1. Organic Chemistry Unit IX

2. I Introduction A. Definition study of carbon compounds forms thousands more compounds than inorganic elements do carbon has four bonding sites and can bond to itself in chains, rings or networks

3. B. Properties Insoluble in water Nonelectrolytes Low melting points Slow to react High activation energies

4. C. Bonding Carbon has 4 valence electrons Forms 4 covalent bonds  Tetrahedral structure  Bond angle = 109.5° 3 different ways to bond to itself Single bonded  saturated Double bonded  unsaturated Triple bonded  unsaturated

5. D Representations Molecular Formula ex. C 4 H 10 Structural Formula HHHHHHHH HCCCCHHCCCCH HHHHHHHH Condensed Structural Formula CH 3 CH 2 CH 2 CH 3

6. II Hydrocarbons A hydrocarbon is a compound that contains carbon and hydrogen only A. Types of hydrocarbons Alkanes Single bonded carbons in compound  C n H 2n+2 Alkenes A double bonded carbon in compound  C n H 2n Alkynes A triple bonded carbon in compound  C n H 2n-2

7. CH 3 CH 2 CHCH 2 C = 4 H = 8 8 is 2(4) Alkene CH 3 CCH C = 3 H = 4 4 is 2(3)-2 Alkyne CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 C = 6 H = 14 14 is 2(6)+2 Alkane Classify the following: n=4 n=3 n=6

8. B. Naming Hydrocarbons Count the longest string of carbons Use a prefix to indicate the number of carbons (Table P) Identify the type of hydrocarbon Use a suffix to indicate the type  Alkane-ane  Alkene-ene  Aklyne-yne Indicate any addition to the carbon chain using a prefix indicates amount number to show location  Use the lowest possible number Example 2-butyne

9. What’s the structure? Longest string of carbons  5 5 Prefix from Table P  pent- Type of hydrocarbon  alkene  Suffix for an alkene is -ene Indicate any addition to the carbon chain using a prefix for the amount  one double bond—no prefix a number to show location  4 th carbon or 1 st carbon  1 st carbon is the lowest Use the lowest possible number 1-pentene  which means 5 carbons; double bond in the 1 st position Name the hydrocarbon C 5 H 10 CH 3 CH 2 CH 2 CHCH 2

10. C. Writing Formulas for Hydrocarbons Write the number of carbons according to prefix used Place bonds according to suffix used Place bonds according to number used ex. 3-hexene 3-hexene 6 carbons 3-hexene double bond 3-hexene 3 rd position CH 3 CH 2 CHCHCH 2 CH 3

11. ex. 1,3,5-cyclohexatriene 1,3,5-cyclohexatriene 6 carbons 1,3,5-cyclohexatriene Double bond 1,3,5-cyclohexatriene Three double bonds 1,3,5-cyclohexatriene In a ring 1,3,5-cyclohexatriene Double bonds are on the 1 st, 3 rd and 5 th carbons

12. D. Complications Occurs when another substance replaces hydrogen Can be halogens ( Group 17) or carbon groups -F  fluoro -Cl  chloro -Br  bromo -I  Iodo -CH 3  Methyl group -CH 3 CH 2  Ethyl group

13. Prefixes used to represent number of replacement elements Mono Di Tri Tetra When writing formulas or naming Build the hydrocarbon name or formula Determine the new pieces How many Where are they located

14. Examples CH 3 CHClCH 2 CH 2 Cl CH 3 CH(CH 3 )CH 2 CH 2 CH 3

15. Examples 1,1,1-tribromo 3-hexene 2,2-dimethylpropane

16. III Functional Groups Replace hydrogen Change the properties of the substance Groups found on Reference Table R Attach to end of chain (terminal) Connect two carbon chains (bridge) Join anywhere on the carbon chain (floater)

17. A. Terminal Functional Groups Types Aldehydes CH 3 COH ethanal Organic Acids CH 3 CH 2 COOH propanoic acid

18. B. “Floaters” Functional group can go anywhere on the carbon chain Types Halide CH 3 CHClCH 3  2-chloropropane Alcohol CH 3 CH 2 CH 2 OH  1-propanol Amine CH 3 CHNH 2 CH 2 CH 3  2-aminobutane

19. C. “Bridges” Connects two carbon chains Types Ketone CH 3 CH 2 CH 2 COCH 3  2-pentanone Ether CH 3 OCH 2 CH 3  Methyl ethyl ether Ester CH 3 CH 2 CH 2 COOCH 2 CH 3  Ethyl butanoate

20. IV Isomers Isomers are compounds with the same molecular formula and different structure C 4 H 8 C 2 H 6 O

21. V Reactions Substitution Alkane + halogen CH 3 CH 2 CH 3 + Cl 2 → CH 2 ClCH 2 CH 3 + HCl Addition Alkene or alkyne + halogen or hydrogen CH 3 CHCH 2 + Cl 2 → CH 3 CHClCH 2 Cl CH 3 CHCH 2 + H 2 → CH 3 CH 2 CH 3

22. Esterification acid + alcohol → ester + water CH 3 COOH + CH 3 OH → CH 3 COOCH 3 + H 2 O Saponification fat + base → soap + glycerol Fermentation sugar → alcohol + CO 2 C 6 H 12 O 6 → 2 CH 3 CH 2 OH + 2 CO 2

23. Combustion hydrocarbon + O 2 → CO 2 + H 2 O C 6 H 12 + 9 O 2 → 6 CO 2 + 6 H 2 O Polymerization monomers → polymer n CH 2 → (-CH 2 -) n  ex. 7 CH 2 → (-CH 2 -) 7 Condensation Polymerization -OH containing monomer produces polymer and water Addition Polymerization Unsaturated monomer makes a saturated polymer