Organic Chemistry

By Alex Chan and Billy Lordi

NomenclatureHydrocarbons- 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

Aromatic Hydrocarbons Defined by a ring

structure Usually a six-

member ring with alternating single and double bonds

Volatile

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

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

Geometric Isomers Two molecules with

identical connectivity and separate geometries Cis- “Same Sided”

(a) Trans- “Opposites”

(b)

AP QUESTION 1Dehydration 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.

Answer

1) trans-3hexene2) cis-3hexene3) trans-2hexene4) cis-2hexene

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

AP QUESTION 2 Consider the hydrocarbon pentane, C5H12(molar mass 72.15 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.

Answers and Explanation (A)A) C5H12 + 8O2 yields 5CO2 + 6H2O

Fairly obvious Simple equation and balancing Combustion- include O2

Answers and Explanation (B) B) 2.5gC5H12 * 1 mol C5H12 / 72.15 g

C5H12 yields 0.0347 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

Answers and Explanation (B) iiPv=NRT

Use it0.137molCO2 x 0.0821 l*atm/mol/K *

298K / (785/760) yields 4.10Liter CO2 0.137 was found in Part B 0.0821 is the constant for R 298 K is standard, 273+25C (25C given) 785/760 is the pressure

Answers and Explanation (C) 5.00g C5H12 * 1 mol C5H12 / 72.15g C5H12

= 0.0693mol C5H12 When in doubt, find the moles! Easiest thing to

do… take the grams given and find the moles of Pentane

243 KJ / 0.0693mol 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

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

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

Answer and Explanation (E)

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