Organic Chemistry I Structure and Reactivity
Organic Chemistry I Structure and Reactivity CHEM 30A
Popular in Course
Popular in Chemistry
verified elite notetaker
This 54 page Class Notes was uploaded by Michael Reilly on Friday September 4, 2015. The Class Notes belongs to CHEM 30A at University of California - Los Angeles taught by Staff in Fall. Since its upload, it has received 1094 views. For similar materials see /class/178001/chem-30a-university-of-california-los-angeles in Chemistry at University of California - Los Angeles.
Reviews for Organic Chemistry I Structure and Reactivity
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 09/04/15
Chem 30A Fall 2001 Prof Garrell Goals TOPICS GOALS Clrjsgngi izs Week 01 Synopsis and Assignments for Sept 26 through Oct 5 2001 REDOX and ELECTROCHEMISTRY Oxidation reduction and oxidation numbers oxid states intro to formal charge Balancing redox reactions by the halfreaction method AG and electrochemical work Galvanic amp electrolytic cells The cell potential under standard conditions E O Faraday s law Relationship between free energy and cell potential The Nernst equation Applications of the Nernst equation determining Q and concentrations Yields from electrolytic processes Be able to assign oxidation numbers states for atoms in inorganic and organic molecules and ions and to use them to identify species that are oxidized and reduced Be able to recognize the correlation between the number of CH and C0 bonds and the oxidation state of carbon Be able to write oxidation and reduction halfreactions and combine them to obtain a balanced chemical reactions b the net cell reaction and c the standard potential E 0 Understand cell diagrams and be able to identify the oxidation and reduction half reactions and determine whether the cell is galvanic or electrolytic Be able to use Faraday s law to quantify the products of electrolysis or determine the current or time necessary to produce a speci ed amount of product Be able to calculate the free energy change and equilibrium constant from the cell potential AGO nFE AG nFE AGO RTan Understand the Nernst equation E E 7 111 Q Be able to apply the Nernst equation to determine cell potentials under non standard conditions eg concentration cells Be able to use the Nernst equation to determine Q and concentrations including pH Understand how the cell potential is related to the spontaneity of a process and the proximity of a reaction to equilibrium Be able to determine the product yield from an electrolytic process RedoxElectrochem synopsis p 2 of 3 Review Session Chem 30A Spring 08 Jesus C 1 Based on the molecule shown below answer the following questions a b C d e V A B How many chiral centers are present in the molecule Assign the right configuration for the carbon labeled as B carbon 2 How many atoms are sp hybridized Is the ether in the cyclohexane ring axial or equatorial What is the name of the functional group assigned as D What is the degree of unsaturation What does the abbreviation OEt stand for as shown in group A How many sp3 hybridized carbons are present in ring C What is the common name of group E How many pi bonds are present in the molecule Review Session Chem 30A Spring 08 Jesus C 2 Draw two reasonable structures for the molecule shown below Show curved arrows on each structure that is required to convert each resonance form into the other Show all formal charges and lone pairs as appropriate 8 o o 7 b OH lt 9 3 Consider the three conformers and answer the questions below CH3 C a at H ire H 77LWXCHQ3 H T EICHQE H C 3 2H5 H In a The conformation with least torsional strain is Answer is b The conformation with the most torsional strain is Answer is c If the rerrbntyl group is replaced by 3 CN cyanide group the torsional strain of the new molecule increases or decreases or remains tuichanged Answer is Review Session Chem 30A Spring 08 Jesus C 4 For each pair shown below choose the best answer that indicates the relationship between the compounds A Equivalent representations of the same compound B Conformers T Constitutional isomers D C is and trans isomers E None from AD As answers place the letter of the temis in the square boxes to the right of each pair 1 M f r s m 7 x 1quot 4 z quot J X quot7 quot 39 J quotquot1 J 1 i If 3 J Va A quot In sh NK lt A x 39w if quot J 39 II EZHJCHu f I r g H J I K I quot H Xvquot quotall CH3 5 I I g 44 xquot r 7 X 391 quotK r quot739 39 r wv39 391 u E f 5 H gr CH CH u m f xv lt k an x m V i 139 L f 4 1 1 Review Session Chem 30A Spring 08 Jesus C 5 Draw curved arrows to indicate the acidbase reaction and draw the products for the compounds shown below j CH3C OOH lgt Wacwcmww 6 Circle the alkyl halide among each of the given pairs that you expect to react faster by an 5N2 mechanism Explain the reasons of your choice Br Bquot i A 3 14 to E in 539 E 3 A 39 Bi V Br Br 4 A B J 1 A7 Bx CB BI UN Review Session Chem 30A Spring 08 Jesus C 7 Give the major product of the following reactions Show stereochemistry where appropriate H35 Lim ar catalyst gt 1 HgOAc2 H30 WWMMWW y 2 NaBm i d 4 H L gt BYEquot H30 Review Session 8 Chem 30A Spring 08 Jesus C Give the major product of the following reactions Show stereochemistry if it is needed Br terr BuO K fe37B110K C3H50 39N gt ethanol CH 31 isopropanol C13H50Na ethanol f TBIXO39K gt rerr BuOH C2H 0Na gt ethanol ww F J k i l x l J VW WW rquot Review Session Chem 30A Spring 08 Jesus C 9 Stereoisomers of 2amino 3hydroxybutanoic acid are shown below COOH COOH H H NH2 H NH2 HOOC NH2 HO H H OH HO H CH3 CH3 CH3 3 What is the relationship between structures and Il b What is the relationship between structures II and III c Which structure represents the 2R3R isomer 10 Provide a reasonable mechanism to account for the transformation shown below Include all the appropriate intermediates curved arrows and the corresponding formal charges a H 80 2 H20 2 4 gt C10quot12002 Terpin Limonene Chem 30A Fall 2001 Prof Garrell W3 13 1 Week 2 Synopsis Reaction Energetics amp Molecular Structure Goals and Assignments for Oct 5 through Oct 12 TOPICS Reaction energetics Gibbs free energy change activation energy reaction mechanism Overview of reaction types Representations of molecules from molecular formulas to threedimensional visualization 0 Functional groups 0 Bond and molecular dipoles polarity 0 Lewis structures and resonance 0 Molecular orbital theory recap GOALS I 0 Understand the following terms Gibbs free energy transition state homolytic exergonic activation energyEa AGI heterolytic endergonic reaction rate radical exothermic reaction intermediate nucleophile endothermic ratedetermining step electrophile potential energy diagram reaction mechanism functional group reaction coordinate carbocation 0 Review topics and oldnew skills practice Lewis structures formal charges resonance VSEPR model Molecular orbitals Electronegativity dipole moments polarity TE We willfollow the chemists convention by writing dipole moment arrows with the tip ofthe arrow at amp and the crossed tail ofthe arrow at amp This is the sign convention used in Munowitz and in BrownFoote but is opposite ofthe convention used in OxtobyNachtriebGillis Hybrid orbitals sp spz sp3 and their geometries arrow notation single double Become familiar with the various two and threedimensional representations of molecules and the structural information each reveals 0 Begin learning how to use the computer software ChemDraw and Chem3D and perhaps Spartan to build and manipulate molecular structures visualize geometries and charge distributions etc Be able to identify functional groups their polarities and hydrogenbonding abilities o Beginto 39 39 39 quot and 39 39 quot Molecular Geometry and Hybrid Orbitals Suggested Reading 39 Review general chemistry textbook as needed 39 Brown and Foote 5Lh edition Chapter 1 Sections 4 6 and 7 39 Klein 2quotd edition Chapter 4 All Lecture Supplement Handouts From lecture also available at course web site Molecular Geometry and Hybrid Orbitals Suggested Text Exercises 39 Brown and Foote 5Lh edition Chapter 1 1712 14 20737 41750 and 646 39 Klein 2quotd edition Chapter 4 All Related Tutorials webchemuclaeduNhardingtutorialstutorialshtml Molecular Model Kit Optional Web Site Reading Hybrid Orbitals wwwwwnorton comCOLLEGEchemistrygi1berttutorialsch7 htm Use Your Molecular Model Kit Visualizing and understanding threedimension structure is critical to mastering organic chemistry Your model kit is a very useful tool perhaps the most useful tool for working with and ultimately conquering concepts based on aspects of the threedimensional structure of organic molecules Pauling Watson and Crick among others made frequent use of molecular models during the Nobel Prizewinning research Therefore it is in your best interest to make your model kit your best chemistry buddy and use it frequently You might even want to give your model kit it s own name Concept Focus Questions 1 Provide precise yet concise definitions for the following terms a Bond angle e Pi bond i spz hybridization b Broken wedge f Sigma bond 139 sp3 hybridization c Electron group g Solid wedge k VSEPR d Hybrid orbital h Sp hybridization 2 In general terms how is molecular geometry controlled by potential energy E What is the fundamental idea of VSEPR theory 4 What is the meaning of solid wedges and broken wedges in a molecular structure Illustrate using the threedimensional structure of methane 5 Describe in one or two words and also sketch the molecular geometry of the following molecules Hz C02 BH3 CH4 and PC15 Give the bond angles for C02 BH3 and CH4 Molecular Geometry anal Hybrid Orbitals 1 6 Rank the following in order of increasing size Hydrogen uorine bromine lone pair and a methyl group 7 Why are all of the HiCiH bond angles of methane equal 8 Why are the HiNiH bond angles in ammonia less than the HiCiH bond angles in methane 9 What must overlap to form a covalent bond 10 Describe the threedimensional arrangement of carbon s valence shell atomic orbitals 11 What is orbital hybridization Why is hybridization necessary for bonding in methane 12 Describe the molecular geometry of ethylene Make a drawing that shows how hybrid orbitals overlap to achieve this geometry Your molecular model kit may be useful to help you visualize the geometry anal orbitals 13 Describe the molecular geometry of acetylene Make a drawing that shows how hybrid orbitals overlap to achieve this geometry Your molecular model kit may be useful to help you visualize the geometry anal orbitals 14 State a general rule concerning the number of electron groups attached to an atom and the hybridization used to achieve it Concept Focus Questions Solutions 1 Illustrated de nitions can be found at the Illustrated Glossary of Organic Chemistry available at the course web a Bond angle The angle formed by three sequentially bonded atoms AiB where B is the vertex of the angle A bond angle is always less than 180 b Broken wedge A bond symbol quotquotquotquotquot indicating the bond is receding away from the viewer and into the plane of the paper or video screen c Electron group An area of electron density a lone pair single atom or group of atoms Also called an electron domain d Hybrid orbital An orbital formed by mathematical combination of atomic orbitals s p py and p2 e Pi bond A covalent bond formed from two or more adjacent parallel p orbitals that overlap perpendicular to the line connecting the nuclei ie the sigma bond axis Also called a 313 bond f Sigma bond A covalent bond formed by the overlap of two orbitals along the line connecting the two nuclei ie the bond axis Also called 0 bond 2 Molecular Geometry anal Hybrid Orbitals E g Solid wedge A bond symbol indicating the bond is pointing out to the Viewer projecting out of the plane of the paper or Video screen h s2 hybridization A hybrid orbital formed by the combination of one s orbital with one p orbital It consists of two equally sized lobes colinear and pointing in opposite directions along the aXis of the p orbital used to create them i 322 hybridization A hybrid orbital formed by the combination of one s orbital with two p orbitals It consists of three equally sized lobes at 1200 angles lying in the plane of the two p orbitals used to create them i 323 hybridization A hybrid orbital formed by the combination of one s orbital with three p orbitals It consists of four equally sized lobes at 109470 angles pointing to the corners of a tetrahedron k VSEPR Valence Shell Electron Pair Repulsion theory A theory that proposes molecular structure is determined by placing electron groups at equal distance from each other in threedimensional space thus minimizing electron pair repulsion The geometry of any molecule is a result of electron group arrangement in a way that minimizes electron repulsion and therefore giving the lowest potential energy The fundamental idea of VSPER theory is that the best ie lowest energy geometry of a molecule places the electron groups are as far apart as possible to minimize potential energy A solid wedge indicates that the bond is projecting out of the plane of the paper or Video screen toward the Viewer The broken wedge indicates that the bond is receding into the plane away from the Viewer This H is in the plane of the paper CuIHltI This H is behind the plane ofthe paper This H1s1n the plane ofthe paperlgtH Hlt This H is In front ofthe plane ofthe paper H H c1 109470 c1 0 VC 0 H B 120 Cl Pquot HH 0 cuIIH 180 H I C1 H H c1 Linear Linear Trigonal planar Tetrahedral Trigonal bipyramid H F lt lone pair bromine lt methyl group In general H F lt lone pair Cl Br I lt groups of atoms Smallest Largest The HiCiH bond angles of methane are all equal because the electron groups the four hydrogen atoms are of equal size and therefore repel each other equally Molecular Geometry and Hybrid Orbitals 3 8 0 A lone pair is larger than a hydrogen atom so a lone pair repels adjacent hydrogen atoms more strongly than the hydrogen atoms repel each other This causes the HiNiH bond angles in ammonia to be compressed compared to the H iH bond angles in methane Ammonia is a distorted tetrahedron HiNiH bond angles 107 whereas methane is an undistorted tetrahedron all HiCiH bond angles 109470 M l 39 M l More repulsmrf ore rep 510 More republfo ore repu Slon N IIIH N quotH I 39 H HjLess repulsion H HJLess repulsion Less repulsion Less repulsion Orbitals must overlap to form a covalent bond Carbon has four valence shell atomic orbitals 23 2px 2py and 2p The ZS orbital is a sphere around the nucleus Each p orbital consists of two lobes that meet at the nucleus The three p orbitals are mutually orthogonal and lie along the X y and zaxes Each p orbital is named for the corresponding axis For example px lies along the XaXis S px py Pz Px py Pz ll Orbital hybridization is the mathematical blending of atomic orbitals to provide new orbitals that point in a different direction than the atomic orbitals Carbon s p orbitals are orthogonal so overlap of each p orbital with a hydrogen ls orbital would cause the H iH bond angles to be 90 This is illustrated below with orbital overlap for two of the four CiH S 3 orbitals bonds In the case of methane hybrid orbitals are necessary to account for fth to the the molecule s tetrahedral geometry and H iH bond angles of 109470 Somers Ofa Thus we combine s p py and pz to form a set of four different orbitals tetrahedron called SP3 orbitals that point to the comers of the tetrahedron 39 Without carbon hybridization 39 HCH bond angle90 2 X 0 gt y Carbon 2py 2pz 2px omitted for clarity Hydrogen ls Two perpendicular CH bonds 4 Molecular Geometry and Hybrid Orbitals With Sp3 hybridization of carbon H 4 X 0 an H Carbon Sp3 orbitals Hydrogen lS Four CiH bonds pointing to the comers of the tetrahedron 12 Each carbon of ethylene HzCCHz bears three electron groups two hydrogens and the rest of the molecule The lowest energy arrangement for three electron groups around a central atom is trigonal planar which requires sz hybridization for the carbon The sz hybrid is formed by miXing the S px and py orbitals causing the sz orbitals to lie in the Xy plane The unhybridized pz orbital is perpendicular to the Xy plane In ethylene as in methane a CiH sigma bond is formed by overlap of a hydrogen lS orbital with a carbon hybrid sz orbital The carboncarbon sigma bond portion of the double bond is formed by overlap of one sz orbital from each carbon This leaves a pair of parallel adjacent pz orbitals one on each carbon which overlap to form the 7 component of the double bond Side view Top view 2 Carbon sz orbitals W W y Ethylene E Each carbon of acetylene HiCECiH bears two electron groups one hydrogen atom and one carbon atom The lowest energy arrangement for two electron groups around a central atom is linear which requires Sp hybridization for the carbon The Sp hybrid is formed by mixing the S and px orbitals causing the Sp orbitals to lie along the XaXis The unhybridized py and pz orbitals are orthogonal to the Sp orbital set In acetylene as in methane and ethylene a CiH sigma bond is formed by overlap of a hydrogen lS orbital with a carbon hybrid Sp orbital The two carboncarbon sigma bond portions of the triple bond are formed by overlap of one Sp orbital from each carbon This leaves two sets of parallel adjacent p orbitals one px and one py set on each carbon which overlap to form the two 7 components of the triple bond Molecular Geometry and Hybrid OrbitalS 5 Side View Top View Carbon Sp orbitals szszw C H o C H o Acetylene CplCM 14 Electron groups Geomet Hybridization Two Linear Sp Three Trigonal planar spz Four Tetrahedral SP3 OWLS Problems 1 Label the hybridization of each atom except hydrogens in the following molecules HO OCH2CH3 b W F CH2CH2CH3 2 Estimate the HiNiH HiNiC and CiCO bond angles in neutral glycine HzNCHzCOOH and zwitterionic glycine H3NCH2COZ39 What is the hybridization of the atoms at the apex of each of these angles 3 Draw a picture that clearly shows how atomic or hybrid orbitals overlap to form formaldehyde HzCO Include all lone pairs Specify the geometry and hybridization of each atom Practice Problems 1 What is the hybridization of the oxygen atom and the approximate C707C bond angle in ethyl methyl ether CH30CH2CH3 6 Molecular Geometry and Hybrid Orbitals 2 Estimate the O iC and H iC bond angles of allyl alcohol HzCCHCHzOH Brie y explain your reasoning in each case 3 Brie y compare the molecular geometry of methane the methyl carbanion 39CH3 and the methyl carbocation CH3 4 For each molecule given below label the geometry of its nitrogen atom as one of the following tetrahedron distorted tetrahedron trigonal planar or distorted trigonal planar Molecules NH3 39NHz NH4 CH33N CH32NH CH33NH 5 Which molecule has a greater bond angle NH3 or NBr3 6 Draw the structure of SF5 clearly showing the molecular geometry Very brie y explain why SF5 adopts this molecular geometry 7 Construct a diagram that shows all the orbitals involved in the bonding of the formic acid HCOOH the simplest carboxylic acid Because of a phenomenon called conjugation which we do not study in Chem 30A the OH oxygen is trigonal planar All of the other atoms in glycine bond as expected based upon what we have learned so far Practice Problem Solutions 1 The hybridization of the oxygen atom is SP3 Because the methyl CH3 and ethyl CH3CH2 groups are larger than the oxygen lone pairs the C iC bond angle is predicted to be greater than 109470 2 The OiCiC bond angle is slightly more than the normal tetrahedral angle of 109470 while the H bond angle is slightly less than the normal tetrahedral angle of 109470 The oxygen atom at the vertex of this bond angle has four things attached to it H atom C atom and two lone pairs Because electrons repel the most stable molecular geometry will be that which places these four things as far away from each other as possible yet still maintains the bonds This is best achieved by pointing the four bonds at the comers of a tetrahedron providing bond angles of 109470 However a lone pair occupies more space than a bonding pair and thus repels more strongly than a bonding pair This results in a compression of the H70 bond angle to a value that is a few degrees less than the normal tetrahedral angle of 109470 3 Methane CH4 The carbon atom is attached to four hydrogen atoms and is therefore SP3 hybridized Because all attachments to the carbon atom are identical they all repel each other equally and the C H HCHbond bond angles are all equal at 109470 All CiH bond lengths are H H e qual H gp3 carbon Molecular Geometry and Hybrid Orbitals 7 4 6 Methyl carbanion 39CH3 The carbon atom has four attachments three hydrogen atoms and one lone pair and is therefore SP3 hybridized However the attachments are not all identical so the bond angles cannot all be identical either A lone pair occupies more space than a bonding electron pair so the lone pair repels the hydrogen atoms more strongly than the hydrogen atoms repel each other This means the HiCiH bond angle is a bit smaller than a perfect tetrahedron less than 109470 5f carbon Cquot39quot H HCHbond H Lquot H angle lt 10947 Methyl carbocation CH3 The carbon atom has three attachments three hydrogen atoms and is therefore spz hybridized A pz orbital szcarborl is also present perpendicular to the spz orbital plane but because this pz orbital is vacant it does not cause any electron repulsion The H Ci quot attachments are identical so their repulsions are equal and all three 0 H HiCiH bond angles are equal at 120 The HiCiH bond lengths are all equal as well In all cases but NH4 the nitrogen geometry is a distorted tetrahedron because the nitrogen has four electron groups which are not all equal Only in the case of NH4 is the nitrogen atom tetrahedral because it has four equal electron groups In both cases the nitrogen has four electron groups one lone pair and either three hydrogens of three bromines making the nitrogen geometry a distorted tetrahedron with bond angles close to 10950 In the case of ammonia hydrogen is smaller than a lone pair so the HiNiH bond angles are predicted to be less than 10950 The actual HiNiH bond angle in NH3 is 107 In the NBr3 case bromine and a lone pair have about equal size so the BriNiBr bond angles are very close to 10950 In simple terms the structure of any molecule is such that placing the various atoms as far apart as possible yet still maintaining the bonds minimizes electron repulsion For a central atom surrounded by siX atoms the best way to achieve this maximum spread is by use of octahedral geometry Flth F S F F F Lewis structure Molecular Geometry and Hybrid Orbitals Bonding and Structure of Organic Molecules Suggested Reading 39 Review general chemistry textbook as needed 39 Brown and Foote 5 11 edition Chapter 1 Sections 1 3 and 5 39 Klein 2 d edition Chapter 1 Sections 5 and 6 Lecture Supplement Handouts From lecture also available at course web site Bonding and Structure of Organic Molecules Optional Web Site Reading More on Lewis Structures wwwtowsoneduladonlewishtml Suggested Text Exercises 39 Brown and Foote 5 11 edition Chapter 1 1 12 14 20 37 41 50 and 64 66 39 Klein 2 d edition Chapter 1 33 68 Related Tutorials webchemuclaeduhardingtutorialstutorialshtml 39 Drawing Lewis Dot Structures 39 Formal Charge 39 Curved Arrows 39 Molecular Model Kit Common Questions About Organic Chemistry Problems I don39t have time to do all these problems There are many problems available to enhance your understanding of organic chemistry including the Concept Focus Questions CFQ in this Thinkbook the Practice Problems PP also in this Thinkbook as well as suggested problems from the text and other sources Working all these problems can be time consuming but this task is essential to your understanding of the course material Rarely do students who skimp on problem solving get good grades Most students who study a topic for the first time really do need to work all the problems Problem solving is an excellent way to reinforce the concepts in your mind If you can honestly say to yourself that you have a firm grasp on the concept then feel free to skip the problem On the other hand if you are not 100 sure then work the problem Even students who have a firm conceptual grasp should work a few quotobviousquot problems every now and then to stay sharp Which problems are most important Here is a suggested order of priority 1 CFQ Always read these before lecture and then again when you study the textbook They outline the fundamental concepts presented in lecture and are the most important problems in this course 2 OWLS and PP Written by the same person who writes the exams OWLS problems are designed to stimulate discussion during discussion section meetings Bonding and Structure of Organic Molecules 1 All other problems including the optional ones Different students get the most out of different problem sources although most students seem to get more out of the textbook problems than from the other sources Organic Chemistry as a Second Language and the course web site tutorials focus on specific topics and should be examined if you are having trouble with these specific topics Wikipedia Another Good Organic Chemistry Learning Tool The Wikipedia wwwwikipediaorg is a huge online encyclopedia of knowledge created and updated by thousands of people all over the world including college students It does contain a few errors here and there and shouldn t be used as the only source of information for a research paper but it is a great place to start learning about a topic Concept Focus Questions 01 Provide precise yet concise definitions for the following terms a 6 d Electronegativity g Lone pair b 6 e Formal charge h Orbital c Covalent bond f Functional group i Polar covalent bond Draw the Lewis dot structure for CH3OH How does electronegativity influence electron distribution within bonds Illustrate with CHSOH as an example Why are functional groups important to a systematic study of organic chemistry Prepare a functional group list that includes the name and basic structure of each functional group as well as an example molecule that contains exactly six carbons Include all lone pairs and formal charges Try to avoid using the same example molecules as given in the lecture supplement Concept Focus Questions Solutions 1 Illustrated definitions can be found at the Illustrated Glossary of Organic Chemistry available at the course web site a E An atom that bears this symbol has a slight electron excess but not enough to give it a full negative formal charge b E An atom that bears this symbol has a slight electron deficiency but not enough to give it a full positive formal charge c Covalent bond A chemical bond formed between two atoms by orbital overlap and sharing of an electron pair Bonding and Structure of Organic Molecules N 9 4 d Electronegativity The measure of an atom s attraction for electrons in a chemical bond e Formal charge The charge on an atom in a Lewis structure if the bonding was perfectly covalent and the atom has exactly a half share of the bonding electrons f Functional group A group of atoms whose bonding is the same from molecule to molecules g Lone pair A pair of electrons assigned to just one atom Also called a nonbonded pair h Orbital The mathematical description of a volume of space in which there is a certain probability of finding an electron of a certain energy i Polar covalent bond A covalent bond between two atoms of different electronegativity resulting in an uneven sharing of the bonding electron pair H HCZOZH H The distribution of bonding electron density is influenced by the electronegativity of the atoms that comprise the bond A more electronegative atom attracts more electron density toward itself Thus end of the bond with the more electronegative atom will have a small negative charge 6 The less electronegative atom loses electron density as so has a small positive charge 6 In the C 0 bond carbon is less electronegative than oxygen so the carbon has a 6 charge and the oxygen atom has a 639 charge The reasoning applies to the 0 H and C H bonds The combination of short bond length and low electronegativity difference causes C H bonds to be nonpolar H 5 6 5 H C O H 1L Nonpolar bond A functional group is a set of atoms bonded together that gives a molecule particular chemical and physical properties Because chemical reactions involve changes in electron distribution and bonds functional groups with similar electronic structures will react in similar ways This is just one reason why the study of functional groups is so critical to a systematic study of organic chemistry Bonding and Structure of Organic Molecules 3 Narne Alkene Alkyne Benzene ring Alkyl halide Alcohol Phenol Ether Epoxide Thiol Disqu ide Amine Generic Structure CC m gtlt m mple Q CEC CH3CH2CH2 CEC CH3 CC Br C C C C l gt4 1 X F Cl Br orI C 39o39 H 6H 9 00 O O O m 2 1 0 OCH3 N CH2CH2CH2CH3 Bonding and Structure of Organic Molecules Name Generic Structure Example I M Amine c ii N CH2CH2CH2CH3 l we I ill I Ketone TCT OER I x 39o Aldehyde CCH H I II Carboxylic acid c c 39o39 H W OH Ester c c 39o39 c ltgt lt I I O CH3 Amide c ii I I R CH3 I II Acid chloride W c1 I II II I o 0 Acid anhydride c c o c c i I l l CH3CH2 o CH2CH3 CH3CH2 Imine Cixi N CH3CH2 CH3 Nitrile CEN OCEN 0 39 N1tr0 N N 9 o Bonding and Structure of Organic Molecules Be prepared to name or draw functional groups but you do not memorize specific examples of molecule that contain functional groups OWLS Problems Solutions to OWLS problem sets are posted on the course web site approximately one week after the material is covered in lecture In this problem set you will explore the structure of glycine the simplest amino acid Amino acids are the small molecules that are the basic building blocks of proteins In aqueous solution neutral amino acids exist in equilibrium with their zwitterionic isomers An incomplete representation of this equilibrium is shown HZNCHZCOOH H3NCH2CO2 Neutral form Zwitterionic form 1 Select the structure for glycine Explain why the others were eliminated 1 F I i If If 1 a H CC O H c H 1I c o e H Iy f c o H H N H H H H H If 1 i 13 b H c H c o H d H IiI IZ IZ H H N H H H O 2 Complete the bond line structure selected in the previous question by including all lone pairs Assume all of atoms carry a formal charge of zero 3 Zwitterion means hybrid ion Verify the ionic nature of zwitterionic glycine by drawing its bond line structure it will be very similar to your answer to question 1 adding lone pairs and calculating all formal charges 4 Functional groups are the centers of reactivity in organic molecules Their systematic study essentially eliminates the need for extensive memorization of organic reactions Therefore it is critical that you learn to rapidly identify and draw functional groups a Using the best structure selected in problem 1 label the functional groups of glycine b Draw an isomer of glycine containing two functional groups not mentioned in problem 4a Label these functional groups 6 Bonding and Structure of Organic Molecules Practice Problems 1 N 9 4 01 Complete the following drawings by adding correct formal charges H H 0 CH3 CC H b 6 quotF39 CC N H a H CO c H NH2 CH3 For the molecules in question 1 add or subtract hydrogens so that each atom has a formal charge of zero Do change any atoms other than hydrogen For each structure in question 1 rank the atoms in order of decreasing electronegativ ity Include the electronegativity values Sulfuric acid H2804 is a common reactant in organic chemical reactions Draw a Lewis structure for sulfuric acid Brie y describe how the polarity of a chemical bond is estimated by using only the Lewis structure of a molecule Illustrate your answer with the 8 0 bond in sulfuric acid For each of the following molecules a Write the chemical formula ie C6H1206 b Label the most polar bond ie largest 6 and 6 charges c Label all functional groups 0 Cl 24 DP a herbicide a chemical used to kill plants Nepetalactone the active component of catnip Bonding and Structure of Organic Molecules 7 N 0 Y Xenical orlistat a weight loss drug that H work by irreversibly blocking stomach and O O 0 O intestinal enzymes involved in fat metabolism CH3CH210 CH2 CH24CH3 6 Draw a single molecule that contains the listed functional groups Do not use any abbreviations such as R or X and do not include any functional groups not listed Label the functional groups in your example molecules Include all lone pairs a Ester amide and alcohol b Carboxylic acid alkene benzene ring and ether c Alkyl chloride nitrile ketone and aldehyde 7 Why do we expect many similarities in chemical and physical properties of alcohols and water 8 A carboxylic acid reacts with LiAlH4 to form a primary alcohol We can therefore conclude that the four membered ring of Xenical question 5 which contains a functional group called a lactone reacts in a similar manner Brie y explain why we can make this conclusion 0 LiAlH4 Carboxylic ac1d reaction A gt ROH HOH R OH 0 HO OH O LiAlH4 Lactone reaction gt R R39 R R39 9 Which two of the following three functional groups has the most similar chemistry ester amide and alcohol Brie y explain Practice Problems Solutions 1 Formal charges of zero are usually not written H H 0 CH3 CC H b39 L a H COI b 8H2 c F CC T H H NH2 CH3 If this problem was not easy you should review the formal charge tutorial available in the Tutorials section of the course web site 8 Bonding and Structure of Organic Molecules H H o CH3 CC H HO 2 a H C9 b CH3 0 F CC T H J NH2 CH3 3 U a 0 EN 35 gt C EN 25 gt H EN 21 b 0 EN 35 gt N EN 30 gt C EN 25 gt H EN 21 c F EN 40 gt N EN 30 gt C EN 25 gt H EN 21 Any legitimate Lewis structure for sulfuric acid is acceptable in this case Formal charges are considered a part of any Lewis structure and must be included as well Bond polarity is a result of uneven distribution of the electron density within a covalent bond This is analyzed by considering the electronegativities of the bonded atoms Oxygen EN 35 is more electronegative than sulfur EN 25 so the bonding electron density is shifted a bit toward the oxygen The slight negative charge that results on oxygen is designated with the 6 symbol Sulfur is on the losing end of the unequal sulfur oxygen bond electron sharing so sulfur is a bit electron deficient This slight positive charge on sulfur is indicated by the 6 symbol 0 39o39 5 lay 5 539 5 H O S 0 H or H o s 39o H 0 0 a When the atom already has a formal charge it is not necessary to show 6 or 6 charges on that atom This is illustrated in the second Lewis structure shown above In the chemical formula of an organic molecule we write the number of carbons the number hydrogens and then the numbers of all remaining elements in alphabetical order The most polar bond has the largest electronegativity difference between its atoms Remember that we do not say that a more complex functional group contains a simpler functional group For example a benzene ring does not contain an alkene an ester does not contain an ether and a carboxylic acid does not contain an alcohol Bonding and Structure of Organic Molecules 9 o Carboxyuc acid OH Benzene ring H C 2 4 DP 3 Formula CQHBCIZO3 10 Cl Most polar bond O H Ether c1 Nepetalactone Formula CmH1402 Most polar bond C O C20 is shorter and therefore predicted to be less polar than C O Y Ester Xenlcal O 4 O H 39 Formula C29H53NOS Esterl O I 0 Most polar bond C O CH3CH210 CH2 CH24CH3 6 There are an infinite number of acceptable answers as long as they contain the required functional groups and no extraneous functional groups Students often ask if functional groups can share atoms They can in some circumstances and not in others The exact rules are a bit complex for Chem 30A so when in doubt avoid this conundrum by inserting one or more non functional group carbons between the functional groups in question Alcohol Ester Carboxylic acid Alkene Benzenering Ether 10 Bonding and Structure of Organic Molecules Ionic Substitution Reactions SN2 Suggested Reading 39 Review general chemistry kinetics and thermodynamics as needed 39 Brown and Foote sections 91 94 39 OCATSA Ionic Substitution Reactions email instructor to obtain access Lecture Supplement 39 Ionic Substitution Reactions 8N2 Optional Reading 39 1999 Nobel Prize in Chemistry quotPhotographsquot of Transition States httpnobelprizeorgnobelprizeschemistrylaureates1999 39 Klein Chapters 8 and 9 Suggested Text Exercises from Brown and Foote 39 Chapter 9 1 3 11 13 15 19 21 51 and 52 Related Tutorials http webchemuclaeduhardingtutorialstutorialshtml 39 Electrophiles and Nucleophiles 39 8N2 Reaction Visualizing 8N2 Transition State via Molecular Models Concept Focus Questions Questions 1 6 refer to this substitution reaction CH3C1 HO gt CH3OH Cl 1 Draw a curved arrow mechanism Is this a concerted reaction 2 Write the rate expression and reaction name 3 Define quottransition statequot Why is this important in the study of chemical reactions 4 Draw the transition state for the reaction Describe in words what this drawing shows 5 Brie y discuss the stereochemistry of this reaction 6 Draw the energy profile for the reaction Assume the reaction is exergonic Label all of the important features of the energy profile 7 What is the relationship between transition state structure and reaction rate 8 List in order of in uence the five important structural features of the nucleophile that control nucleophilicity Brie y describe how each factor operates 9 Define quotleaving groupquot What structural factors make for a good leaving group Ionic Substitution Reactions SN2 1 10 Define quotsteric effect quot Brie y discuss steric effects in the 8N2 reaction 11Define Dielectric constant polar solvent nonpolar solvent protic solvent and aprotic solvent 12 Brie y discuss the relationship between solvent transition state and 8N2 reaction rate Include both polarity and hydrogen bonding effects 13 List the four fundamental requirements for an 8N2 reaction to occur Very briefly describe how each factor controls the rate of an 8N2 reaction Concept Focus Questions Solutions 1 n Ho H3C Cl Ho CH3 c1 This is a concerted reaction because it proceeds through a single mechanism step without any intermediates All bond making and bond breaking occurs within this single mechanism step It is not usually necessary to include the leaving group as one of the products Its omission is not an error but rather an acceptable simplification However in some reactions the leaving group continues to participate in the mechanism after its departure If you are the kind of student who tends to forget that the leaving group is still present even though it has not been written then get yourself into the habit of drawing the leaving group Does CH3Cl confuse you because it implies that the chlorine atom is bonded to hydrogen instead of carbon By convention the methyl group may be written as CH3X or H3CX Rate k nucleophile electrophile k HO CH3Cl The reaction is nucleophilic substitution with bimolecular kinetics hence it is called 8N2 The transition state is the highest energy structure for each mechanism step in the reaction Every mechanism step has its own transition state A reaction mechanism with more than one step has more than one transition states The transition state is important in the study of chemical reactions because the amount of energy needed to achieve the transition state called the energy of activation controls the rate of a reaction see CFQ 7 Reaction rate is easily observable and thus a useful tool to verify a proposed mechanism and to ascertain the effects of changing various factors in a chemical reaction Ionic Substitution Reactions SN2 The carbon oxygen and carbon chlorine bonds are incomplete and therefore longer than normal sigma bonds The hydroxide ion has begun to share a lone pair with the carbon so it has a partial negative instead of a full negative charge The pair of electrons that were the carbon chlorine bond has begun to shift toward the chlorine giving the chlorine a partial negative charge The carbon has shifted toward sp2 or perhaps dsp3 hybridization as the best geometry to accommodate the five atoms surrounding it The brackets and i symbol are used by convention to symbolize that this structure is a transition state It does not matter from what perspective you draw a transition state unless specified by the problem For an SN2 transition state the partial bonds may be drawn horizontally vertically or at any other angle If you are unsure that your answer and the given answer are equal use your model kit to build and compare The nucleophile attacks the carbon from the backside of the carbon leaving group bond so as to allow maximum overlap between the orbital containing the pair of electrons being donated by the nucleophile and the 0 carbon leaving group antibonding orbital This results in net inversion of absolute stereochemistry in an 8N2 reaction one pan 0 0 bond 11 URI R1 R1 1 R3 k R3 R2 R3 R2 R2 The exact structure of the nucleophile and leaving group are not specified so we cannot specify their charges 6 Tsl Gibbs Free Energy Ho39 CHgCl 39c1 C1130H 7 Reaction Coordinate 7 For reactants to become products they must have enough energy to overcome an energy barrier called the Gibbs free energy of activation AGI see diagram in previous answer The height of this energy hill is the difference in Gibbs free energy of the reactants and transition state Recall that Gibbs free energy AG is a function of enthalpy AH energy changes due to bond changes and entropy AS organization In equation form Ionic Substitution Reactions SN2 9 AGz AHz TASr At reaction temperatures up to a few hundred Kelvin the entropy factor is small perhaps one kcal mol l In this temperature range TASr is small compared to AH and so it can often be neglected when making simple predictions or estimations Thus we can approximate AGr as AH This assumption is valid only when enthalpy changes are more important than entropy changes in a reaction This is a convenient approximation because we cannot measure AGr in the laboratory Instead we can determine a closely related energy value called the energy of activation En E is related to AHr by the equation E AHr RT E can be measured in the laboratory because of its relationship to reaction rate something we can easily measure in the lab by the by the Arrhenius equation k Ace EaRT So for our introductory level discussion of reaction rates and variables that influence them we can discuss transition state energy and not have to be concerned about whether it is AGr or E The Arrhenius equation reveals that a lower transition state energy smaller hill leads to a faster reaction Each structural feature is analyzed to see how it influences the ability or driving force of the nucleophile to donate electron density to an electrophile Four factors can be listed in order from most influential resonance to least influential inductive effect Resonance Resonance usually but not always decreases electron density at the atom that shares electrons with the electrophile Atomic radius An atom is driven to share an electron pair in order to decrease its charge density electron density per unit of surface area Everything else being equal smaller atoms have a higher charge density and thus a stronger drive to form new covalent bonds Electronegativity Electronegativity is defined as an atom s attraction for or resistance to sharing electron density Higher electronegativity means the atom has a greater attraction for or is less inclined to share electron density Since the job of a nucleophile is to share electron density with a carbon atom higher electronegativity decreases nucleophilicity Inductive effect This is the non resonance influence that one part of a molecule has on the electron density elsewhere in the molecule It can be an electron withdrawing or electron donating effect and can operate both through the sigma bond framework and through space For a nucleophile it is the electron density effect of atoms other than the atom that is sharing electron density with the electrophilic carbon Remote Ionic Substitution Reactions SN2 gt0 H H atoms can increase or decrease the electron density on the atom sharing electrons with the carbon and thus in uence nucleophilicity The previous four factors are listed in order of decreasing in uence Formal charge also plays a significant role in determining nucleophilicity but its relative in uence is not fixed For example sometimes formal charge carries more in uence than atomic radius and other times its in uence is less Formal charge A formal negative charge indicates excess electron density In the case where all other factors are equal an atom with a formal negative charge makes the nucleophile stronger than one that is neutral Formal charge is of variable in uence for example sometimes it has more in uence than resonance sometimes less so it does not have a fixed position in this sequence of relative in uence Because nucleophiles and bases both share electrons there is a strong although imperfect parallel between the structural factors that control them A thorough review of acids and bases and the structural features that in uence them can be found in the Supplemental Reading link at the course web site Leaving group is the portion of the molecule that departs along with the pair of electrons that was the bond between the leaving group and some other atom This means the leaving group gains electron density The best leaving groups are those that can accommodate and disperse this extra electron density most effectively by resonance atomic size and inductive effects Leaving group efficacy is also enhanced when the leaving group atom that accepts the electron pair has high electronegativity or a positive formal charge In general the best leaving groups are also poor nucleophiles but there are exceptions For example iodide ion is a good nucleophile and superior leaving group A steric effect occurs when a chemical phenomenon is in uenced by van der Waals repulsion In an 8N2 reaction steric effects can in uence the reaction rate by slowing or preventing approach of the nucleophile As the number andor size of substituents on the carbon bearing the leaving group LG increases it becomes increasingly more difficult for the nucleophile to reach this carbon The 8N2 reaction rate decreases as the number or size of substituents increases Relative rates CH3 LG methyl gt RCHZ LG primary 1 gt RZCH LG secondary 2 gtgt RSC LG tertiary 3 A tertiary carbon is so highly hindered toward backside attack that tertiary substrates do not react by the 8N2 mechanism Dielectric constant Dielectric constant e is a measure of a substance39s ability to insulate opposite charges from each other In other words it is a measure of the ability of a substance to separate ions Polar solvent A polar solvent has a high dielectric constant Solvents with s of approximately 20 or more are labeled as polar Ionic Substitution Reactions SN2 5 H H Nonpolar solvent A nonpolar solvent has a low dielectric constant Solvents with s of approximately 20 or less are labeled as nonpolar Protic solvent A protic solvent is a hydrogen bond donor In common organic solvents the hydrogen bond donor is an O H bond Review hydrogen bonding from Chem 14C if necessary Aprotic solvent An aprotic solvent does not have a hydrogen atom that can be shared by hydrogen bonding Reaction rate is controlled by the energy difference between the reactants and transition state If the solvent decreases this gap the reaction is faster If the solvent increases this gap the reaction is slower For example the solvent might stabilize the reactants more than the transition state causing a larger energy gap and slower reaction Most 8N2 reactions involve a negative charged nucleophile and uncharged electrophile This results in a transition state with 6 charges on the nucleophile and leaving group For the fastest reaction we select a nonpolar solvent as this provides less stabilization to the nucleophile than to the transition state However nonpolar solvents do not dissolve most anionic nucleophiles so we are forced to choose a polar solvent instead A protic solvent surrounds a nucleophile with a hydrogen bonding shell This hydrogen bonding provides some stabilization Some of the solvent molecules must move away in order for the nucleophile to come within bonding distance of the electrophile This desolvation decreases the stability of the nucleophile and the transition state Because the nucleophile has greater charge concentration than the transition state the nucleophile suffers a greater decrease in stability than does the transition state Therefore 8N2 reactions are slower in protic solvents than in aprotic solvents of equal polarity Overall the best solvent choice for most 8N2 reactions is polar aprotic Polar protic is usually acceptable but the reaction is slower Nonpolar solvents are usually not useful for 8N2 reactions Moderate or better leaving group Better leaving groups accelerate an 8N2 reaction by allowing for more bonding between the nucleophile and carbon atom accepting the nucleophile This increased bonding stabilizes the transition state thus lowering the activation energy and increasing the reaction rate Good nucleophile Better nucleophiles accelerate an 8N2 reaction by increasing the degree of bonding between the nucleophile and carbon undergoing substitution in the transition state This increased bonding stabilizes the transition state thus lowering the activation energy and increasing the reaction rate Ionic Substitution Reactions SN2 The carbon bearing the leaving group cannot be tertiarv Steric 39 J slows the approach of the nucleophile and thus reduces the reaction rate When the carbon is tertiary the steric hindrance is sufficient to shift the reaction to a different substitution mechanism Solvent A polar aprotic solvent is best Polar protic may be acceptable but the reaction is slower Nonpolar solvents are usually not useful for 8N2 reactions These factors may interact For example if the leaving group is superior then reaction can occur with a poorer nucleophile There are no exceptions to the quotnot tertiaryquot requirement OWLS Problems Please bring your molecular model kit to discussion section this week 1 N 9 Resonance is a common phenomenon in organic molecules Therefore it is useful for you to recognize its presence to be able to draw resonance contributors rapidly and and to J how influences the structure and reactions of organic molecules a Draw all of the significant resonance contributors for each of these ions 0 b Brie y describe how resonance influences each molecule s nucleophilicity and leaving group ability a Select the best nucleophile and leaving group among these four species 0 HOH H O H3C 9 F3C O o o b Write an 8N2 reaction including all reactants and products that utilizes both the best nucleophile and the best leaving group from part a Your 8N2 reaction must occur at a carbon of the electrophile that is both secondary and a stereocenter c Write the mechanism and transition state for your reaction from part b Provide the missing products reactants or starting materials in the boxes Naquot 391 Cl a acetone Ionic Substitution Reactions SN2 7 4 0 gt1 O39 K HOV DMSO C CH3gtCgt OSOZCH3 CH3 ltgtquotIHHHSCH2CH3 d o O H b FCl H3 Is reaction 3d a reasonable 8N2 reaction Explain Consider the four reactions of question 3 What one feature is common to all the nucleophiles What one feature is common to all of the electrophiles Consider this general statement Protic solvents decrease nucleophilicity a Give two examples of protic solvents b Is the statement true Brie y explain c Write a pair of 8N2 reactions that illustrate this solvent effect Select the faster Consider this reaction sequence reaction 3113 2113 CHz Nal gt 1 305mm uuullll quotquotquotquotCEN a Evaluate each step in the sequence Are these reasonable 8N2 reactions b An 8N2 reaction proceeds with inversion so why does the first step not occur as shown below CH3 N31 gt Cl acetone c Considering the given reaction sequence evaluate this statement In an 8N2 reaction the term inversion means that if the electrophile has the R configuration the product always have the S configuration Kt 39CEN gt ethanol MIME uIII Practice Problems 1 For a reaction to occur molecules must collide with sufficient and correct Ionic Substitution Reactions SN2 4 U 0 9 gt0 Give at least one significant reason that the transition state is the highest energy structure between reactants and products in a mechanism step KBr Consider this reaction C1 gt Br DMSO a Write the rate expression for this reaction b Write a curved arrow mechanism for this reaction including the transition state and all lone pairs Write the 8N2 mechanism including the transition state for the following reaction Clearly show product stereochemistry Br I NaI a gt 9 CHgOH H c H 3 Build models of the transition states for the following 8N2 reactions a H3C Cl H0 gt H3C OH Cl b Cst39 139 a a 1 H H SCH C Q NH 0 quotv H3C quot Br H3C NH3 Br The product of reaction c is a salt in which RNH is the cation and Br is the anion We normally do not write a plus sign between the cation and anion of a salt even if they are reaction products Example 2 Na Clz gt 2 NaCl not 2 Na 2 Cl39 CH3 Prov1de the products of this reaction ac stone quotWICl What is the single most important factor more fundamental even than resonance that controls the nucleophilicity or basicity of any molecule or ion Select the poorest nucleophile HO CH3CO2 and CH3O Select the best and worst nucleophiles in an aprotic solvent CH3CHZS CH3CH20 CH3CH20H and CF3CH20H Brie y explain your reasoning Ionic Substitution Reactions SN2 9 10 Select the strongest nucleophile in aprotic solvent in each set List the most important factors that in uenced your decision Illustrate each nucleophile in a reasonable SN2 reaction using a different electrophile in each case a 1 Br F e CF3CH2S CH3CH2S CH3CHZO b HS HSe HO f CH3CH2CH20 CHSCFZCHZO CF3CH2CH20 c NH3 H20 CH3OCH3 g H20 H2S HO HS d HS HO HZN h CFSCOZ CISCOZ CHSCHZCOZ 11 Cyanide ion CEN is an excellent nucleophile a Suggest the structural features that account for its high nucleophilicity b Select the major product of the following reaction and brie y explain your reasoning FCEN HC I gt Hc CEN H3C NC Acetonitrile Methyl isonitrile 12 Select the slowest reaction Brie y explain your choice of Na 39SCH of gt7 Nat 39OCH Br gt SCH or Br gt OCH CHOH CHOH 13 Select the best leaving group CH3S CH3O CF3S or F Illustrate the best leaving group in a reasonable SN2 reaction at a secondary carbon 14 Select the poorest leaving group F I or CH3 15 Label these leaving groups as best middle or poorest CF3SO3 CH3SO3 and CH3C02 16 Select the compound that reacts slowest in an SN2 reaction CH3I CH3CH21 CH32CHI 17 Select the slowest SN2 reaction Brie y explain your choice Na 39SCH O Na rsCH gt or Br gt SCH Br CHOH SCH3 CHOH 18 Without using any reference material such as a table of solvents or your lecture notes assign the dielectric constants of 80 33 and 25 to these compounds CH3CH20H H20 and CH3OH 10 Ionic Substitution Reactions SN2 19 Select the molecules that react fastest and slowest by the 8N2 mechanism F Br H3CI A OH 20 Select the molecule that reacts the slowest in an 8N2 reaction gt7Br H3C OH 030201 CH3CH2 Br 21 For each pair of 8N2 reactions shown below i Decide which reaction is faster Brie y explain your choice ii Write the mechanism of the faster reaction iii Design a reaction that is similar to but faster than the faster reaction of part i Brie y explain why your designer reaction is faster NaI NaCl a gt versus gt I Cl H CH3OH H 1 Cl H CH3OH H 01 LiSH LiSH b gt versus gt C1 CH30H SH 1 CH30H SH 1 I KSCH3 S KSCH I c gt versus gt DMF DMF 1 1 1 SCH 22 For the 8N2 reactions shown below i Write the 8N2 mechanism transition states and product ii By adding subtracting or transmuting at most four atoms of the electrophile or nucleophile rewrite the reaction so that it is obviously slower Nascu3 HI xCl I a Br gt c quot gt CHEOH H3O H acetone 0 a A I I H CH3 0 DMF 23 For the reaction shown below a Write the product and reaction mechanism including all transition states Ionic Substitution Reactions SN2 11 b By changing only the electrophile write a reaction including product that is clearly slower Brie y explain why your new reaction is slower c Changing only the nucleophile write a reaction that is clearly faster Brie y explain why your new reaction is faster OjBr CH3 CHgs39 Na gt CH30H Rank the 8N2 reaction rates of the following molecules with NaSCH3 in ethanol 2 iodo 2 methyl propane 2 bromopropane and 2 chloropropane O O Consider this intramolecular substitution reaction NaOH Br HO H20 a Write two mechanisms for this reaction but avoid carbocations b Which of these mechanisms is most reasonable c Write the transition state for the carbon oxygen bond forming step of the most reasonable mechanism Use a model to visualize the transition state d Write the rate expression for this reaction Is it an 8N2 reaction e Draw another 8N2 product that might be formed in this reaction Select the major product of this reaction Write a mechanism and brie y explain your choice of major product 0 NaI gt 0ng or CF31 H3CO S CF II acetone 0 Many haloalkanes alkyl halides can cause damage to DNA abbreviated DNA NH2 because they can alkylate a free amine by an 8N2 reaction Because of this many haloalkanes are carcinogens DNA NHZ w I a Write the product mechanism and all transition states for the 8N2 alkylation of DNA by R 2 iodobutane reaction shown above b Brie y explain why tert butyl iodide is not a very powerful carcinogen c Methyl uorosulfonate causes DNA mutations more readily than the iodide in the reaction shown above Provide two brief reasons for this Methyl uorosulfonate is also called magic methyl because it is such a powerful methylating reagent 12 Ionic Substitution Reactions SN2 o H Methyl fluorosulfonate H3C O S F ll Magic methyl 0 28 Methylation is an important step numerous biosynthetic pathways This question deals with one step in the biosynthesis of creatine a molecule that is essential for the functioning of muscles a Phosphate ions are used extensively in biosynthesis as leaving groups Why is triphosphate a good leaving group PO Triphosphate ion Common biological leaving group 0 b Write a mechanism that shows the conversion of ATP adenosine triphosphate structure shown below and methionine into SAM S adenosyl methionine NH2 0 N o ii o P o ii o lt j N N o o o O HO OH ATP NH2 CH3 N N I HN S H3N SCH3 3 w N N If 00 O H 90239 HO OH SAM Methionine c Why is methionine alkylated on the sulfur and not the oxygen atom even though the oxygen bears a negative charge and the sulfur is neutral d Write a mechanism that shows how SAM reacts with guanidoacetate to form creatine Ionic Substitution Reactions SN2 13 L JNE 1 H2N TACO H2N TACO H CH3 Guanidoacetate Creatine The mechanism step shown below contains a significant error Very brie y describe the error then write the corrected version of the mechanism step Try to keep your corrected version as close to the original as possible If nothing is wrong with the step write quotOKquot Hint the error is not quotmissing curved arrowsquot Curved arrows are used for mechanisms and are not normally used when the only concern is showing the product of a reaction Br KCN gt CH3CH20H Ibuprofen is an analgesic and anti in ammatory found in many over the counter drugs such as Motrin Imagine that you have inherited a factory that manufactures ibuprofen You are faced with many problems to improve the factory and its product H3C W COOH Ibuprofen A key step in the synthesis of ibuprofen in your factory is the reaction of chloroalkane A with cyanide ion quotArquot is a common abbreviation for an aromatic ring Chemical common sense tells us it is not an abbreviation for argon because argon is a noble gas and does not form chemical bonds H35 Nat 39C EN H3O s H Hi5 91 3 Ar Ar Cl CH3CH2OH Ar CN Ar CN w A B C a Select the major product b Draw the 8N2 mechanism including all transition states c Is this a reasonable 8N2 reaction Brie y explain Clearly state all assumptions d Haloalkanes such as compound A present environmental hazards if spilled As a concerned factory owner you want to avoid this problem and at the same time make the ibuprofen synthesis faster Draw an analog of haloalkane A that does not 14 Ionic Substitution Reactions SN2 have a halogen atom but undergoes a faster 8N2 reaction than A Brie y explain your reasoning e One of your employees approaches you with an idea to make an analog of ibuprofen with an extra methyl group A key step in the synthesis of this analog is the 8N2 reaction shown below Is this employee s idea reasonable or should this employee be fired due to a lack of chemical common sense Explain Ar Cl CH3CHZOH Ar C N D E 31 What is the major product formed when each of the following molecules is reacted with one equivalent of sodium cyanide in ethanol Br r a 0 WE H3C H 32 Design an 8N2 reaction in which the molecule bearing the leaving group has a stereocenter but the product does not 33 In your studies of organic and biochemistry you will encounter reactions that you have never seen before for which you need to figure out the mechanism You can do this by considering the bond changes necessary to convert the starting material into the products and deciding on a reasonable set of curved arrows In addition consider the reactants what have you seen them do in the past Suggest a mechanism for the following reaction which produces only the product shown H3CS H3CS H3CS ac etone osozcr3 I Only product None formed 34 Unlike most other 8N2 reactions the reaction of CH3I with CH3ZS is actually faster in a protic solvent than in an aprotic solvent Explain Practice Problems Solutions 1 For a reaction to occur molecules must collide with sufficient energy and correct orientation 2 The most significant reason that the transition state is the highest energy point between reactants and products along a mechanism step is the presence of significant van der Waals repulsive forces created between the nucleophile and electrophile Ionic Substitution Reactions SN2 15 3 a Rate k R7Cl Br H30 cii3 Your transition state drawing must clearly show the threeidimensional spatial arrangement ofthe groups attached to the carbon that bears the leaving group 1 4 B a 3 8 gtlt l Br 3 5 In the following models lone pairs are depicted with orbital paddles If your model does not appear to be the same as the answer shown here try changing it39s perspective Make sure your model has the correct absolute configuration at the carbon with the nucleophile and leaVing group The computeregenerated models may help you interpret the model photographs 3 b c 16 Ionic Substitution Reactions 8N2 gt1 CH3 Note that inversion occurs only at the carbon attacked by the nucleophile and not at any other acetone carbon IC1 I The role of a nucleophile or base is to share an electron pair to form a new covalent bond with the electrophile or proton so the single most important factor that controls nucleophilicity or basicity is the ability desire or driving force to share an electron pair The structural factor that has the most in uence over nucleophilicity is resonance Acetate ion CH3C02 has resonance that reduces the electron density on the oxygen atoms that share electrons with an electrophile Hydroxide ion HO and methoxide ion CH3O do not have this resonance dilution of charge Alternately upon reaction with an electrophile acetate ion loses resonance stabilization whereas the other two do not Acetate ion is therefore worst nucleophile in this group The role of a nucleophile is to share electrons with an electrophile Using the factors discussed in class we can evaluate the relative nucleophilicity of the species in this question The first four factors discussed below are presented in order of influence Resonance Not factor here not possessed by any of these species Atomic radius Sulfur is larger than oxygen so in an aprotic solvent the nucleophile that donates electrons from oxygen is the best one Electronegativity Of the three remaining nucleophiles all have an oxygen atom as the electron source Because there is no difference in electronegativity between oxygen atoms this is not a useful criterion here Inductive effect Trifluoroethanol CF3CHZOH is a weaker nucleophile than ethanol because the three fluorine atoms are withdrawing electron density from the oxygen Formal charge When everything else is equal a molecule bearing a negative charge is be more nucleophilic or basic than one without Thus ethoxide ion CH3CH20 is more nucleophilic than its conjugate acid ethanol CH3CH20H This analysis suggests the following ranking CH3CH20 gt strongest nucleophile gt CH3CHZS gt CH3CH20H gt CF3CH20H weakest nucleophile The best nucleophiles are in hold a CH3Cl F gt CHSF Cl atomic radius b CH3Br HO gt CH3OH Br atomic radius Ionic Substitution Reactions SN2 17 11 c CH3I NH3 gt CH3NH3 I electronegativity d CH3OSOZCF3 HZN gt CH3NH2 OSOZCF3 atomic radius electronegativity e CH3CH21 CHSCHZO gt CH3CH220 1 atomic radius inductive effect f H2CCHCH2Br inductive effect CHSCHZCHZO gt H2CCHCH20CH2CH2CH3 Br g PhCHZBr H0 gt PhCHZOH Br formal charge atomic radius h H2CCHCH2Cl CHSCHZCOZ gt H2CCHCH202CCH2CH3 Cl inductive effect a Recall that nucleophilicity is strongly influenced by charge density at the atom that shares electrons to form the new covalent bond Consider how various structural features influence the charge density of cyanide Formal charge The carbon end of cyanide bears a formal negative charge where as the nitrogen end is neutral The formal charge enhances the nucleophilicity of cyanide ion and suggests that the carbon end is more nucleophilic than the nitrogen end Resonance Cyanide has just one significant resonance contributor in which each atom has a lone pair and carbon bears a formal negative charge The lack of resonance dispersion of negative charge enhances the nucleophilicity of cyanide ion Atomic radius Carbon is a second row element and thus bears a high electron density This enhances nucleophilicity Electronegativity Carbon s electronegativity is modest 25 This does not significantly enhance or decrease the nucleophilicity of cyanide ion Inductive effect Nitrogen is more electronegative than carbon so nitrogen is pulling some electron density from carbon This dispersion of charge reduces the nucleophilicity somewhat Steric effects In an 8N2 reaction steric effects can influence electrophiles as well as nucleophiles Bulky nucleophiles cause transition state destabilization in the same way as bulky electrophiles Cyanide ion is thin like a spear so it provides very little steric hindrance This enhances the ions nucleophilicity Hybridization The carbon atom of cyanide is Sp Electrons in an Sp orbital are held more tightly than electrons in spz or SP3 orbitals The orbital becomes more 18 Ionic Substitution Reactions SN2 diffuse with increasing p character Electrons in more diffuse orbitals are further from the nucleus and thus not held as tightly This attenuates nucleophilicity b The major product depends upon which end of cyanide ion is more nucleophilic The carbon end bears a formal negative charge whereas the nitrogen end is neutral Carbon is less electronegative than nitrogen Taken together these observations suggest that the carbon end of cyanide ion is more nucleophilic than the nitrogen end so we predict the major product is acetonitrile 12 The only difference between the reactions is the atom of the nucleophile that forms a bond with the carbon bearing the leaving group Oxygen has a smaller atomic radius than sulfur Because this is an atomic radius effect on nucleophilicity we must also consider the solvent CH3OH which is protic A protic solvent forms hydrogen bonds with the nucleophile The strength of these hydrogen bonds is controlled by charge density Oxygen is smaller than sulfur so oxygen has greater charger density and therefore forms stronger hydrogen bonds These hydrogen bonds are broken when the solvent moves out of the way as the nucleophile approaches the electro phile The stronger hydrogen bonds formed by CH3O cost more energy to break than the hydrogen bonds to CH3S Therefore the CH3O activation energy is higher than the CH3S case and the CH3O reaction is slower 13 A good leaving group is stable after departure because it can easily accommodate the lone pair that was once the electrons that once were the carbon leaving group bond The three electron withdrawing fluorine atoms of trifluoromethanethiolate CF38 help stabilize the negative charge by the inductive effect making the negative charge of trifluoromethanethiolate more stable than the negative charge of methanethiolate CH3S Therefore trifluoromethanethiolate is predicted to be the best leaving group Illustration ltgt7scr3 139 gt 01 scr3 14 Carbon and fluorine are both second row elements so their atomic radii are about equal Fluorine is more electronegative than carbon so fluorine can accommodate the formal negative charge more effectively than carbon We know that fluorine is a very poor leaving group so methanide CH3 must be even worse than very poor 15 Best Trifluoromethanesulfonate ion CF3SO3 also called triflate has three resonance contributors and the inductive effect electron withdrawing of the CF3 group resulting in very effective dispersion of the negative charge Middle Methanesulfonate ion CH3SO3 also called mesylate has three resonance contributors as well but is destabilized by weak inductive electron density donation by the CH3 group Ionic Substitution Reactions SN2 19 H H H H N Poorest Acetate ion CH3COZ has only two resonance contributors to delocalize the negative charge as well as the same methyl group electron donation as in mesylate ion The rate of an 8N2 reaction is in uenced by the degree of steric hindrance at the carbon undergoing substitution Increasing the degree of substitution at this carbon makes the reaction slower In isopropyl iodide CH32CHI this carbon is secondary 2 and thus isopropyl iodide reacts more slowly than ethyl iodide CH3CHZI primary or methyl iodide CH3I methyl If the reaction occurs by the 8N2 mechanism the first reaction is the slowest because the carbon bearing the leaving group is more sterically hindered A reaction rate of Zero is less than a reaction rate of slow Dielectric constant e is controlled by molecular structure A molecule with a greater percentage of polar bonds versus nonpolar bonds has a greater dielectric constant Water consists of two polar O H bonds and no nonpolar bonds Methanol has two polar bonds C 0 and H 0 and three nonpolar bonds C H and thus a smaller 8 than water Ethanol has the same number of polar bonds as methanol but more nonpolar bonds and thus ethanol has a smaller 8 than methanol So CH3CH20H s 25 H20 8 80 and CH3OH s 33 It is not necessary to have memorized dielectric constants to answer this question In fact you are strongly discouraged from wasting your time by memorizing such data However the more problems you do the more often you will see this kind of data and the more familiar with it you will become 8N2 reactions work best on substrates that are less hindered Thus we rank CH3I methyl as fastest We might rank isopropyl uoride secondary second fastest but fluoride ion a very poor leaving group so poor that 8N2 reactions in which fluoride is the leaving group are very slow and therefore exceptionally rare The other two molecules are tertiary at the reacting carbon so they do not undergo 8N2 reactions Thus CH3I is fastest and the other three are slower because they don39t react at all the reaction rates are zero These molecules differ in leaving group as well as the degree of substitution at the carbon attacked by the nucleophile Increasing the steric hindrance at the carbon bearing the leaving group slows the 8N2 reaction Based on this criterion the molecule that reacts the slowest in an 8N2 reaction is the benzylic triflate From the perspective of leaving group Br and CF3SO3 are both sufficient leaving groups whereas H0 is a very poor leaving group Based on the leaving group criterion CHSOH reacts the slowest Either answer is acceptable A rate of zero ie no reaction is slower than any other rate including very slow 20 Ionic Substitution Reactions SN2 21 a Iodide ion has a greater atomic radius than chloride ion and in a protic solvent CHSOH is more nucleophilic so the first reaction is faster 1 Mechanism a gt c1 C a a c1 H H Changing chloride to a better leaving group iodide makes the reaction go faster b Iodide ion has a larger atomic radius than chloride ion so iodide ion is a better leaving group The second reaction is faster Mechanism SH H m SH Changing the 2 halide to a 1 halide makes an 8N2 reaction go faster because of reduced steric hindrance to nucleophilic attack NI LiSH SH CH30H N c An 8N2 reaction is sensitive to steric crowding The second reaction is substitution of a 3 iodide which does not proceed at all by the 8N2 mechanism The first reaction substitution at a 1 carbon is therefore the faster one ID scH3 Mechanism gt I 39 CstV 1 1 Changing to a less polar solvent increases 8N2 reaction rate when the nucleophile bears a formal negative charge and the electrophile is neutral The new solvent must still be sufficiently polar to dissolve K SCH3 an ionic compound Acetone s 21 is less polar than DMF s 37 but still dissolves many ionic substances I SCH KSCH3 gt acetone I I l393r 6 t In E 22 a Br quot39L39g H 07mm Br39 SCH 5 cn3 Ionic Substitution Reactions SN2 21 NaSCH3 Slower reactlon C1 gt SCH3 CH30H This reaction is slower because chloride ion is a poorer leaving group than bromide ion N w39 0 t r0 CH3 5 EACH 0 CH3 b CH CH 3 2 CH Q1 0 ECH3 3 o I a 3002 0 CF Slower reaction quot ta 1 DMF 9 O I This reaction is slower because tri uoroacetate ion CF3COZ is a poorer nucleophile than acetate ion H C Nal 3911 Slower reaction 39 quot gt NR 3H3 CH3 acetone This reaction is slower because a tertiary alkyl halide is more sterically hindered than a secondary alkyl halide The reaction is slow that it does not give any detectable amounts of product hence the label NR which means no reaction CH3s 5 H 39 23 a W n CH3 H lquot Br 39 CH3 H CH3 CH3 Br 6 same as same as Br SCH CH3 CH3 b Any change in the electrophile that slows the reaction is acceptable For example changing Br to F a very poor leaving group or adding steric hindrance to the 22 Ionic Substitution Reactions SN2 carbon undergoing nucleophilic attack changing a secondary carbon into a tertiary carbon shuts down the 8N2 mechanism uF CH3SNa gt No reaction CH30H CH3 c Any change in the nucleophile that makes the reaction faster is acceptable Replacing methanethiolate with a nucleophile that suffers less hydrogen bonding iodide ion is one solution Changing the solvent may change nucleophilicity but it does not change the nucleophile itself Br I NaI gt CHgOH CH3 CH3 24 We need to consider the differences between the three reactions and how these differences influence the rate of an 8N2 reaction In 2 iodo 2 methyl propane also called tert butyl iodide the leaving group is bonded to a tertiary carbon whereas in 2 bromopropane and 2 chloropropane the leaving group is bonded to a secondary carbon Increasing steric hindrance at the carbon bearing the leaving group retards the rate of an 8N2 reaction A tertiary carbon is so highly hindered to 8N2 attack that no 8N2 reaction occurs Thus 2 iodo 2 methylpropane the slowest of the three 2 Bromopropane and 2 chloropropane differ in their leaving groups Bromide ion is a better leaving group than chloride ion because bromide ion has a larger atomic radius and can disperse the charge more effectively Thus 2 bromopropane reacts faster than 2 chloropropane Therefore the rate ranking is 2 bromopropane gt 2 chloropropane gt 2 iodo 2 methylpropane 25 a Mechanism possibilities gt o gt JIon V o B WE 5 C O gt gt Iliv I o HROH b Due to formal charge an alkoxide ion RO mechanism A is more nucleophilic than an alcohol mechanism B Therefore mechanism A occurs more quickly ie is the pathway most of the reactants follow to become the given product Ionic Substitution Reactions SN2 23 d This is a two step mechanism in which the first step the proton transfer is much faster than the substitution second step Therefore the second step is rate determining We will discuss this issue in greater depth during our studies of the SNl reaction Therefore the rate expression is rate k RO Even though the rate expression is unimolecular it is still categorized as an 8N2 reaction because it is ionic substitution at an SP3 carbon without the intermediacy of a carbocation e Intermolecular 8N2 with HO can form 14 butanediol HOCHZCHZCHZCHZOH The reaction products arise from attack by the same nucleophile iodide ion at a different site on the electrophile There are two fundamental differences between these reaction pathways Leaving group Triflate CF3SO3 gains three significant and energetically equal degenerate resonance contributors upon departure Methyl sulfite CH3OSOZ gains three significant but energetically unequal resonance contributors upon departure s OCH3 lt gt s OCH3 lt gt OCH3 O mO Recall that degenerate resonance contributors provide more stabilization than do resonance contributors that are not degenerate Review the resonance tutorial on the course website if needed Therefore CF3SO3 delocalizes its negative charge more evenly and is thus more stable than CH3OSOZ Triflate is the better leaving group Steric effects Fluorine van der Waals radius 1355 is a bit larger than hydrogen van der Waals radius 12 A so attack at the CF3 group suffers a bit more steric hindrance than attack at the CH3 group The difference is much smaller than the difference between a CH3 and a primary carbon 24 Ionic Substitution Reactions SN2 o o Mechanism II C r CF3 gt I CH3 39o Isl CF3 0 O DNA NH 5 1 2 DNA 1le2 H H 5H N 27 a CH3CH2 39 CH DNA 3 I DH I a H3C b Assuming the DNA alkylation reaction is an 8N2 process tert butyl iodide is unreactive because it is a tertiary haloalkane and thus too sterically hindered to allow the nucleophile to approach the backside of the carbon iodine bond gt2 I c Reason 1 The carbon in magic methyl that bears the leaving group a methyl carbon is less sterically hindered than the corresponding carbon in 2 iodobutane a secondary carbon Reason 2 Fluorosulfonate ion F803 is a better leaving group than iodide ion due to a combination of resonance and the inductive effect of the uorine atom 28 a The negative charge gained upon departure is delocalized by resonance This makes triphosphate ion more stable and thus a better leaving group 0 o o o b R S CH3 gt SAM 39o iJ o iJ o ii o39 I L J 4 ATP Methionine c Alkylation at oxygen disrupts the resonance stabilization of the carboxylate ion RCOZ Alkylation at sulfur does not detract from the existing resonance NH2 NH2 H2N NAcof ch SR2 gt HZN NOW gt Creatlne H B 1391 ch HquotB Guanidoacetate SAM B is a base of some sort most probably a nitrogen containing functional group such as primary amine within the enzyme active site where this reaction occurs 29 This reaction cannot occur as written because the carbon atom that gains the nitrile group CN has eight electrons to begin with Forming a new bond to the nitrile group without losing an existing bond results in a pentavalent carbon ten electrons The reaction also cannot occur as written because hydride ion HT is not a leaving group Ionic Substitution Reactions SN2 25 except in the Chichibabin reaction One way to rewrite this as a valid reaction is to show the Br leaving an 8N2 reaction KCN Br gt GEN CH3CHZOH 30 a An 8N2 reaction proceeds with inversion of configuration so the correct answer is product C H c H3C l H c b 3 i 57 3 NC NEC 39c1 Ar c1 6 Ar NC Use models to convince yourself that the product shown here is the same as product C The only di erence is the perspective from which the molecule is viewed will 3901 I 2 c An 8N2 reaction requires 39 Moderate or better leaving group chloride is a moderate leaving group 39 Good nucleophile cyanide ion is an excellent nucleophile because the negative charge is localized on carbon and because it is a skinny ion that can readily fit into tight spaces 39 The carbon that accepts the nucleophile cannot be tertiary it is secondary in this case and 39 Solvent The solvent is sufficiently polar ethanol 8 25 It is also protic Hydrogen bonding hinders but does not necessarily prevent cyanide ion from being a nucleophile Thus we conclude this is a reasonable 8N2 reaction ch d Structure of compound A analog Argt0CH3 II o Methanesulfonate ion CH3SO3 is a better leaving group than chloride ion due to resonance stabilization of the negative charge Other answers may also be acceptable e An 8N2 reaction does not occur at a tertiary center so this employee should be fired for suggesting this 8N2 reaction 26 Ionic Substitution Reactions SN2
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'