CHM 25500 Exam 3 Study Guide
CHM 25500 Exam 3 Study Guide CHM 25500 - 001
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CHM 25500 - 001
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This 5 page Study Guide was uploaded by Gayatri on Wednesday November 18, 2015. The Study Guide belongs to CHM 25500 - 001 at Purdue University taught by Christopher H Uyeda in Fall 2015. Since its upload, it has received 199 views. For similar materials see Organic Chemistry in Chemistry at Purdue University.
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Date Created: 11/18/15
CHM 25500 Exam 3 Study Guide Chapter 7 Alkynes contain a triple bond between two carbons 0 Bond length 12 Angstroms shortest 0 Linear structure with sp hybridized orbitals l sigma 2 pi bonds I Pi bonds comes from P2 and Py Nomenclature for Alkynes 0 Same rules as alkanes and alkenes wit yn 0 Number for alkyne is minimized Deprotonation of Alkynes occurs with terminal alkynes have a H attached to triple bond with strong bases 0 Weaker acidbase is favored in equilibrium 0 Deprotonated alkynes are bases and nucleophiles I Act as nucleophiles in SN2 reactions with primary alkyl halides I Act as bases in E2 reactions with tertiary and secondary halides Alkynes can be made from two types of Dibromides o Vicinal dibromides bromines attached to adjacent carbons o Geminal dibromides bromines attached to the same carbon 0 Occurs through two beta eliminations NaNH2 is added twice I Step 1 NH2 acts as a base and attacks a H on the alkane while a bromine on the other carbon leaves the alkane the electrons from the CH bond go toward the CC bond and an alkene is formed along with NH3 and Br I Step 2 The same reaction occurs again this time making an alkyne Electrophillic Addition Reactions 0 Adding Br2 halogenation I Adding one equivalent of Br2 to an alkyne gives a trans alkene I Adding two equivalents of Br2to an alkyne gives an alkane 0 Adding HCl HBr hydrohalogenation I Forms more stable more substituted cation 9 Markovnikov addition I Adding one equivalent of HBr gives an alkene I Adding two equivalents of HBr gives an alkane geminal bromide 0 Adding H20 H2S04 HgS04 hydration I Since alkynes are more dif cult to hydrate than alkenes an extra catalyst is needed Mercury Sulfate HgS04 I Follows Markovnikov addition an OH is added on the more substituted carbon making an enol I Enol is tautomerized into a ketone through resonance and internal proton transfer Electrons from the 0 go to C0 bond 9 C0 and electrons from the CC bond go to the other C 9 CC The CC attacks a H on the 0 and takes it away making the ketone which is favored I Tautomers a subclass of constitutional isomers differ by position of a proton 0 Adding BH3 H202 Na0H Hydroboration I Follows AntiMarkovnikov addition BH3 reacts with alkyne OH is added on less substituted carbon making an alkeneH and BH2 are cis I Alkene reacts with H202 Na0H and BH2 is replaced with an 0H making an enol I Enol goes through resonance and proton transfer and makes an aldehyde Reduction of Alkynes o Hydrogenation adding two equivalents of H2 Pd to al alkyne gives a cis alkane o Semihydrogenation achieved by using Lindlar s catalyst Pd on Ca2C03 I Adding H2 Lindlar s catalyst to an alkyne gives a cis alkene 0 Reduction to a trans alkene from an alkyne is achieved by adding Na NH3 MultiStep Synthesis using known reactions to get to desired products from given starting materials Chapter 8 Nomenclature for alkyl halides 0 Same rules as alkanes and alkenes 0 Additional groups can be written with pre xes ouro chloro bromo iodo etc o R and S centers are speci ed if needed 0 Z and E sameopposite side of high priority groups are speci ed if needed Alkyl Halides are made in a variety of different ways 0 From alkenes with HX or X2 I Hydrochlorination I Bromination 0 From alkanes with X2 and heat or light I Involves radical intermediates formed by homolytic bond cleavage which is shown with movement of 12 arrows each arrow moves one electron heterolytic bond cleavage as seen in previous chapters involves movement of pairs of electrons with full arrows Alkane halogenation reaction mechanism 0 Step 1 Initiation nonradicals are converted to radicals I Br Br 9 Br Br Requires heat or light 0 Step 2 Chain Propagation radicals are used to convert other nonradicals into radicals 39 C2H6 BIquot 9 CH3C39H2 l H BI 39 CH3C39H2 l Br Br 9 CH3CBI H2 l BIquot in solution lots of excess Br Br oats around in excess 0 Step 3 Chain Termination radicals are converted back to nonradicals I Br Br 9 Br Br 39 CH3C39H2 l BIquot 9 CH3CBI H2 39 CH3C39H2 4 CH3C39H2 9 C4H10 39 CH3C39H2 CH3C39H2 9 C2H6 C2H4alkene Formed because radicals on carbon weaken C H bonds Relative stability of carboncentered radical intermediates 0 Allyl gt Tertiary gt Secondary gt Primary gt Methyl o Radicals are also stabilized through resonance I Allyl radicals radicals with a double bond on one side Nbromosuccinimide a molecule which prevents electrophilic addition from occurring gives a radical bromination product Chapter 9 Nucleophilic substitution a reaction where one nucleophile substitutes for another 0 Two mechanisms I SNl Unimolecular nucleophilic substitution Three step mechanism 0 Step 1 leaving group comes off strong electrophilic carbocation is created 0 Step 2 a nucleophile attaches to strong electrophile creating one molecule with a formal charge on one atom an acid 0 Step 3 a base comes in and protonates the acid Similar to electrophilic addition mechanism First order reaction 0 Rate dCH3 3CBrdt kCH3 3CBr Depends only on electrophile SN2 Bimolecular nucleophilic substitution One step mechanism 0 Nucleophile comes in on opposite side of leaving group Stereochemistry is inverted Second order reaction 0 Rate dCH3Brdt kCH3BrOH39 Determining whether a reaction will proceed by an SNl or SN2 mechanism 0 Structure of the electrophile I Less hindered free access for nucleophile is preferred by SN2 reactions Methyl and primary electrophiles Blocked access for nucleophile is preferred by SNl reactions Tertiary electrophiles Secondary electrophiles can undergo either SNl or SN2 Steric hindrance at nearby carbon centers decreases rates of SN2 reactions More substituted beta carbon 9 slower reaction 0 Identity of the leaving group I The more stable the anion the better leaving group Better leaving group 9 faster reaction bene ts both SNl and SN2 Neutral leaving groups are better than anionic ones because they form neutral products instead of unstable anions H20 gt OH The most stable anions have strong conjugate acids 0 I39gt Br39 gt Cl39 H20 gtgt F Similar to acidity trends More stable anions are better same as acidity trends Check for Electronegativity of element with charge 0 F39 gt OH39 gt HzN39 Size of element with charge 0 I39gtBr39gtCl39gtF39 Resonance 0 Double bonds make atom more stable because of their ability to delocalize electrons esp with CO bonds Inductive effects 0 Electronegative elements nearby pull electron density away stabilize anions 0 Identity of the nucleophile I Strong nucleophiles usually anionic favor SN2 Weak nucleophiles neutral favor SNl Strong nucleophiles are often poor leaving groups opposite trends Anionic nucleophiles are better than their neutral conjugate acids More stable anions are weaker nucleophiles o In aprotic solvents the strongest base is the strongest nucleophile I F39 gtCl39 gtBr39 gtI39 o In protic solvents the trend is reversed I I39gtBr39gtCl39gtF39 0 Reaction conditions solvent I Polar Protic has acidic protons is a hydrogen bond donor Eg Water methanol ethanol Protic solvents favor SN I Polar Aprotic has no acidic proteins Eg DMSO DMF acetone Eliminates the possibility of Hbonding Still polar molecules Aprotic solvents favor SN2 Evidence of SN1 vs SN2 mechanisms 0 Rates of reactions I SN1 first order Rate dCH3 3CBrdt kCH3 3CBr I SN2 second order Rate dCH3Brdt kCH3BrOH39 o Stereochemical outcomes I SN1 Generates a 11 racemic mixture Since the carbocation intermediate is planar the nucleophile can come in from either side giving R or S con guration I SN2 Stereocenters are inverted Nucleophile comes in on opposite side of the leaving group 0 Rearrangements 12hydride and alkyl shifts I Occur with SN1 reactions due to formation of carbocations Nucleophiles and leaving groups 0 Strong nucleophiles unstable anionic prefer SN2 reactions 0 Strong nucleophiles make poor leaving groups 0 More stable anions are weaker nucleophiles Solvation of anions o In polar protic solvents the most stable anions are stronger nucleophiles I Stability of the anions is determined by size of element bearing the negative charge 0 l39gtBr39gtCl39gtF39 I This is because a large diffuse ion is weakly solvated in a protic solvent such as water because the charge is dispersed widely whereas in a small ion the charge is concentration making the ion strongly solvated SN1 and SN1 Reaction Conditions Summary Type of Alkyl Halide SN1 SN2 Methyl X Favored Primary Rarely occurs Favored Secondary Favored in protic solvents H20 Favored in aprotic solvents MeOH EtOH other OH s DMSO DMF acetone with with poor nucleophiles neutral good nucleophiles anionic molecules unstable Tertiary Favored X Substitution at a chiral center ll Racemic mixture forms Configuration is inverted 3Elimination reaction a reaction in which a small molecule HCl HBr H20 is removed from two adjacent carbons to form an alkene o If multiple products are possible if 2 beta carbons exist around the alpha the more stable alkene is formed with more substituents 0 Two mechanisms exist I E2 Bimolecular elimination One step mechanism occurs with the H and leaving group aligned anti 180 on Newman projection and coplanar sharing the same plane Rate kRLv Base SN2 vs E2 Favored in reactions with a strong base OH39 OR39 El Unimolecular elimination Three step mechanism 0 Step 1 Rate limiting step leaving group comes off electrophilic O O carbocation is created Step 2 CH30H acts a base and takes a proton from the carbocation which acts as the acid The electrons from the CH bond go to the CC bond and an alkene is created along with CH30H2 Step 3 CH30H2 reacts with Br39 to make HBr Reverse of electrophilic addition Rate kRLv Favored in reactions with more substituted carbocations o Bulky alkyl halides bulky nucleophilesbases and strong bases favor E2 0 If ambiguous a mix of both products is formed SNl vs E1 0 The two mechanisms often compete forming mixtures 0 Weak bases favor elimination Summary of SNl SN2 El E2 Reactions Halide Reaction Methyl Primary Secondary Tertiary SNlEl J in protic solvents such as H20 MeOH EtOH other OH s with poor nucleophiles such as neutral molecules J with poor nucleophilesweak bases if the solvent is polar protic SN2 J with good nucleophilesweak bases such as 139 Cl39 CH3COO39 J with basesnucleophiles Where pKa lt 11 such as I39 and CH3COO39 E2 J with strong bulky bases such as CH33CO J with basesnucleophiles Where pKa gt 11 such as OH39 and CH3 CHzO39 J with strong bases such as HO39 and R0
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