes Physical properties Nomenclature Alkene reaction mechanisms

Introduction to Addition Reactions The characteristic reaction of alkenes is addition the bond is broken and two new bonds are formed. Alkenes are electron rich, with the electron density of the bond concentrated above and below the plane of the molecule. Because alkenes are electron rich, simple alkenes do not react with nucleophiles or bases, reagents that are themselves electron rich. Alkenes react with electrophiles. 4

Introduction to Addition Reactions Because the carbon atoms of a double bond are both trigonal planar, the elements of X and Y can be added to them from the same side or from opposite sides. 5

Introduction to Addition Reactions 6

Hydrohalogenation Electrophilic Addition of HX Two bonds are broken in this reaction the weak bond of the alkene and the HX bond and two new bonds are formed one to H and one to X. Recall that the H X bond is polarized, with a partial positive charge on H. Because the electrophilic H end of HX is attracted to the electron-rich double bond, these reactions are called electrophilic additions. 7

Hydrohalogenation Electrophilic Addition of HX To draw the products of an addition reaction: 8

Hydrohalogenation Electrophilic Addition of HX The mechanism of electrophilic addition consists of two successive Lewis acid-base reactions. In step 1, the alkene is the Lewis base that donates an electron pair to H Br, the Lewis acid, while in step 2, Br is the Lewis base that donates an electron pair to the carbocation, the Lewis acid. 9

Hydrohalogenation Markovnikov s Rule With an unsymmetrical alkene, HX can add to the double bond to give two constitutional isomers, but only one is actually formed: This is a specific example of a general trend called Markovnikov s rule. Markovnikov s rule states that in the addition of HX to an unsymmetrical alkene, the H atom adds to the less substituted carbon atom that is, the carbon that has the greater number of H atoms to begin with. 10

Markovnikov s Rule In the ionic addition of an acid to the carbon-carbon double bond of an alkene, the hydrogen of the acid attaches itself to the carbon atom which already holds the greater number of hydrogens. Them that has, gets! The richer get richer! (V. W. Markovnikov -- 1838-1904) 11

Hydrohalogenation Markovnikov s Rule The basis of Markovnikov s rule is the formation of a carbocation in the rate-determining step of the mechanism. In the addition of HX to an unsymmetrical alkene, the H atom is added to the less substituted carbon to form the more stable, more substituted carbocation. 12

Hydrohalogenation Reaction Stereochemistry Recall that trigonal planar atoms react with reagents from two directions with equal probability. Achiral starting materials yield achiral products. Sometimes new stereogenic centers are formed from hydrohalogenation: A racemic mixture 13

Hydrohalogenation Reaction Stereochemistry The mechanism of hydrohalogenation illustrates why two enantiomers are formed. Initial addition of H + occurs from either side of the planar double bond. Both modes of addition generate the same achiral carbocation. Either representation of this carbocation can be used to draw the second step of the mechanism. 14

15

Hydrohalogenation Reaction Stereochemistry Nucleophilic attack of Cl on the trigonal planar carbocation also occurs from two different directions, forming two products, A and B, having a new stereogenic center. A and B are enantiomers. Since attack from either direction occurs with equal probability, a racemic mixture of A and B is formed. 16

Hydrohalogenation Summary 17

Hydration Electrophilic Addition of Water Hydration is the addition of water to an alkene to form an alcohol. 18

Hydration Electrophilic Addition of Water 19

Halogenation Addition of Halogen Halogenation is the addition of X 2 (X = Cl or Br) to an alkene to form a vicinal dihalide. 20

Halogenation Addition of Halogen Halogens add to bonds because halogens are polarizable. The electron rich double bond induces a dipole in an approaching halogen molecule, making one halogen atom electron deficient and the other electron rich (X + X ). The electrophilic halogen atom is then attracted to the nucleophilic double bond, making addition possible. Two facts demonstrate that halogenation follows a different mechanism from that of hydrohalogenation or hydration. No rearrangements occur Only anti addition of X 2 is observed These facts suggest that carbocations are not intermediates. 21

Halogenation Addition of Halogen Carbocations are unstable because they have only six electrons around carbon. Halonium ions are unstable because of ring strain. 22

Halogenation Reaction Stereochemistry Consider the chlorination of cyclopentene to afford both enantiomers of trans-1,2-dichlorocyclopentane, with no cis products. Initial addition of the electrophile Cl + from (Cl 2 ) occurs from either side of the planar double bond to form a bridged chloronium ion. 23

Halogenation Reaction Stereochemistry In the second step, nucleophilic attack of Cl must occur from the backside. Since the nucleophile attacks from below and the leaving group departs from above, the two Cl atoms in the product are oriented trans to each other. Backside attack occurs with equal probability at either carbon of the three-membered ring to yield a racemic mixture. 24

ADDITION OF BROMINE B r B r C C C C l 4 SLOW C + C B r C C B r B r B r : B r + - B r alkene polarizes bromine 25

ANTI ADDITION H B r B r H B r syn syn - anti + H H B r anti open carbocation would give both cis and trans H H H B r B r B r B r cis compound NOT OBSERVED H trans compound ACTUAL PRODUCT 26

Halohydrin Formation Treatment of an alkene with a halogen X 2 and H 2 O forms a halohydrin by addition of the elements of X and OH to the double bond. 27

Halohydrin Formation Alkenes Even though X is formed in step [1] of the mechanism, its concentration is small compared to H 2 O (often the solvent), so H 2 O and not X is the nucleophile. 28

Halohydrin Formation Alkenes Although the combination of Br 2 and H 2 O effectively forms bromohydrins from alkenes, other reagents can also be used. Bromohydrins are also formed with N-bromosuccinimide (NBS) in aqueous DMSO [(CH 3 ) 2 S=O]. In H 2 O, NBS decomposes to form Br 2, which then goes on to form a bromohydrin by the same reaction mechanism. 29

Halohydrin Formation Because the bridged halonium ion is opened by backside attack of H 2 O, addition of X and OH occurs in an anti fashion and trans products are formed. With unsymmetrical alkenes, the preferred product has the electrophile X + bonded to the less substituted carbon, and the nucleophile (H 2 O) bonded to the more substituted carbon. 30

Halohydrin Formation Alkenes 31

Halohydrin Formation Alkenes 32

Calculating Degrees of Unsaturation An acyclic alkene has the general structural formula C n H 2n. Alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon. Cycloalkanes also have the general formula C n H 2n. Each bond or ring removes two hydrogen atoms from a molecule, and this introduces one degree of unsaturation. The number of degrees of unsaturation for a given molecular formula can be calculated by comparing the actual number of H atoms in a compound to the maximum number of H atoms possible for the number of carbons present if the molecule were a straight chain alkane. This procedure gives the total number of rings and/or bonds in a molecule. Possible structures for C 4 H 6 : 33