Organic Exam1 StudyGuide
Organic Exam1 StudyGuide CHEM350
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This 8 page Study Guide was uploaded by Caroline Emery on Tuesday September 15, 2015. The Study Guide belongs to CHEM350 at University of Tennessee - Knoxville taught by Dr. John Bartmess in Fall 2015. Since its upload, it has received 99 views. For similar materials see Organic Chemistry 1 in Chemistry at University of Tennessee - Knoxville.
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Date Created: 09/15/15
Exam 1 Study Guide Chapter 1 Functional groups Hydroxyl group OH 0 Alcohols Amino group NH3 0 Amines o Hydrogens aren t necessary can be bonded to carbons instead Carbonyl CO o Keytone attached to a center Carbon 0 Aldehyde attached to an end Carbon Carboxyl COOH carbonyhydroxy o Carboxylic Acid Carboxylic Ester o OCOC 0 Similar to carboxylic acid has a carbon group instead of a hydrogen Carboxylic Amide o OCN 0 Similar to carboxylic acid has an amine instead of the OH group Hybridization Bonds overlap and combine orbitals to make a hybridized orbital In a tetrahedral molecule lone pairs included the s and p orbitals hybridize and form the quotsp3quot orbitals 0 These bonds have an angle of approx 1095 When the s and two of the three p orbitals combine orbitals are formed with one large lobe and one much smaller lobes these are quotsp2quot orbitals 0 They have bonds at 120 degrees 0 The remaining p orbital shoots out perpendicular to the plane of three If the s orbital and one p orbital hybridize an quotspquot orbital is formed The orbital is one large lobe and one very small Directly across from each other 0 These bonds are planar and at 180 degrees from each other Formal Charge To assign formal charge to an atom in a molecule valence electrons5number of bonding electrons number of unboned electrons formal charge Resonance If a structure has resonance be able to draw all resonance forms and use arrows to show the relationships between the structures Chapter 2 Alkane Nomenclature Meth 1 Eth 2 Prop 3 But 4 Pent 5 Hex 6 Hept 7 Oct 8 Non 9 Dec 10 o nPr n propyl CH22CH3 normal 0 iPr isopropyl CHCH32 o nBu n butyl CH23CH3 normal 0 iBu isobutyl CH2CHCH32 o sBu secbutyl CHCH3CH2CH3 secondary o tBu tert butyl CCH33 tertiary Naming 1 Find the longest carbon chain in the molecule The root name of the molecule will be that alkane and that longest chain is the parent chain 2 Number the parent chain from each end and choose the set of numbers that give the lower number for the rst branch point 3 Sort the remaining groups alphabetically by name iso counts in alphabetization sec tert n don39t 4 Append di tri tetra etc to any group names if the group occurs more than once in the molecule Those number pre xes are not part of the alphabetization Put all group names in their alphabetic order in front of the parent chain name Assign the appropriate numbers to each group based on the alreadyset numbering of the parent chain These numbers are written just before the group and are separated from the group names both before and after the number by hyphens If there are multiple numbers before a group name eg quot224trimethylquot those are separated by commas Carbon Labeling Carbons are labeled as primary secondary tertiary or quaternary based on how many other carbons are attached to it 0 One attached carbon primary 0 Two attached carbons secondary 0 Three attached carbons tertiary 0 Four attached carbons quaternary Newman Projections Looking straight down a bond Staggered o CH bonds of one carbon are as far apart from the CH bonds on the adjacent carbon as possible 0 Anti con rmation When the methyl etc groups are as far apart as possible 180 degrees 0 Gauche con rmation When methyl etc groups are closer together 60 degrees 0 Lowest energy most stable 0 Echsed o CH bonds of adjacent carbons are directly behind each other 0 Higher energy least stable Torsional Strain difference in energy between staggered and eclipsed con rmations Angle strain bond angle is stretched or compressed beyond optimal value Steric strain strain arises when nonbonded atoms are forced too close together Cyclohexanes Chair con rmations Chair confirmation most stable cyclohexane con rmation all carbons staggered bond angles approx 1095 BE ABLE TO DRAW CHAIR AND ITS FLIP 0 Axial bonds hydrogens parallel to the carbons same plane 0 Equatorial bonds hydrogens perpendicular to the carbons in a different plane Boiling Point Order 0 As the number of carbons in a straight chain alkane increases so does the boiling point of the molecule Chapter 3 The more branched a molecule is the lower it s boiling point will be Molecular Relationships Stereochemistry study of 3D arrangements of atoms in molecules lsomers o ConstitutionalStructural lsomers Same molecular formula different arrangement 0 Stereoisomers Same molecular formula and connectivity different orientation Cis groups on the same side Trans groups on opposite side Enantiomers stereoisomers that are NON superimposable mirror images always refers to a pair Atropisomers lack a chiral center but are different because of blocked rotation Stereo center an atom where swapping two attached groups produces a stereoisomer Diastereoisomers stereoisomers that are NOT mirror images of each other always in pairs Chirality handedness o Chiral not superimposable over its mirror image o Achiral superimposable over its mirror image Meso achiral compounds with several chiral centers 0 Plane of symmetry one half is the mirror image of the other 0 Chiral center an atom usually a carbon that is bonded to four different groups type of stereocenter Labeling as R or S 0 Absolute configuration which isomer a given enantiomer is 0 Establishing priority 0 Priority increases as atomic number increases 0 If two immediately connected atoms have the same priority look at what those atoms are bonded to First point of difference 0 If double or triple bonds are involved use phantom atoms to pretend that multiples of the same substituent are single bonded rather than double or triple bonded RS system specifies absolute con guration also known as the quotCahnlngoldPrelog Systemquot Assigning R or S con guration 0 Locate chiral center 0 Assign priority to all attached atoms 0 Orient so the lowest priority atom is facing away from you 0 Read the remaining three groups in order of priority R priority is clockwise 5 priority is counterclockwise
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