can i get the interpretation of the ir and nmr spectra of m-dinitobenzene?
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Key points to look for include:\n\nInfrared\n-NO2 bonds: 1515-1560, 1345-1385\nAlkenes: C-H: 3020-3080 (moderate), 675-1000\n         C=C: 1640-1680(variable) \nIf wet: Large OH hump in the 3500 area... should obliterate most other peaks in the area.\n\nI am not as up to scratch with NMR.\n\nThis is a common compound and an IR and NMR spectra should be easily found on the internet. I recommend chaecking them carefully as jus tlooking for the peaks above will probably not be highly useful.\n\nEven better than the internet is a resource book, any laboratory or university library should have such resource books.\n\nPersonally I recommend that if you are doing much organic chemistry to get 'Organic CHemistry' by Morrison R.T. and Boyd R.N., published by Prentice Hall.\nThis book has an excellent guide for spectra.
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study on intermolecular particals and there behavior
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1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
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this maybe helpful \n\n\n\nwww.springerlink.com/index/R617G25W81437Q01.pdf\n\nhttp://www.springerlink.com/index/R617G25W81437Q01.pdf
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Refer to IUPAC's web site iupac.chemsoc.org or search on Google.
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Thanks NeuroProf.\n\nThe molecule AsF5 is trigonal bipyramidal in shape. First lay out all the bond pairs in that shape and then see if the 'pulling effect' of all the bonds cancel each other. Here, there are 3 bond pairs (of Cl) at the corners of the triangle, 1 bond pair at the bottom and 1 at the top. The Cl at the top pulls electrons and the pulling effect is cancelled by the one at the bottom. Considering the 3 bond pairs left in the middle, 2 are angled at 120 degress and one is at 90 degrees to the top and bottom bond pair. So the pulling effect of the 2 pairs is cancelled by the one at 90 degrees. Overall dipole cancels out and there is no charge distribution, so the molecule is non-polar.\n\nIn BrF4- (with square planar shape), the pulling effect of the lone pair at the top is cancelled by that of the lone pair at the bottom. This leaves 4 F atoms arranged at the corners of the square in the middle. The pulling effect(dipole) of 2 diagonal F atoms will cancel because they are in the same plane and are the same atoms. Hence all the 4 dipoles cancel out and that of the lone pairs too, so the molecule is non-polar.\n\nIn GaI3, the 3 I atoms are arranged in a trigonal planar manner. The dipole of the 2 I atoms at the sides will cancel, but the I atom at the top is not paired with any other atom at the bottom, so the dipole cannot cancel out. Since overall dipole moment is not equal to zero, the molecule is polar.\n\nAre you sure the molecule is SbF6- and not SeF6? If it's SbF6 then the shape will be square bipyramidal or octahedral (similar to BrF4-). The dipole of the 2 F atoms at the top and bottom will cancel and that of 2 diagonal F atoms will cancel and overall dipole will be zero, so the molecule will be non-polar.
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Mercury: Hg\nEthanol: CH3CH2OH\nBenzene: C6H6
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you can search this sit www.chemweb.com
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Hope this helps:\n\nhttp://jchemed.chem.wisc.edu/Journal/Issues/1997/Feb/abs210.html
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migs is correct.\n\nFor just a qualitative identification of functional groups, IR is the best for organic (and some inorganic) compounds. You load it in and, a few minutes later, you have a spectrum that tells you what groups you have. (assuming your compound is reasonably pure)
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Gee I don't know.  Did it look like\n\nri,nmj, opkj'af,00jslk  kjasei,3-  pem2d\n\n?
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It's fairly easy don't worry.\n\nNo need for d level splitting diagrams to answer this since we're dealing with N and F:  use the spectrochemical series to make an assumption about the ligand.  The difference in t2g and eg levels (∆o) determines whether a complex is low or high spin.  If both ligands were the same, we would have to look at the oxidation state of the ligand in the complex.\n\nHalides <  Oxygen ligands  <  Nitrogen ligands < CN- ligands\n\n   ALWAYS HIGH SPIN    ---------------            ALWAYS LOW SPIN\n\nIf ∆o > P, then a complex is low spin.  IF ∆o < P the complex is high spin.
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You can use Aldrich, which will give you a sample spectra of the compound. It would be very useful for identification based on a comparison of the fingerprint region (<900 cm-1)\n\nAny Organic book worth its salt will have typical vibrational frequencies. You can also do a yahoo search to get them.
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It's not an azo, because the nitrogen bond is not a double bond. It's not an amine because it has a nitrogrn-nitrogen bond.\n\nIt's a substituted version of hydrazine (H2NNH2), the famous rocket fuel.
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My best guess is that you have 2 ethyl groups bound to a nitrogen, and a third ethyl group on which the substituent is placed. Otherwise, it's triethyl amine, a relatively common lab reagent.
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Try this URL... it shows some diagrams.  Although it does not have the full set of configurations for d and f orbitals.
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What he said, except the doublets would likely have some fine meta coupling if you have a high enough resolution NMR.  You should be able to see this with 300MHz or higher.  I use a 400MHz and almost always see meta coupling.