Organic Chemistry Week #3 Notes
Organic Chemistry Week #3 Notes CHM 2210
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This 7 page Class Notes was uploaded by Madison Williams on Friday September 9, 2016. The Class Notes belongs to CHM 2210 at University of North Florida taught by Corey Causey in Fall 2015. Since its upload, it has received 5 views. For similar materials see Organic Chemisty 1 in Chemistry at University of North Florida.
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Date Created: 09/09/16
Line-Angle Formulas o The “ends” of lines and the breaks/intersections/bends imply carbons. o Hydrogen atoms on carbons are assumed. o butane (C H 4:10 The bottom left end of this structure is assumed to be CH 3 the top left bend is assumed to be CH ,2the bottom right bend is assumed to be CH , a2d the top right end is assumed to be CH 3 o pentane (5 carbons): o hexane (6 carbons): o heptane (7 carbons): o octane (8 carbons): o nonane (9 carbons): o decane (10 carbons): Constitutional Isomers o example #1: C H 6 14 All of these have the same formula, but they have different structures (constitutional isomers). Hexane is normal and unbranched, whereas the other ones contain branches (points where they diverge) off of the main chain. Nomenclature of Alkanes o The prefix indicates the number of carbons. o The suffix “-ane” indicates that the molecule is an alkane o Unbranched alkanes are named by the prefix + “-ane” suffix. For Branched Alkanes 1. Select the longest continuous chain of carbons – known as the parent/root. Then, name this chain as a normal, unbranched alkane. 2. Give each substituent (any carbon not apart of the main chain) a name and a locant (location of the main chain). 3. Use the same prefix. The “-ane” suffix is converted to a “-yl”. - example #1: CH - “methyl” 3 - example #2: CH -CH 3 “et2yl” 4. Number the parent chain to give the lowest number to the substituents. - example #1: 2-methyl pentane You number this one from right to left so that the substituent is at 2. (If you numbered it the other way, the substituent would be at 4.) 5. If there are multiple identical substituents, use “di-“, “tri-“, “tetra-“, etc. - example #1: 2, 3 – dimethyl butane For this one, it does not matter which direction you number the parent chain because either way there will be a methyl at 2 and 3. 6. For non-identical substituents, list their names in alphabetical order. (If they have prefixes, do not take them into consideration when putting them in alphabetical order.) - example #1: 4-ethyl-2, 2-dimethyl hexane You number this parent chain from left to right because it is better to have substituents at 2 and 4, than at 3 and 5. The substituent at 2 is two methyls because they are separated, unlike the bonded ethyl on 4. 7. If two parent possibilities have the same length, then choose the chain that is more branched. - example #1: 3-ethyl-2-methyl hexane We choose the second one because it has two branches/two substituents. (The first one only has one branch/one substituent.) Common Sub. Names 1. propyl: isopropyl: an isomer of propyl 2. butyl: bonded to parent chain at carbon 1 isobutyl: 3. secbutyl: bonded to chain at carbon 2 4. tert-butyl (t-butyl) Other Nomenclature o Infix: - “-an-“: alkane - “-en-“: alkene - “-yn-“: alkyne o Suffix - “-e”: hydrocarbon - “-ol”: alcohol - “-al”: aldehyde - “-amine”: amine - “-one”: ketone - “-oic acid”: carboxylic acid Cycloalkane o The formula for cycloalkanes is C Hn.2n The Shapes Nomenclature o Name these by adding the prefix “cyclo-“. o For substituents on the ring: 1. If there is only one substituent, then no locant is needed. - example #1: methyl cyclopentane 2. If there are two substituents, you give the lower number to the substituent that is first alphabetically. - example #1: 1, 2-dimethyl cyclohexane - example #2: 1-ethyl-2-methyl cyclohexane Conformations of Alkanes Newman Projections o There will always be three eclipsed (high energy), and three staggered (low energy). o example #1: ethane - This assumes we are looking down a single C-C bond. - O° (eclipsed) note: The back hydrogen atoms are the three that are not connected to the middle carbon. (Only the front three hydrogen atoms are drawn in this way.) - 60° (staggered) - 120° (eclipsed) - 180° (staggered) - 240° (eclipsed) - 300° (staggered) - If you graph these, the eclipsed structures all exhibit the same energy. o example #2: butane – We are viewing this from the C -C bon2. 3 - O° (eclipsed): This position exhibits the highest energy. - 60° (staggered): This position is called gauche (staggered position in which the large groups are 60° apart). - 120° (eclipsed): This does not exhibit energy as high as 0°. - 180° (staggered): This exhibits the lowest energy and is the best case scenario. It is called the anti conformation. - 240° (eclipsed): This conformation is no better than that of 120°. - 300° (staggered): This conformation is also gauche and is no better than 60°. - If you graph these, the eclipsed structures all exhibit the same energy. Cycloalkanes o Cyclopropane is much higher in energy than propane because: - torsional strain: This molecule must be planar and as a result the carbon atoms cannot twist like they can in ethane and butane. In the structure below, all six hydrogen atoms are eclipsing each other. note: This was drawn in class as just a triangle with the hydrogen atoms (carbon was not shown). - angle strain/ring strain: The hybridization of each carbon atom should be sp 3eaning each angle would be 109.5°; however, in actuality each angle is only 60°. o Cyclobutane does not exhibit as much torsional strain as cyclopropane, but it does possess some angle strain. The image on the right is called a puckered conformation and it helps to remove torsional and angle strain. o Cyclopentane does not exhibit much angle strain because it’s angles are 108° and as an sp3hybridization they want to be 109.5°. However, there torsional strain is present. The image on the right is known as an envelope conformation.
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