Transcript Organic Chemistry

Organic Chemistry
By Alex Chan and Billy Lordi
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
Aromatic Hydrocarbons

Defined by a ring
structure
 Usually a sixmember 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 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.
Answer
 1)
trans-3hexene
 2) cis-3hexene
 3) trans-2hexene
 4) 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)
ii
 Pv=NRT
 Use
it
 0.137molCO2
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)


2 = SqrRoot (72.15g mol^-1 C5H12 / MM)


Solve
4 = (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
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