Fund of Organic Chemistry I
Fund of Organic Chemistry I CHEM 3331
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Date Created: 09/19/15
Chapter 3 The Structure and Stereochemislry of Alkanes Alkanes simplest class of the quot saturatedquot hydrocarbons contain only single bonds We have talked about the simplest member methane Higher quothomologsquot involve replacing H with C H H H HC H Methane H cl3 3 Ethane ll H H H H Hdddapmm All TTTT H C C C C H Sameas CH2CH3 CH8 l HI l l Butane Table 31 nAlkanes LEARN NAMES OF FIRST 10 Think about next 10 CH4 methane CH3CH3 ethane CH3CH2CH3 propane CH3CH22CH3 butane pentane General formula 2 CnH2n2 hexane heptane Homologs differ only by number octane of CH2 groups nonane CH3CH28CH3 decane 30 Are these the same CH3 CH339CH2 CH239CH3 Both are C4H10 No they are not the same One is a branched chain isomer of n butane isobutane Isomers are compounds With the same molecular formula but different chemical structures Butane 2 isomers Heptane 9 draw them Decane 75 C15H32 4347 C40H82 6249117880583l isomers We have already considered the 2 isomers of butane Look at pentane CH8 l Hsc C CH3 CHeCH2CH2CH2CH3 HSC CH CH2CH3 C3H n39pentane isopentane neopentalie We need a simpler naming system to designate isomers in the higher hydrocarbons IUPAC Nomenclature Rules for Alkanes 1 Find the longest contiuous carbon chain and use this name as the base a If two chains of equal length are present choose the one With the more branch points as the parent C C Ht C C 2 Number the longest chain beginning With the end nearest a substituent number structure above 3 Name each substituent group and assign it a number a If there are 2 substs on the same carbon they both get the same number 2Methyl C C e c Cl3 4Methyl 4Ethyl O 0 13 0 Alkyl Groups alkanes in which a hydrogen has been replaced by another group CH3 Cl methyl chloride change quot anequot to quot ylquot CH3 methyl CH2CH3 ethyl CH2CH2CH3 n propyl 32 CH8 HSC CH Isopropyl CH339CH2 CH239CH2 HSC CH CH2CH3 secbutyl iHS HSCCHCH2 isobutyl CH3 HSC tenbutyl or t butyl AHS CH3 HeC c 2 neopentyl CH3 R any akyl group 4 Write out the name substituents come in alphabetical order use di lri letra for multiples of the same subst 5 Name any complex substituent C C G C C C C C C C C C 23dimethyl6 2 methylpropyIdecane 33 Nomenclature Examples CH8 CH2CH3 CH30H2CH2CH CH3 2 methylpentane 3ethy1pentane CH8 CH8 CH3 CH3 224 trimethylpentane CH3 CH3 CH3 HSC EI CH2 EI CH20HTEI CH3 CH8 CH3 CH2 247 tr1rnethyloctane not 257 H H2 CH2 CH HSC CHS 33 diethyl 5 isopropyl 4 methyl octane Complex Substituents H3 C3H3 f CH2 g CH3 2 Methyl 5 12 dimethylpr0py1n0nane The prefixes sec and lerl refer to the degree of alkyl substituion at the carbon in question 34 H H R C H primary carbon 1 R C H secondary 2 one substituent l H R R R R C H tertiary 3 R C R quarternary 4 R R T so if we talk about a tertiary alcohol we mean R CID OH R What is the degree of neopentyl alcohol Properties of Alkanes Table 31 p 84 lists some alkanes and their mp bp and densities Note these vary in a regular fashion see figs 3 3 3 5 bp increases 25 30 carbon Why van der Waals forces Note Branched chains have lower mpbp than straight chains Solubility hydrophobic quotwater hatingquot Good lubricants and corrosion inhibitors Occurences of Alkanes Methane comes from the anaerobic without air decomposition of vegetable matter under water quotmarsh gasquot Natural Gas 2 CH4 and CH3CH3 some propane and butane LPG liquified petroleum gas 2 propane 35 Petroleum 2 mixture of hydrocarbons mostly alkanes and cycloalkanes derived from crude oil by distillation or fractionation bp ie natural gas C1 C4 lt 20 C petroelum ether C5 C6 30 60 li groin C7 60 90 straight run gasoline C5 C11 30 200 kerosene C8 C16 160 230 diesel oil C10 C18 200 320 heating oil C14 C25 275 400 lubricating oil greases paraffin C16 C24 gt400 asphalt Catalytic cracking Si Al catalyst at 400 500 C breaks the Cll Cl4 fraction into smaller molecules which are recombined into C7 C10 alkanes Catalytic reforming converts straight chain alkanes into toluene and benzene aromatics which have a higher octane rating Octane Rating measures the antiknock properties of a fuel nheptane 0 CH3 CH3 H I isooctane 100 224 trimethylpentane Reactions of Alkanes Sometimes referred to as paraffins from Latin quot parum affinisquot slight sffinity Relatively chemically inert l Combustion is the reaction With oxygen CH4 202 gt C02 2H20 213 kcaImole2 Halogenation 36 CH4 Cl2 gtCH3C1 HCl etc Structure and Conformations shapes of Alkanes Methane CH4 tetrahedral replace H with CH3 to get ethane Ethane CH3CH3 H Sawhorse drawing H H There is free rotation around all single bonds for C H bonds no Change for C C bonds Changes H H H Sawhorse H rotate 60 H Formulas H H H H H H H xQ Staggered Conformation Eclipsed 37 CH CH3 CH CH CH3 H 3H H H 3 H H80 3 H CH8 l ii gH 3 1 l QH I l i H H H H H H CH8 HSC H H H HH H H Anti Eclipsed Gauche Eclipsed Gauche 1 mirror images I Gauche Anti Ec p39sed Glauche Eclipsed E E I 09 kcalmo 39 45 kcalmole 38 kcal mole c Potential Energy 5 I o I n I I U f Rotation gt All the above are conformational isomers can be interchanged by rotations about C C single bonds Structural isomers cannot be interconverted by rotations about single bonds ie n butane and isobutane branched chain alkane are structural isomers Branched chain alkanes can also have conformations Consider isopentane 2 methyl 2 butane one gauche H3O CH3 H3 CH3 2 gauche We can continue this treatment for all the higher alkanes The number of conformations becomes very large Energy costs for interactions in alkanes H H Eclipsed 10 kcal mole H CH3 quot 14 CH3 CH3 quot 25 CH3 CH3 gauche 09 Cycloalkanes are alkanes that contain rings of carbon atoms Many important naturally occuring molecules are cyclic mNI VWCOOH HOCHQ W O H OH Prostaglandin E1 regulatory hormone 1mm Corti sone Steroidal hormone Table 3 3 p 105 gives the properties of simple cycloalkanes mp bp density 39 Nomenclature IUPAC 1 Use cycloalkane as base name unless side chain contains CHZCHZCHZCH more carbons than ring 1 cyclopropylbutane CH3 methylcyclopentane 2 Number substituents to arrive at the lowest sum of numbers 13 dimethylcyclohexane 3 not When 2 different alkyl groups are present number alphabetically CH3 1 ethy1 2 methylcyclopentane Halogen substituents are treated like alkyls CH3 1 bromo 2 methylcyclobutane Br 40 When the acyclic portion of the molecule contains more carbons than the cyclic part or When it contains an important functional group the cyclic part is named as a cycloalkyl substituent give example Cis Trans Isomerism use models Note A cyclic structure has less conformational mobility T Hp m Cannot rotate around C C bond HH V eyeball Thus these rings may be considered to have 2 different faces Br Br Br H H H Q l2 dibromocyclopropane trans 12 k Stereoisomers d Other rings show the same effect 41 wow Cyclobutane Cyclopentane Cyclohexane We will study the shape of these rings later Remember cis and trans When naming CI H H CI l bromo 3 methylcyclopentane trans 14 dichlorocyclohexane Conformations Shapes of Cycloalkanes 1 Baeyer Strain Theory Adolf von Baeyer Univ of Munich 1885 Tetrahetral carbon implies that the angle between any 2 substituents 1095 a 6054 and so on Rings smaller than cyclopentane suffer from compressed angles and those larger from suffer from stretch angles How does this theory agree With facts Stability of Cycloalkanes Ring Strain 42 How can we measure the amount of ring strain in kcal mole associated with various size rings Compare heats of combustion The more strain higher energy a cmpd contains the more heat is released upon combustion CH2 32 02 gt C02 H20 Heat To be fair we should look at AH per CH2 unit See Table 34 p 108 CycloCH2n n AH CH2 Total Strain Energy kcal mole 3 1666 kcal mole 276 4 1640 264 5 158 7 6 5 6 1574 0 gt choose as strain free model 7 1583 6 3 8 1586 96 12 1576 24 Strain energy calc as follows For n3 1666 1574 2 92 diff in AH CH2 3 carbons gt total strain 2 3 X 92 276 kcalmole So Baeyer was wrong for large rings Why Not flatll Nature of Ring Strain 1 Angle strain due to expansion or compression of bond angles 2 Torsional strain due to eclipsing of neighboring bonds 3 Steric strain due to repulsive interaction of atoms approaching too closely 43 Let s look at the three smallest cycloalkanes 1 Cyclopropane HJ six pairs of eclipsing interactions A H V T 115 lt increased over 1095 H more pcharacter weaker bond less efficient overlap outside CC Normal CC bond aXiS efficient overlap along x more Mke CC axis A h p bit quot Bent Bondsquot ence or S are easier to break SP439SP5 more reactive 2 Cyclobutane H H H A H infactisslightly bent H H sacrifices some angle strain to relieve some eclipsing interactions 3 Cyclopentane interior angles of a pentagon 2 108 close to 1095 H H 0 lie in a plane quot flap carbon lies out of plane and can rotate from one carbon to the next pseudorotation 4 Cyclohexane most important of the cycloalkanes can adopt a conformation that is essentially strain free no bond angle strain and no eclipsed hydrogen strain H H H H CH I H H H H H H 39 39 K equatorial H CH2 H OIquot H 139 H bond H H lgt39 39 b axial bond Ring Flip Ring Flip interchanges axial amp barrier 108 kcalmole equatorial q H H H H H CH2 H H CH H H H Show how to draw cyclohexanes axial amp eq bonds 45 Substituted cyclohexanes When we put a substituent on a cyclohexane the two chair conformations become non equivalent a ie Methylcyclohexane Gauche interaction CH3 K CH3 C H2 H CH2 H H H CH2 CH3 CH3 CH2 I H H no gauche interaction at 25 C 95 18 kcalmole Note one axial HCH3 more stable interaction 09 kcalmole What is the relationship What is the relationship between the energy difference between two conformers and the relative amounts present of each one at a given temperature AG RT an AG 2 energy difference 46 R 2 gas constant T temp K equil constant 18 kcalmole 1987 X 103 kcal mole K 298 K In K solve for In K an304 K 21 955 Isomer ratio AE kcalmole 25 C 5050 0 6040 024 7030 0502 90 10 1302 982 2306 99 901 4 092 Table 45 p 118 lists some energy differences betw axial and equatorial for different groups on cyclohexane For a t butyl group with 2 such interactions AG 2 2 X 27 54 kcal mole essentially all equatorial H m Conformational Analysis Stereoisomerism in cyclic cmpds consider 12 dimethylcyclohexane 47 CH3 H different chemical cmpds H CH3 Look further CH3 CH3 trans CIS 1 cis12 axialeq eq39aXia39 equal energy 1 GaUChe CHa39CHs 09 1 Gauche OHSCH3 09 2 OHSH 13dIaXIaI 18 2 OHS H 1 3diaxia 18 AE 27 kcalmole AE 27 kcalmole In fact the two conformers are non superimposable mirror images of one another hence we have conformational enantiomerism later 2 trans12 Ring Flip 2 H3 axialaxial eqeq 4 CHsH 13diaxial 36 kcalmole 1 Gauche OHSCH3 09kcallmole AE 27 kcalmole How about the other dimethylcyclohexanes 48 13dimethyl trans Ring Flip 39 CH3 2 CH3 axialeq eqaxia identical 2 compound same energy cis Ring Flip H z 3 CH3 eqeq more stable axax 14dimethyl trans CH3 CH3 H H3O Ring Fllp eqfeq axax CH3 cis 49 CH3 CH3 R39 FI39 H30 mg no CH3 eqax axeq same energy identical Boat Conformation A second higher energy conformation H CH2 H H CH H H H H H Flagpole interaction 4 pairs of ecilpsed bonds sterlc strain mange strain About 7 kcalmole higher E than chair conformation Boats can be interconverted through lower energy quottwist boatquot 55 kcalmole Bicyclic and Polycyclic Molecules 50 Bicyclo221heptane Bicyclo110butane Decalin 2 fused cylohexane Iings Bicyclo440decane H C13 can be cis or trans H QUE Steroids show this kind of stereochemistry at sites of ring fusion trans Q m 51 cholesterol CH3 CH3 H HO H H H H CHEMISTRY 3331 ORGANIC I L G Wade 3rd Edition January 1999 Dr Randy Thummel 2 Chapter 1 Introduction and Review ORGANIC CHEMISTRY is the study of the compounds of carbon 1770 Swedish Chemist Torbem Bergman first distinguished between quotorganicquot and quotinorganicquot compounds Organic cmpds are derived from living organisms quotVital forcequot theory 1828 Friedrich Wohler disproved this theory NH4 39OCN gt H2NCONH2 Birth of Synthesis ammonium cyanate urea Synthesis is one of the three branches of organic chemistry 1 Synthesis involves making and sometimes breaking bonds between simple molecules to make more complex ones 2 Mechanism understanding how chemical reactions occur 3 Structure understanding how atoms are arranged attached to form molecules We will begin with this last subject STRUCTURE What makes carbon unique among the 103 elements Its ability to bond almost infinitely with itself This makes the number of organic compounds essentially limitless For example plastics polymers dyes medicines fabrics food additives pesticides and herbicides your body Let us begin our consideration of organic chemistry on a very simple atomic level 3 Consider a simple picture of an ATOM Nucleus consists of Mass Charge Protons 1 1 N Neutrons 1 0 V Electron ca 0 1 Isotopes of an element have different numbers of neutrons in their nucleus For example 12C has 6 protons 6 neutrons 13C has 6 protons 7 neutrons 14C has 6 protons 8 neutrons t12 5730 years tell ages up to 50K yrs This picture for the nucleus is OK but not so good for the electrons They cannot be represented as point charges This problem was solved by the birth of quantum mechanics 1926 Quantum Mechanics theory was proposed to explain the quotmotionquot of electrons in mathematical terms Hence the Schrodinger wave equation was developed We never know the exact postion of an electron Heisenberg Uncertainty Principle but a volume of the space in which it is likely to be found We call this the orbital The size and shape of an orbital depend upon the energy level of the electron which occupies it Shapes of Atomic Orbitals 3 types s p d 1s orbital is lowest in energy and closest to the nucleus Z S orbital spherically symmetrical 4 p orbitals have a node at the nucleus and two lobes which extend on either side Porbital has a node at the nucleus Y Y a a 2px 2py 2pz d orbitals are more complex we don t use them much in organic chemistry Rules for Filling Atomic Orbitals l Aufbau Principle orbitals of the lowest energy are filled first 2 Pauli Exclusion Principle only two electrons can occupy the same orbital and they must have the opposite spins 3 Hund s Rule if two or more empty orbitals of equal energy are available one electron is placed in each until all are half full Energy Levels of Atomic Orbitals Energy Levels of Atomic Orbitals High 4d 53 1 4p 3d I Energy 43 These 24 orbltals can account for the rst 48 elements Gs 2P 28 Low 13 Let s use our rules to describe the GROUND STATE lowest energy electronic configuration of some simple elements Hydrogen 1H 13 71777 Sodium 11Na as 71777 2 27 4 Neon 10Ne 2p 7 U W 4quot 25 7 outer 2shells complete very stable Carbon 60 2p 7 777 Jr 777777 Chemical Bonds are the forces that hold atoms together in a molecule Note energy is always released when a bond is formed ie a more stable lower energy situation results 1915 G N Lewis proposed the quotoctet rulequot Atoms will tend to transfer or share electrons to acquire a stable complete outer shell configuration similar to He Ne and Ar This leads us to the two types of chemical bonds 1 Ionic Bonds result from the transfer of electrons creating oppositely charged species which then attract one another Ionization Energy tendency of an atom to loose an electron quotElectropositivequot elements Le Na sodium have low LE and hence can easily loose an electron to become positively charged Electron Affinity tendency of atom to gain an electron quotElectronegativequot elements Le F fluorine have low large negative EA and hence can easily add an electron to become negatively charged 6 11Na gt 1e Na F39 ioniccompound 9F 1e39 gt How about carbon Carbon must gain or loose 4 electrons to establish complete octet thus becoming C4 or C394 which is very difficult Rather carbon forms 2 Covalent Bonds result from the sharing of electrons to give a complete outer shell configuration Designate this process using Lewis Dot Structures H H gt H 39 H 2electrons1 bond H H methane CH4 3 3 c 4H gt New H C H H H39 u u 5 39 39o39 2H H9H water H20 note comple e HOH outer shell configurations H methanol H O H These are called I Kekul structures H Drawing chemical structures Showing every bond can be awkward Instead use line drawings each line C C bond add H s to complete valency of 4 for C Butane CHgCHQCHQCHg 1butene CHFCHCHQCHg CH3 CH3 Zbutene C 0 H H cyclopentane 11butylcyclohexene gomz Molecular models Help us to see things in 3 dimensions Show sample Formal charges When an atom looses an electron in forming a covalent bond it acquires a When an atom gains an electron it acquires a a consider the protonation of water A 0 now has only 5 electrons H l r HOH H gt H O H Has lost1 e39 O has 6 electrons Notice how we use curved arrows to designate the movement of 2 electrons These arrows do not show the movement of atoms VERY IMPORTANT b Nitromethane CH3N02 H f both oxygens I d have only seven e T 0 IO 0 with one valence is H C39 N 0 H3 I 0quot N with 4 valences is c Acetonitrile oxide I 3 HSC C N o H C C N O The electrons that contribute to an atom s charge are 1 All its unshared electrons 2 12 of the bonding electrons it shares With other atoms Formal Charge 2 Grp number non bonding electrons 12 unshared pairs Polarization Bonds between non identical atoms are very often polar ie the electron distribution in the bond is unsymmetrical electrons are unequally shared by both nuclei This polarity is due to the intrinsic electronegatiVity of atoms on the right side of the periodic table 6 5 xx XZ Y X Y symmetric polar ionic covalent bond covalent bond bond Look at specific example chloromethane H 6 H C Cl indicates direction of polarity H gt example of inductive effect chlorine draws electron towards itself measure of this polarity is called di ole moment a vector quantity We can determine the net dipole moment by vector addition l gt Net Dipole Moment 39 C gt Net Dipole Moment HQCI How about CCI4 Resonance Structures Some cmpds are not adequately represented by a single Lewis structure Two or more valence bond structures which differ only in the arrangement of electrons are called resonance forms structures H H H 6 6 93 4 N C N f H Resonance structures Resonance Hybrid H H H H H 10 The positive charge is delocalized over C and N We say that the C is resonance stabilized Dispersal of charge is stabilizing and concentration of charge is destablilizing Resonance hybrid is a mixture half assed analogy Look at acetic acid Resonance theory explains the stability of the conjugate anion and hence the Wilingness of HOAc to loose a proton 0 H 0 zoo ch C 2 H3C C 4 ch c Hac C N 0 H o 0 Hybrid General Rules for Resonance 1 All resonance structures must be valid Lewis structures 2 Structures differ only in arrangement of electrons Atoms and bond angles cannot change 3 Number of unpaired odd electrons must remain the same 4 Major resonance contributor is one With lowest energy Neg charge more stable on electroneg atom O o c 039 HH H H minor contributor no octet on carbon i 5 The more resonance forms the better CH Analysis molecular amp empiurical formulae Combustion of organic compounds in the presence of 02 leads to C02 H20 ie 06H12 902 6002 6H20 absorbed absorbed on KOH on CaCI2 Example 055g of an organic compound gave 066 g of H20 1037 g of C02 Wt of H 2 fraction of H in H20 Wt of H20 2 i 2016 066 00749 H 18016 H wtofH 4 X1OO1344H wtof sample 55 wt of C fraction C in C02 wt of C02 12o1 1037 283gC 4401 283 x 100 5147c 55 Difference from 100 3509 0 Determine atomic ratios by dividing element atomic wt 5147 1201 428 1344 1008 1333 Divide by 2193 3509 1600 2193 QIQ Empirical formula C2H60 Molecular formula can be any multiple of this To determine we must know molecular weight MW 2 46 C2H60 MW 2 92 C4H1202 etc Acids and Bases 1 Arrhenius theory defines acids as substances that dissolve in water to give H30 ions LCD 12 CH3COOH H20 gt H3O CH3COO Water autoprotolysis constant Kw H3O OH l X 103914 Hence acidic H3O gt 107 M and basic 2 H3O lt 107 M Furthermore pH log H3O 2 Bronsted Lowry Definition Acid 2 proton donor Base 2 proton acceptor K HA H20 2 H3O A39 Conjugate base of HA Acid Base H30A39 K H3OA39 HA H20 HA Large constant Acidity Constant pKa log Ka The stronger the acid the lower the value of pKa Explain Table 15 lists some typical pKa s Acid strength pKa ConjBase Base Strength Weak Ethanol CH3CH20H 16 CH3CH2039 Strong HECN 92 39CN H F 32 F39 Strong H NO3 43 N0339 Weak Lewis De nition Acid Electron pair acceptor This is more general and covers the Base Electron pair donor Bronsted def but also includes others Ir H or HOH H O H Acid Base quot Empty F CH3 F CH orbital 39 39 39 3 F q FO CH3 F CH3 Acid Base In general Compounds with N O are good Lewis bases because they have lone pair electrons which they can donate Note Another name for Lewis acid is electrophile electron loving Another name for Lewis base is nucleophile nucleus loving Structural Effects on Acidity l ElectronegatiVity A more electronegative element bear an neg charge more easily gives a more stable anion and hence affords a stronger conjugate acid ElectronegatiVity C lt N lt O lt F gt increases Stability CH3 lt NH2 lt OH lt F gt increases Acidity H CH3 lt H NH2 lt H OH lt H F gt increases Basicity CH3 gt NH2 gt OH gt F lt increases 2 Size The negative charge of an anion is more stable if it is spread over a larger area of space 14 Stability F lt Cl lt Br lt I Acidity H F lt H Cl lt H Br lt H I gt 3 Resonance stabilization Acetic acid methanesulfonic acid 0 o H C i z s H3O C O H 0 0 z H3C ISI O39 gt size increases increases H Chapter 2 Structure and Properties of Organic Molecules We have used atomic orbital A0 theory to explain how electrons are to be arranged in atoms When 2 AO s combine to form a covalent bond we have a new molecular orbital MO 6 gt 104kcanole 15 1s H2 molecular orbital MO cylindrically symmetrical 6 bond Note energy is released a lot This energy is the bond strength of the H H bond This much energy is required to split H2 gt 2 H Energy level diagram 39 o Antibonding MO empty 0 W V s 1s 1s s Energy 104 kcalmole Tlrki o Bonding MO 0 Since we start with 2 AO39s we must get 2 MO39s The 2nd one is higher in energy and normally empty Look at the hydrogen molecule gtI 0743lt Internuclear Distance Bond length 1 A 108 cm How about overlap of D orbitals Node Linear Overlap Cgt i X I 39 o x pp sigma antibonding MO d l H 03 t pp sigma bonding MO Node Sideways Overlap i I I I I 8 8 quot n antibonding MO lb c nbonding MO Let s consider the structure of methane CH4 the simplest organic molecule First consider the atomic configuration of carbon C H bonds involve 2 different atoms What kind of AO will each atom use What kind of MOS will results from overlap of these AO s We will see that we need to modify our concept of orbitals to accomodate bonding to carbon Shape of new sp3 hybrid 17 7 How can carbon form 4 bonds 7 if it has only 2 unpaired electrons to share mix together 3 2p and 1 2s orbital to get 4 new orbitals we call Sp 2p 4 l hybridize mix 53 lg t e Shape of new Sp3 hybrid gt Arrange four of these so that they all originate from a common point and are oriented as far apart as possible They point to the corners of a regular tetrahedron I I Structure of ethane Attach 2 sp3 hybridized carbons together by overlapped orbitals Add 6 H s I Sawhorse drawing H C C H E CH30H3 H H 39 Note What happens if we continue to next higher homologue propane W Sawhorse H 39i i i I39 drawing H 0 H H H E H C Cl CI H E CH3CHZCH3 H H H Molecule is not really linear Ethylene sp2 Hybridization The carbons in ethylene C2H4 are each connected to 3 other atoms The fourth valence of the ethylene involves a CC double bond erw 2LL P Pro mote 2S J39 1S 177 Hybridize 2 These three new sp orbitals sz Wii 4777 iALii Jrii p pornt as far apart from each 13 ML other as possible 7 i also represent as CH2C H2 Propylene CH2CH CH3 draw this Acetylene sp Hybridization The carbons in acetylene C2H2 are each bonded to 2 other atoms The third and fourth valences involve a CEC triple bond H CEC H How do we explain this structure with orbitals 6C 2 l39 2pjquot4 7 Promote 2s 1s 15 Hybridize These two new sp orbitals point SP2 ll 4quot Aquot Alli P 1s H as far apart from each other as possible These new sp orbitals point as far apart from each other as possible linear molecule cylindrically symmetrical around CC bond 20 Table Consider the bond lengths and strengths of these new bonds triple gt double gt single sp gt sp2 gt sp3 degree of s Character Molecule Bond Bond Strength Length CH4 Csp3 H 104 kcalmole 110 A Csp3 H 98 110 CH2CH2 CspzzCspz 152 133 Csp2 H 107 1076 HCECH CspECsp 200 12 Csp H 131 106 Hybridization of Other Atoms H H N 1 Nitrogen Consider Ammonia NH3 H 7 Three possibilities 1 Use p orbitals 2 Rehybridize to sp2 N 221 4 4 1s lt HNH 90 2 1 ltHNH120 9 9 4 120 WA measured lt HNH 1071 close to tetrahedral 2 Oxygen Consider water H20 0 H angle 1045 1S 2 lone pairs of electrons RehybridiZeS sp3 3 Boron Consider boron trifluoride BF3 22 empty orbital 5 l B 2p 7 77 7777 i7i 2 2s 7 7 hybridize 2 74 74 iii p 1 s 7 77 Sp2gt 1s J i vacant p orbital can accept a pair of electrons Rotation about Bonds Isomers There is free rotation about the central C C bond of ethane the C H bonds on one carbon may be eclipsed or staggered with respect to one another H H H H H g H H H H CONFORMERS OF H H SAME MOLECULE Staggered Eclipsed Structures which differ only by rotations about single bonds are called conformations There cannot be rotation about the central double bond of ethylene this would cause the n bond to break hence there are no conformations of ethylene However substituents may be on the same or different side of the CC These are sometimes called geometric isomers or stereoisomers differ in the way their atoms are oriented in space H H STEREOISOMERS H DIFFERENT MOLECULES H3 3 H3 H3 cis2butene trans2 butene 23 When molecules have the same molecular formulas but different structures they are called structural or constitutional isomers CH3 CH3 Hac CHz CH2 CH2 CH3 H3C CH CH2 CH3 H3C C CH3 CH3 npentane Isopentane neopentane Dipole moments and the forces that attract molecules We have learned about the plarity of bonds and how to compute the dipole moments of moleculesHow can intermolecular forces attractions or repulsions help us to understand the 3 states of matter gas liquid solid Dipoledipole interactions cause attractions between polar molecules These forces must be overcome in melting or boiling For example CH3C OOOQ What about non polar molecules Why do larger molecules have higher BP and MP This is due to van der Waals forces These result from temporary dipole moments due to the uneven distribution of electrons CH3 CH3 H3CCH2 CH2 CH2 CH3 H3C CH cH2 CH3 Hac A CHa CH3 npentane isopentane b 28 neopentane bp 36 p bp 10 Hydrogen Bonding A hydrogen atom can participate in H bonding if it is bonded to O N F Since no molecules contain H F we consider only 0 H and N H O H O C worth about 5 kcalmole 24 CONSIDER HscCH2OH ch O CHg ethanOI dimethyl ether bp 78 C bp 25 C Consider also trimethylamine bp 35 C ethylmethylamine bp 37 C propylamine bp 49 C Polarity amp Solubility Like dissolves Like Four cases Polar solvent polar solute energy to break up polar solvent provided by solvation Nonpolar solvent polar solute solvent cannot solvate Nonolar solvent nonpolar solute weak solute solute interactions overcome by solvent solute Polar solvent nonpolar solute polar sovent does not want to break intermolecular bonds CLASSES OF ORGANIC COMPOUNDS Functional Groups HYDROCARBONS MOLECULES WITH MORE THAN JUST CH Hydrocarbons Alkanes alkenes alkynes arenes aromatic hydrocarbons 1 Alkanes contain only CH and only single bonds can be cyclic 2 Alkenes contain only CH and CC For example C is characteristic of Alkenes Ethylene CH2CH2 is the simplest alkene 25 Cholesterol is a alkene it is also an alcolhol HO Both ethylene and Cholesterol undergo similar reactions Le bromination 3 Alkynes contain only CH and CEC ethyne l butyne 2 butyne 4 Aromatic Hydrocarbons Benzene Note I the Carbonyl group C is very important found in 6 different fuotional groups Functional Groups 1 alkane CC no functionality I 2 alkene CC 3 alkyne CEC 4 Arene 26 5 Halide X XFCIBrI 6 Alcohol OH 7 Ether C 0 8 Amine CN l 9 NItrIIe ccEN I 10 NItro CN 039 1 Sulflde CSC P 12Sulfoxide 0 O s also written C c39 gt 13 Sulfone CS C 14 Thiol I 15 Aldehyde C C H 16 Ketone C C C 17 Carboxylic Acid C C OH 18 Ester 19 Amide I o I 3 20 Carboxylic c c c Acid Chloride I 9 9 I 21 Carboxylic CC O C C Acid Anhydride I These functional groups can be grouped into three general categories 1 Groups with CC multiple bonds 28 CC CEC Alkenes Alkynes Arenes 2 Groups with carbon singly bonded to an electronegative atom II I C X C OH C g C I II I II I I Alkyl halide Alcohol Ether l CI Nlt Hr Amine Thiol Sulfide 3 Groups with a CO carbonyl group I 9 I E I CI C CH C C C C C OH Aldehyde Ketone Carboxylic Acic I 3 I I CI 9 c c o c C C N CCC Ester Amide Acid chloride
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