1. By Alex Chan and Billy Lordi
Organic Chemistry
By Alex Chan and Billy Lordi

2. Nomenclature
Hydrocarbons- chemical compounds that only contain two elements (carbon and hydrogen)
Classified by the presence of single, double, or triple bonds
Alkanes- C(n)H(2n+2) single bond
Alkenes- C(n)H(2n) double bond
Alkynes- C(n)H(2n-2) triple bond

4. Aromatic Hydrocarbons
Defined by a ring structure
Usually a six-member ring with alternating single and double bonds
Volatile

5. Functional Groups
Chemical properties are determined by “Functional Groups”
Alcohols- Contain an oxygen atom with a single bond to a hydrogen
Carboxylic Acid- most weak acids are this; exist as an acid and its corresponding base
Ketones/Aldehydes- Chemistry of the human body
Esters- Consist of a carbon double bonded to one oxygen and single bonded to another
Amines- Consist of a nitrogen atom with single bonds to one or more carbon atoms

7. Isomers A chemical formula with several different Lewis Structures
Each structure can contain different physical properties
Consider Pentane, C5H12
Neopentane has a BP of +9C
N-Pentane has a BP of +36C
Isopentane has a BP of +28C

9. Geometric Isomers
Two molecules with identical connectivity and separate geometries
Cis- “Same Sided” (a)
Trans- “Opposites” (b)

10. AP QUESTION 1
Dehydration of 3-hexanol yields a mixture of four isomers each with the molecular formula C6H12. Draw structures of the four isomers and name each of them.

11. Answer
1) trans-3hexene
2) cis-3hexene
3) trans-2hexene
4) cis-2hexene

12. Explanations
1) The molecule C6H12 is an alkene having the formula CnH2n; containing one double bond
2) Draw a double bond between the carbon 3 atoms to form the first
3) It is a geometric isomer, so there exists a Cis and a Trans
4) Draw another double bond on the carbon 2 atom to form the second isomer
5) Repeat step 3

13. AP QUESTION 2
Consider the hydrocarbon pentane, C5H12(molar mass g ).
A) Write the balanced equation for the combustion of pentane to yieldcarbon dioxide and water.
B) What volume of dry carbon dioxide, measured at 25 C and 785 mmHg, will result from the complete combustion of 2.50 g pentane?
C) The complete combustion of 5.00 g of pentane releases 243 kJ of heat.On the basis of this information, calculate the value of ∆H for the completecombustion of one mole of pentane.
D) Under identical conditions, a sample of an unknown gas effuses into avacuum at twice the rate that a sample of pentane gas effuses. Calculate themolar mass of the unknown gas.
E) The structural formula of one isomer of pentane is shown below.Draw the structural formulas for the other two isomers of pentane. Be sureto include all atoms of hydrogen and carbon in your structures.

14. Answers and Explanation (A)
A) C5H12 + 8O2 yields 5CO2 + 6H2O
Fairly obvious
Simple equation and balancing
Combustion- include O2

15. Answers and Explanation (B)
B) 2.5gC5H12 * 1 mol C5H12 / g C5H12 yields mol C5H12
Basic conversion to moles for Pentane, the molar mass is given so take the grams and multiply it by mols/grams
0.0347molC5H12 * 5molCO2/1molC5H12 yields 0.173molCO2
Since you need the volume of carbon dioxide, use the formula from part A to ratio 5:1 to find moles of CO2 from moles of Pentane

16. Answers and Explanation (B) ii
Pv=NRT
Use it
0.137molCO2 x l*atm/mol/K * 298K / (785/760) yields 4.10Liter CO2
0.137 was found in Part B
is the constant for R
298 K is standard, C (25C given)
785/760 is the pressure

17. Answers and Explanation (C)
5.00g C5H12 * 1 mol C5H12 / 72.15g C5H12 = mol C5H12
When in doubt, find the moles! Easiest thing to do… take the grams given and find the moles of Pentane
243 KJ / mol C5H12 = 3.51 x 10^3 KJ/Mol, ∆H would be –3.51x10^3 KJ/Mol
Take the KJ given and divide it by the moles of C5H12 you found
Take the negative of your answer because this reaction is exothermic

18. Answers and Explanations (D)
Rate unknown = 2 * rate C5H12
Relationship between rate and velocity needs to be recognized
Rate unknown / rate C5H12 = SqrRoot (72.15g mol^-1 C5H12 / MM)
Formula to be used, recognize rate of unknown over rate of pentane is the square root of the molar mass

19. Answers and Explanations (D) ii
2 * rate C5H12/ rate C5H12 = SqrRoot (72.15g mol^-1 C5H12 / MM)
In this way, we can cancel out “rate C5H12” for the top and bottom; if we substitute this in
2 = SqrRoot (72.15g mol^-1 C5H12 / MM)
Solve
4 = (72.15g mol^-1 C5H12 / MM)
Solve for MM
MM = 72.15g mol^-1 C5H12 / 4 = 18 g mol

20. Answer and Explanation (E)

21. Answer and Explanation (E) ii
Using Chem-draw for first structure, ms Paint for the second- 3 dimensional structures were present
3 Dimensional structures were not mentioned, so the first one can be accepted if it has the following structure:
CH3-CH-CH2-CH3 |
CH3

22. Finally, The END!