I only found that the effective configuration is L-Dopa... what is L-Dopa in terms of R or S? I figured it out by analyzing the only chiral carbon on the molecule... is that okay to do?
136948
The "L" in L-Dopa refers to "levorotatory", meaning that the molecule rotates plane polarized light counterclockwise. The "L" is sometimes written as (-). The opposite is dextrorotatory ("D" or "+"), which rotates plane polarized light clockwise.\n\nYou cannot use "L" to predict whether the molecule is an "S" or an "R" enationmer. They are seperate things.\n\nThe "R" and "S" refer to the arrangement of the atoms around the chiral center. The "L" and "D" refer to the direction that the molecule rotates plane polarized light. All chiral carbons will therefore carry an "R/S" and a "D/L" distiction, but both indicate different things.\n\nYou did the correct thing in determining the R/S configuration by assigning priorities to the 4 groups connected to the chiral carbon and so on and so forth. The L/D distinction can only be determined experimentally.
491823
Note that we're not taking energy differences into account.  In cases where the difference in orbital energy is greater than 15eV you get negligible overlap.  Now onto the symmetry-allowed overlaps.\n\n1s and 2s are both spherical and overlap to form a sigma bond.\n\n1s and 2p(x) form a sigma bond.  You can tell because if you draw the 2p(x) lobe you'll see it overlap with the corresponding 1s.\n\nThe 2p(x) and the 2p(y) do not overlap.  There is no possible bond.  Draw the 2p(x) extending toward the other atom as positive or shaded (remember: positive does not denote charge).  The other atom has its 2p(y) pointing up and down (perpendicular) and there is equal positive and negative overlap, thus canceling both of them out.  No net overlap.\n\n3p(y)-3p(y) form a pi bond.  Once again, draw these out and see it for yourself.  Draw the 3p(y) orbitals as rather fat so you can see how they overlap.  Since they are not cylindrically symmetric they are not a sigma bond.  Pretend you were to rotate one of the atoms about the x-axis... you would change orientation and there'd be less overlap.  For the sigma bonds above, rotation does not affect the overlap.\n\n2p(x)-2p(x) have positive overlap.  Draw them and see why.  They point toward one another.  Moreover, if you rotate one of the 2p(x) orbitals about the x-axis you keep the same structure/orientation since the 2p(x) is along the x-axis.  Thus, you can tell that this gives a sigma bond.
310266
See if this article has a diagram\n\nhttp://jchemed.chem.wisc.edu/Journal/Issues/1999/Apr/abs542.html
449049
Chiral compunds dissolved in liquids, if not a racemic mixture, will rotate the plane of polarisation of polarised light.\n\nBeyond that this is a very complex question!!!!
405697
1. lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion. VSEPR theory. a rough rule of thumb is that with same number of electron pairs, the more lone pairs you have with respect to bond pairs, the smaller the angle between bond pairs.\n\n2. the theory i just mentioned is not obeyed exactly by formaldehyde. you would expect the molecule to be trigonal planar with 120° for all bond angles because there are only bond pairs. however, the oxygen atom is electronegative and has a tendency to pull electrons from the double bond towards itself. the other bond pairs (both C-H) thus are not repelled as much by the C=O bond electrons and so can take a wider angle.
553400
I'm very surprised your book doesn't have information on VSEPR.  The acronym "VSEPR" stands for "valence shell electron pair repulsion".  Please double-check the list of topics in the back of your text.  \n\nA dipole moment in the context that you're considering is an incomplete sharing of electrons. In a covalent bond, one of the atoms involved is better at attracting & holding electrons than the other (more electronegative), so that instead of the covalent bond being perfectly shared it gets distorted to one side.  If a molecule has a dipole moment, it has more electrons (higher electron density) in one direction than in others.  One example of this is the covalently bonded acid, H-Cl.  Hydrogen is not very electronegative, but chlorine is very electronegative, so the chlorine atom does a better job holding onto the two shared electrons in the H-Cl bond.  \n\nI hope this gets you started in the right direction.  If you really can't find the info in your text, please go to a library and look at another chemistry book.
488870
Racemic mixture: an equimolar mixture of two enantiomers, designated (+/-), e.g., (+/-)-alanine\nbecause a racemic mixture contains equal numbers of dextrorotatory and levorotatory molecules, its specific rotation is zero degrees\nResolution: the separation of a racemic mixture into its enantiomers\nSome chiral drugs are sold as racemic mixtures, others as a single enantiomer\nWhen chemicals are synthesized in the lab you get both enantiomers. The body usually uses the Levo form but can convert the Dextro form to Levo through enzymatic reactions.  Because some drugs cannot be converted to the biological useful form, the chemists will seperate them before being sold.
522709
wow... you really do know your chemistry - sorry again, can't help you but I wish you ALLLLLLL the best, and must say the folks who answered - if they are infact correct, very nice.\n\ngood luck in whatever you are trying to figure out.
437693
At physiological pH most of the -COOH of amino acids are deprotonated (-COO-) and -NH2 protonated (-NH3+).\nIf you increase the pH then the -COO- will remain as such but the -NH3+ will lose the proton. -NH3+ can act only as an H-donor for an H-bond so such a bond will be broken. However if the geometry and number of interacting partners allow it it could be replaced by another H of the same -NH2 group.\n\nTyrosine has pKa around 10, so if you go to extremely basic pH it will be deprotonated and thus it will not be able to act anymore as an H-donor for a hydrogen bond and the bond will break. If however it was acting as a lone pair donor, the bond will become even stronger.\n\nGenerally when the pH change (extremely basic or acidic) causes changes in the charge of the group can but not necessarily breaks H-bonds. An important note: even if the H-bonds are not completely broken there will be rearrangement (some bonds might break and others form as the some H-donors become H-acceptors and vice versa) changing the conformation/folding of the protein.\n\nFor sure the effect of pH on ionic bonds is far greater than that on H-bonds.
69247
The occurence of Glycine in alpha helices is slightly less than tyrosine and much less than tryptophan and phenylalanine, which are the aromatic amino acids. Glycine is mostly found in beta turns of proteins and the aromatic ones are found in beta conformation secondary structures of proteins. So if a protein has many glycine residues, it probably has many beta turns and not so many alpha helices where are one with many aromatic amino acid residues is likely to have some alpha helical structures with a predominant beta conformation.\n\nP.S. The Ramachandran plot conveys this information.
484363
This would be (d).\n\nThe presence of an amide bond (R-NH-CO-R) is the main part of a peptide bond.\n\nB may look like this too, however in a peptide bond, but the amino group should be on the alpha-carbon of an amino acid.
452210
ICl2 + should be bent like water since I will have two lone pairs and two Cl's on it.\n\nI3- will be linear\n\nSbCl6- will be octahedral
429913
For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
178591
GHB got a bum rap.  it's just more nonsense hysteria to justify the bloated budgets of the police state.\n\nThat aside, you don't convert beta to gamma.  suggesting an elimination failed to consider that the double bond formed will be more stable in conjugation with the carbonyl group; you'll get crotonic acid.  Subsequent hydrolysis would give you a mixture of alpha and beta.\n\nyou'd have much better luck trying to figure out a way to do it starting with GABA.\n\nOf course, if you have to ask, you lack the skilz.
433401
Both CH3F and CF4 are tetrahedrons, but due to the substitution each has a different symmetry (C4v and Td respectively), as a rule the lower the symmetry the higher the polarity of the molecule. Actually, any Td (fully substituted tetrahedron), Oh (fully substituted octahedron), D4h (fully substituted square planar), D3h (fully substituted trigonal bi-pyramidal) will be non-polar molecules. This is due to the cancellation of the dipole moments.\n\nThus in order of the strongest dipole moment the answer is\n\nCH3F > CF4 (C3v < Td)\nXeO3 > XeO4 (C2v < D4h)\nSO2 > CO2 (C2v < D(inf)H)\nCOCl2 > BF3 (C2v < D3h)\n\nIf you are having a hard time understanding the Schoedinger group types, try a 3D modeling program (eg Spartan Build or ChemDraw 3D) to see the differences in the symmetry of the molecules.\n\nI hope that this helps. Please feel free to contact me if you have any further questions.\n\nKen
30904
The answer has to do with symmetry of the molecule and hydrogen bonding.  That should get you going in the right direction
164792
B.  A catalyst speeds up the reaction, as said in the question.  It's mostly just narrowing it down.  Obviously it provides a lower energy pathway if it speeds the reaction up, so it's not C or D.  It's not A, obviously.  Why would you want something that speeds the back reaction from products to reactants up? Well, unless specifically looking for it, I guess.   \nSo, B.