Week 5 Reading/Lecture
Week 5 Reading/Lecture CHEM 2321
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This 8 page Class Notes was uploaded by Hayley Lecker on Friday September 25, 2015. The Class Notes belongs to CHEM 2321 at University of Texas at El Paso taught by Dr. James Salvador in Fall 2015. Since its upload, it has received 48 views. For similar materials see Organic Chemistry I in Chemistry at University of Texas at El Paso.
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Date Created: 09/25/15
Organic Chemistry Week 5 Important Information Professor s Email isalutepedu Class Website organicutepeducourses2324 Class Code Ebook utep232Xfall2015 112 Acvclic Alkanes with 1 Chiral Center RS nomenclature reference pages 540545 112 Nomenclature of Setereoisomers CahnIngoldPrelog Convention a method that designates each stereoisomer with a letter R or S To determine whether to use R or S 1 Assign priority number to each atom or group of atoms bonded to stereocenter A center will have four different groups bonded to it so if it has two methyls or two hydrogens that is NOT the center Assign the numbers based on the atomic number you may have to look at what the atoms themselves are bonded to to assign their number 2 Redraw into wedge and dash method The lowest priority will be pointing away and the 3 other groups will be in the plane of the paper 3 Draw arrow from 1st priority group past 2quot l to the 2rd f arrow circles clockwise call the stereocenter R if counterclock wise call S Example below Determining Priority Numbers Look at the atomic number of the 4 groups The higher the atomic number the higher the priority number So bromochlorofluroromethane has Br Cl F and H attached to the carbon Assign 1 to Br because the atomic number is 35 2 goes to Cl with 17 3 to F with 19 and 4 to H with 1 If atoms are isotopes of the same element prioritize by atomic weight the heaviest gets priority So 3H comes before 2H and finally 1H If there is a tie the atoms bonded to the tied atoms determine priority Example 2bromo 2methylpentane Br gets 1 and H gets 4 but 2 and 3 are tied because of carbon The first carbon is bonded to 2 hydrogens and 1 carbon The second carbon is bonded to 1 hydrogen and 2 carbons so the second carbon takes higher priority In a cyclic compound start one group with the atom immediately to the right of the stereocenter and start another group with atom immediately to the left of the stereocenter If there is a tie continue walking around the ring until you can find a difference First point of di srsucs With a double or trible bond treat multiple bonds as if broken into single bonds with corresponding atoms at each end of the multiple bonds Thus EC CK becomes C C H t CC H H C c C C H l l l becomes C C H and C 0 becomes C 0 Thus for l C C 0 C The two enantiomers of serine have different structures COOH Him C C NHZ HZN A HOHZC CHZOH Natural swims Synthetic serius f groups are bonded to stereocenter are long look at chains to find difference to give priority Ex C1 CH3 You need to rotate the bond between the sixmembered ring and the carbon hearing the Cl group Doing so helps to see that the atom has an R configuration 5H an The carbon hearing the OH group also has an R con guration Flipping the molecule i800 helps to more clearly see the con guration of this carbon atom C1 H3C HO lR R Rl3l1Chloroethylcyclohexanol Determine whether the following molecule has any stereogenic centers Mark all asymmetric carbon atoms and determine whether they have an R or S con guration 311 Solution This molecule has three stereogenic carbon atoms The carbon bearing the OH group the one bearing the Cl and the one bearing the CHCCH3 group These three are starred in the following structure The carbon atom bearing the CHClCH3 has an R con guration Another example of naming stereocenter lo Assign the priority to the four groups this way 3 F I I H Br 4 1 Then redraw with the lowest priority group back This structure has an S con guration because the arrow curves counterclockwise The molecule is S 2bromobutahe 116 Acyclic Alkanes with 2 or more Chiral Centers reference pages 555560 116 Molecules with Two Stereocenters Enantiomers are stereoisomers that are mirrors images of each other All stereoisomers that are not enantiomers are diastereomers When a molecule contains more than one stereocenter its stereoisomers exist in pairs with an enantiomeric relationships They also exist in pairs with a diastereomeric relationship Example of this C2 and C3 are stereocenters same bonding but different in spatial arrangement thus they are stereoisomers In the image to the right there are not enantiomers because they are not mirror images C2 is S in both structures but C3 has R configuration in the left structure and S in the right C3 is a mirrored image but C2 isn t The 2n rule allows you to calculate the maximum number of stereoisomers possible This is always true for all molecules except 23Butanediol does not it has 2 but the Zn rule says 4 Example on next page CH3 CH3 CH3 CH3 H OH HU H H OH HO H H OH HO H HEO H H OH CH3 CH3 CH3 CH3 As you can see the enantiomers pair on the left is just a different representation of the same molecule CH3 CH3 ng H OH HO H Rotate 1 800 H r CH3 CH3 of the page CH3 Identical to the Apparent pair ofenantiorners I structure on the left In a Fisher projection of these there is an internal plane of symmetry CH i plane H0H of symmetry 11113 Compounds in which exhibit an internal plane of symmetry and are symmetrical even though they have stereocenters are called meso compounds Meso are symmetrical structures with 2 halves having identical substituents but opposite configurations Symmetrical diastereomer is meso diastereomer CHI3 H Plane of symmetry 3 This compound does have an internal plane of symmetry making it a meso compound with no optical activity The top stereocenter is S and bottom one is R When naming molecule with 2 stereocenters consider each center separately Common way to name use erythro and threo Names are derived from 234trihudrocybutanol commonly known are erthyrose and threose CHEOH Erythrose H OH CHQOH Thire use Use erythro when similar groups are adjacent to each other When groups are identical use meso Use threo when similar groups are opposite each other When groups are identical they are dl then use dl These names are not UPAC COOH H OH H 0H COOH maso Tartarlc acid CH3 CH3 H C1 C1 H H OH CH3 CH3 3chlom 2butanol COH H OH HO H COH C OH dBTammie acid CH3 CH3 H II Cl H H 0H CH3 CH3 Threo chloro 2 b1utanol Lecture Notes as Sublimation Cmttle aatio fleljioaitio Emporatimt 1 1q111c Freezing RElti g On his website so will be important Melting points don t change based on location atm Impurities will lower the melting point if the compound has a large melting range it is not pure The melting point increases with size and weight It increases with more polar covalent bonds this includes hydrogen bonds The melting point will decrease with branching longer chains The melting point will increase with better crystal packing and higher rotational symmetry The melting point of even number hydrocarbons is higher than odd numbered Boiling points do change with location the lower the elevation the higher the boiling point The boiling point raises with impurities and more covalent bonds including hydrogen bonds size and weight The boiling point will decrease with branching longer chains Below is an example of what we did in class S23dimethylheptane S is counter clockwise In the example below you can see the priority numbers if you get an R conformation and the hydrogen is in the back just flip the methyl upward and it turns it S or you can mirror it however on some of the homeworks it will give you an diasterisomer message How to calculate density by specific gravity 1 Find the molecular weight of the compound 2 Divide by the total number of atom in the molecule 3 Divide by water s specific gravity 183 6 Then round to two significant figures Example C7H60 The molecular weight will be 12Carbon s weight times 7 16oxygen s weight time 1 541 1hydrogen s weight x 6 This equals 106 now divide that by the total number of atoms in the molecule which is 14 This equals 757 now divide that by water s specific gravity which is 6 this gives a final density of 126 and rounded to 13 I If the number is less than 1 the compound will float If the number is equal to 1 or close to 1 the compound will emulsify If the number is greater than 1 the compound will sink
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