Organic Chemistry I
Organic Chemistry I CHEM 2311
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Last Name First Name Chemistry 2311 Homework Problems Chapter 1 1 For each of the following arrangements of atoms a draw a correct Lewis structure and b indicate the formal charge if any associated with each atom H on 9 l 9 e we SD 80 0r quotp sl 0 H c l I 0396 9 6 9 6 H quot sulfite dianion acetate amon 9 n a e 39 39 H NN v 9 hydrazoic acid nitrite anion H quot039 H CC I H H acetaldehyde enolate anion allyl cation Contal s CC PI H H HCC CQH C I H CC C H H H phenol methyl carbocation 2 Draw a constitutional isomer for each of the following using the same style as used for the starting molecule CIH C H H30 CI H30 H H30 H 2 c c c c c c C C H H H H CI H H O OH OH H c o o H3C C H2TC 2 H CH H3C C CH2 HzC 20 2 CH3 E OH H OH H OH Hzc c Hsc c H30 C H2 H H H H H H OH H A H 6 H C C H HI C Cquot Hyc C H H H H H o e H C OH cumH H Hl H H H H H NH2 H H CCIH C H CN H H H quot H CH30HZCCIZCHZCH3 These are pentane derived There are also branched chain isomers CH3CC12CH2CH2CH3 C12CHCH2CH2CH2CH3 CH3CHC1CHClCH2CH3 CICHZCHZCHCICHZCH3 CICHZCHClCHZCHZCH3 C1CH2CH2CH2CHC1CH3 ClCHZCHZCHZCHZCHZCI 3 Write a contributing resonance structures for each of the following using bondline formulas and indicate whether is will contribute equally more or less than the given structure circle Use curved arrows to show the movement of electron pairs required to generate the resonance structure phenoxide M pentadienyl cation 39 25H C O H30 acetic acid equal less 7VQ more because more highly substituted carboc ations are more stable equal no6 no6 O 0 39 C OH C SH 9 H3 H30 less Ag Elm 059 D99 Emu Far Last Name First Name Roster Noincorrect roster number Chemistry 2311 Homework Problems Chapter 8 No grade Use ChemBioDraw software to draw the requested structures within the red rectangles provided Select and then copy the red rectangle for the question and paste it into the ChemBioDraw window Draw the requested structures and then copy the rectangle and its contents in ChemBioDraw and paste them back into MS Word Answer sheet to be printed out one side of page oan only front side of pages will be graded You may collaborate on the development of strategies to answer to these problems HOWEVER all work using must be completed in dependently of others You cannot exchange electronic les with other students or cutandpaste from other students Violators of this instruction will receive a zero for the assignment anal be referred to the Dean of Students for further sanction 1 A B e 9 Na C D anti addition Br sH 3 CH3 rotate 60 degI 1 Li HNEtz about c2 cs 2H H3O H Br anticoplanarHBr dehydrohalogenate meso E Br H3C CH3 Z 2 bromo 2 butene t BuOK CHCI3 HO HO Et HO u H HO H Et Et meso form 3 B H3 H30 H04quot OH Br 39 Br raccmic form racemic form D H C OH 32 H CH3 HO 1 OH CH3 racemic form Last Name First Name Roster Noincorrect roster number No grade Chemistry 2311 Homework Problems Chapter 6 Use MSWord and0r ChemBioDraw softwares to answer all questions except where indicated For structures select and then copy the red rectangle for the question and paste it into the ChemBioDraw window Draw the requested structures and then copy the rectangle and its contents in ChemBioDraw and paste them back into MS Word Answer sheet to be printed out one side of page only only front side of pages will be graded You may collaborate on the development of strategies to answer to these problems HOWEVER all work using must be completed independently of others You cannot exchange electronic les with other students or cutand paste from other students Violators of this instruction will receive a zero for the assignment and be referred to the Dean of Students for further sanction l d 2 a 3 d 4 b 5 b 6 c 7 c 8 c 9 c 10 a 11 a b U I CI 10 reacts faster in 8N2 c d inversion in SNZ reaction CF3 Last Name First Name Roster Noincorrect roster number Chemistry 2311 Homework Problems Chapter 7 No grade Use ChemBioDraw software to draw the requested structures within the red rectangles provided Select and then copy the red rectangle for the question and paste it into the ChemBioDraw window Draw the requested structures and then copy the rectangle and its contents in ChemBioDraw and paste them back into MS Word Answer sheet to be printed out one side of page oan only front side of pages will be graded You may collaborate on the development of strategies to answer to these problems HOWEVER all work using must be completed in dependently of others You cannot exchange electronic les with other students or cutandpaste from other students Violators of this instruction will receive a zero for the assignment anal be referred to the Dean of Students for further sanction l a E 2methylbut2enlol b E3chlor0pent 2ene c 3R6R 36dimethylcyclohex lene if not named as chiral compound then this would be trans36 dimethylcycloheX l ene d RZ2chlor05 u0r03methylheX2ene e ZZ4Ehepta24diene could also be named cistranshepta24diene f 5R7RZ57dimethyln0n 3ene a b c d 3 Alkene stability order IVgtIIgtHIgtI 4 a b 2 unhindered strong base hindered strong base 0 d H H e f this is obviously not a productive synthetic scheme this is no different from 4a g h 5 Fourmembered ring product Fourmembered ring product 9 Fourmembered ring product Fourmembered ring product 9 gt e gt e F ivemembered ring product Fivemembered ring product OeAQ major product Fivemembered ring product h There is a 4Lh possible 5membered prod gooa m When Iwas developing this problem I basically forgot about the exo product because I knew it would be formed in very small amount relative to the endo products But it is de nitely a possible minor product All of the fouimembered ring compounds will be present in very small amounts relative to the major product 6 An anticoplanar arrangement of the bromine and proton on the adjacent carbon can only be achieved for a conformation in which the phenyl groups are also anti Thus only the E product will be produced Br Ph Ph CH3 Ph gt H Ph rs H30 Ph H quot I H 7 a H30 Br Brz cCH2 gt Br H Br 9 9 3 NaNHz llq NH3 c C Na Br 2 NaBr 2 NH3 9 6 Br C C Na H3CTC3H7 NaBr Last Name First Name Roster Noincorrect roster number Chemistry 2311 Homework Problems Chapter 3 No grade Use ChemBioDraw software to draw the requested structures within the red rectangles provided Select and then copy the red rectangle for the question and paste it into the ChemBioDraw window Draw the requested structures and then copy the rectangle and its contents in ChemBioDraw and paste them back into MS Word Answer sheet to be printed out one side of page oan only front side of pages will be graded You may collaborate on the development of strategies to answer to these problems HOWEVER all work using must be completed in dependently of others You cannot exchange electronic les with other students or cutandpaste from other students Violators of this instruction will receive a zero for the assignment and be referred to the Dean of Students for further sanction l a H TCH O H gt H20 NH3 conjugate acid conjugate base b H H3C NiHLC gt H3C NH3 Ce H conjugate acid conjugate base 2 a Methanol acting as a base H3C OH H2804 gt H3C OH2 HSO439 Any acid with a pKa smaller than that of CH3OHZJr 25 Will work conjugate acid conjugate base b Methanol acting as an acid H3COH NaH a H2 Na CH3O39 Any base stronger than the conjugate base of methanol w1ll work conjugate acid conjugate base 3 Bases from list strong enough to deprotonate propanoic acid so that Kgtl aq ammonia aq sodium hydroxide 3 Bases from list strong enough to deprotonate acetylene so that Kgtl none 3 Bases from list strong enough to deprotonate hydrochloric acid so that Kgtl all 3 Bases from list strong enough to deprotonate t butanol so that Kgtl none 3 Bases from list strong enough to deprotonate diisopropyl amine so that K gt1 NaH 4 a Conjugate acid 4 b Conjugate base f HgH 5 a products of acid base equilibrium CH3NH3 CH3COz K Will be gt1 5 b products of acid base equilibrium HCECH NH239 KWiII be lt1 5 c products of acid base equilibrium CHgCOCHz39 H20 K Will be lt1 5 d products of acid base equilibrium H20N02 H804 KWiII be gt1 6 Use ChemBioDraw to draw curved arrows indicating movement of electrons or you may handdraw them H H H r e H20c H Br HZC C CHs Br CH quot 3 H H e T HZC g CH3 Br gt H2 c CH3 I B 7 a Resonance structures of Oprotonated DMF 35 b39H b Resonance structures of N protonated DMF There are no others c Most stable protonated form a or b and why A is the most stable form because of the resonance stabilization Last Name First Name Roster Noincorrect roster number No grade Chemistry 2311 Homework Problems Chapter 5 Use MSWord and0r ChemBioDraw softwares to answer all questions except where indicated For structures select and then copy the red rectangle for the question and paste it into the ChemBioDraw window Draw the requested structures and then copy the rectangle and its contents in ChemBioDraw and paste them back into MS Word Answer sheet to be printed out one side of page only only front side of pages will be graded You may collaborate on the development of strategies to answer to these problems HOWEVER all work using must be completed independently of oth ers You cannot exchange electronic les with other students or cutandpaste from other students Violators of this instruction will receive a zero for the assignment anal be referred to the Dean of Students for further sanction l 2 a RS designation 2 b RS designation 2 c RS designation b c amp d are all S S R R 2 d RS designation 3 RS designation of i 3 RS designation of ii 3 RS designation of iii S S S R 4a l a cl 3 i 4 I n I 39 I I I I 39 l I I I I I I I I I I I I I I I OH 5 HO OH E HO enantiomers enantiomers ORGANIC CHEMISTRY I CHAPTER 3 Reaction Mechanism a description ofthe events that take place on a molecular level as reactants become products Intermediates the chemical species that intervene between each step in a multistep reaction 31 Reactions and Their Mechanisms 0 Virtually all organic reactions fall into one offour categories substitutions additions eliminations or rearrangements o Substitutions are the characteristic reactions of saturated compounds such as alkanes and alkyl halides and of aromatic compounds even though they are unsaturated o In a substitution one group replaces another 0 A Substitution Reaction 1120 o H3C Cl NaOH gt H3C OH Na Cl 0 Additions are characteristic of compounds with multiple bonds 0 In an addition allparts of the adding reagent appearin the product two molecules becomeone o AnAddition Reaction H l CC14 H H C H H c Br Br gt l Br C C H H H Br 0 Eliminations are the opposite of additions o In an elimination one molecules loses the elements ofanother small molecule 0 Give us a method for preparing compounds with double and triple bonds 0 An alkyl halide becomes an alkene 0 An elimination reaction H KOH T H C C H gt C l Br HBr H Cl 0 In a rearrangementa molecule undergoes a reorganization ofits constituent parts 0 A rearrangement CHa H H acid cat I cc gt H300GCH3 Hac c H l on H30 CH3 0 In this rearrangement not only haverthe positions ofthe double bond and a hydrogen atom changed but a methyl group has moved from one carbon to another 31A Homolysis and Heterolysis of Covalent Bonds 0 Reactions of organic compounds always involve the making and breaking of covalent bonds A covalent bond may break in two fundamentally different ways Heterolysis when a bond breaks such that one fragment takes away both electrons ofthe bond leaving the other fragment with an empty orbital o Produces charged fragments or ions and its termed an ionic reaction The broken bond is said to have cleaved heterolytically Heterolytic bond cleavage o A B aA B39lons Homolysis when a bond breaks so that each fragment takes away one ofthe electrons ofthe bond Homolytic Bond Cleavage o AiBaA B Radicals Heterolysis ofa bond normally requires that the bond be polarized o 5AB395aA B Even with a highly polarized bond heterolysis rarely occurs without assistance The reason Heterolysis requires separation ofoppositely charged ions This requires a lot of energy Often it is assisted by a molecule with an unshared pairthat can form a bond to one ofthe atoms 0 Y395AB539JY A B39 32 AcidBase Reactions 0 Many ofthe reactions that occur in organic chemistry are either acidbase reactions themselves or they involve an acid base reaction at some stage 0 Acid base reactions are simple fundamental reactions that will enable you to see how chemists use curved arrows to represent mechanisms of reactions and how they depict the processes of bond breaking and bond making that occur as molecules react 22A Bronsted Lowrv Acids and Bases o Involve the transfer of protons o Bronsted Lowry Acid a substance that can donate or lose a proton o Bronsted Lowry Base a substance that can accept or remove a proton H 5 I gt HquotH H C39 HH 2C1 Base Add Conjugate Conjugate proton P1390t011 Acid Base acceptor donor of H20 of HCl 0 Just as we classified the reactants as either an acid or a base we also classifythe products in a specific way 0 The molecule or ion that forms when an acid loses its proton is called the conjugate base ofthat acid 0 The molecule or ion that forms when a base accepts a proton is called the conjugate acid 328 Acids and Bases in Water 0 Hydronium ion is the strongest acid that can exist in water to any significant extent any stronger acid will simply transfer its proton to a water molecule to form hydronium ions Hydroxide ion is the strongest base that can exist in water to any significant extent Any base stronger than hydroxide will remove a proton from water to form hydroxide ions 0 Spectacular Ions ions that play no part in the acid base reaction 0 Total Ionic Reaction u n 91 Na O H gt 2H 039 Na c139 H H Spectator Ions 0 Net Reaction H O H u H O H H gt 2 H 39o 39 H 33 Lewis Acids and Bases o Acids are electron pair acceptors o Bases are electron pair donors Any electronde cient atom can act as a Lewis acid Many compounds containing group lllA elements such as boron and aluminum are Lewis acids because group lllA atoms have only a sextet of electrons in their outer shell Many other compounds that have atoms with vacant orbitals also act as Lewis acids Zinc and iron Ill halides ferric halides are frequently used as Lewis acids in organic reactions IIBi Br lt Ie Br Br Fe gt I l3r Fe l3lr B I r Br B 39 o c 39 I r Lewis base Lewis acid 33A Opposite Charges Attract o In Lewis acid base theory as in many organic reactions the attraction of oppositely charged species is fundamental to reactivity o Electrostatic Potential the convention in these structures is that blue represents relatively positive areas and red represents relatively negative areas 34 Heterolysis of Bonds to Carbon Carbocations and Carbanions o Heterosis Ofa bond to a carbon atom can lead to either oftwo ions either an ion with a positive charge on the carbon atom called a carbocation or an ion with a negatively charged carbon atom called a carbanion o carbocations are electron deficient They have only six electrons in their valence shell and because ofthis carbocations are Lewis Acids o Carbocations are electron rich They are anions and have an unshaded electron pair Carbanions therefore are Lewis bases and react accordingly e JG 3926 heterolysm C l l Carbocation 9 9229 heterolysis C O leg I l Carbanion 34A Electrophiles and Nucleophiles becausecarbocations are electronseeking reagents chemists call them electrophiles meaning electron loving All Lewis acids are electrophiles Carbon atoms that are electron poor because of bond polarity but are not carbocations can also be electrophiles Carboanionsare Lewis bases Carboanions seek a proton or some other positive center to which they can donate their electron pair and thereby neutralize their negative charge A Nucleophiles is a Lewis base that seeks a positive center such as positively charged carbon atom Electrophiles are also Lewis acids and Nucleophiles are Lewis bases Lewis acids and lewis base are terms that are used generally but when one or the other reacts to form a bond to a carbon atom we usually call it an electrophile or a nucleophile 35 How to Use Cuned Arrows in Illustrating Reactions Up to this point we have not indicated how bonding changes occur in the reactions we have presented but this can easily be done using curvedarrow notation Curved Arrows Show the direction of electron flow in a reaction mechanism 0 0 Point from e source of an electron pair to the atom receiving the pair 0 Always show the flow of electrons from a site of higher electron density to lower electron density 0 Never show the movement of atoms Atoms are assumed to follow the flow ofthe electrons Example Reaction ofWater with Hydrogen Chloride The Use of Curved Arrows p107 o The curved arrow begins with a covalent bond or unshared electron pair a site of higher electron density endpoints toward a site of electron deficiency 0 Solved Problem 32 p 108 36 The Strength of BronstedLowr Acids and Bases Ka and 2KB 0 Acid strength is characterized in terms of acidity constant Ka or pKa values 36A The Acidity Constant Kg 0 Using a generalized hypothetical acid HA the action in water is 0 HA H20 ltgt H3O Aquot 0 And the expression for the acidity constant is O Ka 2 H3 EEK o A large value of K3 means the acid is a strong acid and a small value of K3 means the acid is a weak acid 368 Acidiy and pKa o Chemists usually express the acidity constant Ka as it39s negative logarithm pKa o PKa 09Ka o This is the analogous to expressing the hydronium ion concentration as pH 0 pH logH3O o The larger the value ofe pKa the weaker is the acid o It is understood that a positive pKa is largerthan a negative pKa o The Ka for water o a Luz o The pKa for water 157 26C Predicting the Strength of Bases o The stronger the acid the weaker will be its conjugate base 0 We can therefore relate the strength ofa base to the pKa of its conjugate acid 0 The larger the pKa of the conjugate acid the stronger is the base 32 How to Predict the Outcome of Acid Base Reactions 0 Acid base reactions always favor the formation of the weaker acid and the weaker base Acid base reactions are said to be under equilibriumcontrol and reactions under equilibrium control always favor the formation ofthe most stable lowest potential energy species The weaker acid and weaker base are more stable lower potential energythan the stronger acid and stronger base 0 P 113 27A Water Solubilitv as the Result of Salt Formation o Water insoluble carboxylic acids dissolve in aqueous sodium hydroxide they do so by reacting to form water soluble sodium salts Icl 25 c l om e n o u NaH H o H Na 2 O H 39 Insoluble in Water Soluble in Water due to its polarity as a salt 0 We can also predict that an amine will react with aqueous hydrochloric acid in the following way H e G R N HCOG H C1 gt R lll H C1 39o H H Stronger Base Stronger Acid Weaker Acid weaker Base pKa 174 pKa 910 28 quot 39 39 39 39 between t trurture and Aciditv 0 When we compare compounds in a single column ofthe periodic table the strength ofthe bond to the proton is the dominating effect 0 Bond strength to the proton decreases as we move down the column increasing its acidity o Proton acidity increases as we descend a column in the periodic table due to decreasing bond strength to the proton Comparing the hydrogen halides with each other H F is the weakest acid and H l is the strongest This follows from the fact that the H F bond is byfar the strongest and the H l bond is the weakest Acidity increases from left to right when we compare compounds in a given row of the periodic table o Proton acidity increases from left to right in a given row of the periodic table due to increasing stability of the conjugate base 0 The electronegativity ofthe atom in question affects acidity in two ways It affects the polarity ofthe bond to the proton and it affects the relative stability ofthe anion conjugate base that forms when the proton is lost 38A The Effect of Hybridization o The protons of ethyne are more acidic than those of ethane which in turn are more acidic than those of ethane Ethyne m Ethane pKa 25 pKa 44 pKa 50 0 We can explain this order of acidities on the basis ofthe hybridization state of carbon in each compound 0 Electrons of 25 orbitals have lower energy that those of 2p orbitals because electrons in 25 orbitals tend on the average to be much close to the nucleus than electrons in 2p orbitals With hybrid orbitals having more 5 character means that the electrons of O the anion will on average be lower in energy and the anion will be more stable The sp orbitals ofthe C H bonds of ethyne have 50 s character because 0 they arise from the combination of one s orbital with onep orbital the sp2 orbitals of ethene have 333 5 character and those ofthe sp3 orbitals of ethane have only 25 5 character 0 Relative Acidity ofthe Hydrocarbons HC E CH gt HZC CEgt H3C CH3 0 Relative Basicity of the Carbanions H3C CH2 gt HZC CH gt HC E C 388 Inductive Effects 0 The C C bond of ethane is completely nonpolar because at each end ofthe bond there are two equivalent methyl groups Inductive Effects are electronic effects transmitted through bonds The inductive effect of a group can be electron donating or electron withdrawing Inductive effects weaken as the distance from the group increases 39 Energy Changes 0 Energy the capacity to do work 0 The two fundamental types of energy are kinetic energy and potential energy 0 Kinetic energy is the energy an object has because of its motion 1 o Ek Emu2 0 Potential energy is stored energy It exists only when an attractive or repulsive force exists between objects 0 Chemical Energy is a form of potential energy It exists because attractive and repulsive eectrica forces exist between different pieces ofthe molecules 0 We usually think in terms of relative potential energy We say that one system has more or less potential energy than another 0 Stability or relative stability the relative stability ofa system is inversely related to its relative potential energy 0 The more potential energy an object has the less stable it is 233A Potential Energy and Covalent Bonds 0 Atoms and molecules possess potentia energy often called chemical energy that can be released as heat when they react Because heat is associated with molecular motion this release of heat results from a change from potential to kinetic energy 0 A convenient way to represent the relative potentia energies of molecules is in terms oftheir reative enthalpies or heat contents H o Enthalpy change the difference in relative enthalpies of reactants and products in a chemical change symbolized by AH o By convention the sign ofAH for exothermic reactions those evolving heat is negative 0 Endothermic reactions those that absorb heat have a positive AH o The heat of reaction AH measures the change in enthapy ofthe atoms ofthe reactants as they are converted to products 210 The Relationshin between the Equilibrium Constant and the quot 39 FreeEnerqv ChangeI AG 0 An important relationship exists between the eq constant Keq and the standard free energy change AH for a reaction 0 AG RT lnE Keq o R is the gas constant and 8314J K391 mol391 and T is the absolute temperature in kelvins K o This equation tells us the following o For a reaction to favor the formation of products when equilibrium is reached it must have a negative value for AG Free energy must be lost as the reactants become products I lfAG is more negative than 13 kJ mol391 the eq constant will be large enough for the reaction to go into completion meaning that more than 99 ofthe reactants will be converted to products when eq is reached 0 For reactions with a positiveAG the formation of products at eq is unfavorable I The eq constant for these reactions will be less than one o The freeenergy change AG has two components the enthalpy change AH and the entropy change AS o AG AH TAS o A negative value for AH will contribute to making AG negative and will consequently favor the formation of products 0 Forthe ionization of an acid the less positive or more negative the value of AH the strongerthe acid will be 0 Entropy changes have to do with changes in the relative order ofa system 0 The more random a system is the greater is its entropy o A positive entropy change from order to disorder makes a negative contribution to AG and is energetically favorable for the formation of products 211 The Aciditv of Carboxvlic Acids o Carboxylic acids are weak acids typically having pKa values in the rage of3 5 o Alcohols by comparison have pKa values in the range of15 18 and essentially do not give up a proton unless exposed to a very strong base 0 Acetic acid and ethanol representative examples of simple carboxylic acids and alcohols O A H OH Acetic Acid Ethanol pKa 475 pKa 16 G 27 kJmol G 908 kJmol o How do we explain the much greater acidity of carboxylic acids than alcohols Considerfirst the structural changes that occur if both acetic acid and ethanol act as acids by donation a proton to water 0 Acetic Acid Acting as an Acid 0 0 H A H o G 0 H quot H 07 0 H H Acetic Acid Water Acetate Ion Hydromum on o Ethanol Acting as an Acid 9 H A A A ltz OH H a a H 39 H 1511131101 Water Ethoxide Ion HydI39ODillm 1011 311A The Effect of Delocalization o Delocalization ofthe negative charge is possible in a carboxylate anion but it is not possible in an alkoxide ion We can show how delocalization is possible in carboxylate ions by writing resonance structures for the acetate ion e 0 lt gt 9 f J Resonance stabilization in acetate ion The structures are equivalent and there is no requirement for charge separation o The two resonance structures we drew above distributed the negative charge to both oxygens atoms ofthe carboxylate group thereby stabilizing the charge This is a delocalization effect by resonance In contrast no resonance structures are possible for an alkoxide ion A rule to keep in mind is that charge delocalization is always a stabilizing factor and because of charge stabilization the energy difference for formation of a carboxylate ion from a carboxylic acid is less than the energy difference for formation of an alkoxide ion from an alcohol 3118 The Inductive Effect 0 The electronegativity ofthese oxygen atoms further helps to stabilize the charge by what is called inductive electronwithdrawing effect 311C Summam and Comparison of Con39ugate Acid Base Strengths o the greater acidity of a carboxylic acid is predominantly due to the ability of its conjugate base a carboxylate ionto stabilize a negative charge better than an alkoxide ion the conjugate base of an alcohol In other words the conjugate base of a carboxylic acid is a weaker base than the conjugate base of an alcohol Therefore since there is an inverse strength relationship between an acid and its conjugate base a carboxylic acid is a stronger acid than an alcohol 211D Inductive Effects of Other Groups 0 The acidstrengthening effect of other electronattracting groups other than carbonyl group can be shown by comparing the acidities of acetic acid and chloroacetic acid 0 0 Cl OH OH pKa 475 pKa 286 o This is an example ofsubstituent effect By adding its inductive effect to that ofthe carbonyl group and the oxygen it makes the hydroxyl proton of chloroacetic acid even more positive than that of acetic acid It also stabilizes the chloroacetate ion that is formed when the proton is lost by dispersing its negative charge 0 Any factor that stabilizes the conjugate base of an acid increases the strength of the acid 212 The Effect of the Solvent on Aciditv o In the absence of a solvent ie in the gas phase most acids are far weaker than they are in solution 0 In the absence of a solvent separation is difficult In solution solvent molecules surround the ions insulating them from one another stabilizing them and making it far easier to separate them than in the gas phase 0 In a solvent such as water called protic solvent solvation by hydrogen bonding is important 0 A protic solvent is one that has a hydrogen atom attached to a strongly electronegative element such as oxygen or nitrogen o Solvation of any species decreases the entropy ofthe solvent because the solvent molecules become much more ordered as they surround molecules of the solute 313 Organic Compounds as Bases If an organic compound contains an atom with an unshared election pair it is a potential base The conjugate acid ofthe alcohol is often called a protonated alcohol although more formally it is called an alkyloxonium ion or simply an oxonium ion Alcohols in general undergo this same reaction when they are treated with solutions of strong acids such as HCl HBr HI and H2504 39 rquot quot H CIJ A gt R Te H 19 H H Alcohol Strong Acid Alkyloxonium Ion Weak Base So too do ethers r R Cl A gt R O H A6 I B R R Ether Strong Acid DiaJkyloxonium Ion Weak Base Compounds containing a carbonyl group also act as bases in the presence of a strong acid 0 n 1 0 H e H A gt Rc A R R R Ketone Strong Acid Protonated Ketone Weak Base Proton transfer reactions like there are often the first step in many reactions that alcohols ethers aldehydes ketones esters amides and carboxylic acids undergo An atom with an unshared electron pair is not the only locus that confers basicity on an organic compound The 11 bond of an alkene can have the same effect As a first step alkenes react with a strong acid by accepting a proton in the following way The pi bond breaks This bond breaks This bond is formed H mgg 9cf 9 0 0 H H Alkene Strong Acid Carbocation Weak Base 0 In this reaction the election pair ofthe 11 bond ofthe alkene is used to form a bond between one carbon ofthe alkene and the proton donated by the strong acid 0 This process leaves the other carbon ofthe alkene trivalent electron deficient and with a formal positive charge 0 A compound containing a carbon ofthis type is called a carbocation an unstable intermediate that reacts further to produce a stable molecule 314 A Mechanism for an Organic Reaction 0 Pages 127129 211 Acids and Bases in nuinnlt o lfyou were to add sodium aminde NaNH2 to water in an attempt to carry out a reaction using the amide ion NH2 as a very powerful base the amide ion would react with waterto produce a solution containing hydroxide ion a much weaker base and ammonia This example illustrates what is called the leveling effect ofthe solvent 0 Water the solvent here donates a proton to any base stronger than a hydroxide ion Therefore it is not possible to use a base stronger than hydroxide ion in an aqueous solution 0 We can use bases stronger than hydroxide ion however ifwe choose solvents that are weaker acids than water 0 Most terminal alkynes alkynes with a proton attached to a triply bonded carbon have pKa values of about 25 therefore all react with sodium amide in liquid ammonia in the same way that ethyne does The general reaction is Last Name First Name Roster Noincorrect roster number No grade Chemistry 2311 Homework Problems Chapter 4 Use MSWord and0r ChemBioDraw softwares to answer all questions except where indicated For structures select and then copy the red rectangle for the question and paste it into the ChemBioDraw window Draw the requested structures and then copy the rectangle and its contents in ChemBioDraw and paste them back into MS Word Answer sheet to be printed out one side of page only only front side of pages will be graded You may collaborate on the development of strategies to answer to these problems HOWEVER all work using must be completed independently of others You cannot exchange electronic les with other students or cutand paste from other students Violators of this instruction will receive a zero for the assignment and be referred to the Dean of Students for further sanction l a 3ethyl5methylheptane b 5ethyln0nane c cis13dimethylcyclohexane d 4methylpentan23diol 0r 4methyl23pentanediol e bicyclo333undecane f bicyclo4200ctane g l3cyclohexadiene h 2methyl2hepten4 yne 0r 2methylhept 2en 4yne i cisl2cyclohexanediol i 23dibr0m023dimethylbutane a b aw r i c d U Ho This was incorrectly shown as 14 on the original answer sheet 6 f g h LA Potential Energy gt I II III IV V VI I 0 60 120 180 240 300 360 Dihedral angle I II III Me Me Me Me Me H Me Me H H Me H Me H Me Me H Me IV V VI Me Me Me H Me H Me Me Me H Me H Me H Me H Me Me 4 a Alternative for 3 is on next page Relative energy 0o 600 1200 1800 2400 3000 3600 Angle of rotation I II 111 Me Me Me Me M e Me Me Me H H Me H Me H Me H H Me IV V VI Me H Me Me Me H H Me Me H Me H Me H Me Me Me Me
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