Unsaturated hydrocarbons Alkenes

Transcription

1 Dr. Seham ALTERARY Chem 145 Unsaturated hydrocarbons Alkenes Chapter nd semester

2 By the end of this chapter you should be familiar with: Definition of Alkenes. Nomenclature of Alkenes. Substituted groups from Alkenes; (Alkenyl) The ybridization of Alkenes. Geometric isomerism among the Alkenes. Cis-Trans isomerism. Physical properties of Alkenes. General Methods of Preparation of Alkenes. Reactions of Alkenes. 2

3 Definition of Alkenes Alkenes: The second class of simple hydrocarbons the alkenes, consists of molecules that contain at least one doublebonded carbon pair. The chemical formula for the simple alkenes follows the expression: C n 2n 3

4 I. Common or trivial name Nomenclature The common or trivial name may be used, often have their origin in the history of the science and the natural sources of specific compounds. Common names used in 1800 s are still in used for some compounds today. Particularly, those of low molecular weight. Example The simplest members of the alkenes series (C 2 and C 3 ) are usually called by their common Names, which are derived from the corresponding alkanes by replacing the (-ane) ending by (-ylene). C 2 C 2 C 3 C C 2 Common Name: Ethylene Propylene IUPAC Name: Ethene Propene

5 II. IUPAC naming II-a. Open-chain Nomenclatures The IUPAC rules for naming alkenes are similar to those used for naming alkanes. The rules are: 1. The longest continuous carbon chain containing the double bond is selected as the parent chain. 2. The name of the parent carbon chain is obtained by replacing the -ane ending of the corresponding alkane by -ene. C 3 C C 2 Propene 3. The postion of the double bond is indicated by the number of the lower numbered doubly bond. 1-Butene C 3 C 2 C C 2 (not 3-Butene) 4. The parent carbon chain is numbered in a manner than will give the doubly bonded carbon atoms the lowest numbers even if it resuls in the substituents getting higher numbers. 5,5-Dimethyl-1-hexene

6 Examples C 3 1 C 3 CC 2 C C Methyl-1-pentene ( not 2-Methyl-4-pentene) 2-ethyl-1-octene Not or 2-Methyl-pent-2-ene 2-Methyl-2-pentene 4-Methyl-3-pentene

7 II-b. Cycloalkenes Nomenclature A cycloalkene or cycloolefin is a type of alkene hydrocarbon which contains a closed ring of carbon atoms, but has no aromatic character. Examples Cyclopropene Cyclobutene Cyclopentene Cyclohexene In cycloalkene, the numbering must begin at the double bond. i.e, the double bond is always found between carbon 1 and carbon 2.! Note that It is therefore not necessary to specify the position of the double bond with a number. C 3 3-Methylcyclohexene (not 1-Methyl-2-cyclohexene)

8 Examples Cl C 3 C 2 3-Chloro-4-ethylcyclobutene (not 1-Chloro-2-ethylcyclobutene) 1,4,4-trimethylcyclobutene ( not 2,3,3-trimethyl cyclobutene) 3 C Br 3 C 1-ethyl-5-methylcyclopentene 3-Bromo-1,3-dimethyl-cyclohexene

9 Alkenyl substituted Group A hydrocarbon group containing carbon-carbon double bond is to be treated as a substituent. In these cases the substituent is named in a similar fashion to simple alkyl substituents. 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. There are two common names that are widely used: 1- Vinyl group (=Ethenyl group) C 2 C 2 2- Allyl group (=Propenyl group) C 2 C 2 C C 2 R CC 3 Example Common Name: C 2 C 2 C R (derived from ethylene) C Br Vinyl bromide (derived from propylene) C 2 C C 2 Cl Allyl chloride

10 Excercise: Write the structural formula of 3,5-dimethyl-4-isopropyl-2-octene. (1) C C C C C C C C C 3 C 3 (2) C C C C C C C C C 3 C 3 (3) C C C C C C C C C C 3 C 3 C 3 C 3 (4) C 3 C C C C C 2 C 2 C 3 C C 3 C 3 10

11 ybridization in Alkenes Example: For the simplest alkene, Ethylene, or Ethene C 2 C 2 2p x1 2p y1 2p z 0 2p x1 2p y1 2p z 1 sp 2 sp 2 sp 2 2p z 1 2s2 2s1 1s 2 1s 2 1s 2 Promotion ybridization Ground state Excited state ybridization These orbitals are formed in the following manner: 2s orbital and only two of the three 2p orbitals hybridize, the result is : 3 equivalent sp 2 hybrid orbitals and one unhybridized orbital 2p z orbital.

12 (a) 2s 2p x 2p y 2p z ybridization The hybridization of 2s, 2p x and 2p y to form sp 2 (b) sp 2 sp 2 2p z sp 2 (or) 120 2p z 2p z 2p z The three sp 2 orbitals get as far away from each other as possible assuming a planer arrangement with an angle of 120 between hybrid orbitals, (trigonal planer). The remaining unhybridized 2p z orbital is perpendicular to the plane of the sp 2 orbitals. 12

13 Bond Formation in Ethylene The two carbons of Ethylene are attached to gather by an Sp Sp 2 = σ bond. The 2p z 2p z overlap resulting the carbon---carbon double bond = π bond p z C sp 2 sp 2 2p z 2p z C 2p z overlap of sp 2 and p orbitals π σ π Equivalent to C C 13

14 The Ethylene molecule has trigonal planer geometrical shape, with a bond angle equal to 120, and bond length 1.34 A Note that; the restricted rotation about the carbon carbon double bond and the planar geometry give type of geometric isomersim. 14

15 Geometric isomerism among the Alkenes. The pi-bond in an alkene does not permit rotation, thus all of the atoms attached directly to the alkene lie in a plane. The restricted rotation about C=C and the planer geometry give rise to a type of isomerism. Called Geometric Isomerism Cis-Trans isomerism. Two different substituents Groups attached to the alkene could be positioned on the same side of the alkene, on opposite sides of the alkene. Br C C Cl or Br C C Cl Cis-1- Bromo-2-chloroethene Trans-1-Bromo-2-chloroethene 15

16 Geometric isomers differ from one another only in the way in which the substituents are arranged in space related to the plane of the C=C bond. Example: the isomers of 2-Butene 3 C C C C 3 3 C C C C 3 Cis-2-Butene (mp= -139 C; bp= 3.7 C) Trans-2-Butene (mp= -106 C; bp= 0.9 C) Cis and Trans are stable molecules have different physical properties and same chemical properties. 16

17 Does this molecule have geometric isomers? 1. Must contain a C=C (or another double bond). 2. On each C of the C=C, there must be 2 different groups. C 3 C 3 C 2 C C 3. The molecule above does not have cis or trans isomers. On the right C of the C=C, there are 2 atoms (must have different groups on each C of the C=C.) Find the 2 groups that are the same on the 2 C of the C=C and circle them; if they re on the same side of the line: cis; different sides: trans. C 3 C 2 C C C 3 Cis C 3 C 2 C C C 3 trans

18 Examples on cis&trans isomers Cl Cl 1- Cl Cl cis-1,2-dichlorocyclohexane trans-1,2-dichlorocyclohexane 2- cis-2-pentene trans-2-pentene Cl Cl Cl 3- cis-1,2-dichloroethene Cl trans-1,2-dichloroethene

19 Exercise: which of the following compound has geometric isomerism? Why? (a) Propene C 3 C C Geometric isomerisim is not possible (b) 2-Methyl-2-butene C 3 C 3 C C C 3 Geometric isomerisim is not possible (c) 2-butenedioic acid OOC C C COO Cis-form (Maleic acid) OOC C C COO Trans-form (Fumaric acid) (d) 1,1-Dichloropropene C 3 C C Cl Cl Geometric isomerisim is not possible 19

20 Physical State: The Physical Properties of Alkenes The first lower member like ethene, propene and butene are colorless gases. Alkenes with five (C 5 ) to seventeen (C 17 )carbon atoms are liquids. and higher ones are solids at ordinary temperatures. Solubility Alkenes are insoluble in water and soluble in nonpolar organic solvents such as benzene C 6 6 or carbontetrachloride CCl 4. Boiling point 1.The boiling points of alkenes gradually increase with an increase in the molecular mass (or chain length). 2. Branched chain alkenes have lower boiling points than the corresponding straight chain isomers. 20

21 General Methods of Preparing Alkenes Alkenes are prepared in the laboratory by one of two general methods. They can be prepared from alcohols or alkyl halides by elimination reactions. 1. Dehydration of Alcohols The General Equation C C O An alcohol acid catalyst ( + ) C C An alkene 2 O When an alcohol is heated in the presence of a mineral acid catalyst, it loses a molecule of water to give an alkene. In Latin De- means= removal of The acid catalyst + ex; 2 SO 4 or 3 PO 4 & is eat. 21

22 Examples C 2 (a) C 2 O IUPAC Name: Ethanol Common Name: Ethyl alcohol + C 2 C 2 2 O + Ethene Ethylene (b) C 3 C C 2 O + C 3 C C O IUPAC Name: 1-Propanol Propene Common Name: n-propyl alcohol Propylene (c) IUPAC Name: Common Name: O O Cyclohexanol Cyclohexene Cyclohexyl alcohol 22

23 Problem 3.7 Write the structure of the product of the dehydration of each of the following alcohols. (a) C 3 C 3 C O C 3 + C 3 C 3 C C 2 (b) O + 23

24 The previous examples of dehydration of alcohols produced a single alkenes as a possible product. Suppose, that the loss of water from adjacent carbon atoms can give rise to more than one alkene as in: C 3 C 2 CC 3 O + C 3 C 2 C C O + Which alkene predominates, The dehydration of 2-Butanol 1-Butene Minor Product C 3 C CC O 2-Butene Major product 1-Butene or 2-Butene 24

25 Saytzeff s Rule The major product is always the alkene with the most substituents attached on the double-bonded carbons. Problem 3.8 Give the major and the minor product of the dehydration of (a) pentanol O 2-pentene Major product 1-pentene Minor product (b) C 3 O + + C 3 C 3 1-methylcylohexene 3-methylcyclohexene 2-methylcyclohexanol Major product Minor product 25

26 2. Dehydrohaloganation of Alkyl halids The second method for preparing an alkene by using dehydrohlogenation reaction, The reaction involves the elimination of and of X from adjacent carbon. General equation C C X + KO alcoholic C C An Alkene + KX + 2 O An Alkylhalide eating an alkyl halide with a solution of KO, in alcohol, yields an alkenes. 26

27 In similar way to that in the dehydration of an alcohol, Saytzeff s rule again applies; that is, the alkene with the most alkyl substituent on the double-bonded carbons Predominates. Br a) C 3 C C C 2 2-Bromobutane (b) 3 C Br Examples + KO C 3 C CC 3 + C 3 C 2 C C 2 Alcoholic 2-Butene 1-Butene Major product Minor product (81%) (19%) + KO Alcoholic C 3 + C 3 1-Bromo-2-methylcyclohexane 1-Methylcyclohexene 3-Methylcyclohexene Major product Minor product 27

28 Problem 3.12 Give the structure of the major and minor products of dehydrohalogenation of: (a) Br C 3 C C 3 C C 3 (b) C 2 C C 3 Cl 28

29 3-Dehaogenation of vicinal dihalides The dehalogenation of vicinal dihalide gives alkene. The reaction takes place in the presence of alcoholic solution of zn at high temperature. For example: Dehalogenation of 2,3-Dibromobutane leads to the formation of 2-Butene in the presence of catalyst. General Equation C C + Zn/Alcoholic C C + ZnX 2 X Example X + Zn EtO Br Br 2,3-dibromobutane 2-butene

30 Reactions of Alkenes All alkenes contain (C=C + a saturated alkyl chain) as part of molecule, except ethylene. C 2 C 2 Ethylene C 2 CC 2 C 2 C 3 Any alkene ex: 1-pentene As the ethylene has no saturated alkyl chain I.E. The chemistry of alkenes can therefore be divided into : 30

31 Reactions of Alkenes Substitution reactions on the saturated alkyl chain alogenation at igh temperature Two types of reactions Addition reactions On ( C=C) Oxidation reactions On ( C=C) 1. Addition of 2 2. Addtion of X 2 3. Addition of X. 4. Addition of 2 O 5. Addition of halohydrin XO 6. Oxidation(visual test for unsaturation) 7. Ozonolysis

32 Alkane vs Alkene the way to react Alkane Alkene Reactivity Less reactive More reactive Reaction condition ard condition -Elevated temperature -Under the influence of UV light -In the gas phas Mild condition -Low temperature -Dark -Solvent

33 1 st Type of Reactions 1. alogenation Substitution reactions on the saturated alkyl chain The product of any reaction is directly affected by factors that known as reaction condition such as temperature, solvent, etc Please see the link below Example substitution vs addition Cl 2 Gas phase Cl C 2 C C 2 3-Chloropropene Allyl chloride C 3 C C 2 Propene Cl 2 / CCl 4 Low temp Cl Cl C 2 C C 2 1,2-Dichloropropane 33

34 2 nd Type of Reactions Addition Reactions on the C=C 1. Addition of ydrogen: Catalytic ydrogenation The C=C of alkene can add a mole of ydrogen in the presence of suitable catalyste such as (Pt, Ni or Pd) at low pressure to give an alkane. This method is called Catalytic ydrogenation. The General equation C C + 2 Pt,Pd, or Ni An alkene Low pressure C C Example C 3 C C 2 Propene (Propylene) 2, Ni Low pressure An alkane C 3 C 2 C 3 Propane 34

35 2. Addition of alogens: alogenation When an alkene is treated at room temperature with a solution of bromine Br 2 or chlorine Cl 2 in carbon tetrachloride, the halogen adds rapidly to the double bond of the alkene to give the correspounding vicinal dihalide. The General equation Examples (a) (b) C C + X 2 CCl 4 C C An alkene X X Vicinal dihalide C 3 C C CCl 2 + Cl 4 2 C 3 C C 2 Propene 1,2-Dichloropropane Cl Cl + Br 2 CCl 4 Br ( X= Cl or Br) Cyclobutene Br 1,2-Dibromocyclobutane Note that, I Iodine is too unreactive and will not add to the C=C of an alkene, F Fluorine is too reactive and reacts explosively making the alkene unstable. 35

36 3. Addition of ydrogen alides: ydrohlogenation Alkenes react with hydrogen chloride, Cl, hydrogen bromide, Br, and ydrogene Iodide, I, to form alkylhalides, RX. This reaction is known as hydrohalogenation As and X is added to the double-bonded carbons. The General equation C C + X C C An alkene X Examples (b) (a) C 2 C 2 + Cl An alkylhalide C 3 C 2 Cl X= (Cl, Br, or I) C 3 C CC 3 + Br C 3 C 2 CC 3 I Br (c) + I 36

37 Addition -X; Case 1: Addition symmetric alkene When hydrogen halide is added to a symmetric alkene such as; 2-Butene [RC=CR], there is only one possible product because the two double-double bonded carbons are equivalent. Example C 3 C CC 3 + Br C 3 C CC 3 Br 2-Butene A symmetrical alkene 2-Bromobutane Only one product possible 37

38 Addition -X; Case 2: unsymmetrical alkenes With unsymmetrical alkenes; such as Propene [RC=CR ] as (R R ), the possibility exists for the X to add in two ways. C 3 C C 2 + Br C 3 CC 3 Br 2-Bromopropane (major product) Propene An Unsymmetrical alkene C 3 C 2 C 2 Br 1-Bromopropane (minor product)

39 Markovnikov s rule In this type of addition of ---X to unsymmetrical alkenes The + of the hydrogen halid adds to the double-bonded carbon that bears the greater number of hydrogen atoms and the X - ion adds to the other double-bonded carbon. Examples The X - part bonds up to the other C, or to the C rich in substituents C 3 C 3 C C 2 + Cl C atoms with greater N of atoms C 3 C 3 CC 3 Cl C 3 + C3CC2Cl 2-methyl propene 2-Chloro-2-methylpropane 1-Chloro-2-methylpropane Major product Minor product

40 4. Addition of Water:( 2 O) ydration When heated with water in the presence of an acid catalyst (usually sulfuric acid),alkenes yield alcohols (RO). The process is called hydration of alkenes because it involves the addition of water across the double bond (C=C). The General equation C C + O + C C Example An alkene O An alcohol 2 O, + C 3 CC 3 C 3 C C 2 O! Propylene Isopropyl alcohol Note that, (Propene) 2-Propanol The addition of O across the double bond is in accordance with Markovnikov s rule. 40

41 5. Addition of OX: alohydrin formation When an alkene is treated with aqueous chlorine or aqueous bromine, the addition product is a halohydrin. (when Cl 2 is used, the product is chlorohydrin; When Br 2 is used the product is a bromohydrin) The General equation C C + X O C C + X An alkene X O A halohydrin The addition of the chloronium ion, Cl + or bromonium ion, Br + and the hydroxid ion, - O, follows the Markovnikov s rule as illustrated in the following example. Example C 3 C C 2 Cl 2, 2 O C 3 C C 2 O Cl Propylene (Propene) Propylene chlorohydrin (1-Chloro-2-propanol) 41

42 6. Oxidation of Alkenes(visual test for unsaturation) A simple visual test can be used to detect unsaturation: Alkenes can easily be oxidized by potassium permanganate (KMnO 4 ) and other oxidizing agents. When alkenes are added to the purple alkaline potassium manganate, there is a color change to brown manganese dioxide (MnO 2 ) as the alkenes are being oxidized. This reaction is called Baeyer test for the unsaturation. The General equation cold, dilute, C C + alkaline KMnO 4 An alkene (Purple) C C + MnO 2 O O (Brown precipitate) A glycol The reaction conditions can be summarized as follows At cold temperatures with low concentrations of oxidizing reagents, alkenes tend to form glycols (diol.) 42

43 Examples (a) C 2 C 2 IUPAC Name: Ethene Common Name: Ethylene cold, dilute, alkaline + KMnO 4 C 2 C 2 + MnO 2 O O 1,2-ethanediol Ethylene glycol (b) + cold, dilute, alkaline O KMnO 4 + MnO 2 O IUPAC Name: cyclopentene Cis-1,2-cyclopentadiol 43

44 7. Ozonolysis (strong oxidation): Oxidation of alkenes by ozone O 3 leads to destruction of both the σ and π bonds of the double-bond system. This cleavage of an alkene double bond, generally accomplished in good yield, is called ozonolysis. The General equation Inert solvent rearrangement C C + O 3 C C O O O C O C O O An ozonide Ozone Molozonide (not isolated) Unstable intermediate Further addition of water in the presence of zinc catalyst results in the formation of two carbonyl groups compounds,(at the position of the C=C double bond was. These products of ozonolysis may be aldehydes and ketones. O R C C zn + 2 O R C O + R C O O O aldehyde ketone An ozonide 44

45 Examples What would be the product of the reaction of each of the following with O 3 followed by Zn/ 2 O? (a) Ethene (b) 1-Butene (c) 2-Butene (d) Isopropylidenecyclohexane The ozonolysis reaction can be summarized by the following equation: R R C C R (1) O 3 (2) Zn, 2 O R R C O + O C R (a) Ethene C C (1) O 3 (2) Zn, 2 O IUPAC Name: Common Name: C O + O 2 equivalents of: Methanal Formaldehyde C 45

46 (b) 1-Butene C C (c) 2-Butene C 2 C 3 (1) O3 (2) Zn, 2 O IUPAC Name: Common Name: C O + Methanal Formaldehyde O C C 2 C 3 Propanal Propaldehyde C 3 C 3 (1) O C C 3 C 3 (2) Zn, 2 O C O + O C 2 equivalents of: IUPAC Name: Ethanal Common Name: Acetaldehyde (d) Isopropylidenecyclohexane C 3 C C 3 C 3 (1) O 3 (2) Zn, 2 O IUPAC Name: Common Name: O Cyclohexanone + C 3 O C C 3 Propanone Acetone 46

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