http://orgo.curvedarrow.com/punbb Curved Arrow Press Forum en http://backend.userland.com/rss http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=521#521 521@http://orgo.curvedarrow.com/punbb Topic: 1-Bromo-1-phenylethane & sodium ethoxide elim rxn to give styrene Message: Yahoo!Answers in chemistry wrote:Reaction between 1-bromo-1-phenylethane and sodium ethoxide in ethanol? This is an inherently ambiguous question. The secondary halide can be involved in several types of reactions, SN1 or SN2-substitution or E1 or E2-elimination reactions. A rule of thumb is that if the pKa of the base (of the conjugate acid) is greater than 12, an elimination reaction will occur. Sodium ethoxide favors elimination. The kinetics for an E1 reaction are proportional to [RX] and for an E2 reaction are proportional to [RX][NaOEt]. The more dilute this reaction is, the more an E1 reaction is favored. High concentrations favor E2. The concentrations are not mentioned, nor is it easy to anticipate which concentrations would favor an E1 over an E2. Because the products of E1 or SN1 reactions can be difficult to predict, I shall predict this is intended to be an E2-elimination reaction to give styrene (1-phenylethene). The mechanism for this reaction is shown below (although with a different halide).        This example is excerpted from A Handbook of Organic Chemistry Mechanisms, available at http://www.curvedarrowpress.com/ or Amazon.com Sun, 29 Aug 2010 22:17:59 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=520#520 520@http://orgo.curvedarrow.com/punbb Topic: Cyclopentadiene synthesis scheme Message: Here is a problem that shows we can't all be prefect: An  organic compound (A) of m.f. C5H6 reacted with ethyl magnesium iodide to give (B).(B) reacted with CO2 to give a salt which on hydrolysis gives (C).(C) when treated with DIBALH gives  (D).(B) reacted with HCHO and on further acidification  gives (E).(E) when oxidised with PCC gives (D).A basic solution  of (A) shows aromatic characters. Find (A) to (E) This is the chemistry that is being suggested. Note that I put the product hoped for between C and D. DiBAL can be used to reduce an ester to an aldehyde, but not an acid. All that would happen is hydrogen would be evolved and a borate ester formed.         There is a second problem with this problem that I didn't address, namely conjugation. Since the starting material is cyclopentadiene and while the anion can be reacted, it is quite difficult to maintain the diene substitution pattern. Again, because cyclopentadiene is quite acidic, adding a carbonyl to it will increase the acidity. It would be very easy and preferred for the double bonds to be in conjugation with the carbonyl group. However, since there is the prior error and it doesn't really seem to address the conjugation-deconjugation, I left it as is. I would very much doubt those would be the products (except possibly for an extremely skilled chemist).[cyclopenta-2,4-dienecarboxylic acid, cyclopenta-2,4-dienecarbaldehyde, cyclopenta-2,4-dien-1-ylmethanol] Tue, 24 Aug 2010 16:38:44 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=519#519 519@http://orgo.curvedarrow.com/punbb Topic: Sigmatropic rearrangement of allyl phenyl ether to allyl phenol Message: This is electrocyclic reaction is a Claisen rearrangement. The scheme shown below shows the bonds that are made and broken in the rearrangement step. Because that intermediate is the less stable form of a ketone (cyclohexadienone) than phenol, a tautomerization reaction is required to complete the reaction. Because the reaction medium will contain phenol, an intermolecular proton transfer is a likely mechanism. For the sake of convenience, I illustrated this reaction by using neutral water as a base followed by protonation of the anion with the hydronium ion.           This example is excerpted from A Handbook of Organic Chemistry Mechanisms, available at http://www.curvedarrowpress.com/ or Amazon.comA better analysis of these reactions is described by orbital symmetry rules. See the answer and links to a question about molecular orbitals here:http://orgo.curvedarrow.com/punbb/viewt … pid=97#p97 Wed, 18 Aug 2010 17:53:50 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=518#518 518@http://orgo.curvedarrow.com/punbb Topic: Cyclohexene with bromine and water Message: Re: unsymmetrical alkenesThe opening of epoxides and bromonium ions can be thought of as paradoxical to the rules for SN1/SN2 reactions. Methylcyclohexene is often used as an example as the stereochemical effects can also be learned. The reaction of methylcyclohexene with Br2/H2O will give the intermediate bromonium ion. This will be opened with water. The opening of the bromonium ring occurs on the tertiary carbon, but this is not an SN1 opening. If it were a classic SN1 opening, then water could attack from either side of the carbocation. If it were a classic SN2 opening, then water should attack at the least hindered side of the bromonium ion. The actual result is an anti opening of the bromonium ion at the tertiary carbon. The result is an SN2-like opening of the bromonium ion at the tertiary carbon with inversion. The net addition is anti to give the trans product.     This example is excerpted from A Handbook of Organic Chemistry Mechanisms, available at http://www.curvedarrowpress.com/ or Amazon.com Wed, 18 Aug 2010 13:50:55 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=517#517 517@http://orgo.curvedarrow.com/punbb Topic: Homolytic, heterolytic bond strengths and lattice energy Message: I saw the following question posted on Yahoo!Answers Yahoo!Answers in chemistry wrote:Bond Strength, Energies...and a bonus question about surface interactions?First off, if you decide to explain these things for me try to give a practical, qualitative explanation along with any mathematical descriptions/crazy jargon you use. Also, please don't answer these questions with vague explanations like "oh, well ionic bonds don't share electrons like covalent bonds do" because this doesn't provide an in depth explanation of how that effects things like bond dissociation energies1) Is it accurate to say that "bond energy" is essentially the average of the "bond dissociation energy" within a molecule. Also, I want to confirm that bond dissociation energy is always referring to the homolytic cleavage of a bond while lattice is referring to heterolytic?2) When people say ionic bonds are weaker than covalent, are they really saying the "Bond dissociation energy" for a covalent compound is generally higher than the "lattice energy" for an ionic compound. IE - (read this carefully) wouldn't homolytic cleavage of ionic compounds be more energy intensive than covalent ones? Analogously, wouldn't the lattice energy for a covalent compound be higher than an ionic one? I'm confused about the interplay between homo and heterolytic cleavage and how that impacts bond strengths in ionic and covalent compounds. 3) This question arises from my confusion in #2.It is often said that the strength of C-F bond is due to a powerful dipole moment between the two atoms that creates a somewhat ionic character. This is also in line with the idea that the flouride ion makes a poor leaving group in organic mechanisms due to its high electron density and small bond length. But how does the ionic character strengthen if ionic bonds are supposedly weaker than covalent?? Wouldn't the fact that flourine likes to hold the negative charge make a C-F weaker not stronger???? As shown below, can you explain why C-F is stronger than C-C? I feel the strong electronegativity of flourine counteracts it's high electron density...the two characteristics seem to go against each other in providing bond strength. And if indeed it is the small size of the flourine atom, why is the C-H bond still weaker than C-F if H is clearly smaller than F.C-H 98 kj/molC-F 117 kj/molC-C 85 kj/molH-H 104 jk/molF-F 158 kj/molIt is shown that C-F has a higher bond dissociation energy, which to me says that homolytic cleavage is costly. But in organic mechanisms, Flourine ALSO makes a poor leaving group which would be a heterolytic cleavage. Does that mean that the C-F bond has high "lattice" energy too?This also begs the question, people say the iodine makes a good leaving group due to it's large size and lack of density. The bond dissociation energy is, appropriately, very low compared to C-F which is charge dense. BUT if BDE refers to homolytic cleavage, how does it apply to leaving groups in organic mechanisms when leaving groups break by heterolytic cleavage. This is a $64,000 question (old tv game show). I have been trying to answer this question also (and many chemists have also tried to provide an answer). I don't think you will find an good answer or if you do, please contact me. Homolytic bond cleavage yields radicals, R-R -> R•  +  •RHeterolytic bond cleavage yields ions, R-X -> R(+) + X(-)Bond dissociation energies measure homolytic bond strengths. Lattice energy is used to calculate the energy to separate a mole of a solid into a gas of its ions. So, the poster is paying attention to what is being said, therefore he arrives at his questions. How can this be? Herein lies the problem, "How can you measure what you think you want to measure?" Is heating sodium chloride the best way to determine how much energy to takes to separate NaCl into a sodium and chloride ions?  Heating may measure lattice energy, but it doesn't measure bond energy. How do they differ? If you make a bridge out of toothpicks, the bridge can be much stronger than an individual toothpick. The lattice energy of ice is relatively low and the mp is low. If you used organic compounds, the lattice energies would be lower still. However, in each of these cases, breaking the lattice would not require breaking any covalent bonds. These lattices are made up of covalent and other weaker bonds, hydrogen bonds or London forces. When an object is broken, what you measure is the weakest force. That is the analogy with a chain. You will find the weakest link. In a lattice of organic compounds, the weakest link are weak forces, like hydrogen bond, London forces, dipole-dipole interactions, etc. Furthermore, the organic lattice, once broken, becomes a solvent to further dissolve the components making up the lattice. That is, what happens upon melting. The solution becomes a good medium to further dissolve the solid. On the other hand, compare this with melting of sodium chloride crystals. Arguably, the "ionic bonds" making up the crystal are stronger bonds than those making up organic solids. Furthermore, sodium chloride probably is not a good solvent for dissolving sodium chloride crystals. If you added a small amount of water, you would find water is much more effective in breaking the lattice. If one wished to use use lattices to compare the strength of ionic and covalent bonds, then one should compare NaCl and diamond. The bonds in NaCl are ionic and the bonds in diamond are covalent. In both cases, the lattice is made up the the same types of bonds. This is different from other organic compounds in which a relatively small number of weak bonds may be contributing to the crystal lattice, such as cholesterol, benzoic acid, acetanilide, etc.Let's try to make simple comparisons. If we wished to compare lattice energies, then we should compare NaCl to diamond. In melting NaCl, we are breaking the lattice. If we attempt to do the same with diamond, we must also break the lattice. Because four covalent bonds are in diamond, it is very difficult to break and diamond is one of the hardest compounds known and much harder than salt crystals. If we wished to break NaCl into its ions, the easiest way to do so would be with water. That is a very easy reaction. If we attempted the same reaction with chloromethane, nothing would happen because the bonds are too strong and the ions too unstable. If we used t-butyl chloride, then a slow ionization would take place. This should indicate that ionic bonds are weak and covalent bonds are strong (to use the chemistry vernacular). It should also indicate that homolytic cleavage is more difficult than heterolytic cleavage (into ions). However, because this ionization involves participation of solvent, no clear and predictable method of calculating bond strengths has arisen from solubility or other methods to detemine ionic or heterolytic bond strengths. The bond dissociation energies measure the homolytic cleavage of the bonds and not the heterolytic strength. The values are important from a conservation of energy point of view, but not necessarily for estimating heterolytic bond strengths. Again, let us return to NaCl. If you calculate the bond energies, it is an exothermic reaction. Sodium donates electrons to chlorine. Think about it. If the affinity of sodium for its valence electrons is so weak, then it readily gives them up and chlorine has such a high affinity for an additional electron, that it readily accepts one, then how great should the attraction of a sodium cation be for the electrons of a chloride anion? Note that I used the term "chemistry vernacular" in referring to ionic and covalent bonds. I posit those are not good terms, but simply describe weak versus strong bonds. Ionic bonds are generally weak and covalent bonds are strong. If we take that definition, then some bonds that break heterolytically could be weak or strong. Thus, silver chloride may form strong bonds that break into ions and trityl chloride may form weak bonds that also break into ions. We may begin to recognize that atoms that have a higher affinity for electrons may hold them more strongly than those with a weak affinity. The metals as a class have a low affinity for electrons and as we increase the number of electron around atoms, we see a trend toward metallic character. The other bond strength we should consider is homolytic bond strengths. Generally, when we break bonds using the least amount of energy possible, we find heterolytic bond cleavage. However, we do find some molecules that can also break bonds homolytically at relatively low energy levels. The halogens and peroxides are examples. In characterizing the reaction path of these low energy radicals, they generally engage in reactions to form stable two electron bonds. That is, unpaired electrons are generally less stable than paired electrons.Fluorine or how is fluorine different? Fluorine is more like oxygen than it is like chlorine. Although fluorine holds its electrons tightly, hence strong bonds, it doesn't hold them as tightly as chlorine, hence less likely to form ions. Fluoride is more basic, HF is a weaker acid than chloride. Fluorine is less dense than chlorine. The electron density of fluorine is the least of the halogens. The volume of atoms is proportional to the volume of its electron clouds. Iodine is more dense than bromine>chlorine>fluorine. Iodine holds its electrons most tightly. It has the largest number of protons to increase the Coulombic attraction. Since iodine holds its electrons the most tightly, the attraction to a proton is weak and HI is a strong acid. How can iodine be a good nucleophile and a good leaving group? In measuring the acidity of HI, this is a thermodynamic property. It is affected by the inherent ability of iodine to withdraw its electrons. Nucleophilicity is a kinetic property. It is a function of which reaction can be faster. Two explanations have been suggested for the nucleophilicity of iodine. The first is a hydrogen bonding effect. The greater the hydrogen bonding, the less available will its electrons be for reaction. Since the electrons of iodine are held more tightly than the other halogens, it would be less subject to the retardation of hydrogen bonding. I think this is correct, but not sufficient for all of the properties of iodine. The other is that because the outer electrons of iodine are subjected to a Coulombic attraction to the nucleus and a Coulombic repulsion to the inner electrons. Because we don't know the properties of electrons completely, we may think that the electrons may extend further beyond its nucleus but with only a weak force. This weak force may kinetically be sufficient to form a bond. Alternately, we may think that local ionizations take place and that one of these electronic peaks forms a bond. Even though iodine can kinetically form a bond at a higher rate than other halogens, the bond strengths may not be large. That is, if iodine pulls its electrons in more than the other halogens, then the thermodynamic bond strength may be weaker. When we measure the kinetic rate of an SN1 reaction, we are determining the slowest or rate limiting step. If we compare iodine to the other halogens, the rate limiting bond cleavage of iodine is much faster than the other halogens. This is because iodine pulls its electrons toward its nucleus more than the other halogens. Check the densities of the halogens. Fri, 06 Aug 2010 17:39:45 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=516#516 516@http://orgo.curvedarrow.com/punbb Topic: Ketal from cis-1,2-cyclohexanediol and acetone Message: Without drawing the mechanism for this reaction, this is the product that one would get from the title reaction.        Sat, 31 Jul 2010 23:49:22 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=515#515 515@http://orgo.curvedarrow.com/punbb Topic: Grignard addition, ketal hydroysis Message: Yahoo!Answers in chemistry wrote:Mechanism for this reaction?Okay, so I'm studying for an exam and I don't get how this works.      I understand how the PCC and the H30+ works in more simple reactions, but i don't understand how we went from that oxygen ring on the grignard compound to it being open and losing an oxygen or something like that? A: There are really only two reactions here. First is the addition of the Grignard reagent to the ketone. This gives a tertiary alcohol, which remains into the product. Then, an acid catalyzed hydrolysis of the cyclic ketal. The result is the carbon which had two oxygen atoms attached becomes the new C=O. That is the final product. PCC will not change the final product because the alcohol is tertiary and so resists oxidation. This is shown diagramatically below. The mechanism for the ketal hydrolysis involves many steps and so is not detailed. The complete mechanism is provided in A Guide to Organic Chemistry Mechanisms, see http://www.curvedarrowpress.com/      Mon, 26 Jul 2010 11:57:44 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=514#514 514@http://orgo.curvedarrow.com/punbb Topic: Mechanism of Corey-House condensation Message: www.chemicalforums.com wrote:Mechanism of Corey-House reaction Q: What's the role of a carbanion in the last step of Corey-House Synthesis?Specifically:-     R2CuLi + R'X   R-R'And why does R' have to be primary? I don't know the answer to this question either, but let me speculate somewhat. In my book, one of the things I did was to write mechanisms in a way that repeated electron movements of other reactions. That sometimes included writing a new mechanism for a reaction that was unknown or may have been written with another mechanism. Accordingly, I have included a copy of my discussion of a Gilman coupling with a vinyl bromide.        Let's think about this reaction. In the first step (no mechanism) an electron rich copper (negative charge) is inserting into a C-Br bond. We might think that this insertion should be consistent with Grignard reactions and reactivity. That is, iodides might be more reactive than bromides, chloride or fluorides. Similarly, a vinyl bromide might be more reactive than an alkyl bromide as we may think of an sp2-carbon as more electron deficient (electron withdrawing) than an sp3-carbon. Accordingly, alkyl halides would further decrease in reactivity by addition of electron donating carbon atoms. Therefore, primary would be preferred. While the insertion of an electron rich copper is suggested, it would be consistent with other reactions of Gilman or similar organocuprates. As you find in the reactions that follow, copper adds to enones and acyl chloride followed by similar bond forming reactions and loss of copper. You can see in the discussion above that I suggest a two electron process to parallel that of several other reactions.In the discussion in the chemicalform link, someone suggested only primary halides due to steric hindrance. I do not believe that would be correct. If the mechanism is similar to Grignard formation, the mechanism of that reaction is not an SN2 substitution reaction and should not be guided by steric hindrance.This example is excerpted from A Handbook of Organic Chemistry Mechanisms, available at http://www.curvedarrowpress.com/ or Amazon.com Sat, 17 Jul 2010 18:05:12 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=513#513 513@http://orgo.curvedarrow.com/punbb Topic: Hydroboration-oxidation methylcyclohexene ->trans-2-methylcyclohexanol Message: The R-groups of borane can be intermediate alkyl groups or hydrogen, depending on the stage of hydroboration reaction. The complete formation of a trialkylborane is in A Guide to Organic Chemistry Mechanisms. This or similar examples are popular subjects for examinations as it shows the stereochemistry of the hydroboration reaction (syn addition) and the stereochemistry is retained in the oxidation step to give the trans-2-methylcyclohexanol.           This example is excerpted from A Handbook of Organic Chemistry Mechanisms, available at http://www.curvedarrowpress.com/ or Amazon.com Mon, 17 May 2010 00:25:19 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=511#511 511@http://orgo.curvedarrow.com/punbb Topic: D.A. rxn between a butadiene + maleic anhydride or maleonitrile Message: Yahoo!Answers in chemistry, draw the products.The products will be Reaction c), cis-1,2-1,2,3,6-tetrahydrophthalonitrile, a cis-dicyanocyclohexene. Reaction d), the Diels-Alder product of the diene with maleic anhydride.Reaction e), a mixture of 3-bromo-4-deuterobut-1-ene, and 1-bromo-4-deuterobut-2-ene. The latter will be the major product as the internal diene is the more stable thermodynamic product. If the reaction were carried out a very low temperature, the kinetic product can be isolated from the carbocation of the greater resonance contributor.         The mechanisms for these reactions are illustrated in "A Guide to Organic Chemistry Mechanisms" (Amazon.com, Barnes and Noble or http://www.curvedarrowpress.com ) Fri, 07 May 2010 15:39:03 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=510#510 510@http://orgo.curvedarrow.com/punbb Topic: Aromatic substitution of p-nitrotoluene and m-methoxybenzonitrile Message: Yahoo question, predict the products of bromination and chlorination of title compounds.            The product of reaction a) will be 2-bromo-4-nitrotoluene. The location of attack is determined by the most activating substituent in substituted benzene compounds. In this case, it will be the methyl group though the nitro complements the site of reaction as it is a meta-director. The product of reaction b) will be 3-methoxy-4-chlorobenzonitrile. The same rules apply. The methoxy is the most activating group. Because the methoxy is an o/p director, I would also expect to find some 2- and 6-chloro product as well. It is difficult to predict the position, ortho or para for different groups. I am not aware of any definitive rules for this prediction. Some groups seem to favor ortho attack and others para.The mechanisms for these reactions are illustrated in "A Guide to Organic Chemistry Mechanisms" (Amazon.com, Barnes and Noble or http://www.curvedarrowpress.com ) Fri, 07 May 2010 15:14:29 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=509#509 509@http://orgo.curvedarrow.com/punbb Topic: Effect of increasing methane concentration in radical chlorination Message: Yahoo!Answers in chemistry wrote:In the halogenation of methane, if we increase the concentraion of methane then the formation of should increase. However, should not the concentration of the ethane increase because its more stable? (1st step of chain propagation) I agree with part of this thinking about increasing methane concentration increases the amount of chloromethane. It will also increase the amount of ethane that forms, but that will only be a minor effect at best. How do we know that? Let us assume that the ratio of moles forming product in the propagation reaction is 9x that of termination steps (1x). Therefore, 10% of the products forming would occur from termination steps. In the reaction, ethane can only form from two CH3• radicals combining in a termination step. However, the termination can occur with methyl radicals to ethane or chloro radicals to chloromethane. Initiation forms two moles of chloro radicals. A chloro radical can form a methyl radical or be involved in a termination step. Therefore, formation of a methyl radical also forms a chloro radical. Therefore, one might expect that termination would be approximately equally likely to form the chloromethane or ethane. However, that would only occur to 10% of the product or at most 5% ethane. If the propagation/termination ratio is much greater than 9/1, then ethane will smaller still. Increasing the methane concentration or decreasing the chlorine concentration will decrease the propagation/termination ratio. However, it will also slow the reaction. The rate of the reaction will depend on the formation of radicals from chlorine, either propagation or initiation. Should they disappear, the reaction will stop. Therefore, while these conditions can increase the amount of ethane formed, it would be difficult to do so without simultaneously increasing the chloromethane to a greater extent. Fri, 07 May 2010 13:59:05 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=507#507 507@http://orgo.curvedarrow.com/punbb Topic: Elimination of (1S,2R)-1-iodo-2-methyl-1-phenylbutane to E-butene Message: What is the E2 elimination product of the reaction of (1S,2R)-1-iodo-2-methyl-1-phenylbutane with sodium methoxide in methanol?The preferred elimination occurs with anti-stereochemistry. Because this reaction is in methanol, the leaving group is a very excellent leaving group, the solvent is a polar methanol, some mechanistic leakage to an E1-elimination may occur, especially as the concentration of the reactant decreases and if it were to also deplete the methoxide. However, the problem does state to give the E2-elimination product so it should be as shown below.          Similar reactions are detailed in A Guide to Organic Chemistry Mechanisms. Mon, 26 Apr 2010 15:48:39 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=506#506 506@http://orgo.curvedarrow.com/punbb Topic: limonene rxns w HBr, MCPBA, H2, or KMnO4 Message: See http://orgo.curvedarrow.com/punbb/viewtopic.php?id=142        Thu, 15 Apr 2010 00:40:25 -0400 http://orgo.curvedarrow.com/punbb/viewtopic.php?pid=505#505 505@http://orgo.curvedarrow.com/punbb Topic: stereoisomers of hexachlorocyclohexane Message: Meso compounds are a special type of compound with two or more chirality centers. Normally, one would predict that if a compound had two chirality centers, then 4 isomers of it could exist. However, if the compound has a plane of symmetry, then the mirror image of a compound may result in a single compound. If you draw the mirror image of (2), you will get (2). Hence it is a meso compound. If a compound does not contain a chirality center, then the mirror image of it may give the same compound, cyclohexane, benzene, (1), or (6). Thu, 15 Apr 2010 00:34:01 -0400