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Organic Chemistry

= the study of carbon and most carbon compounds. Organic Chemistry. as a class of organic compounds structural features and properties importance of their combustion reactions

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Organic Chemistry

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  1. = the study of carbon and most carbon compounds Organic Chemistry

  2. as a class of organic compounds structural features and properties importance of their combustion reactions How can we represent them? Molecular formulas, structural formulas, use of models, 3D representations, and names Structures and nomenclature: decoding chemical names Resources POGILS Fractional distillation of crude oil organic reactions Molecular Models Activity Our TB online resources and practice quiz AP chemistry study cards for organic chemistry http://www.chemmybear.com/apch23sc.pdf Organic Chemistry - Hydrocarbons

  3. Bonding of Carbon Atoms • Carbon atoms have a tendency to covalently bond with other carbon atoms and form chains. Straight chains Branched chains Ring chains • Carbon atoms are able to form up to four covalent bonds: **Remember: Carbon has 4 valence electrons. • Carbon atoms can engage in single, double, or triple covalent bonds: saturated compounds = contain only single bonds unsaturated compounds = contain at least 1 double or triple bond

  4. Organic Molecules - Hydrocarbons • contain H and C atoms. • may be acyclic molecules (linear or branched) or cyclic molecules. • each C atom has a total of 4 bonds and each H has one bond • all C-H bonds are single covalent bonds (  bonds) but carbon-carbon covalent bonds may be single (  bonds) , double (  bond and a  bond) or triple (  bond and two  bonds) .

  5. Molecular vs. Structural Formulas Molecular Formulas – Structural Formulas– Molecular Formula Structural Formula Condensed Structural Formula CH4CH4 C2H6 CH3CH3

  6. Hydrocarbons = organic compounds that contain only atoms of hydrogen and carbon • Homologous series of hydrocarbons: (a) Alkanes = contain only single covalant bonds - General formula: (b) Alkenes = contain one double covalent bond - General formula: (c) Alkynes = contain one triple covalent bond - General formula:

  7. First 10 Alkanes in Series HydrocarbonMolecular Formula Methane CH4 Ethane C2H6 Propane C3H8 Butane C4H10 Pentane C5H12 Hexane C6H14 Heptane C7H16 Octane C8H18 Nonane C9H20 Decane C10H22

  8. Methane (bp: -162 OC) Ethane (bp: -88.5 OC) Propane (bp: -42 OC) Butane (bp: 0 OC) Pentane (bp: 36 OC ) Hexane (bp: 69 OC) Heptane (bp: 98 OC) Octane (bp: 126 OC) Nonane (bp: 151 OC) Decane (bp: 174 OC) Separation of hydrocarbons by distillation? (ex: fractional distillation of crude oil)Animation: http://www.wwnorton.com/college/chemistry/gilbert2/(Chem tours chapter 12: Fractional distillation)POGIL activity on Fractional Distillation

  9. First 10 Alkenes in Series HydrocarbonMolecular Formula Ethene C2H4 Propene C3H6 Butene C4H8 Pentene C5H10 Hexene C6H12 Heptene C7H14 Octene C8H16 Nonene C9H18 Decene C10H20 Notice: There is no alkene corresponding to the methane of the alkane series. That is b/c there must be at least 2 carbon atoms to form a double bond.

  10. First 10 Alkynes in Series HydrocarbonMolecular Formula Ethyne C2H2 Propyne C3H4 Butyne C4H6 Pentyne C5H8 Hexyne C6H10 Heptyne C7H12 Octyne C8H14 Nonyne C9H16 Decyne C10H18 Notice: There is no alkyne corresponding to the methane of the alkane series. That is b/c there must be at least 2 carbon atoms to form a triple bond.

  11. This double bond between the two carbon atoms makes this an unsaturated fatty acid.

  12. Saturated and unsaturated fatty acids Saturated fatty acid Unsaturated fatty acid oleic acid See Jmol files (Interactive molecular animations) http://wps.prenhall.com/esm_brown_chemistry_11/86/22103/5658536.cw/index.html • stearic acid • See Jmol files (Interactive molecular animations) • http://wps.prenhall.com/esm_brown_chemistry_11/86/22103/5658536.cw/index.html

  13. Aromatic Hydrocarbons Simple Examples Aromatics –contain cyclic arrangements of carbon atoms bonded through both σ and delocalized π bonds. example - benzene and toluene

  14. Naming Organic Compounds • Naming straight-chained hydrocarbons:  Use Reference Table P (Organic Prefixes) and Table Q (Homologous Series of Hydrocarbons) to name & write the formulas.  When naming alkenes & alkynes, indicate where the double/triple bond is located in the molecule. **The carbons are numbered so as to keep the number for the double bond as low as possible** The triple bond is located on the 1st carbon…so its name would be: 1-butyne The double bond is located on the 1st carbon…so its name would be: 1-butene Both compounds have four carbons (use prefix but-) and a double bond (use ending –ene) Both compounds have four carbons (use prefix but-) and a triple bond (use ending –yne) The double bond is located on the 2nd carbon…so its name would be: 2-butene The triple bond is located on the 2nd carbon…so its name would be: 2-butyne

  15. AP and Class Exams – you will need to memorize the root names, functional groups and the naming rules associated with simple organic molecules • Memory aids for the first 5 in the series • mary eats peanut butter pancakes • (methyl-, ethyl-, propyl-, butyl-, pentyl-

  16. Naming Organic Compounds • Naming branched hydrocarbons: 1) Find the longest carbon chain which contains the functional group or multiple bond if present and name it (using Tables P & Q to find correct prefix & ending). 2) Number the longest chain (left to right or right to left) so that the functional group/multiple bond/longest side chain (branch) is on the lowest numbered carbon possible. 3) Name each side group but change the ending to -yl. 4) Use a prefix di-, tri-, tetra-, etc. to denote how many side groups of each length are present. 5) Before naming the side group give the number of the carbon to which the side group is attached. 6) Arrange the side groups in alphabetical order ignoring the prefixes di-,tri-, etc.

  17. Examples: 3.) The side group has only one carbon, so use the prefix meth- and add the ending –yl: methyl. 1.) The longest chain has 5 carbons, so the prefix pent- must be used. 2.) There are only single bonds, so the ending –ane must be used. 4.) Since the side group is right in the middle, the carbons can be numbered from either side. The methyl group is located on the 3rd carbon. Name: 3-methyl pentane 3.) Each side group has only one carbon, so use the prefix meth- and add the ending –yl: methyl. Since there are 3 methyl groups, use the prefix tri-: trimethyl. 1.) The longest chain has 4 carbons, so the prefix but- must be used. 2.) There are only single bonds, so the ending –ane must be used. 4.) Count carbons so that the longest side chain has the lowest #. The first 2 methyl groups are located on carbon 2, and the next methyl group is located on carbon 3. Name: 2,2,3-trimethyl butane

  18. Isomers = **As the # of carbon atoms in a compounds increases, the # of possible isomers also increases.** Example of Isomers: All of these compounds have the molecular formula C5H12

  19. Cis and Trans Isomers • Alkenes exhibit not only structural isomerism but geometric (cis-trans) isomerism as well. In geometric isomers the bonds are the same, but the molecules have different geometries. Geometric isomerism is possible in alkenes because rotation about the C=C double bond is restricted.

  20. Functional Groups = • Halides: = when any of the halogens (F, Cl, Br, or I) replaces a hydrogen atom in an alkane - named by citing the location of the halogen attached to the chain and adding the appropriate prefix (fluoro-, chloro-, bromo-, or iodo-) Note: Table R provides examples on how to recognize and name compounds w/ each of the functional groups!

  21. Note: The –OH group does not dissociate, and therefore alcohols are not bases/electrolytes. However, the –OH group does make alcohols polar molecules. (2) Alcohols: = one or more hydrogen atoms of a hydrocarbon are replaced by an –OH group (called a hydroxyl group) - named by citing the location of the –OH group and changing the ending to –ol. - Classifying alcohols: Monohydroxy alcohol: one –OH group Dihydroxy alcohol: two –OH groups Trihydroxy alcohol: three –OH groups

  22. - Alcohols can also be classified according to the position of their –OH group: PRIMARY (1o): the functional group is bonded to a carbon that is on the end of the chain. SECONDARY (2o): The functional group is bonded to a carbon in the middle of the chain. TERTIARY (3o): The functional group is bonded to a carbon that is itself directly bonded to three other carbons.

  23. (3) Aldehydes: = the carbonyl group (-C=O) is found on the end carbon - named by substituting –al in place of the final –e of the corresponding alkane name

  24. (4) Ketones: = the carbonyl group (-C=O) is found on an interior carbon atom that is attached to two other carbon atoms - named by replacing the final –e from the corresponding alkane with –one; if necessary, cite which carbon atom the carbonyl group is attached to.

  25. (5) Ethers: = two carbon chains are joined together by an oxygen atom bonded between two carbon atoms - named by first naming the two methyl groups, followed by the word ether (when both R groups are the same, use prefix di-)

  26. (6) Organic Acids: = contain the carboxyl functional group (-COOH) - named by replacing the –e in the corresponding alkane name with –oic acid

  27. (7) Esters: = have the type formula R-CO-OR’ (R-CO-O- part of formula comes from an organic acid; the R’ part comes from an alcohol- see Esterification) - named for the alcohol and organic acid that make up the ester

  28. (8) Amines: = formed when one or more of the hydrogen atoms of ammonia are replaced by an alkyl group - named by changing the alkane ending of –e to –amine and then numbering the alkane chain to show the location of the amine group

  29. (9) Amides: = a compound formed by the combination of two amino acids (See Condensation reaction) - named by changing the carboxylic acid acid reactant ending –oic acid with -amide

  30. Organic Reactions **Note: Generally occur more slowly than inorganic reactions. When covalently bonded substances react, they must first break relatively strong existing bonds before making new bonds.** • Combustion: =

  31. (2) Substitution: = (3) Addition: = Ethene Ethene

  32. Organic Acid + Alcohol  Ester + Water (4) Esterification: = (5) Saponification: =

  33. (6) Fermentation: = (7) Polymerization: = (a) Addition polymerization = involves the joining of monomers of unsaturated compounds (b) Condensation polymerization = involves the joining of monomers by removing water from hydroxyl groups and joining the monomers by an ether or ester linkage Polymers = organic compounds make up of chains of smaller units covalently bonded to each other

  34. Addition Polymerization: Condensation Polymerization:

  35. Polymerization • Animations for polymerization reactions • http://www.wwnorton.com/college/chemistry/gilbert2/ • Chem tours chapter 12 Animations for • Polymerization

  36. Lab Activity: Esterification • POGIL – organic reactions (group work)

  37. Molecules that possess nonsuperimposable mirror images are termed _________._______________ are stereoisomers that are mirror images of one another that are not superimposable. Resources • Molecular models • Video clip (our TB online resources for chapter 25) www.chm.bris.ac.uk

  38. Many of the molecules occurring in living systems, such as the amino acids, are chiral and exist in nature in only one enantiomeric form. Example: Alanine www.rikenresearch.riken.jp

  39. Many drugs of importance in human medicine are chiral, and the enantiomers may produce very different biochemical effects.

  40. eTextbook problems - Problem Set 4 • Chapter 25 (in part) • GIST problems p. 1056, 1059 1063, 1069 • VC problems 25.1-25.3, 25.6 (p. 1096-1097) • Within chapter 25, Study the sample exercises and then do practice problems:25.1-25.6 • End of chapter exercises for now: 25.7,.8, .9, .12, .17, .21, .23, .25, .31a, .43, .44, .47

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