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This 8 page Class Notes was uploaded by Lucious Gleichner on Wednesday October 28, 2015. The Class Notes belongs to CHM7292 at Villanova University taught by JosephBausch in Fall. Since its upload, it has received 10 views. For similar materials see /class/230584/chm7292-villanova-university in Chemistry at Villanova University.
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Date Created: 10/28/15
Summary for Determining Point Group for a Molecule Cn proper rotation a rotation axis of symmetry where n is the order of the rotation axis or foldedness The number of degrees of the rotation is related to n by 360n The major axis in a molecule is the rotation axis with the highest order highest value of n Types of Point Groups 1 Cn Cnth Cn v these three do Lot have a 02 axis perpendicular to the major axis in Cm the 39h39 indicates a horizontal mirror plane of symmetry present in On V the 39i indicates a vertical mirror plane of symmetry present vertical mirror planes include the major axis horizontal mirror planes are perpendicular to the major axis if both types are present horizontal takes precedent Dni Dnth Dnd these three dio have a 02 axis perpendicular to the major axis in Dnh the 39h39 indicates a horizontal mirror plane of symmetry present in Dnd there are vertical mirror planes present but no horizontal planes Linear Groups a vaunsymmetric ie H Cl b Door symmetric ie OCO Platonic Solids a tetrahedral Tdie CH4 b octahedral 0 Le SFG c icosahedral lhie C60 B12H12239 quotPlan of Attackquot quotspecialquot group no yes I Td Oh h CW th Is On present no yes s mirror plane present 82 or 82nand IIonly collnear WIth unique 0 yes or highest order On Cs no yes s center of symmetry I present S7 Are there n 02 axes no yes perpendicular to 0 C1 C no yes s oh present s oh present no yes no I yes s 0V present 0 s ov present Dnh on on V no yes no I yes Dn Dnd DdMHou t A mcni cornsfmd Q q pkmt ohf 0r RM EM nth wt was Mara cui Q gumf3 32mm on d mauve M van 0 Man an be a new c qukn b M vasth out Few 3 0 ammova o39cnkm m be enamored 1 CD vo Q m bod gimp 01quot O39Auijndd 5 9w stpmd39a than 93 n s M My of 6m mk m can the mva 04 4c3v u J On nuke 5 36 Hg 9 5 7 moquot mm Mquot Mw39cshv 5 M 5 g a HZMM W 0 1 Mk a ka n n mow0 01m om new dun mus 94 bum 1 lt5 MM 6 Q9 93 Hquot Ngkuf orAw check as Is 6M my is u dcm m a phat 5 53mm but cm H mm m whtv in a amt 61an D E WAfed an quot 18 he u c 6K4 um39xw a 33 5 39G39J 39m H39 HAquot H o39v LC I 9 9 kmmw Pm 39 that am pcvpcumdw q a any ma Us bumm K 039 quot Gk is gnaw Q6 not Ame 4w sh 6 C5 invu m of a on Mr Q may 0 snhmdga r gt I 0 k mch dunan is a m 0quot hwcvsion L 3 bungar cusmcwnc 39 1 lt 0 H nok M lt35 isomcv 4a poms M symmdvg akmhf K G dud cm M9wa 0 N B kt no qticnhd Syntade o mrd m 39if K 5Q ch39dknyna 60 Rd HR 59mm dtmm k is Md 0 impropw o in was S A 0 an 55 mush Mn evMAMLa 3 560qu 53 wb mz ou an afoukms unuu vuag 39 Droime Loni a fury Sgt r then 1 Hindu nN a u pcvrn tmhw 6 Ni m Tho cgclosdmk 4 w 39n FiyVK 33 Atm n ks an S New f km MS hggt he 1 59ml domed is uwamf V Ev 6m Sa 5 ch thoan Who wins the Chemistry Death Match between quotformal chargequot and quotoxidation statequot Well the answer is that both are useful but for different things Each is nothing more than a kind of quotelectron bookkeepingquot Oxidation states aka quotoxidation numberquot were popular in gen chem in dealing with redox reactions Here is an example 4 Fe 3 02 gt 2 FE203 For complete rules for assigning oxidation numbers consult a gen chem text Stripped down rules are as follows 1 an atom in its elemental state has an oxidation number 0 2 the oxidation number of a monatomic ion is the same as its charge ie chlorine is 1 in Cl 3 assume electrons in a covalent bond are quotownedquot by the more electronegative atom in the bond So in the above redox reaction the iron has gone from an oxidation number of zero in elemental Fe to 3 in Fe203 hence an oxidation and the oxygen has gone from zero in molecular oxygen 02 to 2 in Fe203 hence a reduction We also have redox reactions in orgo and oxidation numbers can be employed to allow us to determine which species have been oxidized and which reduced Formal charge is used more frequently in orgo than oxidation states It can be thought of as the charge that each atom in a molecule would possess ifal the atoms had the same electronegativity This is clearly a bit different than oxidation number The formal charge of an atom in a molecule can be computed as follows formal charge of valence e oflonepair e 12 of bonding e Knowing the formal charge of an atom in a molecule can sometimes give us some idea of whether that atom is quotelectron richquot quotelectron poorquot or pretty happy with its lot in life But beware sometimes formal charge can be misleading like in the hydronium ion H30 Here the oxygen is 1 formal charge but the electron poor part of the molecule is actually the hydrogens Summary both formal charge and oxidation numbers are just attempts at quotelectron bookkeepingquot for the purpose of trying to indicate whether an atom in a molecule is electron rich poor or neutral The oxidation number method seems to sit on one end of the spectrum where electronegativity plays too much of a role note that the more electronegative element is assumed to quotpossessquot the electron in a bond even when it is only slightly more electronegative than the other atom but formal charge is at the other end of the spectrum where it treats all atoms as having the same electronegativity when it comes to assigning the charge Hmmm kind of makes you wonder if we maybe need a new method that takes some sort of average of the two
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