Table of contents. 44 and ketones Nucleophilic acyl substitution 48 Substitution reactions of carbonyl

Transcription

1 Table of contents Subject Page # Introduction 1 Introductory concepts 2 Acids and bases 5 Alkane conformations 7 Stereochemistry 10 Elimination and substitution 14 Alcohols, ethers, epoxides 18 Alkenes 22 Alkynes 26 Oxidation and reduction 28 Radicals 32 Diers alders and dienes 34 Electrophilic aromatic substitution 37 Organometallic reagents 41 Nucleophilic addition to aldehydes 44 and ketones Nucleophilic acyl substitution 48 Substitution reactions of carbonyl 51 carbons at the alpha carbon Carbonyl Condensations 55 Amines 59

2 1 How to use my notes: Before using my notes as a study tool, please watch chad's videos or use other materials to learn the basic groundwork information. My notes are intended to cover the harder topics/topics that were not covered in chad's videos/groundwork courses to help increase your scores to a 30. These are a complete set of notes for concepts and reactions organic chemistry. I got a perfect score on organic chemistry and it was the easiest science section for me by a mile. Chad's videos = Sections not included (watch chad's videos on these specific sections he does a fantastic job here and nothing more needs to be added than what he says): 1.1 Nomenclature 5.1 Lab Techniques Extractions 5.3 TLC, Gas Chromatography 5.4 Distillation 5.5 IR spectroscopy 5.6 NMR 6.1 Proteins and Amino Acids 6.2 Carbohydrates 6.3 Lipids and Nucleic Acids Learn Chemical lab tests for alkanes, alkenes, alcohols, and alkyl halides in DAT Destroyer (page 135). If you don't have DAT destroyer, all the information is in a quizlet here: This shows up on the DAT so definitely learn this! To get the most out of my notes, study in the following way: 1) Memorize everything in the notes (MOST IMPORTANT). Memorize the mechanisms that I laid out (understanding the mechanism gives you better understanding of the reactions). If you don't understand a certain reaction, feel free to message me on SDN or use the internet or whatever resources you have available. 2) once you have memorized everything in the notes, begin practice problems. I HIGHLY suggest you buy organic chemistry odyssey and DAT destroyer by orgoman (fantastic problem sets). Whenever you get something wrong, understand why you got it wrong, write a general rule on how to solve that specific example, and then add it to my notes. 3) after you have gone through the practice problems, re-memorize areas of my notes that you slacked on and the new additions you added while doing practice problems. Quick note on acid/bases memorize the pka table given in my notes. I know it may be annoying, but it will make your life 100 times easier when solving problems that ask you for the most acidic molecule/proton. Happy studying!

3 2 Electronegativity Electronegativity is a measure of how much atoms want electrons. Electronegativity differences between two atoms helps tells you which type of bond they participate in. Nonpolar bond has electronegativity differences less than 0.5. Polar, covalent bonds have electronegativity differences between 0.5 and 2. Ionic bonds generally have electronegativity differences greater than 2. Trend: Lewis structures special notes Elements not part of the transition metal block and are in the 3 rd period and below can have expanded octets. Boron is electron deficient, only needed 6 electrons to fill its shell. When adding up the valence electrons in a molecule, if you get an odd number, then it is a radical. Formal charge The formal charge is the charge of an atom in a molecule. By adding all the formal charges of all the atoms in a molecule, you can determine if the overall molecule is positive, negative, or neutral. Equation: Resonance structures A resonance structure is another way to depict the same molecule. Only electrons can move around, not atoms! How to determine importance of resonance structure to the overall structure (most important to least important factors): 1) All atoms have full octets 2) Minimal separation of charge 3) More electronegative atom should have the negative charge whereas the more electropositive atom should have the positive charge. In reality, all resonance structures for the same molecule exist at the same time, all the time. Most important resonance contributor best describes the shape of the molecule whereas the second most important resonance contributor best describes the reactivity of the molecule.

4 3 Bond length Bond length describes the distance of a bond between two atoms. How do rank bond lengths: Bond length follows the atomic radius trend. Larger atoms that bond together will have larger bond lengths for this reason. (H-Br > H-F) single bond > double bond > triple bond higher % s character (sp > sp 2 > sp 3 ) of the atoms participating in the bond, the shorter the bond NOTE: watch for resonance! This is NO 2-, or nitrite. When looking at the N-O bond lengths, one might be tempted in saying that one N-O bond is strictly a double bond while the other is a single bond.. However, this is not true! Note that the molecule can resonate! Because of this, the N-O bond length is actually less than 2, it is around 1.5. When given a bond length problem, look for resonance because that can affect the way you answer the question. Bond strength Bond strength indicates how much energy is needed to break a bond into two equal atoms. General rule: the shorter the bond is, the stronger the bond is How to rank bond strength: triple bond > double bond > single bond Higher % s character (sp > sp 2 > sp 3 ) of the atoms participating in the bond, the stronger the bond strength. This is because the atoms participating in the bonds are in lower energy orbitals and are more tightly bound by the nucleus. NOTE: watch for resonance (just like bond length)! Molecular geometry Determining hybridization: 1) count every single, double, or triple bond around a given atom as 1 2) count every lone pair around the same atom as 1 3) add up the two counts and then use the table to determine hybridization Count Hybridization 2 sp 3 sp 2 4 sp 3 5 sp 3 d 6 sp 3 d 2 NOTE: watch for resonance! The resonance structure with the smallest observed hybridization best describes the overall hybridization! sp 3 = ~109 o bond angles and a tetrahedral arrangement 4 bond sand 0 lone pairs = tetrahedral 3 bonds and 1 lone pair = trigonal pyramidal 2 bonds and 2 lone pairs = bent 1 bond and 3 lone pairs = linear sp 2 = ~120 o bond angles and trigonal planar arrangement

5 4 3 bonds and 0 lone pairs = trigonal planar 2 bonds and 1 lone pair = bent 1 bond and 2 lone pairs = linear sp = ~180 o bond angles and linear arrangement lone pair electrons take up extra space and crunches the bond angles. The more lone pairs that exist around an atom, the smaller the bond angles are. Intermolecular forces Intermolecular forces are forces that occur between neighboring particles. They have big effects on physical properties. In terms of strength: Ion-dipole > hydrogen-bonding > dipole-dipole > van-der-waals Ion-dipole is an interaction between an ion and an oppositely charged dipole. Hydrogen bond is the interaction between a lone pair on fluorine, oxygen, or nitrogen and a hydrogen on another molecule that is directly bound to fluorine, oxygen, or nitrogen. NOTE: hydrogen bonding can occur between atoms within the same molecule. Intramolecular H-bonds prevent intermolecular H-bonds from forming and thus weakens its effect on physical properties. Dipole-dipole occurs from polar covalent bonds. NOTE: dipoles within the same molecule can cancel each other out if the dipoles are in the opposite directions and are of the same magnitude. Factor in the NET molecule dipole. Van-der-waals occurs between all atoms. It is the brief attraction between neighboring molecules due to the random movement of electrons. The larger the molecular weight, the stronger the VDW forces. The bigger the molecule is (more surface area), the stronger the VDW forces. The more branched the molecule is, the weaker the VDW forces. Melting points, boiling points, and solubility Boiling point ranking: 1) The stronger the IMFs, the higher the boiling point Note that for hydrogen bonding, oxygen > nitrogen (you will probably never see a fluorine molecule). 2) The larger the surface area of the molecule, the higher the BP 3) The more polarizable the atom is, the higher the BP Polarizability is described as the ease of distortion of the electron cloud of a molecule by an electric field. Polarizability increases down the column of the periodic table. (I > Br > Cl) NOTE: HF > HI > HBr > HCl for boiling point (note that HF can do hydrogen bonding whereas the other 3 acids cannot) Melting point ranking: 1) The stronger the IMFs, the higher the melting point 2) The more branched the molecule is, the higher the melting point Solubility ranking: 1) similar IMFs between solvent and solute 2) the larger the surface area of the solute is, the less dissolving there is

6 5 Acid and base definitions Bronsted-lowry acid is a proton (H+) donor Bronsted-lowry base is a proton (H+) acceptor Lewis acid is an electron acceptor Lewis base is an electron donor Acid strength MAJOR TIP: memorize pka values as it will make ranking acids much easier. I put a terrific table on the very last page of this chapter: memorize the values. General factors that determines acidity (most important to least important): 1) Acidity of H-A increases left to right across a row and down the periodic table (polar and polarizeable). The opposite trend applies to base strength. 2) Acidity of H-A increases if the conjugate base can resonate NOTE: if a ring, when deprotonated, becomes an aromatic ring, then acidity greatly increases. 3) Acidity of H-A increases with the presence of electronegative atoms in A (inductive stabilization) NOTE: distance matters! When an electronegative atom is closer to the proton, the effect is stronger and acidity increases. 4) Acidity of H-A increases as the % s character of the conjugate base increases (sp > sp 2 > sp 3 ). NOTE: intramolecular H-bonding decreases acidity as the hydrogens are less likely to be deprotonated. Basicity of amines and aromatics Alkyl amines are more basic than NH 3 because of electron donating R-groups. Alkyl amines are more basic than aryl amines. Electron donating groups (refer to EAS chapter) increase basicity/decrease acidity on aryl amines. Electron withdrawing groups (refer to EAS chapter) decrease basicity/increase acidity on aryl amines. In heterocylic aromatic compounds, if the heteroatom's (nitrogen, oxygen, etc.) electrons do not contribute to aromaticity, then it is more basic. Higher % s character (sp > sp 2 > sp 3 ) of the orbital containing the lone pair, the less basic/more acidic the compound is. Determining K eq pka conjugate acid pka acid K eq = 10

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8 7 Newman Projections Newman projections allow us to see the relationship of substituents between the front and back carbons. Staggered conformation is when the substituents are not overlapping with each other. Eclipsed conformation is when the substituents are overlapping with each other. A given molecule will rotate between the two conformation many times per minute. Staggered is more stable than eclipsed. Anti vs. gauche vs. partial vs. full eclipse Anti conformer is a type of staggered conformer where the non-hydrogen substituents are 180 o away from each other. This is the most stable. Gauche conformer is a type of staggered conformer where the non-hydrogen substituents are 60 o away from each other. This is the 2 nd most stable. Partial eclipsed conformation is a type of eclipsed conformation where the non-hydrogen substituents are 120 o away from each other. This is the 3 rd most stable. Fully eclipsed conformation is a type of eclipsed conformation where the non-hydrogen substituents are 0 o away from each other. This is the least stable.

9 8 Since a molecule will rotate between staggered and eclipsed conformations, a molecule with non-hydrogen substituents will rotate between all 4 conformers stated above in a given minute. NOTE: if the two substituents bound to the two adjacent carbon atoms can hydrogen bond, make them gauche. Intramolecular hydrogen bonding increases stability of the molecule! Types of strain Torsion strain arises from eclipsed interactions between hydrogen and a non-hydrogen substituent or between 2 hydrogens. Steric strain occurs when 2 non-hydrogen substituents are gauche or fully-eclipsed. Angle strain occurs when atoms participating in a ring have bond angles lower than the expected amount. Conformations of cyclohexane Some rings, such as cyclohexane, will pucker to relieve strain. How to convert from ring to chair: 1) Number the cyclohexane ring. 1 goes to the top priority substituent ( top priority is defined in the nomenclature chapter) and then number in the direction so that the nearest substituents is at the lowest number. 2) Number the boat counterclockwise or clockwise as long as you are consistent (in the picture below, it was numbered clockwise) 3) wedge always points UP, and dashed always points DOWN 4) chair flip: move numbers down one clockwise and flip every axial substituent to equatorial and every equatorial substituent to axial Axial vs. equatorial Axial positions are the ones that are pointing up and down vertically Equatorial positions are the ones that are pointing horizontally on a slant. MEMORIZE the axial and equatorial positions for each chair Determining the most stable chair Non-hydrogen substituents do not like to be in the axial position. Doing so causes them to bump into other axial substituents, called 1,3-diaxial interactions. No such problems exist for equatorial positions! Therefore, the largest, if not all substituents that are not hydrogen should be equatorial.

10 9 A note on cis and trans If the non-hydrogen substituents are both pointing up or down, it is cis. Look at the cis- 1,2-dimethyl-cyclohexane picture: note that both CH 3 groups are always pointing up If one is pointing up, and the other down, it is trans.

11 10 Labeling R and S stereocenters Please refer to this link to learn how to prioritize substituents. S_Sequence_Rules Given dash, wedge structure: lowest priority group is dashed: number substituents using priority system draw a steering wheel from 1 to 3 if the steering wheel is clockwise, it is R. If the steering wheel is counterclockwise it is S. In the picture below, draw a steering wheel from 1 to 2 and to 3. Wen you do this, you will see that you are going in a clockwise direction. So this stereocenter is S. lowest priority group is wedged: number substituents using priority system draw a steering wheel from 1 to 3 REVERSE! If the steering wheel is clockwise, it is S. If the steering wheel is counterclockwise it is R. In the picture below, NH 2 is priority 1, CH 2 CH 3 is priority 2, CH 3 is priority 3, and H is priority 4. When you draw the steering wheel, you will see a clockwise direction. So the stereocenter is R! Lowest priority group is neither dashed nor wedged: rotate molecule (in either direction) so that the lowest priority group is dashed number substituents using priority system draw a steering wheel from 1 to 3 REVERSE! If the steering wheel is clockwise, it is S. If the steering wheel is counterclockwise it is R. In the picture below, H is priority 4 and is neither dashed nor wedged. So here, rotate the molecule (do so with least amount of turns needed) to get H in dashed. Rotating the molecule counterclockwise does the trick (figure on the right). Then assign priorities, do steering wheel, and REVERSE. The stereocenter here is R.

12 11 Given Fischer projection lowest priority group is horizontal: number substituents using priority system draw a steering wheel from 1 to 3 REVERSE! If the steering wheel is clockwise, it is S. If the steering wheel is counterclockwise it is R. lowest priority group is vertical: number substituents using priority system draw a steering wheel from 1 to 3 if the steering wheel is clockwise, it is R. If the steering wheel is counterclockwise it is S. In the figure below, hydrogen is the lowest priority substituent and it is in a vertical position. Simply number the substituents and steering wheel. The stereocenter here is R as the steering wheel is clockwise. If there are 2+ centers of chirality in the fischer projection, consider each center separately Newman projections Watch this youtube video on how to convert a newman projection into a fischer projection Once you converted into the fischer projection, use the fischer projection rules to assign the stereocenters. Chair structure Convert chair back into ring (remember wedge = up and dashed = down) Find stereocenter with dash, wedge structure you created Definitions Chiral molecules have at least one stereocenter and is not a meso compound or a racemic mixture. Rotates plane-polarized light.

13 12 A stereocenter is a tetrahedrally arranged carbon bound to 4 asymmetric substituents. Total number of stereoisomers = 2 n n = number of stereocenters Enantiomers are non-super imposable images. The way I view enantiomers is that every stereocenter has oppositely assigned R, S values. So if molecule A has R configuration at C1 and S configuration at C2, then the enantiomer would have S configuration at C1 and R configuration at C2. Enantiomers are equal in all of their physical properties except for optical rotation. They rotate plane-polarized light with the same magnitude but in opposite directions (so opposite signs). A racemic mixture is a 1:1 ratio of enantiomers. They are achiral (do not rotate planepolarized light) because the equal ratios of enantiomers cancel out the rotations. A meso compound is a molecule with at least one stereocenter and is symmetrical. Meso compounds are achrial. A diastereomer is anything but an enantiomer. The way I view diastereomers is that at least one of the stereocenters have been changed, but not all of them. In the molecules below, note that only one stereocenter has been swapped and the other two remain the same. Diastereomers are characteristically different in their physical properties and thus can be easily separated by conventional separation techniques.

14 13 Determining cis (Z) and trans (E) 1) Split the alkene in half 2) Determine the highest priority groups bound to each carbon that participates in the double bond. Priority is determined in the same way when assigning R/S stereocenters. 3) If the highest priority groups are on the same side, use the Z notation (think on ze zame zide). If the highest priority groups are on the opposite side, use the E notation. Cis and trans isomers have different physical properties and thus can be easily separated by conventional separation techniques. In general, the trans isomer has the higher melting point whereas the cis isomer has the higher boiling point.

15 14 Alkyl halides An alkyl halide is an alkane with a halogen bonded to a carbon. Types (X = halogen): Methyl halide = CH 3 X Primary alkyl halide = RCH 2 X Seocndary alkyl halide = R 2 CHX Tertiary alkyl halide = R 3 CX Carbocations Carbocations are positively charged carbon atoms. + Methyl carbocation = CH 3 + Primary carbocation = RCH 2 Seocndary carbocation = R 2 CH + Tertiary carbocation = R 3 CX + Tertiary > Secondary > Primary > Methyl in terms of stability Leaving group A leaving group is the species that leaves in a substitution or elimination reaction. The better the leaving group, the faster the reaction. How to determine leaving group strength: The less basic/more acidic, the better the LG LG strength follows the periodic trend for acidity Protic vs. Aprotic solvent A solvent is a liquid that serves as the medium for a reaction. It can serve two major purposes: (Non-participatory) to dissolve the reactants. Polar solvents are best for dissolving polar reactants (such as ions); nonpolar solvents are best for dissolving nonpolar reactants (such as hydrocarbons). Participatory: as a source of acid (proton), base (removing protons), or as a nucleophile (donating a lone pair of electrons). The only class of solvents for which this is something you generally need to worry about are polar protic solvents (see below). Protic solvents have O-H or N-H bonds. These bonds serve as sources of protons. We see that protic solvents favor Sn1 and E1 reactions. Aprotic solvents may have hydrogens on them somewhere, but lack O-H or N-H bonds. We see that aprotic solvents favor Sn2 and E2 reactions.

16 15 Electrophiles vs. Nucleophiles Electrophiles accept electrons whereas nucleophiles donate electrons. For this chapter, the electrophile will always be an alkyl halide. A stronger nucleophile will lead to a faster elimination or substitution reaction. How to determine nucleophile strength: In polar, protic solvents: Nucleophilicity increases as the attacking atom moves down a group of the periodic table. Nucleophilicity increases right to left across a row of the periodic table. In polar, aprotic solvents: Nucleophilicity increases as the attacking atom moves up a group of the periodic table. Nucleophilicity increases right to left across a row of the periodic table. If comparing the same atom species, the more basic molecule = the better nucleophile. Sn1 rate = k[rx] Rate law only depends on the alkyl halide concentration, not the nucleophile concentration. The first step is rate determining step (slow step) Likes to happen in protic solvents Favored by weaker nucleophiles Stereochemistry: get both stereoisomers (retention and inversion), but slightly more retention than inversion occurs. Alkyl halide that reacts the fastest forms the most stable carbocation. Tertiary > secondary > primary > methyl halide Mechanism shown below: Sn2 rate = k[rx][nucleophile] Rate law depends on the concentration of the alkyl halide and the nucleophile Likes to happen in aprotic solvents Favored by stronger nucleophiles

17 16 Transition state is triagonal bipyramidal. Leaving group has a negative charge and the nucleophile has a positive charge. The carbon involved changes from an sp 3 to an sp 2 hybridization. Stereochemistry: inversion of stereocenter (called this backside attack) Beta branching (branching at the beta carbon) slows down the reaction Favored by unhindered alkyl halides: methyl halide > primary > secondary > tertiary The mechanism occurs in a single, concerted step. E1 rate = k[rx] Rate law only depends on the alkyl halide concentration, not the nucleophile concentration. The first step is rate determining step Likes to happen in protic solvents Favored by weaker bases More substituted alkene is favored (zaitev's rule) E2 rate = k[rx][nucleophile] Rate law depends on the concentration of the alkyl halide and the nucleophile Likes to happen in aprotic solvents Favored by stronger bases more substituted alkene is favored (zaitev's rule) There must be an antiperiplanar arrangement of H and X, meaning they must be 180 degrees apart. If H and X are not antiperiplanar, rotate the bond until it is. If you cannot rotate the bond to make H and X antiperiplanar, no reaction will occur. How can I tell if H and X are antiperiplanar? Draw a newman projection! If you are doing an elimination of a cyclohexane ring, draw the chair and see if H and X are 180 degrees apart. Note that you may even need to flip the chair in order to see the antiperiplanar relationship (as shown below) Sometimes, you will only see that one anti-periplanar relationship is possible and that may be the less substituted double bond. For more information, visit

18 17 The mechanism occurs in a single, concerted step. Choosing between E1, E2, Sn1, and Sn2 NOTE: bicyclo compounds, halides attached to aromatic rings or to carbon that participates in a double/triple bond do not react in these schemes. Step 1: determine if you have primary, secondary, or tertiary alkyl halide. If I have a primary alkyl halide: Step 2: determine if the base is bulky If it is, then do E2 If it is not, then do Sn2 NOTE: if the base is a bad nucleophile, then there will be no reaction If I have a secondary alkyl halide: Step 2: determine if the base is bulky If it is, then do E2 If it is not, proceed to step 3 Step 3: determine if the base is strong If it is, then E2 and Sn2 will occur, but E2 will be favored If it is not, proceed to step 4 Step 4: determine if the species is a good nucleophile If it is, then E2 and Sn2 will occur, but Sn2 will be favored If it is not, then Sn1 and E1 will be favored If I have a tertiary alkyl halide: Step 2: determine if the base is strong If it is, then do E2 If it is not, then do Sn1 and E1 What does strong base mean? For our purposes, look for anions (e.g. RO-) that are not sulfur-based. Weak bases will be protonated (e.g. ROH) What is a bad base but a good nucleophile? For DAT, think of sulfur-based nucleophiles and cyanide nucleophiles. Benzylic and allylic halides do Sn1 and Sn2 depending on the solvent. Note that if you can create a conjugated double bond system via elimination, do elimination! A special note on bulky bases: (CH 3 ) 3 CO - K + is hoffman elimination. Form the less substituted double bond! C 2 H 5 O - Na + is standard E2 elimination (zaitev's rule) Alcoholic * KOH (1 mol) + heat gives you E2 elimination whereas Aqueous * KOH (1 mol) + heat gives you Sn2 substitution

19 18 The alkoxide nucleophile An alkoxide anion is RO- It participates in Sn2 with Ch3X or primary alkyl halides It participates in Sn2/E2 (more E2) with secondary and tertiary alkyl halides How to make an aloxide use NaH to deprotonate alcohols (acid-base mechanism). The byproduct, H 2 gas, will bubble out of the solution. Williamson ether synthesis React an alkoxide with an alkyl halide to generate an ether. Mechanism is Sn2 inversion of stereochemistry and as a result, the alkyl halide cannot be hindered (the alkoxide can be hindered though). Only use primary or secondary alkyl halides! The solvent is typically the conjugate acid of the alkoxide. Using NaH to form expoxides If the molecule has an OH group and a halogen 1 carbon away (vicinal), use NaH to form the epoxide. Heating alcohols Heat a primary alcohol to 140 o C to create the ether. Heat a primary alcohol to 180 o C to create the alkene. The more substituted alkene will be favored (zaitev's rule). Elimination of tertiary alcohols using strong acid Use H 2 SO 4, TsOH, H 3 O +, etc. on a tertiary alcohol alcohol to generate the alkene. The acid protonates the -OH group to generate a good leaving group. Mechanism is E1.

20 19 Elimination of secondary alcohols using strong acid Use H 2 SO 4, TsOH, H 3 O +, etc. and heat on a secondary alcohol alcohol to generate the alkene. The heat helps stabilize the secondary carbocation. The mechanism is E1 (exactly the same as the mechanism for tertiary alcohols) NOTE: watch out for carbocation shifts! Elimination of alcohols without carbocation generation Add POCl 3 and pyridine to form the alkene These reagents are advantageous because there is no carbocation generation. If you want to eliminate a primary alcohol, you must use these reagents. These reagents also work on secondary and tertiary alcohols. Mechanism is E2. Remember that an antiperiplanar arrangement is needed! Generating tertiary alkyl halides from tertiary alcohols Use HI, HCl or HBr on a tertiary alcohol to generate the tertiary alkyl halide. Mechanism is Sn1. Remember that there will be retention and inversion! Generating secondary alkyl halides from secondary alcohols Use HI, HCl or HBr on a secondary alcohol to generate the secondary alkyl halide. Mechanism is Sn1. Remember that there will be retention and inversion! Watch out for carbocation shifts! Generating primary alkyl chlorides from primary alcohols Use ZnCl 2 on a primary alcohol to generate the alkyl chloride. Note that there will be no carbocation shifts! Mechanism is Sn2. Remember that there will be inversion at the stereocenter! Make R-Cl from primary or secondary R-OH Use SOCl 2 and pyridine to generate the alkyl chloride. Note that there will be no carbocation shifts! Mechanism is Sn2. Remember that there will be inversion at the stereocenter!

21 20 Make alkyl bromide from a secondary alcohol Use PBr3 to generate the secondary alkyl bromide. Note that there will be no carbocation shifts! Mechanism is Sn2. Remember that there will be inversion at the stereocenter! Convert R-OH into R-Ots Add TsCl + Pyridine to generate the tosylate. The point of this reaction is to turn the alcohol into a molecule with a good leaving group. The tosylate is an excellent leaving group (approximately as good as Br-). Ether cleavages Add HI, or HBr in excess to cleave an ether into alkyl halides. Do NOT use HCl! An ether with a primary or secondary alkyl group attached to the oxygen will undergo Sn2. An ether with a tertiary primary alkyl group attached to the oxygen undergoes Sn1. Sn1 is faster than Sn2 in this situation. (3 o > 2 o or 1 o ) 1 o is faster than 2 o since both alkyl groups undergo Sn2 but a primary alkyl group is better suited for Sn2. (1 o > 2 o ) If the question is an ether cleavage but only one equivalent of reagent is used, the fastest reaction will occur first! (3 o > 1 o > 2 o ) If the ether is directly attached to an aromatic ring, keep the phenol! Epoxide opening under basic conditions Use strong nucleophiles to open an epoxide under basic conditions (RS -, RO -, RCN - ) Mechanism is Sn2. Remember that there will be inversion at the stereocenter! Nucleophile attacks the least hindered side: 1 o > 2 o > 3 o

22 Epoxide opening under acidic conditions Use an acid catalyst (H 2 SO 4, H 3 O + ) and a weak nucleophile (H 2 O, ROH, HBr, HI) Mechanism is Sn2. Remember that there will be inversion at the stereocenter! Nucleophile attacks the most hindered side: 3 o > 2 o > 1 o 21

23 22 General Notes on Alkenes Alkenes are somewhat nucleophilic due to its pi bond. Since alkene carbons have sp2 hybridization, they are more acidic than alkanes. Alkenes tend to undergo electrophilic addition reactions. Degrees of unsaturation (DOU) Given a picture of the molecule: DOU = # rings + # pi bonds Maximum Hs Actual Hs Given a molecular formula: DOU= 2 Maximum Hs = 2n + 2 n = # of carbons in the molecule To calculate number of actual Hs: Add up every hydrogen in the formula given. Add 1 to actual Hs for every halogen. Subtract 1 to actual Hs for every nitrogen. Ignore oxygen in calculations. Hydrohalogenation of alkenes Add HI, HCl, HBr to an alkene to create the corresponding alkyl halide. Stereochemistry: syn + anti addition Markovnikov addition: add H to less substituted carbon Watch out for carbocation shifts! In general, you get syn and anti addition when you form a carbocation! Acid-catalyzed addition of H 2 O or ROH to alkenes Add a strong acid catalyst (i.e. H 2 SO 4 ) + H 2 O or primary alcohol to an alkene to generate the corresponding ether or alcohol. Stereochemistry: syn + anti addition Markovnikov addition: add H to less substituted carbon Watch out for carbocation shifts!

24 23 Halogenation Add Br 2, Cl 2 but NOT I 2 to generate the vicinal 1,2 dihalide. stereochemistry = anti addition In the mechanism, you generate a halonium ion intermediate. Nucleophilic X - attacks the more hindered side of the halonium ion. Hydroboration/oxidation of alkenes Add (1) BH 3 (2) H 2 O 2 & NaOH to generate the alcohol. Stereochemistry = syn addition anti-markovnikov addition: add OH to the least substituted carbon. Oxymercuration/deoxymercuration and alkoxymercuration/dealkoxymercuration Oxymercruation: 1) Hg(OAc) 2 /H 2 O 2) NaBH 4 no carbocation shifts no stereochemistry Markovnikov addition of H and OH to make an alcohol.

25 24 Alkoxymercuration: 1) Hg(OAc) 2 /ROH 2) NaBH 4 no carbocation shifts no stereochemistry Markovnikov addition of H and ROH to make an ether. Halohydrin formation Add Br 2 or Cl 2 in H 2 O solvent to an alkene to form a halohydrin. NBS (N-bromosuccinimide) can be used instead of Br 2. Stereochemistry = anti addition OH- adds to more substituted carbon and the halogen adds to the less substituted carbon. Similar to the reaction up above, you can add an alcohol along with X 2 instead of water to generate the ether. Simmonds-smith reaction Add CH 2 I 2, Zn [Cu] to an alkene to generate a cyclopropane. Stereochemistry: trans in alkene trans on cyclopropane cis in alkene cis on cyclopropane

26 25 Cyclopropane formation with diazomethane Add CH 2 N 2 and heat or light to alkenes to generate cyclopropanes. Stereochemistry: trans in alkene trans on cyclopropane cis in alkene cis on cyclopropane

27 26 General properties of alkynes Nucleophilic due to pi bonds. More acidic than alkanes and alkenes due to sp hybridization. Undergoes electrophilic addition reactions. Alkyne synthesis Double E2 elimination of vicinal 1,2 dihalides or geminal 1,1 dihalides. Add NaNH 2 (2 equiv.) or Potassium Tert-butoxide (2 equiv.) and heat. If two alkynes are possible, you will get a mixture of both. Addition of HX to alkynes Add HCl or HBr (1 equiv.) to generate a vinyl halide. Add HCl or HBr (2 equiv.) to generate a geminal dihalide. NOTE: Do not use HI! Halogens add in markovnikov fashion. Mechanism generates special carbocation intermediates called vinyl cations. Terminal and symmetrical alkynes give 1 product = GOOD Internal and asymmetrical alkynes generate 2+ products = BAD Effect of halogens bound to a carbocation Halogens are electronegative, so they will pull electron density away. This will destabilize the carbocation. Halogens can donate electrons via resonance to generate a full octet. This stabilizes the carbocation. Overall, halogens stabilize carbocations. Addition of Br 2, Cl 2 to alkynes This is anti addition so you get the trans alkene (E). Mechanism goes through a halonium ion intermediate, similar to alkenes. Adding 1 equivalent of Br 2, Cl 2 NOT I 2 gives vicinal E-dihalides.

28 27 Adding 2 equivalent of Br 2, Cl 2 NOT I 2 gives you tetrahalides. Hydration of alkynes via oxymercuration Add catalytic HgSO 4, catalytic H 2 SO 4 and H 2 O solvent to generate the ketone. Markovnikov addition of H 2 O. The reason you don't get an alcohol is because the original product is an enol. There will be a spontaneous tautomerization of enol to keto since keto is more stable. Hydroboration-oxidation of alkynes Add (1) BH3 and (2) H2O2, NaOH to generate the ketone. Anti-markovnikov addition of H 2 O. The reason you don't get an alcohol is because the original product is an enol. There will be a spontaneous tautomerization of enol to keto since keto is more stable. The tautomierzation mechanism is the same as the picture above. Generating acetylide nucleophiles Add NaNH 2 to a terminal alkyne to deprotonate it into an acetylide anion. The acetylide anion is very nucleophilic and basic. It is good for substitution reactions.

29 28 What does it mean to oxidize and reduce in terms of organic chemistry? Reduction = more C-H bonds are created Oxidation = more C-O or C-C bonds are created Catalytic hydrogenation H 2 (2 equiv.) + Pd-C catalyst turns alkynes into alkanes. H 2 (1 equiv.) + Pd-C catalyst turns alkenes into alkanes. H 2 + lindlar's catalyst turns alkynes into alkanes. General mechanism for catalytic hydrogenation: Stereochemistry = syn addition Benzene rings do not hydrogenate. Do not try to hydrogenate if there are carbonyl groups present! Sodium metal reduction 2 Na o + NH 3 reduces an alkyne into a trans alkene. Does not work on alkenes and benzene rings. Do not try to use this if there are carbonyl groups present!

30 29 Reductive opening of epoxides Use LiAlH 4 with H 2 O workup to open up an epoxide at the least hindered position. This is like epoxide openings under basic conditions. Markovnikov addition of H 2 O. mcpba/rco 3 H epoxidation Add mcbpa or RCO 3 H to an alkene to create an expoxide. Stereochemistry: If the alkene is cis the epoxide will be cis (syn addition). If the epoxide is trans, the alkene will be trans (anti addition). Anti vs syn dihydroxylation Anti dihydroxylation step 1) mcpba step 2) NaOH, H 2 O H 2 O workup Syn dihydroxylation Use OsO 4 + NMO or KmnO 4 + cold NaOH + H 2 O

31 30 Permanganate oxidation of alkenes Add KMnO 4 + heat to alkenes to oxidatively cleave it into a ketone or a catalytic acid or carbon dioxide. Will turn into a carboxylic acid if the carbon participating in the double bond has 1 hydrogen bound to it. Will turn into a ketone if the carbon participating in the double bond has 0 hydrogens bound to it. Will turn into a carbon dioxide if the carbon participating in the double bond has 2 hydrogens bound to it. Note that you will never produce aldehydes using this reagent! Oxidative cleavage of diols Add HIO 4 or NaIO 4 to a vicinal 1,2 diol to generate the aldehyde or ketone. If the carbon bound to the -OH group is also bound to a hydrogen, it will turn into an aldehyde. If the carbon bound to the -OH group is not bound to a hydrogen, it will turn into a ketone. Ozonolysis of alkenes and alkynes O 3 + Zn/H 2 O or (CH 3 ) 2 S to alkenes cleaves double bond into C=O O 3 + H 2 O workup to alkynes cleaves triple bond into COOH

32 31 In ozonolysis, molecules with multiple alkenes/alkynes are cleaved into fragments. Cyclic alkenes/alkynes become chains. Oxidation of alcohols by chromium compounds There is no oxidation of tertiary alcohols Secondary alcohols oxidize into ketones. Add any of the following reagents: (1) CrO 3 + H 2 SO 4 /H 2 O (2) K 2 Cr 2 O7 + H 2 SO 4 /H 2 O (3) PCC + CH 2 Cl 2 (4) CrO 3 + Pyridine, CH 2 Cl 40 o C Primary alcohols oxidize into aldehydes or carboxylic acids. If you want an aldehyde, add: (1) CrO 3 + Pyridine, CH 2 Cl 40 o C (2) PCC + CH 2 Cl 2 If you want a carboyxlic acid, add: (1) CrO 3 + H 2 SO 4 /H 2 O (2) K 2 Cr 2 O7 + H 2 SO 4 /H 2 O Reaction of alkyl halides into alkanes Add (1) LiAlH 4 and (2) H 2 O or ether Mechanism is Sn2. There will be inversion at the stereocenter. Less substituted alkyl halides react faster than more substituted ones.

33 32 What are radicals and how do they react? Free radicals are chemical species that contain an orbital with one electron in it. They are neutrally charged and highly reactive. Free radical reactions do not involve the donation or acceptance of electrons but instead operate through homolytic cleavage. This means that bonds break in such a fashion that equal numbers of electrons are distributed to each atom. We use fish-hook arrows to represent these cleavages: Stability of radicals VERY USEFUL LINKS: Tertiary radical > secondary > primary > methyl radical If the radical can do resonance, it is more stable. As %s character increases, stability decreases (alkanes > alkenes > alkynes) Stability increases as electronegativity decreases. Stability increases going from right to left across a row periodic table and from top to bottom down a column of the periodic table. If the radical is adjacent to an electron withdrawing group, stability decreases. This only applies to electron withdrawing groups that cannot donate electrons. Adding Br or Cl to alkanes Add Br 2 or Cl 2 and light (hv) or heat to a tertiary carbon to turn it into an alkyl halide. This rarely is done on secondary alkanes and never to primary alkanes. Stereochemistry: retention and inversion of stereocenter (similar to Sn1) Chloronation is exothermic whereas bromination is endothermic. Radical chain mechanism (done with methane, although unrealistic in reality): Step 1: initiation Heat or uv light cause the weak halogen bond to undergo homolytic cleavage to generate two bromine radicals and starting the chain process. There is a NET INCREASE in free radicals! Step 2: propogation A bromine radical abstracts a hydrogen to form HBr and a methyl radical. There is NO NET CHANGE in the number of free radicals! Step 3: propogation The methyl radical abstracts a bromine atom from another

34 33 molecule of Br 2 to form the methyl bromide product and another bromine radical, which can then itself undergo step 2 creating a cycle that can repeat. There is NO NET CHANGE in the number of free radicals! Step 4: termination Various reactions between the possible pairs of radicals allow for the formation of ethane, Br 2 or the product, methyl bromide. These reactions remove radicals and do not perpetuate the cycle. There is a NET DECREASE in the number of free radicals! Adding Br to an allylic carbon Add NBS and light, heat, or ROOR to replace an allylic hydrogen with bromine. The problem with this reaction is that unsymmetrical substrates gives mixtures of products. Anti-markovnikov addition of HBr to alkenes Add HBr and ROOR, hv, or heat to add bromine in anti-markovniknov fashion to alkenes. Stereochemistry: syn + anti addition

35 34 Conjugated dienes and allylic carbocations A conjugated diene has 2 double bonds separated by 1 carbon-carbon bond. A unique property of conjugation is the ability to absorb light energy at certain wavelengths. The more conjugation, the more absorbing power. An allylic carbocation is stable due to resonance: HBr, HCl addition to a conjugated diene Add Hbr or Hcl to a conjugated diene to generate 1,2 and 1,4 addition products The 1,2 addition forms faster (it is the kinetic product) The 1,4 addition is the more stable (it is the thermodynamic product) Add heat to favor the 1,4 product. Keeping the reaction cool (under 0 degrees Celsius) will favor the 12 product. Below is a table showing the effect on temperature and product ratios (don't need to learn this table. More for general understanding): Mechanism:

36 35 Diels-Alder Reaction Diene + dienophile + heat ring The diene must be conjugated and in the s-cis conformation. The diene also cannot be sterically hindered. 4 possible situations in diels-alder problems: Situation 1 The diene above is a diene in the s-cis conformation. Note that there is a wide gap for the dienophile to attack the diene (it is unhindered). As a result, this diene is a good molecule for diels-alder. Situation 2 The diene above is a diene in the s-trans conformation. In this current conformation, this molecule cannot participate in diels-alder. However, this molecule can rotate around its single bonds and thus attain a conformation similar to the diene in situation 1. Note that there is also a wide gap for the dienophile to attack the diene (it is unhindered). As a result, this molecule is also good for diels-alder. S-cis s-trans Situation 3 The diene above is a diene in the s-trans conformation (left picture). Like situation 2, this diene can rotate around its single bonds to attain a s-cis conformation (right picture). When this molecule is in the s-cis conformation, the methyl groups close the opening. This molecule is hindered and the dienophile will not be able to attack the diene. Therefore, this molecule is BAD for diels-alder. Situation 4 The diene above is a diene in the s-trans conformation. This diene cannot rotate around its single bonds to attain a s-cis conformation because it is part of a ring (rings cannot rotate around bonds). Therefore, this molecule is BAD for diels-alder as it will never attain an s-cis conformation.

37 36 Stereochemistry: trans subsituents on the dienophile lead to anti addition cis substituents on the dienophile lead to syn addition Cis dienophile leads to syn addition of R groups Trans dienophile leads to anti addition of R groups You can use alkyne dienophiles instead of alkene dienophiles. Make sure to include the extra double bond directly across from the double bond formed in the diels-alder reaction! The diene can be a ring. If the diene is a ring, you will generate a bicyclo compound. In bicyclo compounds, the endo product is favored. In the endo product, the R groups are pointing downward, not horizontally outward (exo).

38 37 Huckel's rules for aromaticity For aromatic compounds: 4n+2 pi electrons (2,6,10...) planar, cyclic array overlapping pi bonds and/or on orbitals no sp 3 carbons present in the ring unless it is a carbanion For anti-aromatic compounds 4n electrons (4,8,12...) planar, cyclic array overlapping pi bonds and/or on orbitals no sp 3 carbons present in the ring unless it is a carbanion Count lone pairs (maximum of 1 lone pair per atom) if the heteroatom is not connected to a double bond. If the heteroatom is connected to a double bond, do not count them! Fused aromatic rings can be cyclic. If 2 neighboring rings share a double bond, it counts to both rings. Count triple bonds as 2 electrons, not 4 electrons! Note: if the lone pair of electrons on the heteroatom contribute to aromaticity, they are unreactive. EAS regiochemical considerations Memorize this table: Activating groups (aka. Electron donating groups) = faster EAS Deactivating groups (aka. Electron withdrawing groups) = slower EAS In general: substituents with lone pairs on the atom directly attached to the ring and alkyl substituents favor ortho, para. Substituents without lone pairs directly attached to the ring and are not alkyl favor meta. Special cases: Acetals are ortho/para directors. Halogens are ortho/para directors. Substituents that have charge on the atom directly attached to the ring are meta directors. When there are multiple substituents on a ring, the strongest activating group determines regiochemistry. No substitution between meta substituents occurs on a ring.

39 38 A note on EAS mechanisms For the DAT, a common question asked is to identify the most stable intermediate/resonance structure during a specific EAS reaction. If none of the answer choices have structures in which all atoms have a full octet, then: A meta director will have the intermediate looking like this. A para director will have an intermediate looking like this. An ortho director will have an intermediate looking like this. For more information, please visit the site Fantastic source for EAS! Halogenation of aromatic rings Add FeBr 3 + Br 2 or FeCl 3 + Cl 2 to add chlorine or iodine to an aromatic ring. Note, you cannot use iodine. FeBr 3 /FeCl 3 are lewis acids that are designed to help making Br 2 and Cl 2 better leaving groups. Phenols over-halogenate unless you leave out the FeX 3 catalyst. Anilines over-halogenate unless one of the R groups is a carbonyl.

40 39 Nitration of aromatic rings Add cat. H 2 SO 4 and HN O3 to nitrate an aromatic ring. -NO 2 group is a meta director and is strongly deactivating. Below is the lewis structure. Reduction of NO 2 into NH 2 Add H 2, Pd-C or 1) Sn, HCl and 2) NaOH We reduce to an amine since -NH 2 is an ortho, para director and is strongly activating. In addition, the amine group useful for reactions beyond EAS. Friedel-crafts acylation of aromatic rings Add an acid chloride and AlCl 3 to a benzene ring to add a carbonyl group to the ring. AlCl 3 is a lewis acid and is used in the reaction to help make a strong electrophile. Reduction of carbonyl group into an alkyl group Add H 2 NNH 2, NaOH, heat OR Zn(Hg), HCl Friedel-crafts alkylation of aromatic rings Add secondary or tertiary alkyl halide (rarely primary but possible) and AlCl 3 to alkylate the ring. Must add alkyl halide in excess. Watch for carbocation shifts of the alkyl halide!

41 40 Exceptions to friedel-crafts alkylation and acylation If there is a meta director on the ring, then no friedel-crafts will occur. Annilines (-NH 2 ) do not participate in friedel-crafts. Oxidation of alkyl benzenes Add KMnO 4 to turn the alkyl group into a carboxylic acid group. If a ring is attached to the aromatic ring, each connector point (each point where the ring attaches to the benzene ring) turns into COOH. If there are no allylic hydrogens, then there is no reaction. Sulfonation of aromatic rings Add cat. H 2 SO 4 and SO 3 to add an -SO 3 H group to the aromatic ring. -SO 3 H is a meta director and is deactivating.

42 41 LiAlH 4 vs. NaBH 4 aldehyde primary alcohol (use either to achieve this) ketone secondary alcohol (use either to achieve this) ester primary alcohol (LiAlH 4 only) carboxylic acid primary alcohol (LiAlH 4 only) LiAlH 4 and NaBH 4 cannot reduce alkenes. Stereochemistry: retention and inversion (leads to racemic mixtures). Similar to Sn1 stereochemistry. Organometallic reagents: synthesis and properties To create a grignard reagent, add Mg o with Et 2 O or THF to an alkyl halide. To create an organolithium reagent, add 2 equivalents of lithium metal to an alkyl halide. Organometallic reagents are VERY basic and nucleophilic. Due to their high basicity, organometallic reagents will react in acid-base reactions over anything else if there are protons available, so watch out!

43 42 Addition-elimination reactions Use 1) organometallic reagent and 2) H 3 O + to turn an aldehyde or ketone or ester into a tertiary alcohol. This reaction goes through a ketone intermediate. Use 1) LiAlH 4 + THF and 2) H 3 O + to turn an ester and an aldehyde into a primary alcohol and a ketone into a secondary alcohol. This reaction goes through an aldehyde intermediate. How can we stop at the aldehyde or ketone, given an ester as the starting material? Use 1) DIBAL-H and 2) H 2 O to stop at the aldehyde. Use 1) organocuprate reagent (R) 2 CuLi and 2) H 3 O + to stop at the ketone Alpha-beta unsaturated ketone and aldehyde additions An alpha-beta unsaturated carbonyl is when there is a double bond between the alpha (one carbon away from the carbonyl carbon) and the beta (two carbons away from the carbonyl carbon) carbon of a carbonyl compound. ketone aldehyde These compounds react with organometallic reagents in 1,2- and 1,4-addition reactions. What is does 1,2- and 1,4-addition mean? Note that oxygen is numbered 1 in counting.