Exam3StudyGuide.docx

General • Resonance stabilization is more powerful than substitution stabilization - • HSO i4 worse nucleophile than water Questions • 9.38…look at it!!!!!!!111111111111 kinda like a test question FREE RADICAL CHAIN RXN • Alkane halogenation: refers to chlorination/bromination; no I or F • More connected to S 2N • Radical species = neutral in charge but has an unaired electron  unstable  wants to react • Push single electrons w/ single barbed arrows (not double arrows, they’re for electron pair pushing) • Mechanism defined by initiation, propagation, and termination o Initiation requires light to be shined to start rxn  once rxn’s started, initiation can stop, just needs bit of initiation and then can spontaneously go  But for overall process, needs continuous light supplied, b/c need to make up for loss of radicals from termination steps o Propagation itself doesn’t need light b/c breaking + making bonds net energy = 0  Sndrts and ends w/ radical doesn’t create/destroy electrons  2 propagation step consumes product of 1 propagation step, which consumes radical made in 2 step  Both steps = symbiotic creates cycle/chain chain rxn st  1 step = endothermic, has highest activation energy, but = thermoneutral: overall energy increase = slight  2 step = exothermic, activation energy’s lower, product energy’s lowest st nd  1 transition state > 2 transition state in energy  o Termination: 2 radicals combine (can be any combo, alkane + alkane, halogen + halogen, alkane + halogen); stops chain rxn  not very likely b/c amt radicals produced from propagation kept low not as likely to encounter radical as it is to encounter alkene/dihalogen  more of product’s contributed from propagation steps than termination o Overall rxn = ∑propagation steps  no net radicals o Energy of rxn  Bond dissociation enthalpies (BDE): energy needed to break bond into radicals (not ions b/c energy depends on type of sol’n it’s done in, energy of environment doesn’t matter as much for radicals homolytic steps; can’t use this analysis for S N or ionic rxns)  Breaking bonds = endothermic = + BDE; Forming bonds = exothermic = - BDE  Overall BDE = ∑BDEs of breaking/making bonds in a rxn/step • = selective (prefers) more substituted C-H b/c more substituted radicals = stabilized by hyperconjugation (2 electrons ↓ in energy, 1 electron ↑ in energy net effect = stabilization) o Also prefers more resonance-stabilized radical o Secondary C-H bond = weaker than primary b/c has less BDE  Energy difference reflected in radical products…more substituted = more stable  lower energy o Bromination’s more selective than chlorination (forms higher ratio of favored/unfavored products); prefers secondary C-H even more strongly than chlorination st nd  Chlorination: 1 and 2 step closer to thermoneutral, only little bit energy difference reflected in transition state  Bromination: 1 step = endothermic  greater energy difference to get to intermediate, almost all energy difference reflected in transition state • = selective @ allylic/benzylic (1 C away from benzene ring/double bond) sites b/c both = resonance stabilized o Use NBS here b/c Br can2compete w/ addition to double bond; but NBS doesn’t work if there’s not a double bond or benzene ring next door o Benzylic: resonance doesn’t distribute charge (b/c neutral), but helps w/ electron deficiency o Note: resonance structures have same energy (even if some structures hold more weight/significance) b/c it’s really just 1 molecule, can’t separate out; still, though, halogenation might prefer one of the contributors to resonance structure over other b/c better distributed so same intermediate can go to different products w/ different energies; resonance is NOT equilibrium, it’s just 1 thing o Resonance = more stabilizing than substitution lower BDEs for allylic and benylic prefers resonance over substitution • BDE of certain bond = more reflection of stability of product it can form, not b/c the bond itself is any special • Allylic/benzylic Bromination: want Br @ allylic/benzylic position, but Br al2o adds to double bonds (addition) o Use NBS + trace of R. (free radical rxn) instead of Br p2oduces allylic/benzylic addition of Br exclusively  doesn’t halogenate unactivated (those that don’t = radical) C-H’s  Needs light/other chemical radical (better) source • Can run into problem of multiple halogenations (more than 1 X can add onto chloromethane) o @ beginning of rxn, few CH Cl a3d more methane so Cl radical more likely to react w/ methane o @ end, vice versa, so multiple halogenation likely o how to solve this?  incomplete rxn (methane:chlorine ratio = high), then purify chloromethane from methane/starting material (even here, nothing’s really stopping multiple halogenation, but statistically less likely) o If want multiple halogenations chlorine:methane ration = high o Can do multiple halogenations as long as it’s in a position that’s selective ALKENE RXNS • Markovnikov’s Rule: The addition of a proton acid to the double bond of an alkene results in a product with the acid proton bonded to the carbon atom that already holds the greatest number of hydrogen atoms o Anti-Markovnikov: H goes onto one w/ lesser # H’s (more substituted) o Either way, intermediate should be most stable cation (usually so that electrophile’s bound to less substituted C so that charge goes onto more substituted C for stabilization)….have the charge/radical go onto the more substituted C; 1 thing adds to less substituted, 2 thing adds to more substituted o Happens in regioselective cases in which something’s not equally likely to add to both ends of double bond • When more acidic, favors addition, when more basic, favors elimination, when more water, favors hydration • *Whenever attacking to open ring, has an enantiomer most likely b/c can attack from top or bottom face of ring • water deprotonates better than halides Syn Anti Neither o Formation of Halohydrins (alcohol w/ IonicAddition of Hydrogen Halides a H on adjacent C) n Step 1: Protonation of the pi bond  r O Step 1: Electrophilic attack  halonium carbocation v ion i v k a Step 2:Attack by halide ion  addition M Step 2: Water opens halonium ion; product deprotonation  halohydrin Requires addition of acid Alkenes and halides exist in pH -water’s better base than halide, so determined equilibrium when add water, favors hydration Acidic conditions: alkenes halides Basic conditions: halides  alkenes -diastereoselective (never makes cis hydrogen halide addition is the exact version) opposite of elimination mechanism Oxymercuration-demercuration Acid-catalyzed hydration Oxymercuriation Step 1: Protonation of double bond  Step 1: Electrophilic attack  carbocation mercurinium ion Step 2: Nucleophillic attack by water  Step 2: Water opens ring (attacks more protonated alcohol substituted C)organomercurial alcohol Step 3: Deprotonation  alcohol Demercuration: replaces mercuric fragment w/ H  alcohol Exact opposite mechanism for elimination of alcohols -Sulfuric acid loves water drives rxn to -milder (although organomercurial left compounds = highly toxic), higher -presence of water drives rxn to right yielding than acid-catalyzed hydration, -in presence of strongly acidic catalyst and works w/ alkenes that don’t (but not too concentrated???) undergo acid-catalyzed hydration -rxn = reversible not reliable b/c can’t -no free carbocation no purely make something…little bit of rearrangement/polymerization both -ring is able to form b/c Hg is so big -If different acid used, water still attacks -partial + charges on both C’s in ring, in step 2 b/c better nucleophile but more substituted C has greater -Not stereoselective, goes through flat partial charge carbocation, things can add to either -It’s regioselective side (diastereoselective…doesn’t make both diastereomers) (not enantioselective b/c can attack from top or bottom face of ring, no R/S preference racemic mixture) Alkoxymercuriation-Demercuriation makes ethers Alkoxymercuriation Demercuriation: same as oxymercuriation-demercuriation -Pretty much same process as oxymercuration-demercuration, but mercurinium ion’s attacked by alcohol, not water…final product has OR group rather than OH group creates ethers iydroboration-oxidation Free RadicalAddition of Hydrogen a Halides with HBr and peroxides n (doesn’t happen w/ HI or HCl b/c they = r Hydroboration (single step): BH strongly endothermic) o 3 idds to double bond; B adds to Initiation: Formation of radicals vess hindered, less substituted C; H adds to more substituted C r Alkene + BH .THF borated i 3 nlkene A Propogation:Aradical reacts to generate another radical Step 1: Br radical adds to double bond  alkyl radical on more substituted C Oxidation: magical Borated alkene + H O2,2H, H O 2 alcohol Step 2:Alkyl radical abstracts H from HBr  product + Br radical -It’s enantioselective…why? b/c different starting stereoisomers lead to different stereoisomers of *Peroxides = thermal initiators products *Probably best done in nonpolar solvent *2 step is magical, 1 step is 1 concerted step so don’t form carbocation r heduction: Catalytic Addition of Halogens ElectrophillicAdditions eHydrogenation N Step 1: Electrophilic attack  halonium Step 1:Attack of pi bond on e