why is benzoin molecule not optically active even though it has a chiral center?
460627
The keto is in equilibrium with the enol form. In the enol form, the stereogenic carbon assumes sp2 hybridisation and so loses optical activity.
1002550
IUPAC name:  (6aR,9R)-N,N-Diethyl-7-methyl-4,6,6a,7,8,9-\nhexahydroindolo-[4,3-fg]quinoline-9-carboxamide\n\nNote, the R at 6 and 9 is present because the compound has a chirality center at carbon 6 and 9 and it is in rectus configuration (as opposed to sinister).  The N denotes a "normal" chain or ring.  The chemical formula is C20H25N3O.
836459
The L and D prefixes concern how the molecule will rotate a plane of polarized light as it is passed through a tube containing a solution of the molecule.  \n\nThere is levorotary (left, "+") and dextrorotary (right, "-").  The "R" prefix is not used for optical activity, as there are "R" and "S" designations that pertain to the configuration of the molecule.  The actual physical structure of the molecule CANNOT be used to determine if a molecule is "L" or "D" , as both the "R" and "S" forms can rotate light in either direction depending on the molecule.\n\nIf there is a 50/50 mix of the "L" and "D" forms, then the molecule is said to have a "racemic" mixture and would just be noted by its name, e.g. glutamine.  It should be noted however that in naturally occurring amino acids, only the "L" form is found.  \n\nIf you buy amino acids, it is imperative that you buy the "L" form as the "D" form can have serious consequences including death.  To the best of my knowledge, the "D" form is illegal to sell for human consumption in the U.S.
220155
Are you asking for an explnation for the spectral proprties  or doy ou want the actual spectral report? either way go here....http://orgchem.colorado.edu/hndbksupport/irtutor/estersir.html
961812
According to Zaitzev rule your answer is correct. I don't know if they expected you to say that still you'll get some 3-methyl-1-pentene as a by-product, or if they also wanted you to note that the E (trans) stereoisomer of 3-methyl-2-pentene is the main product.\n\nBy the way, the way you wrote the question people mistake the alkoxy base (RO-) with "dash or"
799054
Consider a primary alcohol that has a structure like C--OH\n\nthen consider a tertiary alcohol that has a structure similar to \n          CH3\n           |\nH3C--C--OH\n           |\n          CH3\n\nIn a primary alcohol such as the one shown above, the distribution of charge is relatively polar due to the presence of only one other carbon atom with one very electronegative oxygen atom. In a tertiary alcohol, there are four carbon atoms versus only one electronegative oxygen atom. Therefore, in the tertiary carbon, the polarity is greatly less because the polarity is balanced by the presence of so many carbon atoms around the molecule.
788034
Once the hydroxyl group leaves the alcohol, the structure would lose its pyramidal shape and flatten into a two-dimensional shape.  The halide ion would add from either side, having a 50% chance of attaching from either side.  Once halogenated, the new molecule would again be optically active but you would have an equal abundance of both R and L enantiomers.  This follows one of the rules of stereochemistry in that optically inactive reactants (in this case HX) always yield optically inactive products.
726821
When trying to figure out what dissolves in what solvent, you have to consider POLARITY and remember the rule "like dissolves like".  \n\nSo a polar molecule will dissolve in a polar solvent.\nA nonpolar molecule will dissolve in ONLY a nonpolar solvent.\n\nWater is POLAR...we know that because of the structure of the water molecule haveing lone pair, being bent, and being able to have hydrogen bonding.  Ethanol is also POLAR because of it -OH (hydroxyl) functional group so it will dissolve in water.\n\nBenzene is NONPOLAR (chemists sometimes call it greasy) because it s made up of only carbon and hydrogen and is entirely symmetrical.  There are no electronegative atoms (such as F, Br, Cl, O, S) to cause a partial negative/positive to exist on the molecule so it WILL NOT dissolve in water.   It should dissolve in some NONPOLAR solvent such as hexane or diethyl ether.
545890
http://www.vuw.ac.nz/staff/paul_teesdale-spittle/organic/chiral_web/main.htm
843775
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!!!!
197637
Actually, there are a few things wrong with that structure. If the Red atoms are indeed oxygen then the formal charge would be (C4H8O2)2-, the double minus change would make this very unstable. \n\nOxygen is only found by when it is double bonded, which it can't do here since carbon can only have four bonds. The red could be a halide, that is something from the F, Cl, Br and I column. \n\nIf the red guys were bromine then it would be \n2,3-dibromobutane
945352
Then defend it... I say that there is no way that a random chance process could have formed life forming proteins and DNA with their unique chirality. If proteins and DNA were formed by chance  each and every one of the components would be a 50/50 mixture of the two optical isomers. This is not what we see in natural proteins or in natural DNA. How can a random chance natural process create proteins with thousands of "L" molecules, and then also create DNA with billions of "R" molecules? Does this sound like random chance or a product of design? Even if \nthere were a magic process to introduce chirality, it would only create one isomer. If such a process existed, we do not know anything about it or how it would work. If it did exist, how were compounds with the other chirality ever formed? Even if there were two magical processes, one for each isomer, what determined which process  was used and when it was used, if this was a random chance natural process? The idea of two processes requires a controlling mechanism, and this kind of control is not possible in a random chance natural process.\nYou are an intelligent man alot smarter than a mere theist so I don't think I need to explain chirality more thoroughly. You just need to defend your statement... Too bad you can't now that you have this informant ion.....   Jim
486317
This might not make much sense, but I think it might be\n2,4-Cyclohexadiene with an ammonia substituent.
822762
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
226789
I would say because the molecules are too big to be effectively dissolved by chloroform. Or the oils are not polar where chloroform would be polar (remember like disolves like).
344456
An enantiomeric ratio (ER) is defined as the (+)-enantiomer concentration of a chiral compound divided by its (-)-enantiomer concentration (Vetter and Schurig 1997):\n\n\n                      ER = (+)-enantiomer / (-)-enantiomer                                     (1) \n\nThe (+)-sign indicates that the enantiomer rotates a plane of polarized light to the right (clockwise; March 1985). The isomer with the (-)-sign rotates the light to the left (counterclockwise). Chiroptical detection is not always included in chiral chromatographic analysis. Also, pure enantiomers or enantiomeric enriched reference materials may not be available; therefore the enantiomer elution sequence cannot be achieved. In such cases, the ER is defined as the concentration of the first eluting enantiomer divided by the second enantiomer, under precisely defined chromatographic conditions (Vetter and Schurig 1997). Nowadays, pure enantiomer or enantiomer enriched standards are often available and the elution sequence can be easily determined (Müller and Buser 1994). \n\n\n\nMany organic compounds contain multiple chiral centres.  These are all essentially the same compound, but exist as mirror images of each other (compare with your left and right hand - they are the same, only your left hand can do different things to your right).  Many durgs contain chiral centres, and often several, and it is necessary to ensure there is a process to prepare a specific mirror image.  Giving a patient the "wrong" enantiomer can have serious side effects, or be completely inactive. (Thalidomide being a famous example)\n\nWhere a compound has two or more chrial centres then enantiomers exist.  For example the (R,R) isomer of one molecule is the opposite enantiomer to the (S,S) isomer of the same.  The structure of each look identical, but they are in fact mirror images.  It is not possible to superimpose one on the other.  Enantiomeric enrichment is where a process is introduced to selectively extract one enantiomer from the other, and therefore increase the amount of one compound over the other.  There are many ways to do this, however a common way is to complex selectively the required isomer with another chiral compound, and crytallise this complex from solution.  \n\nExamples escape me I'm afraid, but I'll try.  If you have 100 blue smarties and 100 red smarties.  These are the same, with the exception of colour (replace this with mirror image in chirality and you might see a tentative link...?).  This is a 50:50 mix.  If you introduce your baby sister to these smarties, and she prefers the blue ones, then she'll eat all of those, leaving the red ones behind.  What you have left therefore is an enriched quantity of red smarties.  If all goes well, then you will have 100% red smarties.  What you have done is enriched your mix.  Enantiomeric enrichment is the same principle.  Keeping what you want, but throwing away what you don't.\n\nAlternatively you can isomerise the undesired enantiomer into the desired enantiomer by chemical means.  This depends upon the relative energy states of each enantiomer and which one is the preferred form.  Either way the amount of desired product you have formed increases and you have enriched your mixture.\n\nI am not sure if any of this makes sense, as it is very difficult to explain without pictures.\n\nI've ahd a look and added a couple of links which may help.
306417
Circular dichroism (CD), is the differential absorption of left- and right-handed circularly polarized light. It is a form of spectroscopy used to determine the optical isomerism and secondary structure of molecules.\n\nAt a given wavelength,\n\nΔA = (AL − AR) \nwhere ΔA is the difference between absorbance of left circularly polarized (LCP) and right circularly polarized (RCP) light (this is what is usually measured). It can also be expressed as:\n\nΔA = (εL − εR) C l \nwhere εL and εR are the molar extinction coefficients for RCP and LCP light, where C is the molar concentration and l is path length.\n\nThen\n\nΔε = (εL − εR) \nis the molar circular dichroism. This is what is usually meant by the circular dichroism of the substance.\n\nAlthough ΔA is usually measured, for historical reasons most measurements are reported in degrees of ellipticity.\n\nThe molar ellipticity is:\n\n[θ] = 3298Δε. \nIn general, this phenomenon will be exhibited in absorption bands of any optically active molecule. As a consequence, circular dichroism is exhibited by biological molecules, because of the dextrorotary (e.g. some sugars) and levorotary (e.g. some amino acids) molecules they contain. Noteworthy as well is that secondary structure will also impart a distinct CD to their respective molecules. Therefore, the alpha helix of proteins and the double helix of nucleic acids have CD spectral signatures representative of their structures.\n\nCD is closely related to the optical rotary dispersion (ORD) technique, and is generally considered to be more advanced. CD is measured in or near the absorption bands of the molecule of interest, while ORD can be measured far from these bands. In principle these two spectral measurements can be interconverted through an integral transform, if all the absorptions are included in the measurements.\n\nThe ultraviolet CD spectrum of proteins can predict important characteristics of their secondary structure. CD spectra can be readily used to estimate the fraction of a molecule that is in the alpha-helix conformation, the beta-sheet conformation, the beta-turn conformation, or some other (random) conformation. These fractional assignments place important constraints on the possible secondary conformations that the protein can be in. CD cannot, in general, say where the alpha helices that are detected are located within the molecule or even completely predict how many there are. Despite this, CD is a valuable tool, especially for showing changes in conformation. It can, for instance, be used to study how the secondary structure of a molecule changes as a function of temperature or of the concentration of denaturing agents. In this way it can reveal important thermodynamic information about the molecule that cannot otherwise be easily obtained. Anyone attempting to study a protein will find CD a valuable tool for verifying that the protein is in its native conformation before undertaking extensive and/or expensive experiments with it. Also, there are a number of other uses for CD spectroscopy in protein chemistry not related to alpha-helix fraction estimation. s\n\nCD spectroscopy is usually used to study proteins in solution, and thus it complements methods that study the solid state. This is also a limitation, in that many proteins are embedded in membranes in their native state, and solutions containing membrane structures are often strongly scattering. CD is sometimes measured in thin films.\nCD has also been studied in carbohydrates, but with limited success due to the experimental difficulties associated with measurement of CD spectra in the vacuum ultraviolet (VUV) region of the spectrum (100-200 nm), where the corresponding CD bands of unsubstituted carbohydrates lie. Substituted carbohydrates with bands above the VUV region have been successfully measured.\n\nMeasurement of CD is also complicated by the fact that typical aqueous buffer systems often absorb in the range where structur