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Chapter Two

by: Maggie Bruce

Chapter Two CHM 1240

Maggie Bruce
GPA 3.86

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These notes are on chapter two.
Organic Chemistry 1
Stanislav Groysman
Class Notes
Organic Chemistry
25 ?




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This 3 page Class Notes was uploaded by Maggie Bruce on Wednesday September 21, 2016. The Class Notes belongs to CHM 1240 at Wayne State University taught by Stanislav Groysman in Fall 2016. Since its upload, it has received 35 views. For similar materials see Organic Chemistry 1 in Chemistry at Wayne State University.


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Date Created: 09/21/16
Chapter Two  Alkanes o Simple group o C-C bonds and C-H bonds  Methane o CH 3 o Tetrahedral  Ethane and newman projections o C2H 6r CH C3 3 o Each carbon is tetrahedral o Because the carbons are held together with a single bond, free rotation around that bond leads to different conformations with different energies  Staggered  Lowest energy  Hydrogens are as far apart as they can be  When looking down the C-C bond (Newman projection) you can see each hydrogen  Eclipsed  Highest energy  Hydrogens are as close as they can be  When looking down the C-C bond (Newman projection) you can only see the hydrogens on the front carbon  The difference in energies is due to the distance between hydrogens called steric overlap o Ethyl compounds  CH 3CH 2  X could be anything  Propane o Third member of the alkane series o C H or CH CH CH 3 8 3 2 3 o Line-angle formula o Staggered and eclipsed Newman Projections  CH 3eplaces a hydrogen on the back carbon  The methyl group is larger so there is more steric overlap o Propyl compounds  CH 3f propane is a methyl group  CH 2s methylene  Replacing a methyl group hydrogen with x gives propyl x  Ex. CH3CH 2H B2 propylbromide  Replacing a methylene hydrogen with x gives isopropyl x  Ex. CH3CHBrCH 3  Butane o C4H 10or CH 3H C2 C2 3 o Line-angle formula o Isomer of butane is Y shaped, called isobutane o Newman projections  Looks at the bond between carbon 2 and 3  Four different possibilities  Two staggered o One has the methyl groups as far away as possible, this is the lowest energy o One has the methyl group next to each other  Two eclipsed o One has the methyl groups in line with hydrogens o One has the methyl groups in line with each other, this is the highest energy o Butyl compunds  Primary  CH CH CH CH X 3 2 2 2  Substituent is attached to a primary carbon  Butyl  Secondary  Ste Number CH 3H C2XCH 3  m of Substituent is attached to a carbons secondary carbon  Meth 1 Sec-butyl  Tertiary - Eth- 2  Substituent is attached to a tertiary Prop 3 carbon  - X But- 5  Tert-butyl  Pentane Pent 5 o Five carbons - Hex- 6 o CH 3H 2H C2 Hept 7 2CH 3 o Line-angle formula o Isomers -  Oct- 8 Isopentane Non- 9  Neopentane  Systematic naming Dec- 10 protocol o Developed by IUPAC o Prefix + stem + suffix o First find the longest chain of carbons to get the stem o Number the carbons to make it so the substituents have the lowest number possible o If there are multiple of the same substituent, use di, tri, tetra, etc. o Arrange prefixes alphabetically  Ring coumpunds o Cyclic o CnH2n o Cyclopentane C H5 10 o Naming  Add cyclo in front  Number the ring to give the substituents the lowest numbers  Arrange substituents alphabetically  Rings can be a substituent on linear alkyl groups  Sterisomerism o Appears in cycloalkanes o There is no rotation around the bonds, so there can be cis and trans compounds  Cis – the most significant groups are on the same side  Trans – the most significant groups are on opposite sides  Reactions of Lewis acids and bases o Any reaction between an empty orbital (acid) and a pair of electrons (base) is an acid-base reaction o Curved arrow formalism is used to show the movement of electrons from base to acid


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