Orgo Exam 2

How does the number of alkyl substituents at the double bond affect the stability of an alkene?

- An alkene is stabilized by alkyl substituents on the double bond. When we compare the stability of alkene isomers, we find that the alkene with the greatest number of alkyl substituents on the double bond is usually the most stable one. - It is the # of alkyl groups more than their identities that governs the stability of an alkene. Example: molecule w 2 smaller alkyl groups on double bond is more stable than isomer with 1 larger group on the double bond. - WHY? We are really comparing a tradeoff. The major effect is that an sp2-sp3 C-C bond is stronger than an sp3-sp3 C-C bond. (Bonds to H have similar but smaller effects). Increasing bond strength lowers the heat of formation. - Why are sp2-sp3 C-C bonds stronger than sp3-sp3? Bond strength is directly related to the E of the e-'s in the bond. The lower the E of the bonding e-'s, the stronger the bond. Because s e-'s have lower energy than p e-'s, a bond with more s character involves e-'s of lower energy than one with less s character. Bond strength increases with the fraction of s character in the component hybrid orbitals.

describe the carbon hybridization in alkenes

- C 2s orbital mixed w/ only 2 of the 3 available 2p orbitals. 2pz is unaffected. 1 e- in each hybridized orbital and 1 e- in the p one. - The result is 3 hybrid orbitals and a "leftover" 2pz orbital. Each hybrid orbital has one part s character and 2 parts p character. Called sp2. - Shape of sp2 is much like that of sp3, but the difference is that the e- density within an sp2 orbital is concentrated slightly closer to the nucleus. - Reason: larger amount of s character in an sp2 orbital. (33% sp2 vs 25% sp3) - Axes of 3 sp2 orbitals lie in the xy plane, oriented at max angular separation of 120 - "Leftover" 2p orbital is 2pz so it's axis is the z axis, perpendicular to the sp2 plane

why do we know that bromonium ions rather than carbocations are reactive intermediates in bromine addition? and why is it more stable than a carbocation?

- First: rearrangements are not typically observed in Br addition. They are in HBr however. - Second: bromonium ions have been isolated under special circumstances (though can't isolate under usual rxn conditions bc they're unstable) - Finally; there is compelling stereochemical evidence for bromonium ions we'll consider later So WHY is it more stable than the corresponding carbocation? - The bromonium ion has more covalent bonds than a carbocation, and every atom has an octet :)

describe how ethylene forms using hybridization

- Form hybrid orbital model by bonding 2 sp2-hybridized C atoms and 4 H atoms - Sp2 orbital on 1 C containing 1 e- overlaps with an sp2 orbital on another to form a 2 e- sp2-sp2 C-C sigma bond. Each of the 2 remaining sp2 orbitals, each containing one e-, overlaps with a H 1s orbital, also containing 1 e-, to form a 2 e- sp2-1s C-H sigma bond. - These orbitals account for the 4 C-H bonds and 1 of the 2 C-C bonds of ethylene, which together comprise the sigma-bond framework of ethylene. - Whenever a main-group atom has trigonal planar geometry, its hybridization is sp2!

what is heat of formation and why is it helpful

- Heat of formation (delta H(f)) is heat change that occurs when the compound is formed from its elements in their natural state at 1 atm pressure and 25 deg C. (equation is in notebook) H(P) - H(R) = delta H(rxn) - Exothermic rxn: heat is liberated, neg - energy lower, molecule stabilized - Endothermic rxn: head is absorbed, pos Heats of formation are used to determine the relative enthalpies of molecules - which of the 2 molecules has lower energy. - EXAMPLE: how do you calculate the standard enthalpy diff b/w cis and trans isomers of 2-butene? - Subtract cis-2-butene heat of formation from trans-2-butene heat of formation. - Hess's law of constant heat summation: chemical reactions and their associated energies can be added algebraically. A direct consequence of the 1st law of thermodynamics: requires that E diff between 2 compounds doesn't depend on the path (or rxns) used to make the measurement. We actually just added their forward and reverse reactions. Because they are isomers, the elements from which they are formed are the same and cancel in comparison! - This is analogous to measuring the relative heights of 2 objects by comparing their distances from a common reference, such as the ceiling.

how do we tell if somethings a meso compound?

- It is a meso compound if it can be divided into halves that have the same connectivity. - The corresponding asymmetric atoms in each half of the molecule must have opposite stereochemical configurations (one is R and the other is S) Shortcut: if you can find any conformation of a molecule with asymmetric C's (even an eclipsed conformation) that is achiral, the molecule is meso. So it works, even if the compound does not exist in the eclipsed formation.

priest's optical purity tips

- Mixture of enantiomers exists in a 75:25 ratio = 50% e.e. - Mixture of enantiomers exists in 80:20 ratio = 60% e.e. - Mixture of enantiomers exists in a 95:5 ratio = 90% e.e. Racemic mixture, 50:50 = 0% e.e R = -13 and S = +13, so (.50)x(-13) + (.50)x(+13) = 0 The same R & S enantiomers at 60% have an optical rotation of -7.80. Why? (.80)x(-13) + (.20)x(+13) = -7.80 (.60)x(-13) = -7.80 If we have a 90% e.e. favoring the S enantiomer, then the optical rotation of the mixture would be (.95)x(+13) + (.05)x(-13) = +11.7 So what if a mixture had an optical rotation of +6.90? What would the e.e. be? What is the % of each isomer? (X)x(+13) = +6.90 so X = 53% 53% e.e. = 76.5% S and 23.5% R.

IUPAC nomenclature for alkanes

- Replace the the ane suffix in the name of the corresponding alkane with the ending ene and specifying the location of the double bond with a number. - (Exception: simplest alkene is ethylene rather than ethene) - C's are numbered so that the double bond receives the lowest #. - The names of alkenes are derived from their principal chains. In an alkene, the principal chain is defined as the C chain containing the greatest # of double bonds, even if this is not the longest chain. If more than one candidate for the principal chain have equal #'s of double bonds, the principal chain is the longest of these. It's numbered from the end that results in the lowest #'s for the C's of the double bonds. - ONLY DIFFERENCE FROM ALKANE: if alkene contains an alkyl substituent, the position of the double bond is what determines the numbering of the chain, not the position of the branch. But double bond is cited after the name of the alkyl group. - If a compound contains more than 1 double bonds, put di before en, if three then tri before ene, and so one. So 1,5-hexadiene - NOTE: PLEASE put an A after the parent name. Like heptene -> heptadiene. - If the name remains ambiguous, the principal chain is numbered so that the lowest numbers are given to the branches at the first point of difference. Substituent groups may also contain double bonds. Some widely occurring groups of this type have special names that should be learned. - These are numbered from the point of attachment to the principal chain. - Call them like "butenyl"

describe how bonding and MO of the pi bond works

- The 2 2pz orbitals not used in sigma-bond formation overlap side-to-side to form the second bond of the double bond. Each 2pz orbital contributes 1 e- to make an e-pair bond. - MO theory: 2 AO used, 2 MO formed. Formed by additive + subtractive combos of the 2pz orbitals. Bonding MO results from additive overlap of the 2 C 2p orbitals: pi MO. Has a nodal plane - plane coincides w/ the plane of the ethylene molecule. Antibonding MO results from subtractive overlap, called pi MO. 2 nodes: 1 is in plane of molecule, other is plane b/w 2 C's, perpendicular to the plane of the molecule. - Bonding MO lies at lower E than the isolated 2p orbitals, whereas the antibonding MO lies at higher E. By the aufbau principles, the 2 2p e-'s occupy the MO of lower E - the pi MO. The antibonding MO is unoccupied. - The filled pi MO is the pi bond. Unlike a sigma bond, a pi bond is not cylindrically symmetrical about the line connecting the 2 nuclei. The pi bond has e- density both above and blow the plane of the ethylene molecule, with a wave peak on one side and a wave trough on the other, and a node in the plane of the molecule. The pi bond is 1 bond with 2 lobes, just as a 2p orbital is 1 orbital with 2 lobes. - 2 types of C-C bonds: sigma bond with most of its e- density relatively concentrated b/w the C atoms, and a pi bond with most of its e- density concentrated above and below the plane of the ethylene molecule - Shows why ethylene is planar: if the 2 CH2 groups were twisted away, the 2p orbitals could not overlap to form the pi bond. Thus, the overlap of the 2p orbitals and consequently the very existence of the pi bond require the planarity of the ethylene molecule.

information about the hydration of alkenes

- The alkene double bond undergoes reversible addition of water in the presence of moderately concentrated strong acids such as H2SO4, HClO4, and HNO3 - The addition of the elements of water is in general called hydration. The addition of water to the alkene double bond is called alkene hydration. - Hydration does not occur at a measurable rate in the absence of an acid, and the acid is not consumed in the rxn. Hence, alkene hydration is an acid-catalyzed rxn. Because the catalyzing acid is soluble in the rxn solution, it is a homogeneous catalyst. - Notice this rxn, like the addition of HBr, is regioselective! Just like with the addition of HBr, the H adds to the C of the double bond with the smaller # of alkyl substituents. The more electronegative partner of the H-OH bond, the OH group, like the Br in the HBr addition, adds to the C of the double bond with the greater number of alkyl substituents.

describe transition states in reactions

- The rate of a chemical rxn: the number of reactant molecules converted into product in a given time. - Theory used: as reactants change into products, they pass through an unstable state of maximum free energy, called the transition state. Has a higher energy than either the reactants or products and therefore represents an energy barrier to their interconversion. This energy barrier is shown graphically in a rxn free-energy diagram. - Progress of reactants to products: reaction coordinate. That is, the reactants define one end of the reaction coordinate, the products define the other, and the transition state is at the energy maximum somewhere in between. Energy barrier delta G double dagger called: standard free energy of activation: equal to the difference between the standard free energies of the transition state and the reactants. - The size of the energy barrier determines the rate of a reaction: the higher the barrier, the smaller the rate. - In the same sense that relative free energies of reactants and products determine the equilibrium constant, the relative free energies of the transition state and the reactants determine the rxn rate. NOTICE: a rxn and its reverse have the same transition state.

let's apply hammond's postulate to HBr addition

- The rate-limiting step of the HBr addition to an alkene was the first step: protonation of the alkene by HBr to give a carbocation. This protonation could occur in 2 diff and competing ways: protonation of double bonds at 1 C gives the tert-butyl cation as the unstable intermediate; protonation of the double bond at the other C gives the isobutyl cation. - We apply Hammond's postulate by assuming that the structures and energies of the transition states are approximated by the structures and energies of the unstable intermediates - the carbocations - themselves - Because the tertiary carbocation is more stable the transition state leading to the tertiary carbocation should also be the one of lower energy. Transition state has smaller free energy and is thus the faster of the 2 competing rxns. NOTE: the stabilities of the carbocations themselves do not determine which reaction is faster; the relative free energies of the transition states for carbocation formation determine the relative rates of the two processes. - We need this postulate because the structures of transition states are uncertain, whereas the structures of reactants, products, and reactive intermediates are known. Therefore knowing that a transition state resembles a particular species helps us to make a good guess about the transition-state structure.

information about the energy barrier in reactions

- The relationship between rate and standard free energy of activation is an exponential one. Reaction rates are VERY SENSITIVE to their standard free energies of activation - Where do they get the energy to overcome the energy barrier? From thermal motions! - Maxwell-Boltzmann distribution: shows the energy of a collection of molecules. The rate of a rxn is directly related to the fraction of molecules that has enough energy to cross the energy barrier. (The smaller the barrier, the greater the rxn rate will be.) For a given rxn under a given set of conditions, we cannot control the size of the energy barrier: it is an intrinsic property of the rxn. Some are intrinsically slow, others are intrinsically fast. However, we can sometimes control the fraction of molecules with enough energy to cross the barrier. We can increase this fraction by raising the temperature. - The Maxwell-Boltzmann distribution is skewed to higher energies at higher temperature, and as a result, a greater fraction of molecules have the energy required to cross the barrier. In other words, rxns are faster at higher temperatures. - Very rough rule of thumb: rxn rate doubles for each 10 deg C inc in temp. NOTE: equilibrium constant tells us ABSOLUTELY NOTHING About its rate.

why is there a dipole moment in alkenes? and why are they greater than that of alkanes?

- Turns out that sp2-hybridized C's are somewhat more electronegative than sp3 hybridized C's. As a result, any sp2-sp3 C-C bond has a small bond dipole in which the sp3 C is the positive end of the dipole and the sp2 C is the negative end. - Reason: Sp2-hybridized C has an electron in a 2p orbital. E- density in a 2p orbital is not evenly distributed around the nucleus in all directions. That is, if a 2p orbital is oriented along the z axis, a 2p e- does not screen the nucleus from other e-'s situated in the xy plane. Therefore, the relatively unscreened positive charge of the atomic nucleus in an sp2 C pulls the e-'s in the sp3-sp2 C-C bond toward itself and creates a small partial pos charge on the attached methyl groups, and thus creates the bond dipole. (a similar effect occurs in C-H bonds in ethylene, for example) - The dipole moment of cis-2-butene is the vector sum of all the H3C-C and H-C bond dipoles. Although both types of bond dipole are oriented toward the alkene C, there is good evidence that the bond dipole of the H3C-C bond is greater. Would be expected from the greater length of the C-C bond. (dipole moment for a given charge separation increases with the length of the dipole). Thus cis-2-butene has a net dipole moment. IN SUMMARY: bonds from alkyl groups to trigonal planar sp2-hybridized C's are polarized so that e-'s are drawn away from alkyl groups toward the trigonal C. This means that a C-C double bond, when viewed as a substituent group, exerts an electron-withdrawing polar effect!

describe catalytic hydrogenation of alkenes

- When a solution of alkene is stirred under an atmosphere of H, nothing happens But if the same solution is stirred under H in presence of a metal catalyst, H is rapidly absorbed by the solution. H is consumed bc it undergoes an addition to the alkene double bond! - Catalytic hydrogenation: an addition of H to an alkene in the presence of a catalyst. One of the best ways to convert alkenes into alkanes! - Hydrogenation can be carried out at room temp and pressure or, for especially difficult cases, at higher temp and pressure in a "bomb" (a closed vessel designed to withstand high pressures) - Bc hydrogenation catalysts are insoluble in the rxn solution, they are examples of heterogeneous catalysts. Can be reused and in small amounts! How does it work? - Both the H and the alkene must be adsorbed on the surface of the catalyst for the rxn to occur. The catalyst is believed to form reactive metal-C and metal-H bonds that ultimately are broken to form the products and to regenerate the catalyst sites. Beyond this, the chemical details are not thoroughly understood! Curved-arrow mechanism cannot be written!!!!! - NOTE: benzene rings are inert to conditions under which double bonds react readily. They can be hydrogenated, but only with certain catalysts under conditions of high temp and pressure. NOTE: syn addition! They're added on the same molecule plane side bc they're added at the same time PRIEST: the metal creates H radicals

what is the regioselectivity of the hydrogen halide addition?

- When alkene has an unsymmetrically located double bond, 2 constitutionally isomeric products are possible. ONLY ONE of the 2 is formed in a significant amount - Generally: the main product is that isomer in which the halogen is bonded to the more substituted C, H to the less - When this happens, the rxn is said to be a regioselective rxn. Hydrogen halide is highly regioselective because it gives only 1 of the 2 possible products. - NOTE: this is regardless of size! Small alkyl group on one C, large on other, nearly equal amounts of both products!

why does regioselectivity occur in hydrohalogenation?

- When double bond of alkene is not located symmetrically within the molecule, protonation of the double bond can occur in 2 distinguishable ways to give 2 diff carbocations - These 2 rxns are in competition: one can only happen at the expense of the other bc the 2 rxns compete for the same starting material. The one where H goes to the less substituted C occurs much faster than the other. - The regioselectivity of hydrogen halide addition is due to the formation of only 1 of 2 possible carbocations. - To understand why it's faster, you need to understand factors that influence reaction rate: relativity stability!

fun priest notes

- You will NEVER make a primary carbocation - We prefer strong electrophile and weak nucleophile or vice versa- not both strong or both weak - You don't have to say cis/trans for cyclic rings below 8 C's because it must always be cis. But once we hit 8 then you have to - Benzene's are NOT the same as alkene's and don't react the same way - A carbocation one away from a benzene ring is called a benzylic cation! It is more stable than the identical secondary carbocation one more carbon away from the benzene ring.

what are the 2 distinct steps in the reaction of ozone (O3) with alkenes

1) Alkene reacts w/ ozone to form addition product called an ozonide 2) Ozonide treated with oxidizing or reducing agents to form various products

what are the rules for rearrangements

1) Rearrangement almost always occurs when a more stable carbocation can result 2) Rearrangement that would give a less stable carbocation generally does not occur 3) The group that migrates in a carbocation rearrangement comes from a C directly attached to the e-deficient, pos charged C of the carbocation 4) The group that migrates in a rearrangement is typically an alkyl group, aryl group, or H 5) When there is a choice b/w the migration of an alkyl group (or aryl group) or a H from a particular C, a hydride migration typically occurs because it gives the more stable carbocation

uses of knowing bond dissociation energies

1) You can use these energies to estimate the delta H of a nonradical rxn by using Hess's law to treat it as the sum of fictitious radical reactions. The sum of the delta H values of the radical rxns, obtained from bond dissociation energies, provides the delta H of the overall rxn. - The lower a bond dissociation energy, the lower the temperature required to rupture the bond in question and form free radicals at a reasonable rate. 2) An important use of bond dissociation energies is to calculate or estimate the delta H of free-radical reactions. Why in the second initiation step for the free-radical addition of HBr is that H abstracted from HBr rather than the Br? The reason lies in the relative enthalpies of the 2 rxns. To calculate the delta H for a rxn, subtract the bond dissociation energies (BDE) of the bonds formed from the bond dissociation energies of the bonds broken! Delta H = BDE (bonds broken) - BDE (bonds formed) - This works because BDEs are the enthalpies for bond dissociation - Exothermic (favorable) has - value and endothermic has positive. And why doesn't HCl and HI have a peroxide effect? The first propagation step is endothermic for them! Remember that the propagation steps of any free-radical chain rxn are in competition with recombination steps that terminate free-radical rxns. The energy required for an endothermic propagation step represents an energy barrier that reduces the rate of this step. In effect, only exothermic propagation steps compete successfully with recombination steps. In summary, you can use it for anything where ions are not created or destroyed

what are the 2 ways carbocations can react

1) react with nucleophile 2) rearrange to more stable carobcations

what are the steps of the hydrogen halide addition?

2 successive steps. Step 1: e- pair in the pi bond of an alkene is donated to the proton of the hydrogen halide. The e-'s of the pi bond react rather than the e-'s of the sigma bonds bc pi e-'s have the highest E. As a result, the C=C is protonated on a C atom. The other C becomes positively charged and e-deficient. - The species with a positively charged, e-deficient C is called a carbocation. The formation of the carbocation from the alkene is a Bronsted acid-base rxn, in which the pi bond acts as a Bronsted base toward the Bronsted acid HBr. The pi bond is a very weak base. Nevertheless, it can be protonated to a small extent by a strong acid such as HBr. Step 2: The resulting carbocation is a powerful e-deficient Lewis acid and is thus a potent electrophile. In the second step, the halide ion (Lewis base/nucleophile) reacts w/ the carbocation at it's e-deficient C atom. This is a Lewis acid-base association rxn. NOTES: - The carbocations involved in hydrogen halide addition to alkenes are examples of reactive intermediates or unstable intermediates: species that react so rapidly that they never accumulate in more than very low concentration. Most carbocations are too reactive to be isolated except under special circumstances. Thus, can't be isolated from these rxns. - Mechanism of reaction: complete description of a rxn pathway, including any reactive intermediates such as carbocations. To summarize: 1. C of pi bond is protonated (Bronsted acid-base rxn) 2. Halide ion reacts w/ resulting carbocation (Lewis acid-base association rxn)

what are racemates?

A mixture containing equal amounts of 2 enantiomers is encountered so commonly it's given a special name: racemate or racemic mixture. - Referred to as: racemic 2-butanol, (+/-)-2-butanol, or d,l-2-butanol Have diff properties from those of pure enantiomers. - Melting point of either enantiomer of lactic acid is 53 deg C, but the melting point of racemic lactic acid is 18 deg C. The reason: the crystal structures differ! - The melting point largely reflects interactions b/w molecules in the crystalline solid. (Like packing left and right shoes in a box). The optical rotations of enantiomers and racemates are another example of differing physical properties. - The optical rotation of any racemate is zero, bc they contain equal amounts of 2 enantiomers whose optical rotations of equal magnitude and opposite sign exactly cancel each other out. The EE is 0 Process of forming a racemate from a pure enantiomer: racemization - Mix or do chemical rxn Bc enantiomers have the same boiling points, melting points, and solubilities, the separation of enantiomers poses a special problem - Separation of enantiomers (enantiomeric resolution) requires special methods

what are the catalysts and support in catalytic hydrogenation of alkenes

A number of noble metals (platinum, palladium, and nickel (Ra/Ni) and Rh in a giant complex called Wilkinson's catalyst) are useful as hydrogenation catalysts, and are often used in conjunction with solid support materials such as alumina (Al2O3), barium sulfate (BaSO4), or activated C NOTE: Pt/C is read as "Platinum supported on C" or "Platinum on C". Is a finely divided platinum metal that has been precipitated, or "supported" on activated charcoal.

describe the formation of the ozonide on an alkene

Addition of ozones to alkenes occurs at low temp and breaks the C-C pi bond to give an unstable addition product. The spontaneous conversion of this addition product into an ozonide breaks the second C-C bond. - The rxn of an alkene w/ ozone to yield products of double-bond cleavage is called ozonolysis. (-lysis describes bond-breaking processes) First step: addition rxn of the alkene pi bond. Central O of ozone is a positively charged electronegative atom and therefore strongly attracts e-'s. The curved-arrow notation shows that this O can accept an e- pair when the other O of the O=O bond accepts pi e-'s from the alkene. - This rxn results in the formation of a ring because the 3 O's of the ozone molecule remain intact. Additions that give rings are called cycloadditions. Furthermore, the cycloaddition of ozone occurs in a single step - so it's a concerted mechanism - This product is unstable and spontaneously forms the ozonide. The remaining C-C bond of the alkene is broken. The creation of the ozonide starts with a cyclic e- flow to form an aldehyde and an aldehyde oxide. An O-O bond (a very weak bond) is broken in the process. - The aldehyde flips over, and a second cycloaddition, much like the ozonolysis itself- but this time to the C=O bond - completes the formation of the ozonide

what is the mechanism for OXYMERCURATION of an alkene

Bears a close resemblance to halohydrin formation. The first step of the rxn mechanism involves the formation of a cyclic ion called a mercurinium ion. - mercuric acetate adds to C, one OAc is leaving group, forms mercurinium ion - Like Br addition, oxymercuration does not involve carbocations bc carbocation rearrangements are not observed. Consequently, the mechanism can be viewed as a one-step process. - Just as the bromonium ion reacts with the solvent water, which is present in large excess, the mercurinium ion also reacts with the solvent water - Of the 2 C's in the ring, the rxn of water occurs at the C w/ the greater number of alkyl substituents, just as in the rxn of water with a bromonium ion. A difference b/w oxymercuration and halohydrin formation, however, is the degree of regioselectivity. In oxymercuration, the rxn of water occurs almost exclusively at the C w/ more alkyl substituents, even if that C has only 1 alkyl substituent. In halohydrin formation it was only regioselective if one of the alkene C's had 2 alkyl branches. The addition is completed by the transfer of a proton to the acetate ion formed from the first step

describe the process of hydroboration

Borane (BH3) adds regioselectively to alkenes so that the B becomes bonded to the C of the double bond with fewer alkyl substituents, and the H becomes bonded to the C with more alkyl substituents. - Because borane has 3 B-H bonds, 1 borane molecule can add to 3 alkene molecules - Addition of BH3 is called hydroboration. Hydroboration product of an alkene is a trialkylborane - Ethers used as solvents in the hydroboration reaction: diethyl ether, THF, diethylene glycol dimethyl ether (diglyme). Borane-ether complexes are the actual reagents involved in hydroboration rxns. For simplicity, the simple formula BH3 often is used for borane. Hydroboration is believed to occur in a single mechanistic step because carbocation rearrangements are not observed and stereochemical evidence - Is said to occur by a concerted mechanism because everything happens "in concert" or simultaneously. Despite the evidence against carbocation intermediates, the concerted mechanism is consistent with the regioselectivity of the rxn only if some degree of e- deficiency is built up on the tertiary C in the transition state of the rxn. - Just as alkyl substitution at the e-deficient C stabilizes a carbocation, alkyl substitution at a partially e-deficient C stabilizes a transition state. Thus, hydroboration occurs w/ the regioselectivity that places partial positive charge on the C with more alkyl substituents. Priest: H and B add at the same time in syn addition (cuz its concerted mechanism) Overall result is the H and B add :)

let's compare HBr addition with and without peroxides

Both reactions begin by attachment of an atom to the C of the double bond with fewer alkyl substituents. In the absence of peroxides, the proton adds first to give a carbocation at the C with the greater number of alkyl substituents. The nucleophilic reaction of the bromide ion at this C completes the addition. In the presence of peroxides, the free-radical mechanism occurs; a Br atom adds first, thus placing the unpaired e- on the C with the greater number of alkyl substituents. A H atom is subsequently transferred to this C.

describe the most common mechanism for bromine addition? (halogenation)

Bromonium ion: species that contains a Br bonded to 2 C atoms; the Br has an octet of e-'s and a pos charge. Formation of the bromonium ion occurs in a single mechanistic step involving 3 curved arrows 1) One C of the double bond acts as a nucleophile toward Br using the pi e-'s to form a C-Br bond 2) The other Br acts as a leaving group to form bromide ion 3) The electrophilic Br also acts as a nucleophile toward the other C of the double bond to form the other C-Br bond NOTE; analogous cyclic ions form in Cl and Iodine addition! Q: how can Br2 act as an electrophile if it's a nonpolar molecule? Bc e-'s are always moving and sometimes they create a partial pos charge Bromine addition is completed when the Br ion donates an e- pair to either one of the ring C's of the bromonium ion (and the C-Br bond breaks and e-'s go to bromine, making it neutral) - This is another electron-displacement rxn, in which the nucleophile is the bromide ion, the electrophilic center is the C that accepts an e- pair from the nucleophile, and the leaving group is the Br of the bromonium ion. (The leaving group doesn't actually leave the molecule, bc it is tethered by another bond) - This rxn occurs bc the positively charged bromine is very electronegative and readily accepts an e- pair. Also because three-membered rings are strained; opening it releases considerable energy NOTE: it adds on TRANS. Because the bromonium ion takes up the entire plane, so the other Br has to add to the other side!

what are double-bond stereoisomers and is interconversion possible

Butenes exist in isomeric forms. - In butenes with unbranched C-C chains, the double bond may be located either at the end or in the middle of the C chain. Isomeric alkenes that differ in the position of their double bonds are further example of constitutional isomers - 1-butene or 2-butene There are 2 separable, distinct 2-butenes, each with its own characteristic properties. - Cis-2-butene or (Z)-2-butene: CH3 on same side of double bond. Higher BP too (bc higher dipole moment so more ionic) - Trans-2-butene or (E)-2-butene: CH3 on opposite sides of the double bond Have identical atomic connectivities, but differ in the way their constituent atoms are arranged in space. Called: stereoisomers! The interconversion (isomerization) of cis- and trans-2-butene requires a 180 degree internal rotation about the double bond. - For such to occur, the 2p orbitals on each C must be twisted away from coplanarity: that is, the pi bond must be broken. Bc bonding is energetically favorable, lack of it is energetically costly. It takes more energy to break the pi bond than is available under normal conditions; thus, the pi bond in alkenes remains intact, and internal rotation about the double bond does not occur. In contrast, internal rotation about the C-C single bonds of ethane or butane can occur rapidly, because no chem bond is broken! Double-bond stereoisomers (cis-trans stereoisomers or E,Z stereoisomers): compounds related by an internal rotation of 180 about the double bond. Related by interchange of the 2 groups at either C of a double bond. - When an alkene can exist as a double-bond stereoisomer, both C's of the double bond are stereocenters. An atom is a stereocenter when the interchange of 2 bonded groups gives stereoisomers. Also stereogenic atom or stereogenic center.

how can we visualize the transition state?

Can visualize it as a structure. - Consider step 1 of addition of HBr to an alkene. H-Br bond and C-C pi bond are partially broken, new C-H bond is partially formed, and the new charges are only partially established. Can use dashed lines to be partial bonds and make partial charges. Looking at this event frozen in time. - It's the point of highest potential energy. You're never there for more than an instant

what are hydride shifts

Carbocation rearrangements are not limited to the migrations of alkyl groups. Hydride shift: migration of an H with its 2 bonding e-'s Hydride migrates instead of an alkyl group because the rearranged carbocation is tertiary and thus is more stable than the starting carbocation. Migration of an alkyl group would have given another secondary carbocation.

describe the structure and stability of carbocations

Carbocations are classified by the degree of alkyl substitution at their e-deficient C atoms. Primary: 1 R group. Secondary: 2 R groups and so on. - Alkyl substituents at the e-deficient C strongly stabilize carbocations! - Stability of carbocations: tertiary > secondary > primary Why? The e-deficient C of the carbocation has trigonal planar geometry and is therefore sp2-hybridized, like the C's involved in double bonds - however, the 2p orbital on the e-deficient C has no e-'s. - The explanation is in part the same as the explanation for the stabilization of alkenes by alkyl substitution - the greater number of sp2-sp3 C-C bonds in a carbocation w/ a greater number of alkyl substituents. But this is CONSIDERABLY greater in stabilizing carbocations than in alkenes. - Additional factor: phenomenon called hyperconjugation: overlap of bonding e-'s from the adj sigma bonds with the unoccupied 2p orbital of the carbocation. (In the diagram in the book, the sigma bond that provides the bonding e-'s is a neighboring C-H bond). - The energetic advantage of hyperconjugation is that it involve additional bonding. The e-'s in the C-H bonds participate in bonding not only with the C and H, but also with the e-deficient C. Additional bonding is a stabilizing effect! It's like if you made a resonance structure where C-H bond became pi bond between C=C and H was just floating with partial pos charge. It hasn't actually moved, it's just showing that CHARACTER! - So each alkyl group provides additional hyperconjugation and thus more stabilization! priest: hyperconjugation is that C's have e-donating stabilization effect

what are meso compounds?

Certain compounds containing 2+ asymmetric C's are achiral (no enantiomer). - 2,3-butanediol has (2S, 3S), (2R, 3R), (2S, 3R) (2R, 3S) - The 2S,3S and 2R,3R structures are noncongruent mirror images, and are thus enantiomers - But 2S,3R and 2R,3S have a center of symmetry and are actually identical molecules - Hence, there are only 3 stereoisomers of 2,3-butanediol, bc 2 of the possibilities are identical. And this 3rd one is achiral and thus optically inactive (the light twists one way and then back as it goes through the molecule) Example of a meso compound: an achiral (and therefore optically inactive) compound that has chiral diastereomers. In virtually all examples we'll encounter, it will have at least 2 asymmetric centers. (so alkene C's are not bc they don't have asymmetric C's) - Notice difference between meso compound and racemate. Both are optically inactive, but a meso compound is a single achiral compound, and a racemate is a mixture of chiral compounds- specifically, an equimolar mixture of enantiomers. The existence of meso compounds shows that some achiral compounds have asymmetric C's. Thus, the presence of asymmetric C's in a molecule is an insufficient condition for it to be chiral, unless it has only 1 asymmetric C. - If a molecule contains n asymmetric C's, then it has 2^n stereoisomers unless there are meso compounds. If there are meso compounds, there are few than this

what causes chirality?

Chiral molecules lack certain types of symmetry Plane of symmetry (internal mirror plane): divides an object into halves that are exact mirror image - A molecule or other object that has a plane of symmetry is achiral - Chiral molecules and other chiral objects do not have planes of symmetry Center of symmetry (point of symmetry): you can reproduce molecule by forming its mirror image and rotating it by 180 deg about an axis perpendicular to the mirror. It's a point through which ANY line contacts exactly equivalent parts of the object at the same distance in both directions. In summary: If a molecule has a single asymmetric C, it must be chiral - The most general way otherwise is to build 2 models/draw 2 perspective structures and test for congruence - If congruent, achiral; if not, chiral

describe the conversion of organoboranes into alcohols

Converts into alcohols with hydrogen peroxide (H2O2) and aqueous NaOH You don't need to know the mechanistic details Priest: the -OH oxidizes the B off! The important thing to notice is the replacement of the B by an -OH in each alkyl group. The -OH group comes from the H2O2. - Typically, the organoborane product of hydroboration is not isolated, but is treated directly with the alkaline hydrogen peroxide to give alcohol. The addition of borane and subsequent rxn with H2O2, taken together, are referred to as hydroboration-oxidation We find that the net result is addition of the elements of water (H, OH) to the double bond in a regioselective manner so that the -OH ends up at the C of the double bond with the smaller number of alkyl substituents!! - So this rxn is an effective way to synthesize certain alcohols from alkenes! And because carbocations are not involved in either rxn, the alcohol products are not contaminated by constitutional isomers arising from rearrangements

what is D and L?

D-sugars and L-amino acids, but not the other kind!

how do the physical properties of enantiomers compare?

EVERYTHING identical: melting point, boiling point, densities, indices of refraction, heats of formation, standard free energies, and many others So how do we tell them apart? A compound and its enantiomer can be distinguished by their effects on polarized light

describe the geometry of alkanes and information about it's bond length relative to others

Each C has trigonal planar geometry. 120 deg b/w each surrounding atom - Ethylene is a planar molecule! - For alkenes in general, the C's and atoms directly attached lie in the same plane C=C shorter than C-C. - Shows relationship of bond length + bond order C-C single bond of propene is shorter than C-C single bond of propane. Likewise, bonds to H's attached to C=C are shorter than attached to C-C - Shortening of all these bonds is a consequence of the particular way that C atoms are hybridized in alkenes

give the general overview of enantiomeric resolution

Enantiomers have same properties, so we cannot exploit these for the separation of enantiomers as we might for other compounds. How are they separated then? - The resolution (separation) of enantiomers takes advantage of the fact that diastereomers, unlike enantiomers, have different physical properties. The strategy used is to convert a mixture of enantiomers temporarily into a mixture of diastereomers by allowing the mixture to combine with an enantiomerically pure chiral compound called a resolving agent. - The principle of enantiomeric differentiation: the separation or differentiation of of enantiomers requires that they interact with an enantiomerically pure chiral agent Example: hand must try on L and R hand gloves to see which one is which. Hand is pure chiral agent 3 ways we talk about - chiral chromatography - diastereomeric salt formation - selective crystallization

enzyme catalyst info

Enzymes are usually homogeneous catalysts. However, can be other too. Important example of enzyme-catalyzed addition to alkene: hydration of fumarate ion to malate ion with the use of fumarase enzyme!

asymmetric nitrogen: amine inversion

Ethylmethylamine has 4 diff groups around the N: H, ethyl group, methyl group, and e- pair. Bc geometry is tetrahedral, ethylmethylamine appears to be a chiral molecule - it should exist as 2 enantiomers. They cannot be separated, bc they rapidly interconvert by a process called amine inversion. In this process, the larger lobe of the e- pair seems to push through the nucleus to emerge on the other side. It is not simply turning over; it is actually turning itself inside out! This process occurs through a transition state in which the amine nitrogen becomes sp2-hybridized. Process occurs at room temp, so it's impossible to separate the enantiomers. Ethylmethylamine is a mixture of rapidly interconverting enantiomers. An example of racemization!

what is the postulation of the tetrahedral carbon?

First chemical substance in which optical activity was observed was quartz - When cut a certain way + exposed to polarized light along particular axis, the plane of polarization of the light is rotated. Exists as both levorotatory and dextrorotatory - Clearly associated the chirality of a substance w/ optical activity phenomenon Some substances have optical activity, some do not. Bc optical activity can be displayed by compounds in solution, it must be a property of the molecules themselves. - First observation of enantiomeric forms of same organic compound involved tartaric acid - Racemic acid was separated by human hands (enantiomeric resolution. First organic compound shown to exist as enantiomers - object and noncongruent mirror image. One was identical to (+)-tartaric acid, other was previously unknown. How can the existence of enantiomers be used to deduce a tetrahedral arrangement of groups around C? - Planar molecule is congruent, so enantiomers impossible. - In pyramidal geometry, 2 diastereomers would be known (negative evidence)

what makes the transition state of the alkene to a carbocation so unstable? what is hammond's postulate?

First: bonds undergoing transition are neither fully broken nor fully formed. The unstable bonding situation is why the transition state lies at an energy maximum. Additionally: a significant contribution to the high energy of the transition state comes from the same factors that account for the high energy of the carbocation. - Carbocation has one bond fewer than HBr and alkene. Bc bonding releases E, this fact alone means that the carbocation has a considerably higher energy than starting materials (or products). - Other factor is separation of pos and neg charge. The electrostatic law tells us that separation of opposite charges requires energy. Conclusion: the structure and energy of the transition state can be approximated by the structure and energy of the carbocation intermediate. Can be generalized by: - Hammond's Postulate: for a rxn in which an intermediate of relatively high E is either formed from reactants of much lower energy or converted into products of much lower energy, the structure and energy of the transition state can be approximated by the structure and energy of the intermediate itself Priest note: things of similar energy have similar geometry

how do you tell if a molecule has a double-bond stereoisomer?

Flipping a C. It should be a diff molecule from what it was originally. Should be the same for the other one. If you flip it and it's actually the same molecule, it's not a stereoisomer.

what are diastereomers

For a pair of chiral molecules with 1+ asymmetric C to be enantiomers, they must have opposite configurations at every asymmetric C (S,S and R,R or vice versa) - Stereoisomers that are not enantiomers are called diastereoisomers, or diastereomers. NOT mirror images. Diastereomers differ in all of their physical properties. - Diff melting points, boiling points, heats of formation, standard free energies - Can in principle be separated by conventional means: fractional distillation or crystallization - If they happen to be chiral, they can be expected to be optically active, but their specific rotations will have no relationship

describe the initiation of free-radical chain reactions

Free radicals are formed from a free-radical initiator: a molecule that undergoes homolysis with particular ease. The initiator is the source of free radicals. Peroxides are frequently used as free-radical initiators. First initiation step: homolysis of the peroxide - (Peroxides are not the only type of free-radical initiators. Another widely used initiator is azoisobutyronitrile (AIBN). Readily forms free radicals bc the very stable molecule dinitrogen is liberated as a result of homolytic cleavage) - Sometimes heat or light initiates a free-radical rxn. This usually happens bc the additional energy causes homolysis of the free-radical initiator - or in some cases, the reactants themselves- into free radicals. - The effects of initiators provide some of the best clues that a reaction occurs by a free-radical mechanism. If the rxn occurs in the presence of a known free-radical initiator but does not occur in its absence, we can be fairly certain that the rxn involves free-radical intermediates. Second step: the removal of a hydrogen atom from HBr by the tert-butoxy free radical that was formed in the first initiation step. - Example of a common type of free-radical process, called atom abstraction: a free radical removes an atom from another molecule, and a new free radical is formed (Br free radical).

which hydrogen halides undergo addition to give alkyl halides?

HF, HCl, HBr, HI undergo addition to C=C to give products called alkyl halides, compounds in which a halogen is bound to a saturated C atom HBr and HI preferred. HCl is slow. HF is HAZARDOUS!

what are vicinal halides and which halogens do halogenation

Halogens are added across a double bond. The product are called vicinal dihalides: vicinal means "on adjacent sites" - thus, are compounds with halogens on adj C Br and Cl are the 2 halogens used most frequently; F is so reactive that it not only adds to the double bond but also rapidly replaces all the H's with F's, often with considerable violence. Iodine adds to alkenes at low temp, but most diiodides are unstable and decompose to the corresponding alkenes and I2 at room temp. Bc Br is a liquid that's more easily handled than chlorine gas, many halogen additions are carried out with bromine

what is bond dissociation energy?

How easily does a chemical bond break homolytically to form free radicals? The question can be answered by examining its bond dissociation energy. The bond dissociation energy of a bond b/w 2 atoms X-Y is defined as the standard enthalpy delta H of the rxn. Note that this always corresponds to the enthalpy required to break a bond homolytically, not to the heterolytic process where they become ions rather than atoms. A bond dissociation energy measures the intrinsic strength of a chemical bond.

comparison of methods for the synthesis of alcohols from alkenes

Hydration: useful industrial method for preparation of a few alcohols, but not a good laboratory method. - An industrial method typically works well in the specific case for which it was designed, but it cannot necessarily be applied to other related cases. Bc chemical industry has gone to great effort to work out conditions that are optimal for preparation of particular compounds of commercial significance, using reagents that are readily available and inexpensive. Often require high temps, pressures, or elaborate reactors. Hydroboration-oxidation and oxymercuration-reduction are both general laboratory methods for the preparation of alcohols from alkenes: they can be applied successfully to a wide variety of alkene starting materials. - The choice hinges on the difference in their regioselectivities - For alkenes that yield the same alcohol by either method, the choice is arbitrary

notes about optical activity and enantiomers

If plane-polarized light is passed through 1 enantiomer of a chiral substance (either pure or in solution), the plane of polarization of the emergent light is rotated. A substance that rotates the plane of polarized light is said to be optically active. Individual enantiomers of chiral substances are optically active. Optical activity is measured by: polarimeter, basically the system of two polarizers. Sample is placed in the light beam b/w 2 polarizers. Because optical activity changes with the wavelength (color) of the light, monochromatic light is used to measure optical activity. Yellow light from sodium arc is often used. - An optically inactive sample (air or solvent) is placed in a beam. Light polarized by the first polarizer passes through the sample, and the analyzer is turned to establish a dark field. This setting of the analyzer defines the zero of optical rotation. - Next, sample whose optical activity is to be measured is placed in the light beam. The number of degrees (alpha) that the analyzer must be turned to reestablish the dark field is the optical rotation of the sample. - If the sample rotates the plane of the polarized light in the clockwise direction, the optical rotation is (+). Is said to be dextrorotatory. If it rotates in the counterclockwise direction, the optical rotation is (-) and said to be levorotatory. Enantiomers are distinguished by their optical activities bc enantiomers rotate the plane of polarized light by equal amounts in opposite directions. Lower-case prefix d or l are like dextrorotatory and levorotatory. +-2-butanol can be d-2-butanol. Very important point: there is no general correspondence b/w the sign of the optical rotation and the R or S configuration of a compound. The only way to determine optical rotation is to measure it experimentally

what is heterolytic vs homolytic, what is fishhook notation, and how does this relate to peroxides

In a heterolytic process, e-'s involved "move" in pairs. - Heterolysis means bond breaks and results in 2 ions In a homolytic process, e-'s involved "move" in an unpaired way - Homolysis means bond breaks and results in 2 atoms Peroxides, because of their very weak O-O bonds, are prone to undergo homolytic cleavage - Priest: why does O bond form radicals? Bc it's a symmetrical molecule so why would they favor going both one way! So if R and H on either side, won't form! - (EXCEPTION: hydrogen peroxide (H2O2) is not commonly used as a source of initiating free radicals. The reason is that the O-O bond is considerably stronger than in other peroxides and is therefore harder to break homolytically. The cleavage of organoboranes by hydrogen peroxide (hydroboration-oxidation) is not a free-radical reaction.) - Any species with at least 1 unpaired e- is called a free radical

in-depth explanation for why HI and HCl don't have the peroxide effect

In most cases, only exothermic propagation steps (steps with favorable or negative delta H values) occur rapidly enough to compete with the recombination rxns that terminate free-radical chain processes. Both of the propagation steps in the free-radical HBr addition to alkenes are exothermic, and they both occur readily. However, the first propagation step of free-radical HI addition, and the second propagation step of free-radical HCl addition, are quite endothermic. For this reason, these processes occur to such a small extent that they cannot compete with the recombination processes that terminate these chain rxns. Consequently, the free-radical addition of neither HCl nor HI to alkenes is observed.

describe what carbocation rearrangement is

In some cases the addition of a hydrogen halide to an alkene gives an unusual product. The minor product is the result of the ordinary regioselective addition of HCl across the double bond. The major product shows a rearrangement has occurred: a group from the starting material has moved to a different position in the product. - Rxn begins like a normal addition of HCl - protonating of the double bond to yield the carbocation with the greater number of alkyl substituents at the e-deficient C. - However, the carbocation can also undergo a second type of reaction: it can rearrange! The methyl group moves with its pair of bonding e-'s from the C adj to the e-deficient C. The C from which this group departs, as a result, becomes e-deficient and pos charged. That is, the rearrangement converts 1 carbocation into another. This is essentially a Lewis acid-base rxn in which the e-deficient C is the Lewis acid and the migrating group with its bonding e- pair is the Lewis base. The rxn forms a new Lewis acid - the e-deficient C of the rearranged carbocation. The major product of this reaction is formed by the Lewis acid-base association rxn of Cl- with the new carbocation. - WHY: a more stable tertiary carbocation is formed from a less stable secondary one. Rearrangement is favored by the increased stability of the rearranged ion. (Minor one still exists to some extent, just less!)

overview of halohydrins and why it would occur

In the addition of Br2, the only nucleophile available to react with the bromonium ion is the bromide ion. When other nucleophiles are present, they too can react with the bromonium ion to form products other than dibromides. - Common situation: when the solvent itself can act as a nucleophile - For example: when alkene treated with bromine in a solvent containing a large excess of water, a water molecule rather than bromide ion reacts with the bromonium ion, because water is present in much higher concentration than bromide ion - This ion is very acidic, so the solvent H2O can remove this acidic proton to give the product - a bromohydrin: a compound containing both an -OH and a -Br group - Member of a general class of compounds called halohydrins: compounds containing both a halogen and an -OH group. - In the most common type of halohydrin, the 2 groups occupy adjacent or vicinal positions Halohydrin formation involves the net addition to the double bond of the elements of a hypohalous acid, such as hypobromous HO-BR or hypochlorous acid HO-Cl. And although the products of I2 addition are unstable, iodohydrins can be prepared

describe the termination of free-radical chain reactions

In these steps, 2 radicals react to give nonradical products. Typically, termination involves a radical recombination rxn: 2 radicals come together to form a covalent bond. Radical recombination is the reverse of a homolysis. - 2 can happen in the free-radical addition of HBr to alkenes: Br free radicals make covalent bonds between themselves, and the ethyl free radicals can combine. They are present in very small amounts because they are formed ONLY from free radicals, which are also present in very small amounts. - Because each recombination reaction takes 2 free radicals "out of circulation," it terminates 2 propagation rxns and breaks 2 free-radical chains The recombination rxns of free radicals are in general highly exothermic; that is, they have very favorable, or negative, delta H values. They typically occur on every encounter of 2 free radicals: in other words, there is no delta G double dagger for radical recombination. In view of this fact, we might ask why free radicals do not simply recombine before they propagate any chains. The answer is simply a matter of the relative conc of the various species involved. - Free-radical intermediates are present in very low conc, but the other reactants are present in much higher conc. Consequently, it is much more probable for a Br atom to collide with an alkene molecule in the propagation reaction than with another Br atom in a recombination reaction. In a typical reaction, a termination rxn occurs once for every 10,000 propagation rxns. As the reactants are depleted, however, the probability becomes significantly greater that 1 free radical will survive long enough to wander into another free radical with which it can recombine.

describe the propagation of free-radical chain reactions

In these steps, radicals react with non-radical starting materials to give other radicals; starting materials are consumed and products are formed. Propagation steps occur repeatedly. When the propagation steps are considered together, there is no net formation or destruction of any of the radical species involved. This means that if a radical is formed, it must be consumed in a subsequent propagation step and another radical must be formed to take its place. First step: rxn of the Br atom with the pi bond. - Reaction of a free radical with a C-C pi bond is another common process. The pi bond reacts, rather than a sigma bond, because the C-C pi bonds are weaker than the C-C sigma bonds. Second step: another atom abstraction rxn. The removal of a H atom from HBr by the free-radical product of the last step to give the addition product and a new bromine atom. - This Br atom in turn can react with another molecule of alkene, and this can be followed by the generation of another molecule and etc. It's a chain reaction. These 2 propagation steps continue in a chainlike fashion until the reactants are consumed. Because no net destruction of free radicals occurs, the initial conc of free radicals provided by the initiator, and thus the conc of the initiator itself, can be small. Typically the initiator conc is only 1-2% of the alkene conc. - The free radicals involved in the propagation steps of a chain rxn are said to propagate the chain. - REMINDER: the initiation steps had to occur only once for the subsequent propagation steps to occur repeatedly.

what inert solvents are used in halogenation

Inert solvents such as methylene chloride (CH2Cl2) or carbon tetrachloride (CCl4) are typically used for halogen additions bc these solvents dissolve both halogens and alkenes. The addition of Br to most alkenes is so fast that when Br is added dropwise to a solution of the alkene, the red Br color disappears almost immediately. In fact, this discharge of color is a useful qualitative test for alkenes

what are the steps in the cahn-ingold-prelog system

Involves assignment of relative priorities to the 2 groups on each C of the double bond according to a set of sequence rules. We then compare the relative locations of these groups on each alkene C. If the groups of higher priority are on the same side: Z configuration. If are on opposite sides: E configuration. (E opposite Z zame) Sequence rules: assign relative priorities. Atoms in each group can be organized into levels. - Level 1: atoms directly attached to the bond. - Level 2: atoms attached to level-1 atoms. Sequence rules start with a comparison of atoms at level 1. If no decision is possible, proceed to level 2 and so on. Step 1: Examine level 1 - Rule 1a: Assign higher priority to group containing atom of higher atomic # - Rule 1b: Assign higher priority to group containing isotope of higher atomic mass (Deuterium over H) Step 2: If level 1 atoms are the same, working outwards, consider within each group the set of attached atoms. Apply each rule 2 to each level 2 set - Rule 2: arrange atoms within each set in descending priority order (O,C,H), and make a pairwise comparison of the atoms in the 2 sets. The higher priority is assigned to the atom of higher atomic # (or mass in case of isotopes) at the first point of difference. Step 3: If level 2 sets are identical, then within each set, choose the atom of highest priority (within that last set it would be O). Identify the level 3 set of atoms attached to it. Compare the level 3 sets in each group by applying rule 2. Step 3a: If no decision emerges, choose atoms of next highest priority in level 2 set (It would be C in this case) and repeat the process. Choose atoms of progressively lower priority in the level-2 sets until a difference is found. Steb 3b: If 2+ atoms in any set are the same, decide on their relative priorities for step 3 by continuing to explore outward from each, and choose as the atom of higher priority the one that gives the path of higher priority. Step 3c. If no decision is possible, move away from the double bond within each group to atoms at the next level and repeat step 3. Continue this exploration, level-by-level, until the first difference is found. Sometimes the groups to which we assign priorities themselves contain double bonds. Double bonds are treated by a special convention, in which the double bond is rewritten as a single bond and the atoms at each end of the double bond are duplicated. -CH=O is treated as CHO-OC Notice that the duplicated atoms bear only 1 bond: they have no other groups attached to them. The treatment of triple bonds requires triplicating the atoms involved. -CtriplebondN is treated as -CNN-NCC

what is the isomer flowchart

Isomers have same molecular formula Structural (Constitutional): same formula but different atom connectivity - Positional: differ in location of functional group - Functional: have different functional groups Stereochemical: differ in 3D orientation - Enantiomeric: non-superimposable mirror images. (R: clockwise, S: counterclockwise) - Diasteromeric: non-mirror image - Conformational: can be interconverted by conversions - Geometric: different structure?? (cis/trans)

why is oxymercuration-reduction better than hydration sometimes

It gives the same overall transformation as the hydration rxn. However, oxymercuration-reduction is much more convenient to run on a laboratory scale, and it is free of rearrangements and other side rxns that are encountered in hydration, because carbocation intermediates are not involved in oxymercuration. The absence of rearrangements is one reason that mercurinium ions, rather than carbocations, are thought to be the reactive intermediates in oxymercuration

what are oxidation numbers (priest) and then ALL THE RULES

Just take a polar covalent bond and make more electronegative -1 and less +1, and then you can sum it all up. So methane, 4H's are all +1 so the C is -4. Double bonds counted twice 1) any free element in elemental state (H2, O2) is 0 2) monoatomic ions (Na+, Ca+2) = charge on ion 3) H in most compounds is +1 - Exception: (LiH, NaH, is -1) 4) O in most compounds is -2 - Exception: peroxides, being -1 5) sum of all #'s on atoms in molecule must equal apparent charge on molecule. 6) C changes by +1 for each bond to a more electronegative heteroatom 7) C changes by -1 for each bond to a less electronegative heteroatom 8) Bonds between 2 C's are not counted

what is polarized light?

Light is a wave motion that consists of oscillating electric and magnetic fields. The electric field of ordinary light oscillates in all planes, but it is possible to obtain light with an electric field that oscillates in only one plane. - Called plane-polarized light, or simply polarized light. - Obtained by passing ordinary light through a polarizer, such as a Nicol prism (prism made of specially cut and joined calcium carbonate crystals). The orientation of the polarizer's axis of polarization determines the plane of the resulting polarized light. - Analysis of polarized light hinges on the fact that if plane-polarized light is subjected to a second polarization whose axis of polarization is perpendicular to that of the first, no light passes through the second polarizer.

some characteristics of alkenes

Like alkanes, alkenes are flammable, nonpolar compounds that are less dense than, and insoluble in, water. The alkenes of lower molecular weight are gases at room temperature.

priests fun steps for drawing mechanisms

Look at pH Write if acidic (+) or basic (-) - if you're under acidic conditions there can't be any - sign in the mechanism Best nu- goes after best electrophile+ NOTE: you can use general acid/base notation like (B-H/B-)

what is an asymmetric C and some notes about it

Many chiral molecules contain 1+ asymmetric C atoms Asymmetric C atom: C to which 4 different groups are bound - Also: chiral C or chiral center A molecule that contains ONLY 1 asymmetric C is chiral. No generalizations can be made for molecules with 1+ asymmetric C's - Moreover, some have no asymmetric C's at all Any atom can be asymmetric. Atoms called asymmetric centers - Another type of stereocenter, or stereogenic atom - Stereocenter: atom at which the interchange of 2 groups give a stereoisomer Not all C stereocenters are asymmetric C's. The term stereocenter is not associated solely with chiral molecules. All asymmetric atoms are stereocenters, but not all stereocenters are asymmetric atoms

what is the rate-determining step in multistep reactions

Many rxns take place w/ formation of reactive intermediates. Such are: multistep rxns - Bc when intermediates exist, what we commonly express as 1 rxn is rly seq of 2+ Each step has its own characteristic rate and therefore its own transition state - Generally the rate of a multistep rxn depends in detail on the rates of various steps. However, it often happens that 1 step of a multistep rxn is considerably slower than any of the others. The slowest step in a multistep rxn: rate-limiting step or rate-determining step - The rate of the overall rxn is equal to the rate of the rate-limiting step. The rate-limiting step is the step with the transition state of highest free energy. This step has special importance: anything that increases the rate of this step increases the overall rxn rate. Conversely, if a change in the rxn conditions (like change in temp) affects the rate of the rxn, it is the effect on the rate-limiting step that is being observed.

what happens in addition reactions for alkenes

Most characteristic type of alkene rxn: addition at the C=C. C-C pi bond of alkene and X-Y bond of reagent are broken, and new C-X and C-Y bonds are formed

general notes about free-radical chain reactions

Most free radicals are very reactive. They generally behave as reactive intermediates- react before they can accumulate in significant amounts. Most free-radical reactions can be classified as free-radical chain reactions: involves free-radical intermediates and consists of the following 3 fundamental rxn steps 1) Initiation steps 2) Propagation steps 3) Termination steps

what are enantiomers

NOT: Molecules that are congruent to their mirror images: all atoms and bonds in a molecule can be simultaneously superimposed onto identical atoms and bonds in its mirror image Those that are noncongruent to their mirror images are enantiomers - Molecules that can exist as enantiomers are said to be chiral: possess the property of chirality, or handedness. - Enantiomeric molecules have the same relationship as the right and left hands - Molecules that are not chiral are said to be achiral- without chirality

what is the math associated with optical rotation

Optical rotation of a sample is the quantitative measure of its optical activity. Observed optical rotation (alpha), in degrees, is proportional to the # of optically active molecules present in he path through which the light beam passes. - Alpha is proportional to both the concentration © of the optically active compounds in the sample as well as the length (l) of the sample container - Alpha = [alpha] * c * - The constant of proportionality, [alpha], is called the specific rotation. By convention, the concentration of the sample is expressed in grams per milliliter. Path length in decimeters (dm). C is taken as the density in a pure liquid. - 1 dm = 10 cm - Typically, the specific rotation is determined as the slope of a plot of observed rotation alpha against the concentration c. It's independent of c and l, so used as a standard measure of optical activity. Bc the specific rotation of any compounds varies with wavelength, solvent, and temperature, it is conventionally reported with a subscript that indicates the wavelength of light used and a superscript that indicates the temp.

describe what happens in the reduction part of oxymer-curation reduction

Oxymercuration is useful bc its products are easily converted into alcohols by treatment with the reducing agent sodium borohydride (NaBH4) in the presence of aqueous NaOH We won't consider the mechanism of this rxn. The key thing to notice is its outcome: the C-Hg bond is replaced by a C-H bond. The H- comes from the NaBH4! The oxymercuration adducts are usually not isolated, but are treated directly with a basic solution of NaBH4 in the same rxn vessel Priest: this is the reduction step, where H- reduces off (displaces) the mercury

what is percentage yield and how do you find it

Percentage yield: percentage of the theoretical amount of product formed that has actually been isolated from the rxn mixture by a chemist in the lab - Gives rough idea of how free the rxn is from contaminating by-products and how easily the product can be isolated from the rxn mixture 2A + B -> 3C + D should give 3 moles of C for every 1 of B and 2 of A 90% yield of C means that 2.7 moles of C per mole of B were actually isolated under these conditions. The 10% loss may have been due to separation difficulties, small amounts of by-products, or other reasons.

give some more information about pi bonds

Pi e-'s have higher E than the e-'s in hybrid orbitals. Thus, pi e-'s generally have higher E than sigma e-'s, just as p e-'s have higher E than s e-'s. - Consequence: pi e-'s are more easily removed than sigma e-'s. We'll find that electrophiles react preferentially with the pi e-'s in an alkene bc those e-'s are most easily donated. Most of the important rxns of alkenes involve the e-'s of the pi bond, and many of these rxns involve the rxn of electrophiles with the pi e-'s. Pi bond is weaker than typical C-C sigma bonds bc pi overlap which is "side-to-side" is inherently less effective than sigma overlap, which is "head-to-head." Single bonds to an sp2-hybridized C are somewhat shorter than single bonds to an sp3-hybridized C. Bc the e- density of an sp2 orbital is somewhat closer to the nucleus than that of an sp3 orbital, a bond involving an sp2 orbital, such as the one in propene, is shorter than one involving only sp3 orbitals, such as the one in propane. In other words, within bonds of a given order, bonds with more s character are shorter.

peroxide types

R on either side of O's, peroxide One H on one side of O's, hydroperoxide 2 H's, hydrogen peroxide

describe ring expansion/contraction mechanisms

Ring Expansion Mechanism - OH is a bad leaving group so we protonate it - H3O+ now can leave - but it can't leave behind a primary carbocation - So a nearby C of the ring breaks its bond and makes a new one with the C with the leaving group, and the C with which it broke its bond becomes the secondary carbocation Ring Contraction Mechanism - Nucleophile attacks partially positive C in ketone. O has neg charge now. - We want Cl to leave, but we can't have a pos charge bc its a basic rxn - So in a concerted mechanism, C-C bond breaks, O neg charge makes pi bond, and C makes a new bond with where the Cl leaves so there's no pos charge left behind.

how do you do the R,S Cahn-Ingold-Prelog thing

Same Cahn-Ingold-Prelog priority rules as before. A stereochemical configuration at each asymmetric C in a molecule can be assigned using the steps: 1. Identify an asymmetric C and the 4 diff groups bound to it 2. Assign priorities to the 4 diff groups according to the rules (highest atomic #) 3. Put lowest priority group in the back 4. Consider clockwise or counterclockwise order of the remaining group priorities. If they decrease in the clockwise direction, has R configuration. If decrease in counterclockwise direction, has S configuration. (So count priorities 1-2-3) TRICK: Swap any 2 substituents to change configuration (R to S). Switch lowest priority to be in the back this way, figure out configuration, and it'll be opposite of what you were supposed to determine. A stereoisomer is named by indicating the configuration of each asymmetric C before the systematic name of the compound (3S, 4S)-3,4-dimethylhexane

what is a catalyst

Some rxns take place much more rapidly in the presence of certain substances that are themselves left unchanged by the rxn. A substance that inc the rate of a rxn without being consumed is called: catalyst - A catalyst inc the rxn rate. This means that it lowers the standard free energy of activation for a rxn - A catalyst is not consumed. It may be consumed in 1 step of a catalyzed rxn, but if so, it is regenerated in a subsequent step - IMPLICATION: a catalyst that strongly accelerates a rxn can be used in very small amounts - A catalyst does not affect the energies of reactants and products. It does not effect the delta G of a rxn and consequently also does not affect the equilibrium constant. - A catalyst accelerates both the forward and reverse of a rxn by the same factor.

what is "stability" and stuff about it

Stability: has lower energy. But energy can take diff forms, and the energy we use to measure relative "stability" depends on the purpose we have in mind. - Delta G is the energy quantity related to the equilibrium constant. - But if we're interested in the total energy change for a rxn, we use the standard enthalpy change, delta H. Approximates very closely the total energy difference between reactants and products, and it reflects the relative stabilities of bonding arrangements in reactants and products. Delta G = delta H - T(delta S) - Where delta S is the entropy change for the reaction and T is the absolute temperature.

what is the reaction mechanism of hydration for alkenes

Step 1: (rate-limiting step) double bond is protonated so as to give the more stable carbocation. Because water is present, the actual acid is the hydrated proton (H3O). This is a Bronsted acid-base rxn. Bc this is the rate-limiting step, the rate of the hydration rxn increases when the rate of this step increases. The strong acid H3O+ is more effective than the considerably weaker acid water in protonating a weak base (the alkene). If a strong acid is not present, the rxn does not occur because water alone is too weak an acid to protonate the alkene. Step 2: the nucleophile water combines with the carbocation in a Lewis acid-base association rxn. Step 3: a proton is lost to solvent in another Bronsted acid-base rxn to give the alcohol product and regenerate the catalyzing acid H3O+ (or whatever acid is there) NOTES: - It consists entirely of Lewis acid-base association-dissociation and Bronsted acid-base rxns - Although the proton consumed is not the same as the one produced, there is no net consumption of protons - Note that the hydroxide ion is not acting as the nucleophile in Step 2 or the base in Step 3. Although hydroxide ion is more basic, it is not in the solution! And it isn't needed: the carbocation is very reactive- reactive enough to react rapidly with water, and the acid is strong enough to donate a proton to the weak base water! (R-OH2+) - WE CAN SAY: whenever H3O+ acts as an acid, its conjugate base H2O acts as a base. More generally, acids and their conjugate bases always act in tandem in acid-base catalysis. REMEMBER: bc hydration rxns involve carbocation intermediates, some alkenes give rearranged hydration products!!!

how can stereoisomers interconvert by internal rotations

The chirality of gauche-butane shows that some chiral molecules do not contain asymmetric centers. Shows that the existence of an asymmetric atom is not a requirement for chirality. - The 2 gauche conformations of butane are conformational enantiomers: enantiomers that are interconverted by a conformational change. In this case, it's an internal rotation. The anti conformation of butane, in contrast, is achiral and is a diastereomer of either one of the gauche conformations. - Anti-butane and either one of the gauche-butanes are therefore conformational diastereomers: diastereomers that are interconverted by a conformational change Strictly speaking, the terms chiral and achiral can only be applied to a single rigid object. Thus, each gauche conformation of butane is chiral, and the anti conformation is achiral. Butane is a mixture of conformations, however, and is therefore a mixture of "objects." - Chemists have broadened the use of the terms chiral and achiral to molecules that consist of many conformations by introducing the dimension of time into the definitions in the following way: A molecule is said to be achiral when it consists of rapidly equilibrating enantiomeric conformations that cannot be separated on any reasonable time scale. In butane, therefore, rapidly equilibrating conformations are the 2 gauche conformations. We cannot isolate each conformation on any reasonable time scale because the equilibration is too fast. When we think of butane this way, we are in effect considering it as one object w/ a time-average conformation that is achiral. - If we know that the conformational equilibrium is rapid, then the molecule is achiral if we can find ONE achiral conformation - even an unstable conformation such as an eclipsed conformation. This works because, once the molecule is in (or passes through) an achiral conformation, formation of either enantiomeric conformation is equally likely. Thus, recognizing that either the anti or eclipsed conformation of butane is achiral is sufficient for us to know that any chiral conformation of butane must be in rapid equilibrium with its enantiomer and that butane is achiral as a result Like meso compounds: contain at least 1 achiral conformation and rapidly interconverting enantiomeric conformations - but meso compounds differ with presence of asymmetric C's So a molecule is chiral only if it has NO achiral conformations, or equivalently, only if ALL of its conformations (even unstable eclipsed conformations) are chiral

le chatelier's principle and how it relates to hydration of alkenes

The equilibrium constants for many alkene hydrations are close enough to unity that the hydration rxn can be run in reverse. The reverse of alkene hydration is called alcohol dehydration. The direction in which the reaction is run depends on the application of Le Chatelier's principle: if an equilibrium is disturbed, it will react so as to offset the disturbance. - For example: if the alkene is a gas, the rxn vessel can be pressurized with the alkene. The equilibrium reacts to the excess of alkene by forming more alcohol. Neutralization of the acid catalyst stops the rxn and permits isolation of the alcohol. - To run hydration rxn backwards, the alkene is removed as it is formed, typically by distillation. The equilibrium responds by forming more alkene. Alkene hydration and alcohol dehydration occur by the forward and reverse of the same mechanism. Generally, if a rxn occurs by a certain mechanism, the reverse rxn under the same conditions occurs by the exact reverse of that mechanism. Called the principle of microscopic reversibility. - If you know the mechanism of alkene hydration, then you know the mechanism of alcohol dehydration as well. A consequence: the rate-limiting step of each is the reverse of the other.

what's the unsaturation number or DOU

The molecular formula of an organic compound contains "built-in" information about the number of rings and double (or triple) bonds. The presence of rings or double bonds within a molecule is indicated by a quantity called the unsaturation number, or degree of unsaturation, U. It is equal to the total number of its rings and multiple bonds. Also can be calculated like this: C - 1/2 H - ½ X + ½ N + 1

so what is the overall result of oxymercuration-reduction

The net addition of the elements of water (H and OH) to an alkene double bond in a highly regioselective manner: the -OH group is added to the more branched C of the double bond. Conversion of alkene to alcohol is conducted over 2 separate steps, put over the arrow. Writing consecutive rxns in this manner saves time and space. But number the rxns!!! Adding all at once would not give the desired product

why is allene (C=C=C) a chiral molecule

The pi bond on the y axis for the right and middle C's exists, so the pi bond for the left and middle C's needs to be on the Z axis. That means there's a 90 degree rotation and so the left C has substituents in the plane of the page but the right C has them coming in and out of the plane!

WHY are trans alkenes more stable!!!!

Trans has a lower enthalpy of formation. Almost all trans alkenes are more stable than their cis isomers. Reason: in a cis-alkene, the larger groups are forced to occupy the same plane on the same side of the double bond. Experience van der Waals repulsion! No such repulsions occur in a trans isomer. Heats of formation not only suggest the presence of van der Waals repulsions, but also give us quantitative info on the magnitude of such repulsions. Priest: trans alkenes have better IMF than cis alkenes. Can stack! Cis has kinks

what is diastereomeric salt formation?

Used for the enantiomeric resolution of acidic or basic compounds. Particularly well suited for large-scale separations. Example: enantiomeric resolution of the racemate of alpha-phenetylamine - Amines: derivatives of ammonia in which 1+ H's have been replaced by organic groups - Diastereomeric salt formation takes advantage of the fact that amines, like ammonia, are bases, so they react rapidly and quantitatively w/ carboxylic acids to form salts - The resolving agent is an enantiomerically pure carboxylic acid. Gives a mixture of 2 diastereomeric salts - They are diastereomers bc they differ in configuration at only 1 of their 3 asymmetric C's. (Enantiomers must differ at EVERY asymmetric C). Have diff physical properties! - In this case, have significantly different solubilities in methanol, a commonly used alcohol solvent. One diastereomer is less soluble and crystallizes selectively, leaving the other that may be recovered. Once either pure diastereomer is in hand, the salt can be decomposed with base to liberate the water-insoluble, optically active amine, leaving the tartaric acid in solution A simple and convenient rxn, often used for enantiomeric resolution of amines and carboxylic acids PRIEST: derivatize, separate, un-derivatize. While in the previous one you use electrostatic forces, in this one you are using covalent bonds. Have enantiomers with alcohol, puts a chiral group on the alcohol (replace the proton) to make diastereomers that are easy to un-derivatize!

what is selective crystallization

Used to resolve large quantities of chiral compounds that form crystalline solids. Crystallization is often a slow process, and can be accelerated sometimes by adding a seed crystal of the compound to be crystalized. In selective crystallization, a solution of a mixture of enantiomers is cooled to supersaturation and a seed crystal of the desired enantiomer is added. In this case, the seed crystal serves as the resolving agent and promotes crystallization of the desired enantiomer. - The seed crystal only contains molecules of the pure enantiomer of the desired compound. The seed crystal can grow in 2 ways: can incorporate more molecules of same enantiomer, or some of opposite enantiomer. These 2 possibilities generate 2 diastereomeric crystals. - Diff properties -> diff solubilities. It is common that the "pure" crystal has the higher melting point, and thus lower solubility. (Higher melting point = less soluble). - Requires a small amount of the desired pure enantiomer to start with, but this can be obtained, for example, by chiral chromatography. So it provides a mechanisms for "amplification" of an enantiomeric resolution to a larger scale.

what is enantiomeric excess

When 1 enantiomer of a chiral compound is uncontaminated by the other enantiomer, it is said to be enantiomerically pure. However, mixtures of enantiomers occur commonly. The enantiomeric composition of a mixture of enantiomers if expressed as the enantiomeric excess (EE): defined as the difference b/w the percentages of the 2 enantiomers in the mixture: - EE = % of the major enantiomer - % of the minor enantiomer - 80 - 20 = 60. 20 of the (-) enantiomer cancels the rotation of 20 of the (+) enantiomer, leaving a net rotation of 60. Thus, if we know the optical activity of the pure major enantiomer, and if the 2 enantiomers are the only optically active substances present, then EE can be determined from the specific rotation of the mixture EE = 100% x (([alpha]mixture)/([alpha]pure)) - Where [alpha]mixture and [alpha]pure are the specific rotations of the mixture and the pure major enantiomer determined under the same conditions When EE is calculated from optical activities, it is sometimes also called optical purity Finally, the actual percentages of each enantiomer can be calculated from the enantiomeric excess. % minor enantiomer = 100% - % major enantiom

heterogeneous vs homogeneous catalyst

When a catalyst and the reactants exist in separate phases: heterogeneous catalyst - Like a catalytic converter of an automobile is a solid while reactants are gases Or a rxn in solution may be catalyzed by a soluble catalyst: homogeneous catalyst

what is a steric effect?

When a chemical phenomenon (such as a reaction) is affected by van der Waals repulsions, it is said to be influenced by a steric effect. Thus, the regioselectivity of free-radical HBr addition to alkenes is due in part to a steric effect. Other examples of steric effects are the preference of butane for the anti rather than the gauche conformation, and the greater stability of trans-2-butene relative to cis-2-butene.

what are the possible results of ozonides and how do we get there

When ozonide is treated with dimethyl sulfide (CH3)2S, ozonide is split - The net transformation is the replacement of a C=C with 2 C=O groups - If the 2 ends of the double bond are identical, then 2 equivalents of the same product are formed. If the 2 ends of the alkene are different, then a mixture of 2 diff products is obtained - If a C of the double bond in the starting alkene bears a H, then an aldehyde is formed. In contrast, if it bears no H's, then a ketone is formed instead If the ozonide is simply treated with water, hydrogen peroxide (H2O2) is formed as a byproduct. Under these conditions (or if hydrogen peroxide is added specifically), aldehydes are converted into carboxylic acid, but ketones are unaffected.

why does the presence of peroxides reverse the normal regioselectivity of HBr addition?

When the Br atom adds to the pi bond of an alkene, 2 rxns are in competition: the Br atom can react at either of the 2 C's of the double bond to give diff free-radical intermediates What's the difference? Free radicals, like carbocations, can be classified as primary, secondary, or tertiary. And the formation of the tertiary free radical is faster! 2 reasons: 1) when the rather large Br atom reacts at the more-branched C, it experiences van der Waals repulsions with the H's in the branches. These repulsions increase the energy of this transition state. When it reacts with the less-branched C, these repulsions are absent. Bc the reaction with the transition state of lower energy is the faster rxn, rxn of the Br atom at the alkene C with fewer alkyl substituents to give the more alkyl-substituted free radical is faster 2) Has to do with tertiary radical's relative stability. Stability of free radicals: tertiary > secondary > primary. Same as carbocations! However, the energy differences between isomeric free radicals are only about ⅕ the magnitude of the differences b/w the corresponding carbocations. - This can be understood from the geometry and hybridization of a typical C radical. The methyl radical .CH3 is trigonal planar, but other carbon radicals are slightly pyramidal. However, they are close enough to planarity that we can, to a useful approximation, consider them also to be sp2-hybridized with the unpaired e- in a 2p orbital. The stability order implies that free radicals are stabilized by alkyl-group substitution at sp2-hybridized C's. The magnitude of the alkyl-group stabilization of free radicals is very similar to that observed for alkyl substitution at the sp2-hybridized C's of alkenes. - By Hammond's postulate, a more stable free radical should be formed more rapidly than a less stable one. The more stable free radical is formed. - Priest: radicals act like carbocations - they are stabilized through induction hyperconjugation and resonance

inversion of other atoms (into enantiomers)

When the central atom comes from the 2nd period of the periodic table, inversion is very rapid, as it is with amines. (Carbanion, oxonium ion, the like. All inversions are very fast at room temp). Therefore, if one of these atoms is the only asymmetric center in a compound, the compound cannot be resolved into enantiomers and cannot maintain optical activity - However, when the central atom comes from the 3+ periods, inversion is very slow- so slow it doesn't occur at room temp. This means that when the phosphorus atom of a phosphine, or sulfur or whatever, is an asymmetric center, such a compound can be resolved into enantiomers - The reason lies in the hybridization of the central atom. The unshared e- pair on the N of ammonia (and amines) occupies an approx sp3-hybridized orbital. Has 75% 2p character. In the transition state for inversion, the central atom is sp2-hybridized, and the unshared pair occupies a 2p orbital. A relatively small amount of energy is required to add another 25% p character to the unshared pair; so the inversion energy barrier is small, and inversion is fast - If the central atom is from the 3+ greater period, the unshared e- pair occupies an orbital with a high degree of s character. It takes significant energy to convert an e- pair in a 3s orbital to one in a 3p orbital. Therefore, the inversion barrier for these atoms is larger and inversion is slow - Why do the unshared pairs in 3+ period atoms have more s character than in 2 period? The unshared e- pair in a 3rd period atom is held less tightly than it is in a 2nd period atom, and it takes up a lot of space. According to VSEPR theory, repulsion of this unshared pair with the e-'s in neighboring bonds causes the neighboring bonds to compress more than they do in 2nd-period atoms. In fact, the R-S-R bond angles and R-P-R are around 100, whereas the R-N-R angles in amines are typically 110. Remember that bond angle and hybridization are intimately related. Bond angles closer to 90 require that the bonds have a great deal of p character. If the bonds to S and P contain most of the 3p character, the unshared e- pair on the S or P has little 3p character and a lot of 3s character. (this stuff is dumb)

describe the mechanism of halohydrin formation, starting from the bromonium ion

When the double bond of the alkene is positioned unsymmetrically, the rxn of water with the bromonium ion can give 2 possible products, each resulting from the breakage of a diff C-Br bond. The rxn is ONLYYYY highly regioselective when 1 of the C's of the alkene contains 2 alkyl substituents - OH goes to the more substituted carbon - Why? Before water joins, on the tertiary C, the bond b/w C and Br is so long and weak that this species is essentially a carbocation containing a weak C-Br interaction - Priest's explanation: one of the Br bonds has to break, and we'd rather break the C with more substitution. Is basically a carbocation in the transition state. But we DON'T completely break the bond!!!! There is NOT a carbocation!!! Because if there were then the OH would have a 50/50 shot of going in from either side of the trigonal planar C, but it can't bc the Br bond is there! Therefore it is a trans addition

what is the peroxide effect?

When traces of peroxides (compounds of the general structure R-O-O-R) are present in the rxn mixture during HBr addition, the regioselectivity of HBr addition is reversed! This is called the peroxide effect! Causes anti-Markovnikov regioselectivity It was also found that light further promotes the peroxide effect. (UV prolly... put hv over arrow to show light is being shined on it) NOTES: - the peroxide-promoted rxn is faster than HBr addition in the absence of peroxides. Very small amounts of peroxides are required to bring about this effect. - the regioselectivity of HI or HCl addition to alkenes is NOT affected by the presence of peroxides. For these, normal regioselectivity of addition predominates, whether peroxides are present or not. The addition of HBr to alkenes in the presence of peroxides occurs by a completely different mechanism from that for normal addition. Involves reactive intermediates known as free radicals.

why are reaction rates important

Whenever a rxn can give 1+ possible products, 2 or more rxns are in competition. One rxn predominates when it occurs more rapidly than other competing reactions. Understanding why is often a matter of understanding the rates of chemical rxns.

what is anomalous disperson?

You can't apply the R and S system to a molecule until you know the actual 3D arrangement of its atoms - that is, its absolute configuration or absolute stereochemistry. One way to do this is to use a variation of X-ray crystallography called anomalous dispersion. But very expensive. Usually determined by using chemical rxns to correlate them with other compounds of known absolute configurations: stereochemical correlation. Hypothetical example. You know an alkene is (-) or levorotatory. How to find absolute configuration? - Convert alkene into carboxylic acid by ozonolysis! This reaction breaks the double bond but it does not break any of the bonds to the asymmetric C! Converts into (+)-hydratropic acid. - The corresponding groups in the 2 compounds must be in the same relative positions! No bonds to the asymmetric C were broken! - It also follows that one must be R and one must be S because if they have opposite sign opposite rotations they must be both either R or S! (But this is not true in general. It's possible for R,S configurations of reactant and product to be the same.) To summarize: we can determine the absolute configuration of one compound by converting it into another compound whose absolute configuration is known. We could also take the same approach "in reverse" and deduce the configuration of a product from a starting material of known configuration. - Unambiguous when bonds to the asymmetric atom(s) are unaffected by the rxn

enantiomeric drugs

Zopiclone: sold worldwide as racemic mixture - Eszopiclone - S configuration: Lunesta. Dose is exactly ½ that of Zopiclone - Sleep aid Thalidomide, enantiomers responsible for different activities - R-enantiomer: anti-inflammatory activity - S-enantiomer: teratogenic activity (damages DNA, short arms/spontaneous abortion Ibuprofen - S-enantiomer: active one - R-enantiomer: converts into S and then is used! - How? Benzylic proton is epimerizable. C becomes sp2-hybridized, and there's a 50/50 shot of putting H on either face

what is required for the oxymercuration of alkenes

alkenes react with mercuric acetate Hg(OAc)2 in aqueous solution to give addition products in which an -HgOAc (acetoxymercuri) group and an -OH (hydroxy) group derived from water have added to the double bond - -HgOAc group goes to C less substituted, -OH groups goes to more substituted one. The solvent is a mixture of water and THF (tetrahydrofuran), a widely used ether THF is an important solvent bc it dissolves both water and many water-insoluble organic compounds. Its role in oxymercuration is to dissolve both and alkene and the aqueous mercuric acetate solution. (Alkenes are not soluble in water alone!) Water is required as both a reactant, and as a solvent for the mercuric acetate.

what is cycloaddition?

an addition reaction that forms a ring

what is an atropisomer

biphenyl (2 benzenes connected by a sigma bond) can rotate freely around the sigma bond. But once you put substituents near the sigma bond, it becomes locked! Basically only situation where sigma bonds becomes locked - one biphenyl is horizontal and other is vertical. They can't rotate anymore bc the substituents repel. Can't eclipse! Prefers staggered

what is chiral chromatography

chromatographic separation of enantiomers. Stationary phase typically consists of microscopic porous glass beads to which an enantiomerically pure chiral compound has been covalently attached. Serves as a resolving agent. Combo of glass beads and resolving agent: chiral stationary phase, (CSP). Each bead contains many copies of resolving agent. If a mixture of enantiomers is passed through CSP column, each of the 2 enantiomers forms a noncovalent complex with the immobilized resolving agent. Bc the resolving agent is enantiomerically pure, the two complexes differ in configuration at only 1 of their asymmetric C's. In other words, the 2 complexes are diastereomers. In general, they have diff free energies and diff stabilities. For this reason, the equilibrium constants for their formation differ. This means that 1 of 2 enantiomers binds more tightly to the resolving agent than the other, and as a result, the concentrations of the 2 complexes are different How do we know what CSP to use? - We don't know unless someone has done it before. Sometimes matters of trial and error, but experience has led to some principles by which a resolving agent can usually be chosen rationally Important point: enantiomers are separated by their interaction with the CSP by the temporary formation of diastereomers. (Gloves separated, CSP is the hand)

What are stereoisomers?

compounds that have the same atomic connectivity but a different arrangement of atoms in space - Recall the E and Z isomers of an alkene - are stereoisomers

what are fischer projections and what's the tricky thing about it

flat drawing represents a 3D molecule. A chiral carbon is at the intersection of horizontal and vertical lines. Horizontal lines are forward, out-of-plane. Vertical lines are behind the plane. - Easy to draw, easy to find enantiomers, easy to find internal mirror planes TRICKY THING: both the C's in the vertical are going away, but when you're looking at it in a bondline drawing they switch off zig zag and that switches the horizontal groups

what is column chromatography

mixture of compounds is introduced onto a cylindrical column containing a finely powdered solid: stationary phase. Components adsorb (bind) reversibly to the stationary phase. Results from non covalent attractions b/w molecules of the stationary phase and the molecules in solution to be separated. Column is then eluted (washed) continuously with a solvent. If the components of the mixture adhere to the stationary phase with sufficiently different affinities, the component with the smallest affinity for the solid emerges first from the column, and the components with progressively greater affinities for the solid emerge later Graph of concentrations of compounds in mixture against time (or volume of the solvent) is called a chromatogram

priest's carbocation stability chart

primary carbocation secondary and primary that's resonance stablilized tertiary and primary that's doubly resonance stabilized CONCLUSION: resonance stabilization adds a degree of stability

what are the conventions for writing organic reactions

solvent: written under arrow catalysts: written over arrow abbreviate rxns by showing only the organic starting materials and the major organic products. The other reactants and conditions are written over the arrow - by-products are not given and the equation is not balanced

common thread of all addition reactions

the first step of each rxn is the donation of an e- pair from the alkene pi bond, acting as a nucleophile, to an electrophilic center - Electrophilic atom becomes bonded to the alkene C w/ fewer alkyl substituents - A nucleophilic atom that was either part of the original reactant or present in solution completes the addition by donating e-'s to the alkene C with the greater number of alkyl substituents. WHY is it that the less electronegative atom is the electrophile? - The electrophilic atom (H in HBr for example) both accepts and gives up electrons in an electron-pair displacement reaction, Br is the leaving group - Leaving group only accepts electrons. Therefore, electronegativity is more important for the leaving group. So these are called electrophilic additions: begins with the donation of an electron pair from a pi bond to an electrophilic atom

unsaturated vs saturated hydrocarbons

unsaturated: have double/triple bonds, so fewer H's saturated: alkanes :)