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Fundamental Organic Chemistry Prepared By Dr. Essam M. Hussein Assistant Professor of Organic Chemistry Chemistry Department Faculty of Applied Science.

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Presentation on theme: "Fundamental Organic Chemistry Prepared By Dr. Essam M. Hussein Assistant Professor of Organic Chemistry Chemistry Department Faculty of Applied Science."— Presentation transcript:

1 Fundamental Organic Chemistry Prepared By Dr. Essam M. Hussein Assistant Professor of Organic Chemistry Chemistry Department Faculty of Applied Science -Umm Al-Qura University

2 Carbon Compounds and Chemical bonds: Organic chemistry is the chemistry of the compounds of carbon. Carbon compounds include DNA, RNA. DNA deoxyribonucleic acid. RNA Ribonucleic acid. & Proteins & Hydrocarbons and carbohydrates. Compounds are formed by covalent Bonds.

3 Covalent Bonds: Covalent Bonds are formed by sharing electrons by atoms. The first explanations of the nature of chemical bonds were advanced by G. N. Lewis and W. Kossel The atoms in covalent bond achieve noble gas configurations by sharing electrons. Covalent bonds form between the atoms form molecules. Molecules may be represented by electron –dot formulas or by dash formulas where each dash represents a pair of electron shared by two atoms.

4 Examples: (1) H 2 molecule H 1s ↑ + H 1s ↑



7 All the previous formulas are called "Lewis structures". Note: in writing Lewis structures we show only the electrons of the valence shell.


9 Polar Covalent bond and Electronegativity: The electrons of the covalent bond between non identical atoms are attracted to the atom having high electronegativity (higher atomic number) in this case the covalent bond will be polarized (called polar covalent bond). Ex: δ δ Electronegativity: The ability of atoms to attract covalence bond electrons to it’s side.

10 Writing Lewis Structures: When we write Lewis structures (electron-dot formula). We assemble the molecule or from the constituent atoms showing only the valence electrons (the electrons of the outer most shell). By having the atoms share or transfer electrons we try to give each atom the electronic configurations of the noble gas in the horizontal row of the periodic table (octet rule). e. g. give hydrogen atoms two electrons like He structure and C, N, O and F eight electrons like Ne.

11 * The no. of valence electrons can be obtained from the periodic table because it is equal to the group number of the atom e. g. for carbon in IVA equal four and it has four electrons. * Fluorine (F) in group VIIA has seven. * Hydrogen in group IA has one.

12 Example1: Write the Lewis structure of CH 3 F (1) We find the total no. of valence electrons of all the atoms. C + 3H + F 4 + 3(1) + 7 = 14 (2) We use pairs of electrons to form bonds between all atoms that are bonded to each other. We represent these bonding pairs with lines, in our example four pairs of electrons (8 of our 14 valance electrons).

13 (3) We then add the remaining electrons in pairs as to give each hydrogen 2 electrons and every other atom 8 electrons in our example the remaining 6 valence electrons were assigned to the Fluorine atom in three non bonding pairs (unshared electrons). So the Lewis structure of CH 3 F is

14 Example 2: Write the Lewis structure of ClO 3 - (1) We find the total no. of valence electrons of all the atoms and the extra electrons needed to give the ion a negative charge: Cl + O3 + e (6) + 1 = 26 (2) We use three pairs of electrons to form bonds between the chlorine atom and three oxygen atoms.

15 (3) then add the remaining 20 electrons in pairs to give each atom octet. So the Lewis structure of ClO 3 - is Example 3: Write the Lewis structure of C 2 H 6 (1) We find the total no. of valence electrons of all the atoms: 2C + 6H 2(4) + 6(1) = 14 (2) We use three pairs of electrons to form bonds between the carbon and hydrogen atoms.

16 7 X One pair (2 electron) = 14 electrons i. e. only 14 bonded electrons. Formal charges: When we write Lewis structure it is often convenient to assign unit positive or negative charge called "formal charge" to certain atoms in the molecule or ion. Calculations of formal charges is a bookkeeping method for electrical charges, because the arithmetic sum of all of the formal charges equals the total charge on the molecule or ion.

17 Simply to calculate the formal charge for atoms in molecules or ions. (1) Write the right Lewis structure. (2) Use the equation F = Z – S/2 – U for each atom in the molecule. (3) To know the net hole charge of the molecule (should be zero) or ions; calculate the arithmetic sum of the all atoms in molecule or ion. Ex. (1): Ammonia NH 3 * Write the Lewis structure of ammonia N + 3H 5 + 3(1) = 8

18 To calculate the formal charge use the equation F = Z – S/2 – U Where F Formal charge Z group no. of the atom S no. of shared (bonding electrons) U no. of unshared (nonbonding electrons) For NH 3 the Lewis structure is

19 The formal charge of H is F = 1 – 2/2 – 0 = 0 For nitrogen F = Z – S/2 – U F = 5 – 6/2 – 2 F = 5 – 3 – 2 = 0 So the charge on NH3 molecule = sum. Of the formal charge of the constituent atoms = F(N) + F3(H) = 0 + 3(0) = 0

20 Ex. (2): H 2 O (1) Write the Lewis structure of H 2 O 2H + O 2(1) + 6 = 8 F(H) = 1 – 2/2 – 0 = 0 F(O) = 6 – 4/2 – 4 F(O) = 6 – 2 – 4 = 0 So the charge on molecule = F(O) + F(H)= = zero so H 2 O is a neutral molecule

21 Ex. (3): HNO 3 Write the Lewis structure of HNO 3 because the hydrogen atom should attach to oxygen (mor electronegative atom).

22 F(H) = 1 – 2/2 – 0 = 0 F(O1) = 6 – 4/2 – 4 = 0 F(O2) = 6 – 2/2 – 6 = -1 F(O3) = 6 – 4/2 – 4 = 0 F(N) = 5 – 8/2 – 0 = +1 So the charge (net charge) on molecule = F(O1) + F(O2) + F(O3) + F(N) +F(H)= (-1) + (+1)= zero so nitric acid is a neutral molecule Ex. (4): NO 3 - (1) Write the Lewis structure of NO 3 ­

23 F(O1) = 6 – 2/2 – 6 = -1 F(O2) = 6 – 2/2 – 6 = -1 F(O3) = 6 – 4/2 – 4 = 0 So the charge (net charge) on molecule = F(O1) + F(O2) + F(O3) + F(N) = 0 + (-1) + (-1) + (+1)= -1 so NO 3 ­ has one negative charge.

24 Ex. (5): NH 4 + * Write the Lewis structure of ammonia N + 4H – e = 5 + 4(1) – 1 = 8 The formal charges F(H) = 1 – 2/2 – 0 = 0 F(N)= 5 – 8/2 = = 5 – 4 = +1 So the charge on ion = 4(0) + (+1) = 0+ 1 =+1

25 Orbital hybridization structure of alkanes (sp3): sp3 hybridization: According to the quantum mechanism the electronic configuration of a carbon in its lowest energy state-called the ground state. C 6 1s 2 2s 2 2p 2 1s 2 2s 1 2p x 1 2p y 1 2p z 1 Looking only on the valence electrons the carbon atom has tow unpaired electrons i. e. it can only two bonds with hydrogen in its ground state. To account (explain) the four covalent bonds of carbon, we should discuss "The hybridization of carbon orbitals".

26 C 6 1s 2 2s 2 2p 2 1S 2 2S 2 2P 2 PyPy PxPx PzPz Ground state excitation 1S 2 2S 1 2P 3 PxPx PyPy PzPz excited state

27 Hybridization (sp 3 ) i. e. ( one s + 3p) i.e. Hybridization: mixing of orbitals of the same main shell. (the same principle quantium no.). to give new hybridized orbitals with the same energy. Ex. (1): in methane CH 4

28 * In methane (CH4) each C-H σ bond formed by the overlap of sp 3 electron (from carbon) with 1s electron (from hydrogen) * sp 3 by hybridization and their compounds have tetrahedrol structure and the bond angle of H-C-H (in methane) or H-C-C- (in other alkanes) equals i.e. sp 3 compounds are non planer. Ex. (2): for sp 3 hybridization ROH, ROR, H 2 O, NH 3 NH 3 ammonia structure: N 5 1S 2 1S 2 2P 3

29 1S 2 2S 2 2P 2 PyPy PxPx PzPz Ground state excitation and hybridization 1S 2 SP 3 excited state Formation of N-H bonds

30 σ- bonds * The geometry of a molecule of ammonia is a triagonal pyramid. * The bond angles in a molecule of ammonia are 107º * The hybridization of Nitrogen atom in ammonia explains the presence of a lone pair of nonbonding (unshared) electrons which explains the basicity of ammonia.

31 Sp2 hybridization and the structure of thene (ethylene): * Before we going to sp 2 hybridization, we should know that the covalent bonds in carbon compounds are classified into: (1) Sigma bond (σ- bond): Formed by head-head overlap of 1s with 1s electrons 1s with p x electrons 1s with sp 3 electrons 1s with sp 2 electrons 1s with sp electrons p x with p x electrons p x with sp 3 electrons p x with sp 2 electrons p x with sp electrons

32 (2) Pi bond (Л bond): Formed only by sidewise overlap of p y with p y electrons and p z with p z electrons. sp 2 hybridization of ethene:

33 C 1s 2 2s 2 2p 2 1S 2 2S 2 2P 2 PyPy PxPx PzPz Ground state excitation 1S 2 2S 1 2P 3 PxPx PyPy PzPz excitation (promotion of electron) C 1s 2 2s 2 2p 2 1S 2 2S 2 2P 2 PxPx PyPy PzPz 1S 2 2S 1 2P 3 PzPz PyPy PxPx

34 1S 2 sp 2 hybridization PyPy 1S 2 SP 2 PyPy sp 2 1s 1 PyPy PyPy

35 σ- bond Л bond * It is planer molecule Sp hybridization: This type of hybridization explained the band formation of alkynes.

36 C 1s 2 2s 2 2p 2 1S 2 2S 2 2P 2 PyPy PxPx PzPz Ground state excitation 1S 2 2S 1 2P 3 PxPx PyPy PzPz excitation (promotion of electron) C 1s 2 2s 2 2p 2 1S 2 2S 2 2P 2 PxPx PyPy PzPz 1S 2 2S 1 2P 3 PzPz PyPy PxPx

37 1S 2 sp hybridization PyPy 1S 2 SP PyPy sp 1s 1 PzPz PzPz C C PyPy HH PzPz PzPz PyPy Bond formation

38 how many sp 3 -sp σ sp 2 -sp 3 σ sp 2 -sp 2 σ sp-sp σ

39 SP SP 2

40 σ- bond Л bond Cleavage ( Breaking) of bonds: A covalent bond between two atoms can be broken by two ways (1) Homolytic bond fission: i.e. each atom takes one electron of the bond electrons

41 ForCarbon-Carbon δ bond breaking gave free radicals of SP 2 hybridization i.e

42 (2) Heterolytic bond fission: sp 3 -sp 3 σ sp 2 -sp 2 σ sp-sp σ sp 3 -sp 2 σ sp 3 -sp σ sp 2 -sp σ

43 In the carbocations the hybridization is sp 2 ( the carbocations are planer ). In the carbanions the hybridization is sp 3. Types of Reagents: Reagents can be classified into:- (1) Nucleophiles (Nucleophilic) Reagents: They are nucleus seeking species, may be negatively charged ions or groups or neutral atoms or groups and can react with the donating of electrons.

44 Examples of negative charged nucleophiles:- neutral nucleophiles are Lewis bases such as: (2) Electrophilic ( electrophilic) Reagents: They react with accepting a lone pair of electrons, they are Lewis acids.

45 Examples of electrophiles: (3) Free Radicals: They are atoms or groups having one unpaired electrons, they are highly reactive with very short life time. e.g.

46 Reaction Types: There are four general types of reaction which organic compounds can under goes. (a) Displacement (Substitution) Reactions. (b) Addition Reactions. (c) Elimination Reactions. (a)Rearrangement Reactions. (a) Displacement (Substitution) Reactions. It is displacement from carbon, the atom displaced be either hydrogen or another atom or group and can be classified into: (1) Nucleophilic substitution reactions. It is often an atom other than hydrogen that is displaced. e.g

47 ( 2) Electrophilic substitution reactions. It is often hydrogen that is displaced. e.g. aromatic substitution. (Electrophilic substitution)

48 It can be electrophilic, nucleophilic or radical in character depending on the type of species that initiates the process. e.g addition to simple carbon-carbon double bonds is either electrophilic or radical induced e.g. addition of HBr (b) Addition Reactions: Which can be initiated by the attach of either

49 Nucleophilic addition reaction: e.g. the base catalyzed formation of cyanohydrins in liquid HCN

50 (C) Elimination Reactions: The reversal of addition reactions.

51 (d) Rearrangement Reactions: e.g. Pinacol- Pinacolone rearrangement

52 Represntative carbon compounds: (1) Aliphatic hydrocarbons: (a) Saturated hydrocarbons (1) Open chain Saturated hydrocarbons: Ex. CH 4 methane, CH 3 -CH 3 ethane and all alkanes with general formula C n H 2n+2 (2) Cyclic Saturated hydrocarbons: Ex

53 (b) Saturated hydrocarbons (1) Alkenes and cycloalkenes: and all hydrocarbons with general formula CnH2n Alkenes Ex. Cycloalkenes : i.e cyclic hydrocarbons contain double bond Ex.

54 (2) Alkynes: Unsaturated hydrocarbons contain triple bond CnH2n-2.

55 A Represntative Aromatic Hydrocarbons: They are cyclic unsaturated hydrocarbons containing conjugated double bonds, example benzene is six membered ring with alternating single and double bonds called a Kekule structure.

56 In fact, the carbon-carbon bonds of benzene are all the same length (1.39A), a value in between that of carbon-carbon single bond and carbon-carbon double bond. There are two ways of dealing with this problem: with resonance theory or with molecular orbital theory. Based on the principles of resonance theory we recognized that benzene can not be represented adequately by either structure but that, instead, it should be visualized as a hybrid of the two structures. We represent this hybrid by a hexagon with a circle in the middle.

57 Two contributing Kekule Structures for benzene A representation of the resonance hybrid In the molecular orbital explanation. Sp 2 hybridized carbon

58 When the benzene ring is attached to some other group of atoms in a molecule, it is called a phenyl group and it is represented in several ways or C 6 H 5 - or ph or ф The combination of a phenyl group and a –CH 2 - group is called a benzyl group or C 6 H 5 -CH 2

59 Functional groups: A functional group is the part of a molecule where most of its reaction occur. It is the part that effectively determines the compound’s chemical properties and many of its physical properties as well. The functional group of alkanes is C-H and C-C alkenes has carbon-carbon double as a functional group. The functional group of alkynes is carbon-carbon triple bond

60 Alkyl groups and the symbol R: They are groups that would be obtained by removed a hydrogen atom from an alkane Examples:

61 abbreviationAlkyl groupAlkane Me-CH 3 - (methyl group) CH 4 (methane) Et-CH 3 CH 2 - or C 2 H 5 - (ethyl group) CH 3 (ethane) Pr- i-Pr- CH 3 CH 2 CH 2 - (propyl group) or CH 3 CH 1 CH 3 (isopropyl group) CH 3 CH 2 CH 3 (propane)

62 Alkyl Halides or haloalkanes: Alkyl halides are compounds in which a halogen atom ( fluorine, chlorine, bromine or iodine) replaces a hydrogen atom of an alkane. Ex. CH 3 Cl methyl chloride C 2 H 5 Br ethyl bromide Alkyl halides are classified into Primary (1º), or secondary (2 º ), or tertiary (3 º ) and this classification is based on the type of the carbon to which the halogen is directly attached. Examples of primary, secondary and tertiary alkyl halides are the following.

63 1º carbon 1º alkyl chloride 2º carbon 3º carbon 2º alkyl chloride 3º alkyl chloride

64 Alcohols (R-OH): The functional group of alcohols is the hydroxyl (OH) group attached to sp 3 - hybridized carbon atom. is the functional group of an alcohol Examples: CH 3 OH methyl alcohol (methanol), C 2 H 5 OH ethyl alcohol (ethanol) Alcohols may be viewed in two ways (1) as hydroxy derivatives of alkanes. (2) as alkyl derivatives of water.

65 Example ethane ethyl alcohol (ethanol) 109º105º water ● As with alkyl halides, alcohols are classified into three groups Primary (1º), or secondary (2º), or tertiary (3º) alcohols 1º carbon 1º alcohol

66 Geraniol(1º alcohol) Benzyl alcohol (1º alcohol) ● Primary carbon is the carbon atom which is attached to one other carbon. ● 1º carbon also has two other carbon atom attached to it. ● 3º carbon also has three other carbon atom attached to it.

67 2º alcohol 3º alcohol 2º carbon 3º carbon

68 Ethers: Ethers have the general formula R-O-R or R-O-R’ where R, R’ are alkyl groups. ● ethers can be thought as derivatives of water which both hydrogen atoms have been replaced by alkyl groups. General formula for an ether Dimethyl ether (a typical ether) The functional group of an ether

69 ethylene oxide Tetra hydro furan (THF) Cyclic ether Amines: ● Amines may be considered as organic derivatives of ammonia ammonia (an amine) amphetamine (1º amine)

70 Amines are classified as Primary (1º), or secondary (2º), or tertiary (3º) amines, this classification is based on the number of the organic groups (alkyl groups) that attached to the nitrogen atom. 2º amine 1º amine 3º amine Examples: Isopropyl amine 1º amine Diethyl amine 2º amine

71 Tri ethyl amine 3º amine Cyclic 2º amine ● The bond angle C-N-C in tri ethyl amine is 108,7º which is very close to H-C-H in methane, thus the nitrogen atom of an amine can to be sp 3 hybridized and the unshared electron pair occupy an sp 3 orbital. Aldehydes and ketones: Both contain the carbonyl group. the carbonyl group.

72 In aldehydes, the carbonyl group is bonded to least one hydrogen atom, i.e. aldehydes has the follwing general formula. R may be hydrogen atom or alkyl group In ketones, the carbonyl group is bonded to two carbon atom, i.e. aldehydes has the following general formula. Where R and R 1 may be the same or different

73 KetonesAldehydes acetoneFormaldehye ethyl methyl ketoneAcetaldehyde benzophenonebenzaldehyde

74 Aldehydes and Ketones have a trigonal planar arrangement of groups around the carbonyl carbon. The carbon atom is sp 2 hybridized e.g in formaldehyde 121º 118º Carboxylic acids: It have the general formula the functional group Is called the carboxyl group (carbonyl + hydroxyl) Acarboxylic acid

75 The carboxyl group Examples : formic acid(1) (2)acetic acid (3)benzoic acid

76 Amides:- They have the formula Examples: acetamide N-Methylacetamide N,N-diMethylacetamide Esters:- esters have the general formula

77 Examples: it is prepared by this reaction Ethyl acetate

78 Physical properties and molecular structure: (1) Electronegativity: The electron cloud that bonds two atoms is not symmetrical except when the two atoms are the same and have the same substituents. The cloud is necessarily distorted toward one side of the bond depending on which atom maintains the greater attraction of the cloud. Electronegativity is the ability of atoms to attract the electron clouds to it’s side. Electronegativity can measured by Pauling method which is based on bond energies of diatomic molecules.

79 Fluorine is the most electronegative elements in the periodic table. some examples of the electronegativities of some atoms. 1.2Mg2.5I4.0F 0.9Na2.5C3.5O 0.7Cs2.1H3.0Cl 2.1P3.0N 2.0B2.8Br 1.8Si2.6S Bond distances: It the distance between the nucleus which form the molecule


81 Bond distance is determined by X- ray diffraction ( for solid), electron diffraction ( for gases). Bond distances in a molecule are characteristic properties of the molecule which give some information about the chemical properties of that molecule. The bond distance of the vibration of the molecule. Bond distance are measured in angstrom Aº. The effect of the vibration for a bond between two sp 3.

82 Bond length A º C-C bond inBond length A º C-C bond in ± Diamond 1.532t-butyl chloride ± C2H6C2H n- butane to n- hexane ± C 2 H 5 Cl ± 0.001isobutene1.532 ± 0.003C3H8C3H8

83 Table of Bond distances. Typical compds. Length A º Bond type Typical compds.Length A º Bond type C=CC-C Ethylene1.34sp 2 - sp 2 CH 3 CH sp 3 - sp 3 Ketene1.31sp 2 - spCH 3 CHO, toluene 1.50sp 3 - sp 2 C≡CButadiene1.48sp 2 - sp 2 Acetylene1.20sp- spAcrylonitrile1.43sp 2 - sp methane1.11(C-H) sp 3 - H cyanoacetylene1.38sp- sp

84 Typical compds. Length A º Bond type Typical compds. Length A º Bond type C-NC-H Amines sp 3 - NBenzene1.10sp 2 - H C=NHCN, acetylene1.08sp- H Oximes, imines 1.28sp 2 - NC-O 1.38sp 3 - FDimethyl ether1.41sp 3 - O 1.35sp 2 - FC=O 1.27sp- FFormaldehyde1.20sp 2 - O CO sp- O

85 As the percentage of s increas the bond distance is shortened, C-D is shorter than C-H. Bond angle: The angle between two molecular orbitals ( three atom, and central atom). 120º SP 2 S S 109º28` SP 3 109º28`

86 Oxygen, Sulfur, Nitrogen bond angles:- Compd. ValueAngleCompd.ValueAngle (CH 3 ) 2 SC-S-Cwater 104 º.27` H-O-H NH º.46` H-N-Hmethanol 107 º º C-O-H CH 3 NH º H-N-H(CH 3 ) 2 O 111 º.43` C-O-C (CH 3 ) 2 N H 112 º C-N-H(ph) 2 O 124 º.5` C-O-C (CH 3 ) 3 N 108 º.7` C-N-HH2SH2S 92 º.1` H-S-H CH 3 SH 99 º.4` C-S-H

87 Bond Energies: There are two kinds of bond energy. (1) Dissociation energy D Which defined as the energy necessary for cleavage a bond to give the constituent radical. e.g. D for water +118 kcal/mol and D for OH bond in water is +100 kcal/mol ● The average is 109 kcal/mol is taken as bond energy. ● In diatomic molecules D=E.

88 Cal. fromMean value E kcal/mol BondCal. from Mean value E kcal/mol Bond CCl 4 79C-ClH2OH2O O-H C 2 H 5 -SH66C-SCH C-H CH≡CH C≡CNH 3 93N-H (CH 2 =CH C=CH2SH2S82S-H CH 3 I52C-ICH 3 OH85-91C-O C2H6C2H C-C Table: Bond energy E values for some important bonds

89 From the previous Table (1) There is a correlation of bond strengths with bond distances in general, shorter bonds are stronger bonds. Since we have seen that increasing S character shortens bonds bond strengths increase with increasing S character. (2) Bonds become weaker as we move down the periodic table. Compare C-O and C-S and C-Cl and C-I. (3) Double bonds are shorter and stronger than the corresponding single bonds but not twice as strong because Л overlap is less than σ overlap. This means that a σ bond is stronger than a Л bond.

90 The difference in energy between a single bond, say C-C and C=C is the amount of the energy necessary to cause rotation a round the double bond. Polarity of bonds: Two atoms joined by a covalent bond, share electrons and their nuclei are held by the same electron cloud. In most cases the two nuclei do not share the electrons equally; the electron cloud is denser about one atom than the other i.e. one end is relatively negative and the other is relatively positive ( δ- & δ+).

91 i.e. there is a negative pole and a positive pole such bond is called polar bond or to possess polarity. We can indicate polarity using δ- & δ+, δ+ is partial positive, δ- is partial negative. Some time we say delta plus and delta minus. δ+δ+ δ-δ- δ-δ- δ-δ- δ+δ+ δ+δ+ δ+δ+ δ+δ+ δ+δ+ polar bonds We can expect a covalent bond to be polar if it join atoms that differ in electronegativity. The most electronegative elements are those located in the upper right hand corner of the periodic table.

92 Electronegativity can arrange in the following order F >O >Cl, N >Br >C,H Bond polarities are concerned with physical and chemical properties. e.g The polarity of bonds leads to polarity of molecules which affect the melting point, boiling point and solubility. The polarity of a bond determines the kind of reaction that can take place at that bond and even affects reactivity of near by bonds. Polarity of molecules:- They are molecules which contain polar bonds, such a molecule constitutes a dipole ( two equal and opposite chages separate in space).

93 A dipole is often symbolized by where the arraw points from positive to negative. Polar molecules possess a dipole moment μ which is equal to the magnitude of the charge e, multiplied by distance d between the centers of the charge. ∴ μ (Debye) = e (e.s.u) x d (Cm) 1.86CH 3 Cl1.75HF0H2H2 0CCl H2OH2O0O2O2 0CO NH 3 0N2N2 0CH 4 0Cl 2 Table of Dipole moments

94 Molecules like H 2, O 2, N 2, Cl 2 and Br 2 have zero dipole moments i.e. They are non- polar, since each molecule has two identical atoms e is zero hence μ = 0 HF has large μ of 1.75 D, F is very high dectro- negativity, d is small and e is large μ is large too.

95 Melting point:- It is the temperature at which the thermal energy of the particles is great enough to over come the intracrystalline forces that hold them in position. i.e. Solid ∆ liquid Crystalline is arranged in very regular symmetrical way ( highly ordered) More random arrangement

96 Intermolecular forces:- The forces hold neutral molecules to each other seen to be electrostatic in natural involving attraction of positive charge for negative charge. There are two kinds of inter molecular forces. (1) Dipole – Dipole interactions. (2) Van der Waals forces. (1) dipole – dipole interaction is the attraction of the positive end of one polar molecule for the negative end of another polar molecule or

97 ● As a result of dipole – dipole interaction, polar molecules are generally held to each other more strongly than non- polar molecules of comparable molecular weight. Hydrogen bonding: it is strong kind of dipole-dipole attraction in which a hydrogen atom serves as a bridge between two electronegative atom holding one by a covalent bond and the other by purely electrostatic forces. ● For hydrogen bonding both electronegative atoms must come from the group F, O, N. …… 5 kcal/ mol kcal/ mol

98 …… & or ● These three elements owl their special effectiveness to the concentrated negative charge on their small atoms. ● Hydrogen bonding affects the b. p. and solubility properties of compounds and plays a key role in determining the shapes of large molecules like proteins and nucleic acids.

99 Van der Waals forces:- The forces between the molecules of non- polar compound which make them can solidify. Van der Waals forces can explained by the following: The average distribution of charge about say a methane molecule is symmetrical so that there in no net dipole moment. However the electrons move about so that at any instant of time the distribution will probably be distorted and a small dipole will exist. This momentary dipole will affect the second methane molecule near by the negative end of the dipole tends to repel electrons and the positive end tends to attract electrons, the dipole thus induces an oriented dipole in the neighboring molecule

100 Although the momentary dipole and induced dipoles are constantly changing the net result is attraction between the two molecules. Van der Waals forces have a very short range, they act only between the portion of different molecules that are in close contact that is between the surfaces of molecules. Boiling points:- The temperature at which the thermal energy of the particles is great enough to overcome the cohesive force that hold them in the liquid. In the liquid state the unit of a non- ionic compound is the molecule. The weak intermolecular forces of non – ionic liquid is dipole – dipole interactions and van der waals forces are more readily overcome than the strong interionic

101 forces of ionic compounds and boils at very much lower temperature. e.g. the non – polar methane boils at ° and even polar HCl boils at - 85°. liquids whose molecules are held together by hydrogen bonds are called associated liquids and boil at higher b. p. than compounds of there molecular weight and dipole moment. e.g. (1) HF boils at 100 degree higher than the non – associated HCl. (2) H 2 O boils at 160 degree higher H 2 S. Boiling points increase as molecular weights increase.

102 Dipole – Dipole forces:- Most organic molecules are not fully ionic but have instead a permanent dipole moment resulting from a non uniform distribution of the bonding electrons. δ+δ+δ-δ- Solubility : non- ionic solute when a solid or liquid dissolves the structural units ions or molecules become separated from each other and the space in between become occupied by solvent molecules. The energy required to break the bonds between solute particles is supplied by the formation of bonds between the solute particles and the solvent molecules i. e. the old attractive forces are replaced by new ones. δ-δ-δ+δ+

103 The solubility characteristics of non – ionic compounds are determined by their polarity. Non–polar or weakly polar compounds dissolve in non-polar or weakly polar solvents, highly polar compounds dissolve in highly polar solvents “ like dissolves like” it is useful rule. e.g. methane dissolve in carbon tetrachloride molecules to each other is Van der Waals interaction are replaced by very similar forces holding methane molecules to carbon tetrachloride molecules. Neither methane nor carbon tetrachloride is readily soluble in water. The highly polar water molecules are held to each other by very strong dipole – dipole interactions hydrogen bonds. The highly polar organic compound methanol CH 3 OH is

104 quite soluble in water, hydrogen bonds between water and methanol molecule can replace the very similar hydrogen bonds between different methanol molecules and different water molecules. Alcohols: Non polar part like methane hydrophobic (water hating) some times called lipophilic ( solubility in non polar solvents). polar part like water hydrophilic (water loving) ( solubility in polar solvents).

105 Tables of solubility of alcohols in water. Solubility g/100g water alcoholSolubility g/100g water alcohol 7.9CH 3 (CH 2 ) 3 OHXCH 3 OH 2.3CH 3 (CH 2 ) 4 OHXCH 3 CH 2 OH 0.05CH 3 (CH 2 ) 7 OHXCH 3 CH 2 CH 2 OH From the above table it is clear that as the lipophilic increase the solubility in water decrease.

106 Solubility: ionic solutes. Protic and aprotic solvents and ion pairs: In the dissolution of ionic compounds a great deal of energy is necessary to overcome the powerful electrostatic forces holding together in an ionic lattice. Only water or other highly polar solvents that are able to dissolve ionic compounds. Polar molecule has positive and negative ends, consequently, there is electrostatic attraction between a positive ion and the negative end of the solvents molecule and between a negative ion and the positive end of the solvent molecule, these attractions are called ion – dipole bonds. Each ion – dipole bond is relatively weak but in the aggregate they supply enough energy to overcome the interionic forces in the crystal.

107 Ion – dipole interactions, solvated ( hydrated in case of H 2 O) cation & anion To dissolve ionic compounds a solvent must also have a high dielectric constant, that is have high insulating properties to lower the attraction between oppositely charged ions once they are solvated. Water owes its superiority as a solvent for ionic substances in part to its high polarity and high dielectric constant.

108 Cations are attracted to the negative pole of a polar solvent, in water the negative pole is oxygen. δ+δ+ δ-δ- δ+δ+ …… Anions are attracted to the positive pole of a polar molecule, in water the positive poles are an hydrogen. δ+δ+ δ+δ+ δ-δ- ……

109 Hydrogen bonding permits particularly strong solvation of anions. Thus, water owes a large part of its special solvating power to its OH group: it solvates cations strongly through the unshared pairs on oxygen; it solvates anions strongly through hydrogen bonding. CH 3 OH resembles H 2 O having an OH group dissolves ionic compound. CH 3 OH is less polar than water, CH 3 - group is bigger than the second –H of water. Protic solvents: solvents like water and methanol are called Protic solvents containing hydrogen that is attached to oxygen or nitrogen and hence is appreciably acidic.

110 Aprotic solvents: polar solvents with moderately high dielectric constants which do not contain acidic hydrogen. Dimethylsulfoxide DMSO Dimethyl formamide DMF Hexamethylphosphoro- -triamide (HMOT) They dissolve ionic compounds but in doing their action differ from protic solvents, they can not form hydrogen bonds to anions. They are highly polar with dipole moments several times as large as that of water. The negative pole in aprotic solvents is an oxygen atom that just out from the rest of the molecules. Through unshared pairs of electrons on these negatively charged well exposed atoms. Cations are solvated very

111 The positive pole is buried within the molecule and anions are solvated very weakly. water dissolves ionic compounds very well but poor solvent for most organic compounds, this difficulty can be overcome by addition of second solvent like methane, methanol’s hydrophilic –OH makes it miscible with water, through its lipophilic CH 3 it brings about dissolution of organic compounds. CH 3 CH 2 OH does like CH 3 OH dissolve both ionic and non- ionic compounds. Polar covalent bonds: when two atoms of different electronegativities from a covalent bond, the electrons are not shared equally between them.

112 The atom with greater electronegativity draws the electron pair closer to it and a polar covalent bond result. δ-δ-δ+δ+ δ-δ- δ+δ+ ∴ HCl has a partially negative end with charge δ- and a partially positive end with δ+ charge and hence forms a dipole and has + - a dipole moment μ = 1.08 D ∴ (positive end)(negative end)

113 Acids and bases:- The terms acid and base can be defined in a number of ways, According to the Lowry- Bronsted definition acid is a substance that gives up a proton and a base is substance that accepts a proton. Ex c Stronger acid Stronger acid Stronger base Stronger base weaker base weaker base weaker acid weaker acid

114 According to the Lowry definition, the strength of an acid depends upon its tendency to give up a proton and the strength of a base depends upon its tendency to accept a proton. Ex. H 2 SO 4 & HCl are strong acids because they tend to give up a proton very readily. Conversely HSO4 - and Cl - must be weak bases. If aqueous H 2 SO 4 is mixed with aqueous NaOH, the acid H 3 O + gives a proton to the base OH - to form the new acid H 2 O and the new base H 2 O. stronger acid stronger base weaker acid weaker base + -

115 Also for NH 4 Cl and NaOH + - stronger acid stronger base weaker acid weaker base Acids are arranged in the following order: Acid strength: H 2 SO 4, H 3 O + > NH 4 + >H 2 O for the corresponding conjugate bases have the opposite order ∴ Base strength HSO 4 -, Cl - < H 2 O < NH 3

116 -+ oxonium protonated ethyl alcohol Ex. 1 Ex. 2 oxonium protonated diethyl ether +-

117 Lewis definition of bases and acids:- Base is a substance that can furnish an electron pair to form a coordination bond i.e. the base is electron pair donor e. g. Lewis acid is a substance that can take up an electron pair to form a coordination bond i. e. is an electron pair acceptor e. g. equation of lewid acids bases reactions: (1)

118 (2) To be acidic in the Lowry- Bronsted sense, a molecule must of course contain hydrogen and the degree of acidic is determined largely by the kind of atoms that holds the hydrogen and the ability to accommodate the electron pair left behind by the departing hydrogenation. The ability to accommodate the electron pair depends upon. (1) The atom’ s electronegativity. (2) The atom’ s size. Thus within a given row of the periodic table.

119 acidity increases as electronegativity increases H-CH 3 < H-OH 2 < H-OH < H-F < H-SH < H-Cl within a given family acidity increases as the atomic size increase. H-F < H-Cl< H-Br < H-I < H-OH < H-SH < H-SeH Acids and Bases in Nan-aqueous solutions:- Many of the organic acid-base reactions occur in solutions. These acids are much weaker acids than water and the conjugate base of these very weak acids are powerful bases. Reactions of organic compounds are carried out in liquid ammonia. The most powerful base that can employed in liquid ammonia is the amide ion – NH 2.

120 Ethane and all other alkanes are much weaker acids tha ammonia. The general order of the acidity of some weak acids. RH < RCH=CH 2 < H 2 < NH 3 < R-C≡CH < ROH < H 2 O < RCOOH increasing acidity

121 The order of basicity of conjugate bases R: - > RCH=CH: - > H: - > - :NH 2 > RCH≡C: - > RO: - > :OH - > RCO 2 - increasing basicity Acid – base reactions and the preparation of deuterium and tritium labeled compounds: Chemists often use compounds in which deuterium and tritium atoms have replaced one or more hydrogen atoms of the compounds as a method of “ Labeling ” or identifying particular hydrogen atoms D is H & Tritium H. One way to introduce deuterium or tritium into a specific location in a molecule is

122 Through acid – base reaction that takes place when a very strong base is treated with D 2 O or T 2 O. e.g. Isomerism: Isomerism : compounds have the same molecular formula and different physical and chemical properties due to the difference in structural formula.

123 e.g. C 2 H 6 O can be Ethanol H 3 CCH 2 OH b. p. 78° + HI CH 3 CH 2 I + Na C 2 H 5 ONa + H 2 Dimethyl ether H 3 C-O-CH 3 b. p. -24° + HI 2CH 3 I + Na No reaction Toutomersion: it is rapid equilibrium between a mixture Definite compound, it is a proton shifts from one atom of a molecule to another atom. Keto–Enol toutomersion:- It takes place between a carbonyl compounds containing an α- hydrogen and its enol form.

124 The % of enol is affected by the structure, solvent, concentration and temperature. Ethyl acetoacetate is in high % enol form because it is stabilized by internal hydrogen bonding which is unavailable to the keto form.

125 Constitutional Isomers: “isomer” is composed of two part “iso” in Greek words means the “same” and “meros” means part i.e. isomer means the same part. Constitutional (structural) isomers: They are compounds having the same molecular formula but differ in the way of their atoms are bonded together (structural) formula. Ex. 1: urea and ammonium cyanate have the same molecular formula CH 4 N 2 O and different structures.

126 Ex.2: nitro methane and methyl nitrite CH 3 NO 2

127 Hydrocarbons Hydrocarbons are compounds composed of carbon and hydrogen only and they are classified as follows Hydrocarbons Aromatic hydrocarbons ( arenes) Aliphatic hydrocarbons Alkanes CH 3 -CH 3 Ethane Alkenes CH 2 =CH 2 Ethylene Alkynes CH≡CH acetylene benzene

128 Alkanes Alkanes are saturated hydrocarbons and are classified into. Alkanes a cyclic alkanes General formula (C n H 2n+2 ) e.g. CH 4 methane C 2 H 6 ethane C 3 H 8 propane C 4 H 10 butane cyclic alkanes General formula (C n H 2n ) e.g.

129 A cyclic Alkanes They are saturated hydrocarbons having the general formula C n H 2n+2 and they are found as branched and unbranched alkanes. Nomenclature of unbranched Alkanes: They are found as straight carbon chain ( called n-alkanes). e.g. CH 4 methane, CH 3 CH 3 ethane, CH 3 CH 2 CH 3 propane, CH 3 CH 2 CH 2 CH 3 n-butane, CH 3 CH 2 CH 2 CH 2 CH 3 n-pentane, CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 n-hexane, CH 3 (CH 2 ) 5 CH 3 n-heptane, CH 3 (CH 2 ) 6 CH 3 n-octane, CH 3 (CH 2 ) 7 CH 3 n-nonane, CH 3 (CH 2 ) 8 CH 3 n-decan. For writing the accurate structure of sp 3 hybridization e.g.

130 For simple representation we can use the Zigzag structure (bond- line structure). e.g. CH 3 (CH 2 ) 4 CH 3 n-hexane can be shown as. ≡

131 Note: The alkanes take the suffix ane. Alkyl groups: e.g. CH 3 - methyl, CH 3 CH 2 - ethyl, CH 3 CH 2 CH 2 - n-propyl, CH 3 CH 2 CH 2 CH 2 - n-butyl, CH 3 CH 2 CH 2 CH 2 CH 2 - n-pentyl. Classification of carbon atoms: Carbon atoms are classified according to their degree of substitution by other carbons. A primary carbon is one that is directly attached to one other carbon.

132 Secondary carbon is directly attached to two other carbon atoms. A tertiary carbon is directly attached to three carbon atoms.

133 A quaternary carbon is directly attached to four carbon atoms. There is no quaternary alkyl group.

134 Com. Names (trivial) CH 3 - methyl CH 3 CH 2 - ethyl CH 3 CH 2 CH 2 - n-propyl CH 3 CH 2 CH 2 CH 2 - n-butyl ethane methane propane n- butane iso-butane n- pentane iso- pentane neo- pentane iso-butyl

135 Branched Alkyl groups: They are named by using the longest continuous chain that begins at the point of attachment as the base name. i.e. Numbering starts from the carbon where the hydrogen atom was removed. e.g CH 3 CH 2 CH 2 - n- propyl group, IUPAC 1-methyl ethyl group (comm.) isopropyl

136 isobutyl group ≡ tertiarybutyl group

137 n- pentane n- pentyl iso- pentane iso- pentyl

138 neo- pentane neo- pentyl ≡ IUPAC (International Union of Pure and Applied Chemistry) Nomenclature:- e.g. (1) 1- Methyl propyl group 2 1 3

139 Methylpropyl group e.g. 3 e.g. 2 1,1- Dimethyl ethyl group (tert-butyl)

140 Systematic IUPAC Nomenclature of branched Alkanes:- Branched- chain alkanes are named according to the following rules:- (1) Locate the longest continues chain of carbon atoms; this chain determines the base name of the alkane. e.g hexane e.g. 2 heptane (2) Number the longest chain beginning with the end the chain nearest constituent. (give the position of the constituent the lowest).

141 substituent ok (3) Use the numbers obtained by application of rule 2 to designate the location of the substituent group, the base name is placed last, and the substituent group, preceded by the number designating its location on the chain, is placed first. Numbers are separated from words by a hyphen. e.g.

142 ≡ Methylhexane 2- Methylheptane ≡ (4) When two or more substituents are present, give each substituents a number corresponding to its location on the longest chain.

143 4- ethyl-2-methylhexane The groups should be listed alphabetically (i.e. ethyl before methyl). In deciding on alphabetical order disregard multiplying prefixes such as “di” “tri” and disregard structure –defining prefixes that are written in italics and are separated from the name by a hyphen such as sec- and tert- and consider the initial letter (s) of the substituent name. Thus ethyl precedes dimethyl and tert- butyl precedes ethyl but ethyl precedes isobutyl.

144 5- when two or more substituents are present on the same carbon atom use that number twice e.g.. 3- ethyl-3-methyl hexane 6- when two or more substituents are identical, identical this by the use of prefixes di, tri, tetra, …. and so on and then make certain that each and every substituent has a number. Commas are used separate numbers from the other.

145 ≡ ,3- dimethylbutane 2,3,4- trimethylpentane 2,2,4,4- tetramethylpentane

146 7- when two chains of equal length compete selection as base chain, choose the chain with the greater number of substituents ,3,5- tri methyl-4-propyl heptane (four substituents) (three substituents) not 4- sec-butyl-2,3- dimethylheptane

147 8- when branching first occurs at an equal distance from either end of the longest chain, choose the name that gives the lower number at the first point of difference. 2,3,5- tri methyl hexane ( not 2,4,5- tri methyl hexane) Classification of hydrogen atoms:- The hydrogen atoms of an alkane are classified on the basis of carbon atom to which they are attached. A hydrogen atom attached to a primary carbon atom is a primary hydrogen atom, attached to secondary carbon is a secondary hydrogen, and attached to tertiary carbon is a tertiary hydrogen.

148 Example: 2- methyl butane has primary (1°), secondary (2°), tertiary (3°) hydrogens. 1° hydrogen atom 3° hydrogen atom 2° hydrogen atom ≡ 3° hydrogen atom 2° hydrogen atom 1° hydrogen atom

149 Cycloalkanes Nomenclature: Cycloalkanes are also called alicyclic (aliphatic cyclic) hydrocarbons. Cycloalkanes are alkanes in which two carbons are bonded together to form a ring, they are characterized by the molecular formula C n H 2n. Examples: ≡ cyclopropane cyclobutane

150 ≡ cyclopentane cyclohexane Cycloalkanes are named under the IUPAC system, by adding the prefix cyclo to the name of the unbranched alkane with the same number of carbons as the ring.

151 Substituent groups are identified in the usual way. The positions of the substituents are specified by numbering the carbon atoms of the ring in the direction that gives the lowest number to the substituent groups at the first point of difference. Ethyl cyclopentane 3-Ethyl-1,1-di methylcyclopentane

152 ethyl-3-methylcyclohexane (not 3-ethyl-1-methylcyclohexane) (not 1-ethyl-5-methylcyclohexane) When the ring contains fewer carbon atoms than an alkyl group attached to it, the compound is named as an alkane and the ring is treated as cycloalkyl substituent:

153 cyclobutylpentane cyclopropylhexane ,1-dimethyl-2-propylcyclopentane 1-ethyl-2-methylcyclopentane

154 Functional groups in hydrocarbons and substituted alkanes:- A functional group of a compound is the structural unit of this molecule that is responsible for its chemical reactivity under a particular set of conditions. The functional group may be single atom or a set of atom. The functional group of a hydrocarbon is any one of its hydrogens, the reactivity of hydrocarbons may be illustrated by replacement with chlorine.

155 Substituted Alkanes:- The hydrogen atoms of alkanes are not reactive. When a group replace a hydrogen in alkane, the group becomes almost functional group. Examples of functionally substituted alkanes:- ExampleSubst. AlkaneAlkane C 2 H 5 OH Ethanol R-OHR-H C 2 H 5 Cl Ethyl Chloride R-XR-H C 2 H 5 NH 2 Ethyl Amine R-NH 2 R-H C 2 H 5 OC 2 H 5 Diethyl ether R-O-RR-H

156 ExampleSubst. AlkaneAlkane C 2 H 5 CN Propane nitrile R-CNR-H C 2 H 5 NO 2 Nitroethane R-NO 2 R-H C 2 H 5 SH Ethanethiol R-SHR-H Epoxide Ethylene Oxide

157 Example 2 The C 5 H 12 isomers There are three isomeric alkanes having the molecular formula. C 5 H 12 CH 3 (CH 2 ) 2 CH 3 ≡ Comm. Name: n- pentane IUPAC name: pentane ≡ Comm. Name: isopentane IUPAC name: 2-methyl butane

158 ≡ Comm. Name: neo- pentane IUPAC name: 2,2- di methyl propane Sources of hydrocarbons:- Hydrocarbons are the main constituents of petroleum. C 1 -C 4 gas, Heating, cooking, petrochemical, raw material. C 5 -C 12 liquids (Naphtha), Fuel, lighter fraction, petroleum ether b. p °C, laboratory solvents. C 12 -C 15 Kerosene, fuel. C 15 -C 18 Fuel oil, diesel fuel. Over C 40 lubricating oil, greases, paraffin waxes, asphalt.

159 The Biological effects of alkanes:- Gasoline, which is a mixture of alkanes should not be swallows because it contains additives such as tetra ethyl lead which is quite poisonous. Liquid alkanes can cause damage if they get into lungs. They dissolve the liquid molecules in the cell membranes. Liquid alkanes can also harm the skin by dissolving the natural body oils and cause the skin to out. Mixtures of high molecular weight liquid alkanes are used to soften and moisten skin. Example petroleum jel “vaseline” which used to protect skin and it can protect babies from diaper rash caused by the skin’ contact with urine.

160 Conformations of alkanes and cycloalkanes:- The different spatial arrangements that result from rotation about single bond in a particular molecule are called, Conformations, Conformers, rotational isomers or rotamers. (1)Conformational analysis of Ethane: Ethane exists in two conformations known the staggered conformation and the eclipsed conformation and can represented as follow.

161 (a) Wedge- and dash (b) Newman projection (c) Sawhorse Staggered conformation of ehtane Eclipsed conformation of ehtane (a) Wedge- and dash(b) Newman projection (c) Sawhorse

162 The H-C-C-H unit in ethane is characterized by a torsion angle ( dihedral angle). Torsion angle is the angle between the H-C-C plane and the C-C-H plane and H-C-C-H unit. 180° 0°60° Torsion angle = 0° Eclipsed Torsion angle = 60° Gauche Torsion angle =180° Anti

163 The staggered and eclipsed conformations of ethane are interconvertable by rotation about its C-C single bond. The angle between hydrogens and the two carbons of ethane is called the “ torsion angle” or “ dihedral angle”. The torsion angle in staggered conformation is 60° while eclipsed form it is 0 °. The spatial relationship between substituents in the staggered conformation is called “ gauche”. Between the staggered and eclipsed conformation there will be an infinite number of conformations that differ by tiny increments in their torsion angle and are known as “ skew” conformation. Conformational analysis of Butane: * Butane posses two different staggered conformations in the first the two methyl groups in gauche relation. in the second they are anti to each other

164 * In the gauche conformation of butane, a hydrogen form each methyl group will be very close to each other to the extent that they repel each other, the repulsive destabilization of a molecule resulting from crowding of atoms or groups is called “ Van der Waals strain”, “ Steric hindrance” Gauche conformation

165 Anti conformation


167 Conformation of cyclohexane: * Cyclohexane exists in a stable conformation known “chair” conformation, this conformation is free of angle strain becouse all of its bonds are staggered Chair conformation “ more stable”

168 * Newman projection of the chair conformation of cyclohexane.

169 Boat conformation Less stable because a steric strain of two flagpole hydrogens and torsional strain due to eclipsing of the bond on four of its carbon. Torsional strain

170 Synthesis of Alkanes: (1)Hydrogenation of Alkenes General Reaction

171 (2) Reduction of Alkyl halides General Reaction (3) Lithium Dialkylcuprates: The Corey-Posner, Whitesides House synthesis:- The over all synthesis:

172 Reaction steps: Restirictions of the reaction:

173 R-X is any alkyl halide, R’-X is a methyl, 1° alkyl or 2° cycloalkyl halide

174 Outline the synthesis of Using Cory-House synthesis? or

175 Reactions of Alkanes: (1) Halogenation of Alkanes X=F fluorination X=Cl chlorination X=Br bromination X=I iodination The order of reactivity is F>Cl >Br >I * Fluorination is very vigorous exothermic reaction and is not used in the laboratory. * Iodination is an endothermic reaction and alkyl iodides are not prepared by direct iodination of alkanes.

176 Chlorination of methane:- * In presence of heat or light ° ° ° °

177 Mechanism of Chlorination of methane:- Initiation step Propagation step Termination step

178 Structure and Stability of Free Radicals:- >>> Tertiary Free radical Secondary Free radical Primary Free radical methyl Free radical Also carbocations >>> Tertiary carbocation Secondary carbocation Primary carbocation methyl carbocation

179 * Chlorination of methane, ethane, cyclobutane gives only mono halogen derivatives. * Chlorination is less selective process (i.e. Cl 2 is highly reactive and less selective). * Chloriniation of alkanes containing nonequivalent hydrogen affords mixture of mono halogen derivatives (all possible derivatives)

180 * Bromination process is milder than chlorination and more selective. e.g. bromination yields the tert-mono bromoalkane almost exclusively


182 Alkenes Alkenes are hydrocarbons that contain a Carbon-Carbon double. Their functional group is a Carbon-Carbon double bond. They have the general formula when they have one double bond and alicyclic is C n H 2n. Nomenclature of alkenes and cycloalkenes: * There are old names for simple alkenes for example. IUPAC: Ethylene Ethene Propene 2-methyl Propene PropyleneisobutyleneCom.:

183 The IUPAC Rules for naming Alkene: (1) Determine the longer carbon chain that contains the double bond, change the ending of the alkane to ene. (2) Number the carbon chain that include the double bond and begin numbering at the double bond; Designate the location of the double bond by using the number of the first atom of the double bond as a prefix Butene (Not 3-Butene) Hexene (Not 4-Hexene)

184 (3)Indicate the location of the substituent groups by the numbers of the carbon atoms to which they are attached. Examples: 1 2-methyl-2-butene (Not 3-methyl-2-butene) ,5-dimethyl-2-hexene (Not 2,2-dimethyl-4-hexene) ,5-dimethyl-2-hexene (Not 2,5-dimethyl-4-hexene) 1-chloro-2-butene (Not 4-chloro-2-butene)

185 (4) Number substituted cycloalkenes in the way that gives the carbon atoms of double the 1- and2-positions that gives the substituent groups the lower numbers at the first point of difference. The position of the double is not nescessary to specify (mentioned). Examples: 1-methylcyclopentene (Not 2-methylcyclopentene) ,5-dimethylcyclohexene (Not 4,6-dimethylcyclohexene) 2

186 (5) Two frequently encountered groups are the vinyl group and ally group. the vinyl group (ethenyl) the allyl group (2-propenyl) e.g Bromo ethene (vinyl bromide) 3-chloropropene (allyl chloride)

187 (6)Designate the geometry of the double bond of disubstituted alkene with perfixes Cis and trans (cis: two identical groups on the same side, trans if the two identical groups on opposite sides). e.g. Cis 1,2-dichloroethenetrans1,2-dichloroethene

188 Structure and Bonding in alkenes sp 2 1s 1 PyPy PyPy Ethene (ethylene) Ex.

189 Higher alkenes are related to ethylene by replacement of hydrogen substituents by alkyl groups. There are two different types of carbon-carbon bonds in propene, CH 3 -CH=CH 2, the double bond is of the δ+Π type, and the bond to the methyl group is a δ bond formed by sp 3 - sp 3 overlap. SP 2 hybridized carbon SP 3 hybridized carbon

190 Isomerisation in alkene (Cis & trans isomerisation): The alkene hydrocarbon C 4 H 8 can exist in four isomers Cis 2-butene trans 2-butene 1-butene2-methylpropene ( No cis, trans isomerism here) ( two geometrical isomerism) Ex.1

191 Ex.2 Cis -isomer maleic acid trans -isomer fumaric acid Naming stereo isomeric alkenes by the E-Z system: When one of two different substituents on a carbon of the C=C bond is identical with a substituent on the other carbon, it will be easy to describ the isomers as cis-trans Example. Oleic acid (Cis) cinnamaldehyde (trans)

192 When the four substituents on the two carbons of the C=C bond are different, cis-trans description become ambiguity. In such cases we will use E/Z description ( E from German means Entgegen = Opposite, Z means Zusammen = together) For the assignment of E/Z we use the sequence rules, the isomer having the two atoms or groups with high priority on the same side of C=C bond will be described Z and when they are on opposite sides of the molecule the isomer will be E. sequence rules: (1) Rule 1 The atom of higher atomic number getting high priority, if two atoms are isotopes of the same element, the atom of higher mass number has higher priority.

193 e.g the sequence is I > Cl > S > D > H Example1. 1-bromo-1-chloro-2-iodo ethene E E Z Z Example 2: 1-bromo-2-chloro-1-flouro ethene

194 Example 3: 1-bromo-1-chloro propene. Z (2) Rule 2: if the relative priority of two groups can not be decided by rule 1 it shall be determined by a similar comparison of the next atoms in the groups e.g. C 2 H 5 group has higher priority than CH 3 ; (isopropyl) has higher priority than C 2 H 5 & -CH 2 Cl Chloro methyl has higher priority than

195 Precedence is determined at the first point of difference: ethyl C 2 H 5 [-C-( C, H, H)] outranks methyl CH 3 [-C( H, H, H)] & similarly tert-butyl outranks isopropyl and isopropyl outranks ethyl. -C(CH 3 ) 3 > -CH(CH 3 ) 2 > -CH 2 CH 3 -C(C, C, C) > -C (C, C, H) > -C( C, H, H) Ex. lower higher lower E Also -CH(CH 3 ) 2 outranks -CH 2 CHOH [-C-( C, C, H)] outranks [-C( C, H, H)] Ex.

196 lower higher E Also –CH 2 OH [-C-( O, H, H)] outranks –C(CH 3 ) 3 [-C( C, C, C)] Ex. lower higher Z

197 (3) Rule 3: where there is a double or triple bond both atoms are considered to be duplicated or triplicated thus C=A equals &equals e.g. equals e.g. for –OH, -CHO, -CH 2 OH groups the sequence is –OH, -CHO, -CH2OH

198 ≡ ≡ Phenyl C 6 H 5 - e.g. has higher priority than(isopropyl) & -CH=CH 2 (vinyl group) equals takes priority over(isopropyl)

199 i.e. –CHO is treated as if it –C( O, O, H) so -CHO, [–C( O, O, H)] outranks -CH 2 OH [–C( O, O, H)] Ex. The compound lower higher E Also -CH=CH 2 is treated as if it were A vinyl group –CH=CH 2 outranks an isopropyl group

200 Ex. The compound higher lower Z Relative stabilities of alkenes: The heat of combustion of the four isomeric alkenes having the same molecular C 4 H 8 showed that 1-butene cis-2-butene trans-2-butene2-methylpropene order of decreasing heat of combustion and increasing of stability

201 Stabilities and degree of substitution of the alkenes: Stability of alkenes increases with increasing of degree of substitution on the C=C bond i.e > > >> > > Stability

202 The degree of substitution in alkenes can extend in cyclic alkenes e.g * *, the ring carbon indicated by an asterisk counter as a unique substituent on the double bond > > tetrasubstituted alkene trisubstituted alkene disubstituted alkene stability

203 Ex. * This is because the C=C is electron attracting and will be stabilized by the electron releasing effect of the alkyl substituents. Steric factors (steric effect): Bulky substituents attached in a cis configuration to C=C will repel each other as a result of being too Close. In contrast, to same bulky groups will be far a part in the trans configuration. Accordingly cis form is less stable than the trans form > >>

204 Cis isomer trans isomer Synthesis of alkenes via Elimination Reactions: There are three main reactions (Elimination reactions) for the synthesis of alkenes. (1) Dehydrohalogenation of alkyl halides: α ß

205 (2) Dehydration of alcohols: (3) Debromination of vic-Dibromides: Dehydrohalogenation of alkyl halides: E 2 ( Bimolecular elimination Reaction) Beta elimination Reaction.

206 B, the used base (basic catalyst) and may be sodium ethoxide in ethanol (C 2 H 5 -ONa / C 2 H 5 OH ). transition state concerted process

207 Or pot. tert.butyloxide in tert. Butyl alcohol ((CH 3 ) 3 COK / (CH3)3COH); Or pot. hydroxide in ethanol ( alcoholic KOH, KOH / C 2 H 5 OH). * RX should be secondary or tertiary alkyl halide. * Higher temperature favored E 2 reactions. E 2 Reactions: the orientation of the double bond in the product; Zaitsev’s Rule: E 2 occurs to give the most stable, highly substituted alkene. ° EX. 1

208 EX. 2 EX. 3 (b) (a) 2-bromo-2-methylbutane

209 (a)(b) δ δ More stable transition state resembles a tri substituted alkene less stable transition state resembles a di substituted alkene (b) (a)

210 An Exception to Zaitsev’s Rule: The use of bulky base such as pot. tert-butoxide / tert-butyl alcohol favors the formation δ ß ß ß ß ß δ

211 ß δ δ ß ß ß Explain the formation of 4-methylcyclohexene in the dehydrobromination 4-methyl bromocyclohexane of less substituted alkene.

212 2-methyl-1-butene 2-methyl-2-butene Synthesis of alkenes via the dehydration of alcohols (E 1 ): * Dehydration of alcohols is an elimination and is favored at high temp.. * Acids used may be Bronsted acids such as sulfuric acid and phosphoric acid. Also Lewis acids such as Al 2 O 3 is often used

213 * The Dehydration process is closely related to the structure of alcohols and temp. and the conc. of the used acid. Ex.1 85% 80% 84% Ex.2 Ex.3 20%

214 * In some case 1 ry and secondary (2 ry ) alcohols undergo rearrangements of their carbon skeleton during dehydration. Ex. 85% 80% 20% * Thus the relative ease of the dehydration of alcohols can be arranged in the following order. > >

215 i.e Is the starting carbon skeleton and Is the producing carbon skeleton. Mechanism of alcohol Dehydration E 1 (unimolecular elimination) Reaction: Example the dehydration of tert-butyl alcohol

216 Step 1: Protonation of alcohol ( alkyloxonium ion formation) Protonated alcohol ( alkyloxonium ion formation) Step 2: carbocation

217 (b) Dehydration of primary Alcohols(E 2 ): * 1 ry carbocation is not stable enough to be formed. * The acid- catalyzed dehydration of primary alcohols takes place by the following mechanism

218 Step 1: Step 2:

219 Molecular Rearrangements in alcohols Dehydration (carbocation stability): It has been found that dehydration of some alcohols afforded alkenes which were different from the expected ones. A rearrangement process was proposed to account for the formation of the unexpected alkenes (methide shift) Ex. 3,3-dimethyl-2-butanol

220 - H + Methyl shift 2 ry carbocation - H + 3 ry carbocation (unexpected)

221 1,2-hydride shift: In which a hydride ion H - migrates to an adjacent positively charged carbon. ~ * Hydride ion shift usually takes place during the dehydration of primary alcohols Ex.

222 - H + ~H 1 ry carbocation - H + 2 ry carbocation (unexpected)

223 ~ ~ ~ 2 ry carbocation The following examples illustrated the carbocation rearrangements. Ex.1 3 ry carbocation Ex.2 Ex.3

224 Ex.4: rearrangements of carbocations can also lead to a change in ring size as the following. 3 ry carbocation

225 Synthesis of Alkenes by the debromination of vicinal dibromides: Vicinal dihalide (avic-dihalide) geminal dihalide (agem-dihalide) * vic- dibromides undergo debromination with NaI/acetone or with a mixture of Zn dust/CH 3 CO 2 H

226 * The debromination by NaI takes place via E 2 mechanism. Additions to Alkenes: The addition reactions are the characteristic to the C=C bond with the general type shown below.

227 Ex. alkyl halide alkyl hydrogen sulfate dihalide alcohol

228 Two characteristics of the double bond addition reactions: (1) The conversion of one Π bond and one δ into two δ bonds δ bond Π bond 2 δ bond Bonds formedBonds broken (2) The Π electrons are particularly susceptible to the electrophiles Π bond

229 Ex. HX react with alkenes by donating a proton to the Π bond Π bond alkene or electrophile nucleophile

230 Addition of hydrogen halides to alkene: Markavikov’s rule HCl, HBr and HI add to the double bond of alkenes. * The addition is carried out by dissolving HX in acetic acid or CH 2 Cl 2 or bubbling the HX gas in the alkene and the alkene itself as the solvent. * The order of the reactivity of HX is HI > HBr > HCl > HF

231 Markovnikov’s rule: In the addition of HX to an alkene, the hydrogen atom adds to the carbon atom of the double bond that has (already has) the greater number of hydrogen atoms. Ex.2

232 Ex.1 The mechanism of Markovnikov’s addition: Step1. Π complex step2 δ complex

233 Theoretical Explanation of Markovnikov’s rule:

234 Ex.2

235 Modern statement of Markovnikov’s rule: In the ionic addition of unsymmetrical reagent to a double bond, the positive portion of the adding reagent attaches itself to a carbon atom of the double bond so as to yield the more stable carbocation as an intermediate. Example:

236 Addition of sulfuric acid to alkenes: Example:

237 Alcohols from alkyl hydrogen sulfates: Addition of water to alkenes: Acid catalyzed hydration. General equation: Ex.2 Ex.1

238 The mechanism of the hydration of 2-methylpropene. Step 1. Step 2. Step 3.

239 The reaction produces tert-butyl alcohol because step 1 leads to the formation of the more stable tert-butyl cation rather than the much less stable isobutyl cation. for all practical purposes this reaction does not take place because it 1 ry carbocation. The rearrangements associated with alkenes hydrations. 3,3-dimethyl-1-butene 2,3-dimethyl-2-butanol

240 The mechanism for the above rearrangements: ~

241 Addition of Bromine and Chlorine to Alkenes: The reaction of alkenes with Br 2 /CCl 4 is a useful test for carbon-carbon multiple bonds. Alkenes react rapidly with bromine at room temp. and in the absence of light. If we add Br 2 to C=C, the red-bromine disappears. Thus Alkene (colorless) red-brown Vic-dibromide (colorless) Examples

242 1) 2) 3) C° -9 C° Trans-1,2-dibromo cyclohexane - 5 C° (as a racemic form)

243 Π complex δ complex SN2SN2 Vic-di bromide Step 2 Step 1 Mechanism of halogen Addition: Π complex formation and δ complex.

244 Stereo chemistry of halogen Addition to alkenes: The addition of Br 2 to alkenes is anti addition. The product is trans in isomer in the cases of cyclic bromonium cation and the anti addition of Br - to that intermediate via SN 2 mechanism could be illustrated by the following reaction. Ex. cyclopentene Trans1,2-dibromocyclopentane

245 Mechanism of the reaction: Π complex δ complex (bromonium ion) Trans1,2-dibromocyclopentane

246 Halohydrin formation: alkene halhydrin Vic-dihalide The mechanism of Halohydrin formation: Step 1 Π complex Step 2 δ complex (bromonium ion)

247 Step 3 Example

248 Epoxidation of alkenes: Epoxides are cyclic ethers with three membered rings. An epoxide Ex. IUPAC name : Oxirane common name : ethylene oxide Synthesis of (epoxidation):

249 * The used peracids are peracetic and perbenzoic acids, the mechanism of epoxidation is as shown * The epoxidation reaction is a syn addition. Ex.1 Ex.2

250 Acid-catalyzed hydrolysis of epoxide: Ex. 1,2-ethanediol (ethyleneglycol) Anti hydroxylation of alkenes: Ex.

251 Acid catalyzed hydrolysis of cyclopentene oxide yields a trans-diol, trans-1,2-cyclopentanediol.

252 Alcohols from alkenes through Oxymercuration- Demercuration:

253 * Oxymercuration- Demercuration is highly regioselectiv. The net orientation of water H- and –OH is in accordance with Markovnikov’s rule. General example: Specific example: Ex.1

254 Ex.2 Alcohols from Alkenes Through Hydroboration-Oxidation: The addition of the elements of water [ H,OH ] to C=C through the use of diborane (B 2 H 6 ) or THF:BH 3. The addition of diborane to C=C is hydroboration and the second is oxidation and hydrolysis of organoborn intermediate. The method gave the alcohols ( Anti- Markovnikov addition of water) which can not prepared through the acid-catalyzed hydration of alkenes or oxymercuration-demercuration

255 Example: H 2 O / H 3 O + 3 mole THF : BH 3 hydroboration HO 3 moleH 2 O 2 / OH - 3 CH 3 CH 2 CH 2 OH 1- propanol

256 Ex.2 Ex.3 Ex.4

257 The stereo chemistry of the oxidation of organoborans: The oxidation step in the hydroboration-oxidantion synthesis of alcohols takes place with retention of configuration, the hydroxyl replaces the boron atom where it stands in the organoboron compound the net result of the two steps is the syn addition of –H and –OH. Ex.

258 Radical addition to alkenes: The Anti Markovnikov addition of HBr: When alkenes that contained peroxides or hydroperoxides reacted with HBr, Anti-Markovnikov addition of HBr. HBr Absence of peroxide

259 Ex.1 The mechanism of anti Markovnikov’s addition: ( HF, HCl, HI do not give anti Markovnikov’s addition) in the presence of ROOR only HBr gives anti Markovinkov’s addition, the mechanism is a readical chain reaction initiated by peroxides: and involves the following steps. Chain Initiation Step 1 Step 2

260 Chain propagation Step 3 Step 4 Reaction termination:

261 Oxidations of alkenes: Syn Hydroxylation Alkenes undergo the oxidation of C=C bond with KMnO 4 or OsO 4 to give 1,2-diols (glycols) Examples (1) (2)

262 Syn Hydroxylation of alkenes: The mechanism for the formation of glycols by KMnO 4 and OsO 4 involve the formation of cyclic intermediates followed by the cleavage at the oxygen-metal bond to give the syn hydroxylation product.

263 The Syn Hydroxylation can be seen by when cyclopentene reacts with KMnO 4 / OH - or OsO 4 followed by treatment with NaHSO 3 or Na 2 SO 3. The product in either case is cis- 1,2-cyclopentanediol.



266 Oxidative cleavage of alkenes: Hot KMnO 4 oxidize alkene to pot. Salts of carboxylic acids. Ex. The terminal CH 2 group of 1-alkene is oxidized to CO 2 and H 2 O by hot KMnO 4. A disubstituted carbon atom of a double bond becomes the C=O group of a ketone Example:

267 The oxidation cleavage of alkenes has been used to prove the location of the double bond the alkenes chain or ring (via retrosynthetic analysis) Example:

268 Ozonolysis of alkenes: A more widely used method for locoing of the double bond of an alkene involves the use of O 3 (ozone) ( either cis or trans 3-octene) Ozone react with alkenes to form the unstable initial ozonide which rearrange to ozonide. The mechanism of this rearrangement can be depicted as follow

269 Ozones are very unstable and can not isolated but are reduced with Zinc and water. The reduction products are carbonyl compounds

270 The overall process of ozonolysis followed by reduction with Zinc and water accounts to a disconnection of the carbon- carbon double bond in the following fashion A –H attached to the double bond is not oxidized to –OH as it is with KMnO 4 Example:

271 Alkynes They are hydrocarbons containing -C≡C- (carbon-carbon triple bonds). * Noncyclic alkynes have the general formula C n H 2n-2 * We called R-C≡CH monosubstituted or terminal alkynes, R-C≡CR’ are called to have internal triple bonds * R-C≡CH have acidic proton H-C≡C-H acetylene (ethylene) is the simplest alkyne Sources of Alkynes: (1) From lime stone and Coke

272 (1) By thermal dehydrogenation of ethylene: Nomenclature of alkynes: IUPAC rules for hydrocarbons were followed and alkynes take the suffix yne Ex. H-C≡C-H (ethyne), CH 3 C≡CH (propyne), CH 3 CH 2 C≡CH (1-butyne), CH 3 C≡CCH 3 (2-butyne), (CH 3 ) 3 CC≡CCH 3 (4,4- dimethyl-2-pentyne). * -C≡CH is named as ethynyl if it is a substituent * We can use the older nomenclature as follow HC≡CH (acetylene), CH3C≡CH ( Methyl acetylene),

273 Physical properties: * They are resemble alkanes and alkenes in their physical properties such as low density and low water solubility. * They are non polar and dissolve in organic solvents such as alkanes Acidity of acetylene and terminal alkynes: Alkynes are more acidity than alkenes and alkanes ( Acetylene and terminal alkynes are stronger than other hydrocarbons) HC≡CH > H 2 C=CH 2 > CH 3 CH 3 pKa Ka * - OH ion is too weak base to convert acetylene to it’s anion,

274 The position of equilibrium in this equation lies to the left. Amide ion is a much stronger base than acetylide ion and converts acetylene to it’s conjugate base quantitavely. Ka = pKa = 26 Ka = pKa = 36

275 Preparation of Alkynes by the alkylation of acetylene and terminal alkynes: Alkynes are prepared by combining smaller structural units to build longer carbon chains. e.g. Dialkylation of acetylene can be achieved by carrying out the sequence twice

276 Dehalogenation of dihalides: Alkenes can be converted to alkyne e.g. general equation

277 The geminal dihalides and vicinal dihalides can be dehyrohalogenation two reaction steps. General equations: The second dehydrohalogenation step is more difficult than that the first so that rather strongly basic conditions or high temperature are needed to convert dihalides to alkynes

278 Chemical properties of alkynes: (1)Hydrogenation of alkynes: (a) Catalytic hydrogenation

279 (b) Metal-Ammonia Reduction of Alkynes: It gives trans or E alkenes Addition of hydrogen halides to alkynes: The addition of HX to alkynes is regioselective and follows Markovinkov’s rule. Ex.

280 The reaction mechanism is. Step 1 Step 2 is bromide ion captures the alkenyl cation The carbocation formed by addition of a proton to an alkyne is an alkenyl cation. The positively charged carbon of an alkenyl cation is SP hybridized which is more electronegative than its SP 2 hybridization count part thus alkenyl is less stable than alkyl carbocation

281 RCH 2 CH + R’ is more sable than RCH=C + R’ Alkyl cation alkenyl Positive charge is positive charge is On SP 2 hybridized carbon on SP hybridized carbon * In the presence of excess HX, geminal dihalides are formed Ex.

282 * Free-radical addition of HBr to alkynes is a regioselectivity oppositey to Markovinkov’s rule is observed Ex. Hydration of alkynes: Addition of the elements of water to the triple bond ( HO - &H + ) to yield an enol which converted to the most stable form, the keto form.

283 Addition of halogens to alkynes:

284 A dihaloalkene is an intermediate and is the isolated product when the alkyne and the halogen are present in equimolar amounts. The stereochemistry of addition is anti. Ozonolysis of Alkynes: Carboxylic acids are produced when alkynes are subjected to ozonolysis.

285 Ozonolysis is some times used as a tool in structure determination by edifying the carboxylic acid.

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