Chem 2120 – 4 – Flashcards
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Unlock answers| Oxidations of Alkenes: Syn 1,2-Dihydroxylation (and unsaturation test) |
-KMnO4 and alkene react in a cold solution of OH-/H2O to produce 1,2-Ethandiol
Cold temperature and alkaline condition can prevent the glycol to be further oxidized by KMnO4
**Either OsO4 of KMnO4 will give 1,2-diols (glycols) -always will be syn addition (same side) because oxygens add at the same time -Formation of a cyclic intermediate leads to syn addition (** 5 member ring) -cyclic intermediates result from reaction of the oxidized metals -the initial syn addition of the oxygens is preserved when the oxygen-metal bonds are cleaved and the products are syn diols This reaction can be used as a test for unsaturation: -This is chemical test for alkenes in which the purple color of the KMnO4 disappears and forms brown MnO2 residue if akene (or alkyne) is present -the alkene becomes oxidied and Mn(VII) is reduced to Mn (IV) *of course the other way to identify alkenes and alkynes is to brominate the solution at room temperature and in the dark. |
| Oxidative Cleavage of Alkenes |
-Reaction of an alkene with hot KMnO4 results in cleavage of the double bond and formation of highly oxidized carbons -unsubstituted carbons become CO2, monosubstituted carbons become carboxylates and disubstituted carbons become ketones
-only carbons attached to double bond react with an oxygen
**OH- and H3O are involved in this reaction |
| Ozonolysis of Alkenes |
-Cleavage of alkenes with ozone and workup with zinc in acetic acid leads to less highly oxidized carbons than products from cleavage with hot HKnO4 -Unsubstituted carbos are oxidized to formaldehyde(C=O), monosubstituted carbons are oxidized to aldehydes, and disubstituted carbons are oxidized to ketones
-very cold temperatures involved
reaction proceeds: 1) O3, CH2Cl2, -78C 2)Zn/HOAc |
| Oxidative Cleavage of Alkynes |
| -reaction of alkynes with ozone or basic potassium permanganate both lead to formation of carboxylic acids |
| Elimination Reactions of Alkyl Halides |
**If non-primary alkyl halides are used then elimination reactions predominate
dehydrohalogenation is used for the synthesis of alkenes -elimination competes with substitution reaction -strong bases such as alkoxides favor elimination
*the alkoxide bases are made from the corresponding alcohols |
Mechanisms of Elimination: The E2 Reaction
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-E2 reaction involves concerted removal of the proton, formation of the double bond, and departure of the leaving group -both alkyl halide and base concentrations affect rate and therefore the reaction is 2nd order
rate = k[A][B] **transition state must involve both reactants!!
E2 = base Sn2 = nucleophile
* four partial bonds exist in transition state *3 products in the end |
| E1 reaction |
-The E1 reaction competes with Sn1 reaction and likewise goes through a carbocation intermediate
-step 1 is the slow dissociation step of the halide to form carbocation -step two is the base grabbing the hydrogen to form a conjugate acid, thus producing the alkene |
| Substitution vs Elimination |
Sn2 vs E2 primary substrate -if the base is small, SN2 competes strongly because approach at carbon is unhindered
Secondary Substrate -approach to carbon is sterically hindered and E2 is favored
tertiary substrate approach to carbon is extremely hindered and elimination predominates especially at high temperatures
temperature -increasing temperature favors elimination over substitution
size of base/nuc -large sterically hindered bases favor elimination becasue they cannot directly approach the carbon closely enough to react in a substitution -Potassium tert-butoxide is an extremely bulky base and is routinely used to favor E2 reaction
OVERALL SUMMARY CH3X -bimolecular reaction only -gives SN2
R-CH2X -bimolecular reaction only -gives mainly SN2 except with a hindered strong base and then gives mainly E2
RRCHx -Gives mainly SN2 with weak bases and mainly E2 with strong bases
RRRCX -SN1/E1 or E2 -No SN2 reaction. In solvolysis gives SN1/E1 and at lower temperatures SN2 is favored. With strong base E2 predominates
*rarely see E1/E2 with halide (?)
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| Relative Stability of Alkenes, and Synthesis of Alkenes via Elimination Reactions |
-generally cis alkenes are less stable than trans alkenes because of steric hindrance
**-the greater the number of attached alkyl groups (ie. the more highly substituted the carbon atoms of the double bond), the greater the alkene's stability -disubstituted are most stable in order of 1)both R on same carbon, 2)trans, and 3)cis ; -the more electrons around the double bond the happier it is!! ; ; Synthesis -reactions by an E2 mechanism are most useful -E1 reactions can be problematic -E2 reactions are favored by secondary or tertiary alkyl halides, and alkoxide bases such as sodium ethoxide or potassium tert-butoxide -bulky bases such as potassium tert-butoxide should be used for E2 reactions of primary alkyl halides |
| Zaitsev's rule |
When two different alkene products are possible in an elimination, the most highly substituted (most stable) alkene will be the major product. -this is true only if a small base such as ethoxide is used
....if you want the other product... -bulky bases such as potassium tert-butoxide have difficulty removing sterically hinderd hydrogens and generally only react with more accessible hydrogens (primary hydrogens)
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| The stereochemistry of E2 reactions: The orientation of Groups in the transition state |
-all four atoms involved must be in the same plane -anti-coplanar orientation is preferred because all atoms are staggered
-syn coplanar transition state only occurs with certain rigid molecules
*In a cyclohexane ring the eliminating substituents must be diaxial to be anticoplanar
In neomenthyl chloride, the chloride is in the adial position in the most stable conformation -two axial hydrogens anti to chlorine can eliminate, and the Zaitzev is major (Mar 23rd/s30)
-in menthyl chloride the molecule must first change to a less stable conformer to produce an axial chloride -elimination is slow and can yield only the least substituted (Hoffman) product from anti elimination(s31) |
| Making Alkenes from Alcohols |
Acid Catalyzed Dehydration -recall that elimination is favored over substitution at higher temperatures -typical acids aused in dehydration are sulfuric acid and phosphoric acid -the temperature and concentration of acid required to dehydrate depends on the structure of the alcohol
**primary alcohols are most difficult to dehydrate, tertiary are the easiest -need higher temperature and acid concentration for primary
-for E1 dehydration of secondary and tertiary alcohols, only a catalytic amount of acid is required since it is regenerated in the final step of te reaction
Step1: The alcohol accepts a proton from the acid in a fast step Step 2: The protonated alcohol loses a molecule of water to become a carbocation. This step is slow and rate determining. Step 3: Another molecules of alcohol, water, or th conjugate base grabs the hydrogen and forms the double bond.
*tertiary alcohols react the fastest here because they ahve the most stable tertiary carbocation-like transition state in the second step
E2 -primary alcohols cannot undergo E1 because of the instability of the carbocation-like transition state in the 2nd step -in the E2 dehydration the first step is again protonation of the hydroxyl to yield the good leaving group water -first the alcohol accepts a proton from the acid in a fast step, then a base removes a hydrogen from the beta carbon as the double bond forms and the protonated hydroxyl group departs. (The base may be another molecule of the alcoholor the conjugate base of the acid) |
| Predicting major Products |
Always ask 3 questions: 1)What Mechanism? (consider 4 factors: substrate, reagent(Nu or base), leaving group, solvent. 2)What regiochemistry? (not applicable to SN reactions) 3)What stereochemistry? (applicable to all reactions)
RULE OF THUMB -nucleophiles that do not act as bases: halides, CN-, RS-, RSH. NO E when these are used!! -For tertiary alkyl halide, weak base E1 and SN1 can compete, strong base E2 -if high temperature, then mostly E
TYPE STEREO REGIO SN1 racemization N/A SN2 inversion of config. N/A E1 if cis/trans isomers, get both Zaitzev E2 anti coplanar dictates the product *
*Zaitzev if small base, Hoffman if strong bulky base) |
| Synthesis of Alkynes by Elimination Reactions |
-alkynes can be obtained by two consecutive dehydrohalogenation reactions of a vicinal dihalide
-alkenes can be converted to alkynes by bromination and two consecutive dehydrohalogenation reactions
geminal dihalides can also undergo consecutive dehydrohalogenation reactions to yield the alkyne (s53 of Mar23) |
| Reactions of Radicals and Stability of Radicals |
-homolytic bond cleavage leads to the formation of radicals (also called free radicals) -radicals are highly reactive, short-lived species -single barebed arros are used to show the movement of single electrons -homolysis of relatvely weak bonds such as O-O or X-X bonds can occur with addition of energy i the form of heat or light
-Radicals tend to react in ways that lead to pairing of their unpairedelectron -hydrogen abstraction is one way a halogen radical can reat to pair its unshared electron
STABILITY -the relative stabilities of radicals follows the same trend as for carbocations -the most substituted radical is most stable
-alkanes undergo substitution reactions with halogens such asfluorine, bromine, and chlorine in the presence of heat or light |
| Multiple Substitution Reactions vs Selectivity |
-Radical halogenation can yield a mixture of halogenated compounds because all hydrogen atoms in an alkane are capable of substitution -all different kinds of halogenation will be seen
-Monosubstitution can be achieved by usng a large excess of the alkane -a large excess of methane will lead to predominantly monohalogenated product and excess unreacted methane |
| Chlorination (radicals) |
-chlorination of higher alkanes leads to mixtures of monochlorinated product (and poly-substituted products) -chlorine is relatively unselective and does not greatly distinguish b/w types of hydrogens (primary, secondary, tertiatry)
chlorination vs bromination -molecular symmetry is important in determining the number of possible substitution products Reactivity of halogen radicals: F>>Cl>Br>>I **Bromination is milder ad more selective, easier to obtain pure products |
| Chlorination of Methane: Mechanism of Reaction |
-The reaction mechanism has three distinct aspects: chain initiation, chain propagation, and chain termination chain initiaton: chlorine radicals form when the reaction is subjected to heat or light (Cl2 -> 2Cl)
chain propagation -a chlorine radical reacts with a molecule of methane to generate a methyl radical -a methyl radical reacts with a molecule of chlorine to yield chloromethane and regenerate chlorine radical -a chlorine radical reacts with another methane molecule, continuing the chain reaction -a single chlorine radical can lead to thousands of chain propagation cycles (a chain reaction is a stepwise mechanism in which each step generates the reactive intermediate that causes the next cycle of the reaction to occur)
chain termination: -occasionally the reactive radical intermediates are quenched by reaction pathways that do not generate new radicals -the reaction of chlorine with methane requires constant irradiation to replace radicals quenched in chain-terminating steps
IN GENERAL, any mechanism for a chain reaction must have the three reactions: initation, propagation, and termination
** in termination, all reasonable combinations of radicals must be considered |
| Autoxidation and free radicals |
-autoxidation happens when there is light, oxygen, and sometimes a radical initiator
Advantages: Surface film formation as in the hardening of paints and varnishes Disadvantages: The ageing processes athat form fancid fats, deterorated rubber and plastics, skin wrinkles
Anti-oxidants: some phenols and aromatic amines can act as "radical traps" to slow down autoxidation Vitamins C and E play some roles in radical trapping
The reaction: free electron attaches to an O2, which then accepts 2 H+ to become H2O2, then accepts another H to ecomes OH(radical), which can transfer the radical to proteins or lipids -vitamin E or C can take this radical
Factors causing radical formation: 1)UV radiation 2)air pollution 3)tobacco smoke
Defending against Free Radicals -De-oxidant enzyme system (rely on selenium, copper, manganese, and zinc) -Antioxidant Vitamins (E defends lipids, C defends fluids and restores vitamin E -Phytochemicals (nonnutrients, plant-derived foods that have biological activity in the body) |
| Phytochemicals in Disease Prevention |
-phytochemicals are metabolites poduced by plants 1)antioxidant activity 2)phytosterols (slow the growth of certain cancers) 3)lycopene (inhibit the reproduction of cancer cells) 4)flavenoids (preventing heart disease)
*phytochemicals absorb free radicals and become inactive
DEFENDING AGAINST CANCER -de-oxidant enzyme system -antioxidant Vitamins -phytochemicals (soy products are rich in phystosterols, which correlates with low rates ofbreast and prostate cancer....... tomatoes are rich in lycopene, a red pigment, and offer protection against cancers of esophagus, prostate, and stomach)
DEFENDING HEART DISEASE -de-oxidant enzyme system -antioxidant Vitamins -Phytochemicals (flavonoids are found in whole grains, veggies, fruits, herbs, spices, teas, red wine...... lycopene and carotenoids also protect) |
| Radical Additions to Alkenes: The anti-Mark addition of Hydrogen Bromide |
-addition of hydrogen bromide in the presence of peroxides gives anti-Markovnikov addition -the other hydrogen halides do not give this type of anti-Markovnikov addition
Step1: Heat brings about homolytic cleavage of the weak RO-OR bond Step2: The slkoxyl radical abstracts a hydrogen atom from HBr, producing a bromine atom
In step3, the first step of propagation, a bromine radical adds to the double bond to give the most stable of teh two possible carbon radicals (in this case, a secondary radical) -attack on the primary carbon is also less sterically hindered
Step 4 regenerates a bromine radical -bromine radical reacts with another equivalent of alkene |
| Review: Catalytic Hydrogenation |
-an internal alkyne will yield a cis double bond with heterogeneous catalysts -to break only one pi bond, special catalysts need to be used -3 metals do the trick: Ni, Pd, Pt ... these make it go all the way to single..... when with carbon...... with anything else, triple goes to cis double
Another Method -Lindlar's catalyst also produces cis-alkenes from alkynes -Lindlar's catalyst: Pd/CaCO3 |
| Anti Addition of Hydrogen: Synthesis of trans-Alkenes |
A dissolving metal reaction which uses lithium or sodium metal in low temperature ammonia or amine solvent produces trans-alkenes -net anti addition occurs by formal addition of hyddrogen to the opposite faces of the double bond [lithium only has 3 electron... 1 must be unpaired]
THE MECHANISM -the mechanism is a free radical reaction with two electron transfer reactions from the metal Step1: A lithium atom donates an electron tot he pi bond of the alkyne. An electron pair shifts to one carbon as the hybridization states change to sp3. -the radical anion acts as a base and removes a proton from a molecule of the ethylamine Step2: The vinylic anion prefers to be trans and this determines the trans stereochemistry of the product -A second lithium atom donates an electron to the vinylic radical -the anion acts as a base and removes a proton from a second molecule of ethylamine |
| Radical Polymerization of Alkenes: Chain-Growth Polymers (Binyl Polymerization) |
Polymers are macromolecules made up of repeating subunits -the subunits used to synthesize polymers are called monomers
Polyethylene is made of repeating subunits derived from ethylene (cntaining a vinyl group CH2=CH-) -polyethylene is called a chain-growth polymer or addition polymer
THE MECHANISM -A very small amount of diacyl peroxide is added in initiating the reaction so that few but very long polymer chains are obtained -the diacyl peroxide dissociates and releases carbon dioxide gas -alkyl radicals are produced which in turn initiate chains (R43)
-The propagation step simply adds more ethylene molecules to a growing chain -chains propagate by adding successive ethylene units, until their growth is stopped by combination or dispoportionation
-Chain branching occurs by abstraction of a hydrogen atom on the same chain and continuation of growth from the main chain
-the radical at the end of the growing polymer chain can also abstract a hydrogen atom from itself by what is called "back biting". This leads to chain branching. |
| Polymers from other monomers (heat-to-tail polymerization) |
-polystyrene is made using styrene as the monomer -polyvinyl chloride (PVC) is made using vinyl chloride as the monomer. -exposure to viyl choride linked to liver cancer Control of Polymer Properties -irregular arrangement of the halide groups leads to weak polymer -catalyzed polymeriztion creates more orderly arrangement and leads to stronger polymer -copolymerization is one way to modify polymer properties
-cross-linking is another way to modify polymer properties -some are reversible, some are irreversible - unrecyclable
Plastics and Biodegradability -synthetic polymers were unknown 100 years ago -nature does not know what to do with them - not biodegradable -incinceration is one way of displosal -reducing the usage is the best way for now
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| Other important radical reactions |
Nitric Oxide -Synthesized in the body from the amino acid arginine, serves as a chemical messenger in a variety of biological processes, including blood pressure regulation adn the immune response -plays a role in relaxation of smooth muscle in vascular tissues, leading to increased blood flow to certain tissues -drugs such as viagra inhibit the enzyme that counters effects of Nitric Oxide
Combustion of Alkanes -in oil and gas furnaces and in internal combustion engines, alkanes react with oxygen via a radical chain mechanism with chain-initiating and chain-propagating steps: (ultimately producing CO2 and water)
Also CFC action (R53) |
| Carboxylic Acids and Esters (handout 6) |
Nomenclature and Physical Properties -In IUPAC nomenclature, the name of a carboxylic acid is obtained by changing the -e of the corresponding parent alkane to -oic acid -the carboxyl carbon is assigned position 1 and need not be explicitly numbered -the common names for many carboxylic acids remain in use -methanoic and ethanoic acid are usually referred to as formic and acetic acid -carboxylic acids can form strong hydrogen bonds with each other and with water -carboxylic acids with up to 4 carbons are miscible with water in all proportions (completely soluable)
Acidity of Carboxylic acids -the carboxyl proton of most carboxylic acids has a pKa=4-5 -carboxylic acids are readily deprotonated by sodium hydroxide or sodium bicarbonate to form carboxylate salts -carboxylate salts are more water soluble than the corresponding carboxylic acid
Preparation of Carboxylic Acids -by oxidation of alkenes *KMnO4 or O3 -by oxidation of aldehydes and primary alcohols
ESTERS -the names of esters are derived from the names of the corresponding carboxylic acid and alcohol from which the ester would be made -the alcohol portion is named first and has the ending -yl -the carboxylic acid portion follows and its name ends with -ate or -oate -esters cannot hydrogen bond to each other and therefore have lower boiling points than carboxylic acids -asters can hydrogen bondto water and have appreciable water soubility
Esterification -acid catalyzed reaction of alcohols and carboxylic acids to form esters is called Fischer eterification -ester formation is favored by use of a large excess of either the alcohol or carboxlic acid -ester formation is also favored by removal of water
-esterification with labeled methanol gives a product labeled only at the oxygen atom bonded to the methyl grou -handout page 7
*the reverse reaction is acid-catalyzed ester hydrolysis
saponification: base-promoted hydrolysis of esters -reaction of an ester with NaOH results in the formation of a sodium carboxylate and an alcohol -the mechanism is reversible until the alcohol product is formed -protonation of the alkoxide by the initially formed carboxylic acid is irreversible -this step draws the overall equilibrium toward completion of the hydrolysis |
