Organic Chemistry I
Organic Chemistry I CEM 351
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Date Created: 09/19/15
Chapter 1 Structure and Bonding Chapter Outline 1 Atomic Structure Sections 11 13 A Introduction to atomic structure Section 11 1 ALAN Atoms consist of a dense positively charged nucleus surrounded by negatively charged electrons a The nucleus is made up of positively charged protons and uncharged neutrons b The nucleus contains most of the mass of the atom c Electrons move about the nucleus at a distance of about 10 10 m The atomic number Z gives the number of protons in the nucleus The mass number A gives the total number of protons and neutrons All atoms of a given element have the same value of Z a Atoms of a given element can have different values of A b Atoms of the same element with different values of A are called isotopes B Orbitals Section 12 l The distribution of electrons in an atom can be described by a wave equation a T1216 solution to a wave equation is an orbital represented by 1 b P predicts the volume of space in which an electron is likely to be found There are four different kinds of orbitals s p d f a The s orbitals are spherical b The p orbitals are dumbbellshaped c Four of the five d orbitals are Cloverleafshaped An atom39s electrons are organized into shells a The shells differ in the numbers and kinds of orbitals they have b Electrons in different orbitals have different energies 0 Each orbital can hold two electrons The two lowest energy electrons are in the 13 orbital a The 25 orbital is the next in energy Each p orbital has a region of zero density called a node b The next three orbitals are 2px 2py and 2pz which have the same energy C Electron Configuration Section 13 l 2 The groundstate electron configuration of an atom is a listing of the orbitals occupied by the electrons of the atom Rules for predicting the ground state electron configuration of an atom a Orbitals with the lowest energy levels are filled first The order of filling is ls 2s 2p 3s 3p 45 3d b Only two electrons can occupy each orbital and they must be of opposite spin 0 If two or more orbitals have the same energy one electron occupies each until all are halffull Hund39s rule Only then does a second electron occupy one of the orbitals All of the electrons in a halffilled shell have the same spin II Chemical Bonding Theory Sections 14 15 A Development of chemical bonding theory Section 14 1 2 Kekul and Couper proposed that carbon has four quotaffinity unitsquot carbon is tetravalent Other scientists suggested that carbon can form double bonds triple bonds and nngs 2 Chapter 3 4 1 Van39t Hoff and Le Bel proposed that the 4 atoms to which carbon forms bonds sit at the comers of a regular tetrahedron In a drawing of a tetrahedral carbon a wedged line represents a bond pointing toward the viewer and a dashed line points behind the plane of the page B Covalent bonds 1 Q Atoms bond together because the resulting compound is more stable than the individual atoms a Atoms tend to achieve the electron configuration of the nearest noble gas b Atoms in groups 1A 2A and 7A either lose electrons or gain electrons to form ionic compounds Atoms in the middle of the periodic table share electrons by forming covalent bonds The number of covalent bonds formed by an atom depends on the number of electrons it has and on the number it needs to achieve an octet C Covalent bonds can be represented two ways a In Lewis structures bonds are represented as pairs of dots b In linebond structures bonds are represented as lines drawn between two atoms 4 Valence electrons not used for bonding are called lonepair electrons C Val l 2 4 5 6 7 III Hybridiza A SP3 1 2 B sz 1 2 3 Lonepair electrons are represented as dots ence bond theory Section 15 Covalent bonds are fonned by the overlap of two atomic orbitals each of which contains one electron The two electrons have opposite spins Each of the bonded atoms retains its atomic orbitals but the electron pair of the overlapping orbitals is shared by both atoms The greater the orbital overlap the stronger the bond Bonds formed by the head on overlap of two atomic orbitals are cylindrically symmetrical and are called 0 bonds Bond strength is the measure of the amount of energy needed to break a bond Bond length is the optimum distance between nuclei Every bond has a characteristic bond length and bond strength tion Sections 16 110 Orbitals Sections 16 17 Structure of methane Section 16 a When carbon forms 4 bonds with hydrogen one 23 orbital and three 2p orbitals combine to form four equivalent atomic orbitals sp hybrid orbitals b These orbitals are tetrahedrally oriented c Because these orbitals are unsymmetrical they can form stronger bonds than unhybn39dized orbitals can d These bonds have a specific geometry and a bond angle of 1095 Structure of ethane Section 17 a Ethane has the same type of hybridization as occurs in methane b The C C bond is formed by overlap of two sp3 orbitals c Bond lengths strengths and angles are very close to those of methane Orbitals Section 18 If one carbon ZS orbital combines with two carbon 2p orbitals three hybrid 5p2 orbitals are formed and one p orbital remains unchanged The three Sp2 orbitals lie in a plane at angles of 120 and the p orbital is perpendicular to them Two different types of bonds form between two carbons a A 5 bond forms from the overlap of two sp2 orbitals b A 7 bond forms by sideways overlap of two p orbitals c This combination is known as a carboncarbon double bond 4 C sp 1 2 3 Structure and Bonding 3 Ethylene is composed of a carbon carbon double bond and four 0 bonds formed between the remaining four sz orbitals of carbon and the ls orbitals of hydrogen The double bond of ethylene is both shorter and stronger than the C C bond of ethane Orbitals Section 110 If one carbon 23 orbital combines with one carbon 2p orbital two hybrid Sp orbitals are formed and two p orbitals are unchanged The two sp orbitals are 180 apart and the two p orbitals are perpendicular to them and to each other Two different types of bonds form a A 6 bond forms from the overlap of two sp orbitals b Two at bonds form by sideways overlap of four p orbitals c This combination is known as a carbon carbon triple bond Acetylene is composed of a carbon carbon triple bond and two 6 bonds formed between the remaining two Sp orbitals of carbon and the Is orbitals of hydrogen The triple bond of acetylene is the strongest carbon carbon bond D Hybridization of nitrogen and oxygen Section 110 1 2 3 4 Covalent bonds between other elements can be described by using hybrid orbitals Both the nitrogen atom in ammonia and the oxygen atom in water form 3173 hybrid orbitals The lonepair electrons in these compounds occupy sp3 orbitals The bond angles between hydrogen and the central atom is often less than 109 because the lonepair electrons take up more room than the 0 bond Because of their positions in the third row phosphorus and sulfur can form more than the typical number of covalent bonds IV Molecular orbital theory Section 111 A Molecular orbitals arise from a mathematical combination of atomic orbitals and belong to the entire molecule 1 2 3 Two ls orbitals can combine in two different ways a The additive combination is a bonding MO and is lower in energy than the two hydrogen ls atomic orbitals b The subtractive combination is an antibonding MO and is higher in energy than the two hydrogen ls atomic orbitals A node is a region between nuclei where electrons aren t found If a node occurs between two nuclei the nuclei repel each other The number of MOs in a molecule is the same as the number of atomic orbitals combined V Chemical structures Section 112 A Drawing chemical structures 1 2 Condensed structures don39t show C H bonds and don t show the bonds between CH3 CH2 and CH units Skeletal structures are simpler still a Carbon atoms aren t usually shown b Hydrogen atoms bonded to carbon aren t usually shown c Other atoms are shown 4 11 Chapter 1 Solutions to Problems a To nd the groundstate electron configuration of an element first locate its atomic number For oxygen the atomic number is 8 oxygen thus has 8 protons and 8 electrons Next assign the electrons to the proper energy levels starting with the lowest level Fill each level completely before assigning electrons to a higher energy level Notice that the 2p electrons are in different orbitals According to Hand s rule we must place one electron into each orbital of the same energy level until all orbitals are half lled Zplii l 2s as Remember that only two electrons can occupy the same orbital and that they must be of opposite spin A different way to represent the groundstate electron configuration is to simply write down the occupied orbitals and to indicate the number of electrons in each orbital For example the electron configuration for oxygen is 1 232 2p4 Oxygen b Silicon with an atomic number of 14 has 14 electrons Assigning these to energy 3pfl 3s bass as we The more concise way to represent groundstate electron configuration for silicon ls2 232 2p6 3s2 3p2 Silicon c 152 232 2p6 332 3p4 wees as bass as us Sulfur 12 13 14 15 Structure and Bonding 5 Strategy The elements of the periodic table are organized into groups that are based on the number of outer shell electrons each element has For example an element in group 1A has one outershell electron and an element in group SA has five outershell electrons To find the number of outer shell electrons for a given element use the periodic table to locate its group Solution a Magnesium group 2A has two electrons in its outermost shell b Molybdenum is a transition metal which has two electrons in the 4s subshell plus four electrons in its 3d subshell c Selenium group 6A has six electrons in its outermost shell Strategy A solid line represents a bond lying in the plane of the page a wedged bond represents a bond pointing out of the plane of the page toward the viewer and a dashed bond represents a bond pointing behind the plane of the page Solution l C Chloroform Cl cu Cl H H 7 HKC Ci Ethane H H Strategy Identify the group of the central element to predict the number of covalent bonds the element can form Solution a Germanium Group 4A has four electrons in its valence shell and forms four bonds to achieve the noblegas configuration of neon A likely formula is GeCl4 Element Group Likely Formula b Al 3A A1H3 c C 4A CH2C12 d Si 4A SiF4 e N 5A CH3NH2 6 Chapter 1 1 6 Strategy Start by drawing the electron dot structure of the molecule 1 Determine the number of valence or outershell electrons for each atom in the molecule For chloroform we know that carbon has four valence electrons hydrogen has one valence electron and each chlorine has seven valence electrons 4 4x1 H 1x1 1 7 x 3 21 26 total valence electrons 2 Next use two electrons for each single bond H on CI CI 3 Finally use the remaining electrons to achieve an noble gas configuration for all atoms For a line bond structure replace the electron dots between two atoms with a line Solution Molecule Electrondot structure Linebond structure H H a CH013 il CE l b H28 HS H LTJ 8 valence electrons H H 1 H T T c CH3NH2 zc 1H Hec N H l4 valence electrons H H n T d CH3Li H 9 Li H c Lu 8 valence electrons H H 1 7 Each of the two carbons has 4 valence electrons Two electrons are used to form the carbon carbon bond and the 6 electrons that remain can form bonds with a maximum of 6 hydrogens Thus the formula C2H7 is not possible 18 19 110 111 Structure and Bonding 7 Strategy Connect the carbons and add hydrogens so that all carbons are bonded to four different atoms I I I Sp3lllH H C C C H H C C Propane I I I m3 H H H 0 H H39Hlsp3 109 HH HH HH H H H H H H H 6 6 6 I I I I I I C C C H H C C C C C C H Hexane l l l I I I I HH HH HH H H H H H H H H H H I H SP3 2 H C C C CSP Propene I 3 2 H lC H H lC H xspz H H The C3 H bonds are 0 bonds formed by overlap of an SP3 orbital of carbon 3 with an s orbital of hydrogen The C2 H and Cl H bonds are 0 bonds formed by overlap of an sp2 orbital of carbon with ans orbital of hydrogen The C2 C3 bond is a 0 bond formed by overlap of an sp3 orbital of carbon 3 with an sp2 orbital of carbon 2 There are two C1 C2 bonds One is a 0 bond formed by overlap of an sp2 orbital of carbon 1 with an Sp2 orbital of carbon 2 The other is a 1 bond formed by overlap of a p orbital of carbon 1 with a p orbital of carbon 2 All four atoms connected to the carbon carbon double bond lie in the same plane and all bond angles between these atoms are 120 The bond angle between hydrogen and the sp3hybridized carbon is 109 H H I I 2 3 2 1 v H SP C SP C All atoms he 1n the same plane and all bond 4CI sp22I3 sp2H angles are approximately 120 H H 13Butadiene 8 112 113 Chapter 1 Aspirin All carbons are sp2 hybridized with the exception of the C I indicated carbon All oxygen atoms have two lone pairs 0 H of electrons H HSp3SP Sp C C H l3 2 1 H Propyne The C3 H bonds are 0 bonds formed by overlap of an sp3 orbital of carbon 3 with an s orbital of hydrogen The Cl H bond is a 0 bond formed by overlap of an sp orbital of carbon 1 with an s orbital of hydrogen The C2 C3 bond is a 0 bond formed by overlap of an Sp orbital of carbon 2 with an sp3 orbital of carbon 3 There are three C1 C2 bonds One is a 0 bond formed by overlap of an Sp orbital of carbon 1 with an sp orbital of carbon 2 The other two bonds are 7 bonds formed by overlap of two p orbitals of carbon 1 with two p orbitals of carbon 2 The three carbon atoms of propyne lie in a straight line the bond angle is 180 The H C12C2 bond angle is also 180 The bond angle between hydrogen and the sp3 hybridized carbon is 109 3 H H H H I I 3 C C The sp hybr1dized oxygen atom H H has tetrahedral geometry b H301N CH3 Tetrahedral geometry at nitrogen and carbon H30 0 7 Like nitrogen phosphorus has P five outershell electrons PH H l H 3 H has tetrahedral geometry 115 Structure and Bonding d CID H30 CHZCHZCIDHCOH aso Strategy Remember that the end of a line represents a carbon atom with 3 hydrogens a two way intersection represents a carbon atom with 2 hydrogens a threeway intersection represents a carbon with 1 hydrogen and a fourway intersection represents a carbon with no hydrogens The sp3hybridized sulfur atom has tetrahedral geometry Solution a b H OH OH 1H 3H 0 Ho1H NHCH3 0H 2H OH gt HO 1H 1H Adrenaline CgH13NO3 Estrone C18H2202 1 1 6 Several possible skeletal structures can satisfy each molecular formula 3 CSH12 M l 7l 1 C2H7N NH2 NH C C3HBO 7 0 WOH NOH O O O O OH A A D U OH 1 C4H9C39 M Cl Al kCl 4 10 Chapter 1 117 OH PABA H2N Visualizing Chemistry 118 a H H H C H H C N Cx CH2CH2CH3 H I H H I I i H H C C CH3 H H CBH17N c 1N H I H H H b i H INHZ H N H 39 I I o o H C C C O H H30 ff H I I I II O H H 0 C3H7N02 119 Citn39c acid C6H807 contains seven oxygen atoms each of which has two electron lone pairs Three of the oxygens form double bonds with carbon Citric acid Structure and Bonding 11 1 2 0 H H 0 0 H I r H 9 Clo 9 N a H C 9 061 H H O H Acetaminophen All carbons are Sp2 hybridized except for the carbon indicated as sp3 The two oxygen atoms and the nitrogen atom have lone pair electrons as shown 1 2 1 CH3 H l I H H H l l H H0 I C c C C C IN I H C C H I l O H H Aspartame HZN H 0 Additional Problems 1 2 2 Atomic Number of Element Number valence electrons a Zinc 30 2 b Iodine 53 7 0 Silicon 14 4 1 Iron 26 2 in 45 subshell 6 in 3d subshell 1 2 3 Atomic Ground state Element Number electron con guration a Potassium 19 132 2s2 2p6 3s2 3p6 481 b Arsenic 33 1s 2s2 2p6 352 3p64s2 3d 4 c Aluminum 13 15 239quot 2p6 3s2 3p d Germanium 32 ls2 2s 2p6 332 3p6 432 3d 10 4p2 124 a NH20H b AlCl3 0 313202 d CHQO 12 125 126 127 128 129 Chapter 1 H HICICIII N H Acetonitrile In the compound acetonitn39le nitrogen has eight electrons in its outer electron shell Six are used in the carbonnitrogen triple bond and two are a nonbonding electron pair The H3C carbon is Sp3 hybridized and the CN carbon is sp hybridized H iEI CC Vinyl chloride H H Vinyl chloride has 18 valence electrons Eight electrons are used for 4 single bonds 4 electrons are used in the carbon carbon double bond and 6 electrons are in the 3 lone pairs that surround chlorine a b 0 C 39s39 CH3 H30 quot s H30 In molecular formulas of organic molecules carbon is listed first followed by hydrogen All other elements are listed in alphabetical order Compound Molecular Formula 3 Aspirin C9H804 b Vitamin C C6H306 c Nicotine C10H14N2 d Glucose C6H1206 Structure and Bonding 13 1 30 To work a problem of this sort you must examine all possible structures consistent with the rules of valence You must systematically consider all possible attachments including those that have branches rings and multiple bonds Wm M 7 T i H C3 l3 H H cls N H H o p o H H o o p H H H H H H H H H W i T T M i i 8 H H H H H H H W if T H H H C C H H o o H o o I l H H H H TM ii iii ii i f f i l i H l f H H N H N H H H H H H c H H H H H HH H 131 a b 3 C 1 SP3 SP3 SP3 Hg SP2 SP2 Sp 5 pl CH3CH2CH3 3 sp2 Sp HZCCH CECH SP3 CSP2 CCH2 CH3 OH H3C sp3 132 H H H C C H Benzene c All carbon atoms of benzene are sp2 hybridized and all bond angles of benzene are 120 Benzene is a planar molecule 14 133 134 135 136 Chapter 1 a O 2 inf120 0 NH2 Glycine a NH 3 0H 3 H l H i 1 H l H 3 H 0 H l H 1 H 1 0st 0 H C10H11N Examples 21 CH3CH2CH CH2 M b HZCCH CH CH2 2 C H fo H R 0 120 C H 0 109 C I ll 3 H Cf H H1 OH H Pyridine Lactic acid C d O 0 Cl 0 6 OH H 3H 2H 1 m 02H 1H 1H 2H OH OH 0H 2H 0 1H 2H T 1H 2H T 1H OH OH C11H113r02 C9H120 c H2020H c ECH 1 3 7 a The 4 valence electrons of carbon can form bonds with a maximum of 4 hydrogens Thus it is not possible for the compound CH5 to exist b If you try to draw a molecule with the formula C2H6N you will see that it is impossible for both carbons and nitrogen to have a complete octet of electrons Therefore C2116N is unlikely to exist 0 A compound with the formula C3H5Br2 doesn39t have filled outer shells for all atoms and is thus unlikely to exist Structure and Bonding 15 138 3 H C H Ethanol OH 139 H O H O O R R H H O C C C C O H H O C C C C H I O H O O H II H O H C CIE C C O H H 0 140 a b c d H H T T T T T T T H quotH C CEN H C C O C H H C C C C H H H H C l I I C C quot39 H H H H H H H H CC H I H H 141 T H IDO39 K All other bonds are covalent H ionic 142 a SP3 SP3 b sp3 H O CH CH CH OH sp2 Isp2 3122 SP3 SP3 39 2 3 SP EH3 2 HCCCSgOCH2CH2IIH Cl HOCspOspO spzcl 522 02392 05 13 HOCZ 2 HZN chz H sp sp SP 53 CSP IL HO OH Procaine Vitamin C 16 Chapter 1 143 Pyridoxal phosphate The bond angles formed by atoms having sp3hyb1139dization are approximately 109 The bond angles formed by atoms having spzhybridization are approximately 120 1 44 In a compound containing a carbon carbon triple bond atoms bonded to the sphybridized carbons must lie in a straight line It is not possible to form a ve membered ring if four carbons must have a linear relationship 1 4 5 The n bonding molecular orbital in ethylene results from the combination of two p atomic orbitals with the same algebraic sign A second molecular orbital can form by the combination of two p orbitals with opposite algebraic signs but it is an antibonding orbital 146 rtbonds Sp Sp2 039 bond The central carbon of allene forms two 0 bonds and two 1r bonds The central carbon is sp hybn39dized and the two terminal carbons are spZhybridized The bond angle formed by the three carbons is 180 indicating linear geometry for the carbons of allene Structure and Bonding 147 nbonds Carbon dioxide is a linear molecule 148 0 H CH3 H3CquotC N N C I H 0 H Caffeine c CH3 All of the indicated atoms are spzhybridized 17 1 4 9 a The positively charged carbon atom is surrounded by six valence electrons carbon has three valence electrons and each hydrogen brings three valence electrons b The positively charged carbon is sp hybridized c A carbocation is planar about the positively charged carbon 150 9 7 H H a A carbanion is isoelectronic with has the same number of electrons as a trivalent nitrogen compound b The negatively charged carbanion carbon has eight valence electrons c The carbon atom is sp3hybridized d A carbanion is tetrahedral 18 151 Chapter 1 According to the Pauli Exclusion Principle two electrons in the same orbital must have opposite spins Thus the two electrons of triplet spinunpaired methylene must occupy different orbitals In triplet methylene sphybn39dized carbon forms one bond to each of two hydrogens Each of the two unpaired electrons occupies a p orbital In singlet spinpaired methylene the two electrons can occupy the same orbital because they have opposite spins 5 Including the two C H bonds there are a total of three occupied orbitals We predict Sp 152 153 154 155 hybridization and planar geometry for singlet methylene p vacant p orbital 99 eggH 120056 Triplet methylene Singlet methylene linear planar as CH3CHQCH2CH3 CH3CHCH3 The two compounds differ in the way that the carbon atoms are connected H Hc c H H H20 CH CH3 One compound has a double bond and one has a ring CH3CH20H CH3OCH3 The two compounds differ in the location of the oxygen atom CH3 H C C 2 CH3 CH3CHZCHCH2 CH3CH CHCH3 The compounds differ in the way that the carbon atoms are connected and in the location of the double bond