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A tidy laboratory means a lazy chemist ^^
Section 2
Dr. B.E. McCarry
Course Information:
We re going to cover the first 4 chapters (14-17) and then have our midterm.
Dr. J.C. Laundry is the course coordinator (ABB/121) Ext. 22485
Friday February 2nd is the first test
Week A groups start Jan 8. / Week B groups start Jan. 15
Some Chem Cheer =)
2007
In case you guys are studying for midterms and are stressed, here’s a joke to lighten the mood:
Two hydrogen atoms bumped into each other.
One said: "Why do you look so sad?"
The other responded: "I lost an electron."
Concerned, One asked "Are you sure?"
The other replied "I'm positive."
Lame yes, but I know you smirked =P
WASIM’SINTRO TO CHEMISTRY II[1AA3] NOTES
contents:
14: chemical kinetics 2
15: equilibrium 31
16: acid + base 48
11: chemical bonding 72
26: organic chemistry 82
chem
1aa3
a. reaction ratesRate or speed is something that happens in a unit of time. Reaction rates are defined as: the change in concentration of reactants as a function of time. Going back to Calculus, if you remember ^^, as the change in time becomes extremely small, ie as the change in time approaches 0, it is considered the
instantaneous, ie
dxdt rate at time t
Example:
2NO + O2 −→ 2NO2
For every mole of O2 consumed, there are 2 moles of NO consumed and 2 moles of NO2 produced.
Relating this back rate laws, the rate of O2 consumption = 1/2 NO consumption.
Rates are expressed on a per mole basis. At anytime t, we can express the reaction rate in the follow-ing way:
Note rates are considered going “forward” for that reason the reaction rate of O2 and NO are nega-tive, ie they are decreasing (refer to graph). The following is the rate at any time:
Reaction rate in terms of [O2] = −d(O2)
dt = - slope of plot of O2 vs t at time t
Reaction rate in terms of [NO] = −d(NO)
dt = - slope of plot of NO vs t at time t
Reaction rate in terms of [NO2] = +d(NO)
dt = + slope of plot of NO vs t at time t
At t = 0:
Initial rate of reaction for NO:
NO = − 12 (d(NO)
dt ) = −(d(O2)dt )
For the general reaction:
aB + bB −→ cC + dDThe reaction rate is equal to:
= − 1a
d(A)dt = − 1
bd(B)
dt = − 1c
d(C)dt = − 1
dd(D)
dt
Rates are always expressed in terms of a loss of reactants (negative sign needed) or formation of products.
Net Reaction Rate = Forward Reaction Rate - Backward Reaction Rate
The four main factors that effect the the rate of reaction are:
1. Temperature
2. Concentration reactant (surface area)
3. The presence of a catalyst (or inhibitor)
4. The nature of the chemical reaction. ie what the chemical is exactly. Some chemical react more vigorously than others
Remember, the initial rate of reaction is the rate of reaction at t = 0. In the real world scenario, it may not always be possible to determine the rate exactly at t = 0. In this case you can measure the con-centration as soon as possible after mixing. The rate of reaction at this point will essentially be the same as t = 0, as the rate does not increase that much in such a short time interval. For instance, in the decomposition of hydrogen peroxide, the rate of reaction is actually the very close for the first 200 seconds.
For a comparison on both methods, if you drew a tangent at t = 0, the rate law for the decomposition of hydrogen peroxide may be 1.71 x 10-3 M/s , whereas if you took the average for the first 200 sec-onds you may have found the rate to be 1.6 x 10-3 M/s. As you can see the rate is not that far part. Between the two methods, taking the tangent at t = 0 is obviously more accurate, but the second method also works. If using the first method, just make sure that you draw the tangent properly ^^.
b. rate lawsRate laws are experimentally determined equations which describe the relationship between reactant concentrations and reaction rates.
General Reaction Equation:
aA + bB −→ cC + dDQ: Can you predict the rate for a reaction based on the balanced equation? Or in other words can you predict how the concentration of A and B effect the rate?
A: No we cannot. We must determine the rate law experimentally.
However, from the stichometry, you can write down the general equation.
In general:
Rate ∝ [A]x[B]y
Rate Laws take the form:
− 1a
d(A)dt = k[A]x[B]y
[Note the negative sign is because d(a)/dt from the last section was defined as negative]
a is the stichometric coefficient
x and y are call the order of reactions. If x is equal to one, we say that the reaction is first order in A. If y is equal to 2, it is second order in B
x = order of the reaction with respect to reactant A
y = order of reaction with respect to reactant B
x+y = overall order of reaction (ie sum of individual reactant orders)
k is the proportionality constant. It relates the rate of reaction to reactant concentrations. The larger the value of k is, the faster a reaction goes. The value of k is dependent on the specific reaction, the presence of a catalyst and the temperature.
Remember that x and y are generally positive whole numbers. However in some cases it may be zero, fractional and/or negative. We don t have to worry about these though, because we re not going to be tested on them, but we should be aware that they exist though =)
Also remember that although a catalyst affects the value of k, you never include the components of the catalyst in the rate law.
The procedure for determining orders of reaction:
1. Find the initial rate for the series of reactions in which [A] is varied, [B] & T are held constant
2. From initial rate data, we determine exponent x which fits data as Rate ∝ [A]x
3. Repeat this procedure for every other reactant, ie B
Rate ∝ [B]y determine y
Are you wondering why we use initial rates to determine the order of the reaction? Scientists use the initial rates so they don t have to follow the reaction to the end. This is especially true for very slow reactions. However you don t specifically HAVE to use initial rates. If you have a graph of concentra-tion vs. time you can arbitrarily set any point to t = 0 and make it initial technically. However what s important though is that the rates and concentrations must be measured at same time for all species.
Example: What is the rate law in terms of loss of NO for the reaction below, given the following data.
2NO(g) + H2(g) −→ N2O(g) + H2O
Experiment # [NO] [H2] Initial Rate
1 0.3 M 0.70 M 2.0
2 0.15 M 0.70 M 0.5
3 0.30 M 1.40 M 4.0
You would use Experiment 1 and 2 to find NO, because if you notice in those to experiments, the concentration of NO changes, however H2 stays constant. Conversely, to find H2, you would use Ex-periment 1 and 3.
Rate Law has the form:
Rate = − 12
d(NO)dt = k[NO]x[H2]y
Rate dependance on [NO]:
Rate1 = k[NO]x1
Rate2 = k[NO]x2
You divide the above equations and you get:
Rate1Rate2
= k[NO]x1k[NO]x2
The k s cancel out and become:
Rate1Rate2
= ( [NO]1[NO]2
)x
Then you plug in numbers. In this case we decided to use experiment 1 and 2, and so you just enter the data into their respective places. You get the equation:
2.00.5 = ( 0.3
0.15 )x
So you just solve for x, and you get:
x = 2The same procedure is followed for finding the rate dependance on H2
Rate dependance on [H2]:
Rate1 = k[H2]y1
Rate2 = k[H2]y2
Divide both equations:
Rate1Rate2
= k[H2]y1
k[H2]y2
The k s cancel out and becomes:
Rate1Rate2
= ( [H2]1[H2]2
)y
Like you did with NO, you plug in the respective experimental data
2.00.4 = ( 0.7
1.4 )y
Then solve for y:
y = 1After you have found the exponents for all species, you plug the values into the rate law equation. For this reaction the rate law is:
− 12
d[NO]dt = k[NO]2[H2]
In this experiment the exponents happen to be the same as the coefficients in the stichometric for-mula. But don t get too ahead of your self, you were just lucky =P. You have to do the whole working out every time.
The important thing to understand in this example is that when the initial concentration of the a par-ticular reactant is doubled, the initial reaction rate doubles in first order reactions and quadruples in second order reaction.
In relation to this a zero order reactant would have no effect on the initial rate of reaction and a third order reactant would increase the rate of reaction 8 fold.
c. first order reactions: half livesFirst order reactions are those that have a rate law in which the sum of the exponents (x + y) is equal to 1. A very common first order reaction is a decomposition reaction.
First order reactions / Half Lives:
Example:
N2O5 −→ 2NO2 + 12O2
Rate Law (remember this was experimentally determined):
−d(N2O5)dt = k[N2O5]
This is a first order decomposition of N2O5. This reaction is first order in N2O5 and first order overall
Let c0 = [N2O5] at time t = 0
Let c = [N2O5] at any time t
The rate law becomes: −dc
dt = kc (differential form of rate law)
This can be written as:
dcc = −kdt
∫ c
c0
dcc = −k
∫ t
t0dt
ln c]cc0= −kt
ln c − ln c0 = −kt
ln c = −kt + ln c0
If you look closely, the equation is actually in the form y = mx + b, where -k is the slope and ln c0 is the y intercept.
How do you test for first order reactions? The plot of ln c vs. t is always linear for first order reactions.
Wondering how to use the integrated rate law? Take the following example:
H2O2 (aq) has an initial concentration of 2.32 M. What will the concentration of H2O2 be at t = 1200? Use k = 7.30 x 10-4 s-1
ln[H2O2] = −kt + ln[H2O2]0= −(7.30 ∗ 10−4s−1x1200s) + ln 2.32
= −0.034
[H2O2]t = e−0.034 = 0.967MMany body and biological processes follow first order kinetics. An example in biology is the removal of chemical waste. However other applications include physics and even astronomy.
The reason why scientists use a straight line instead curved is because it s much easier to do tests on straight lines. Sure there are methods for curves such as regression and all that, but none are as ac-curate, or as easy as testing for straight lines.
Half Life of first order reactions:
The notion of half life only comes out of first order reactions. Half life is defined as the time required for the concentration of a reactant to decrease 2-fold (or concentration of a product to increase 2-fold). This is only true for first order reactions.
When:
t = 0 c = c0When:
t = t 12
c = 12c0
When:
t = 2 ∗ t 12
c = 12 ( 1
2c0) = 14c0
When:
t = 3 ∗ t 12 c = 1
2 ( 14c0) = 1
8c0 = ( 12 )3c0
In general:
t = n ∗ t 12
c = ( 12 )nc0
The relationship between t1/2 & k:
What s the relationship between k and the thing we have here?
when
t = t 12
c = 12c0
ln( cc0
) = −kt
ln(12 c0
c0) = −kt 1
2
ln(12 ) = −kt 1
2
−ln(2) = −kt 12
t 12
= ln(2)k
Thus if we know k we can calculate t1/2 and vice versa.
A graphical representation of the concept of half-lives
The whole notion of a half life is that the concentration to half takes the same amount of time.
d. collision theory and reaction ratesWe now want to talk about temperature effects. In order for molecules to react, collision theory pre-dicts that :
1. molecules must collide with proper alignment (with respect to each other) for reaction to occur.
2. The molecules must have sufficient energy to continue through the reaction. (i.e. must have more than minimum energy)
An example of proper alignment in collision:
A. If the collision is correctly aligned, and there is sufficient energy, a reaction will take place
B. If the collision is not aligned, the reaction does not take place.
According to the collision theory of gases, as temperature increases, the kinetic energy of molecules increases and the rate of collisions between molecules increases.
In gas phase, collision rate ∝ T1/2
[Note in case you have a brain block this second, reminder, that the power of 1/2 means square root]
What this basically means is that, when T increases from typical room temperature 300K to 310 K (a 10 degree increase). What s noted is that
1. Collision rates in gas phase increases by only about 2%. So yes the collisions increase, but be-cause of the square root sign, the collisions only increase by a little bit.
2) Reaction rates however are known to increase by 50% to 300%, typically about 200%
as T increases, the number of “effective” collisions increases much more rapidly than the total number of collisions. If you think about it, although the number of collisions don t increase much, the amount of energy available is much more. and therefore the amount of effective collisions are much more.
It was probably in the late 19th century when Maxwell-Boltzman got together and realized that the distribution of kinetic energies had the following form:
[See figure 14-8 in the text]
Comments:
1. Molecules exhibit broad range of Kinetic energy.
2. Area under the curve is ∝ to the number of molecules in the system
3. Areas under parts of curve ∝ number of molecules in it.
4. In order to react, molecules must have kinetic energies greater or equal to some minimum E value
5. As T increase, # of molecules .. Kinetic energies >= E minimum, increases much more rapidly than total # of collisions.
[DEMO]
This demonstration allows you to see and hear a chemical reaction as it produces from start to finish down a glass tube.
The tube is filled with nitric oxide, NO, a toxic gas
Observations:
Professor put some sort of liquid into the tube. Then he unscrewed the cap of the tube to release pressure commenting that the ends may blow off off.
Just for memory, this was the demo, in which that really loud and bright blue flame screamed across the tube =). He also did it on the last day of class, my goodness, such a cool demo =D
e. temperature dependance of k [arrhenius equation]
Arrhenius s equation is:
k = Ae−EaRT
Where Ea = activation energy for the reaction (Emin)
T = absolute temperature (ie in Kelvin)
R = gas constant (8.3145 J/mol/K)
A = pre-exponential factor.
The term A contains terms to account for:
1. Collision geometries
2. Collision frequencies ( ∝ T1/2)
3. Masses of colliding species
There is little dependance of T on A.
Methods to determine Ea s:
You can either do this graphically or mathematically
[see lab experiment]
k = Ae−EaRT
If we ln both sides we get:
ln k = lnA − EaRT
and then we simply take this equation, and put it in the form y = mx + b, to form
ln k = (−EaR ) 1
T + lnA
−Ea
R is the m and
1T is the x ln A is the b (the intercept) and ln k is the y
Procedure
1. Determine k s experimentally at various temperatures
2. Plot ln k vs 1/T
This is the way to find k experimentally.
2. The second way is the calculation method.
First you must determine k1 & k2 at temperatures t1 and t2. Then find Ea by calculation
ln k2 = lnA − EaRT2
ln k1 = lnA − EaRT1
Then we subtract both equations. Note: We have to assume that ln A does not change over the tem-perature range of interest. Because we made this assumption the ln As in both equations cancel out and so you re left with:
ln k2 − ln k1 = EaRT2
− EaRT2
You can factor out common terms and then you end up with :
ln(k2k1
) = (Ea
R )( 1T1
− 1T2
)Now you can do some more rearranging and you can solve for Ea:
Ea =R ln(
k2k1
)1
T1− 1
T2We have 5 variables in this equations. The 5 variables must know any 4 to calculate the 5th. Often rate rations are used:
Rate2Rate1
∝ k2k1
Example:
CH3CH2 − I + OH− −→ CH3CH2 − OH + I−
Q: On raising temperature from 25 degrees C to 60 degrees C, the reaction rate increases 133-fold. Calculate Ea
A: Which means:
Rate(60◦C)Rate(25◦C) = 133 = k2[CH3CH2−I][OH−]
k1[CH3CH2−I][OH−] = k2k1
the Iodo-ethanes cancel out, as do the OH- and so basically you end up with:
133 = k2k1
You know the equation for Ea (we defined it before).
Ea =R ln(
k2k1
)1
T1− 1
T2
As you can see, in the previous step you calculated k2/k1 , which can be plugged into the equation, along with the remaining variables:
Ea = 8.315 ln(133)1
(25+273)− 1(60+273)
Ea = 115 kJmol
Note: Be careful of unit: J/mole vs kJ/mol
Note: The temperature is in absolute values (ie in Kelvin)
On tests and exams you may be asked variants of this question. You could be given something like: “Given the activation energy, how much will the rate increase with an increase of temperature from so to so”
How does k vary with temperature? and with activation energy?
We know that:
k = Ae−EaRT
k ∝ e−1T
k ∝ e−Ea
Therefore
1) As T increase, 1/T decreases, e(-1/T) increases
2) As Ea increases (e-Ea decreases) and k decreases
f. reaction progress diagramsWhat are they? They are a way of diagraming energetic reactions as they progress with time. But most of the time we don t use time, and instead something else that is measurable.
They are plots of enthalpy ( H) vs a “reaction coordinate” - eg. degree of bond breaking, or bond for-mation ect.
see fig: 14-10
1. Single Step Mechanism
A + B −→ C Endothermic by 25Kj/mole
2. Two-step mechanism:
Step 1: A + B −→ I
Step 2: 2I −→ C
In this mechanism, A + B reacts to form I. Then I goes onto another reaction to form C. I is called a reaction intermediate. It is a stable, but reactive, species that is formed in one step of a reaction mechanism & consumed in a later step.
The enthalpy difference between reactants and the highest point on a reaction diagram is the overall Ea for the reaction. It is this Ea that is determined by calculation and by find the slop of the ln k vs 1/T graph.
Transition States are defined as extremely short lived. They are also called an activated complex (a transient, short-lived association of reacting species). They shouldn t be confused with intermediates.
Transition states are hills and intermediates are valleys.
In this example, step one is the slowest step. It requires the most activation energy and is called the Rate Determining step.
g. reaction mechanismsThe real crux of kinetics is to get a mechanism.
A reaction mechanism is a set elementary steps or reactions by which an overall reaction occurs. This is called a reaction mechanism.
The overall reaction is the sum of the individual elementary steps.
For any Elementary Steps you can write down the rate law for elementary steps BUT NOT for the overall reaction.
So for any elementary step, we can immediately write down a rate law for that elementary step. How-ever, you CANNOT write a rate law based on the overall stichometric reaction.
The molecularity of any of the elementary steps, is equal to the number of reacting species in that step.
This is made more clear below:
When you talk about the molecularity of a step, you can start to associate the step as being a unimolecual, bimolecular or termolecular step
Examples of elementary steps and their molecularities:
1. First order step [unimolecular step] Molecularity = 1
O3 −→ O2 + O Rate = k[O3]
2. Second order step [bimolecular step] Molecularity = 2
O + O3 −→ 2O2 Rate = k[O][O2]
3. Third Order Step [termolecular step] Molecularity = 3
O + O2 + N2 −→ O3 + N2
Rate = k[O][O2][N2]
Note that termolecular steps are usually very rare.
Stichiometry of the overall reaction is related to the sum of the individual elementary steps.
Rate of Overall reaction: equals the rate of the slowest elementary step. It is also called the rate de-termining step.
Rate law of overall reaction - equals to the rate law of for the slowest step [rate determining step ]
IMPORTANT: It is possible to PREDICT the rate law for a reaction based on a proposed mechanism and is identified by the slowest step or the rate determining step.
Example: The rate law for the reaction:
2NO2 + F2 −→ 2NO2Fis first order in NO2 & F2 . Derive the rate law based on following mechanism & show whether it is consistent of inconsistent with observed rate
Rate Law:
*From experiments* Rate = k[NO2][F2]
Mechanism (proposed)
1. NO2 + F2 −→ NO2F + F ( slow )
2. NO2 + F −→ NO2F (fast)
Rate determining step (RDS) = slower step = step 1
RDS = Rate overall reaction = k1[NO2][F2]
Rate law derived from the proposed mechanism is identical to observed rate law, and therefore the proposed mechanism is consistent with the observed rate law.
h. reaction mechanisms involving equilibria
For some reaction mechanisms the reactive intermediate species are produced (& sometimes con-sumed) in equilibrium steps in the mechanism. If such intermediates appear in rate laws, then the concentration of these species must be expressed in terms of reactants or products.
Basically, you re going to get a situation where the reaction intermediate is going to appear in the rate equation, which is a problem
Example:
2NO + O2 −→ 2NO2
Observed Rate Law: Rate = k[NO2]2[O2]
Proposed mechanism:
1. NO + O2 � NO3 (fast)
2. NO3 + NO � 2NO2 (slow)
Overall reaction rate = rate determining step = rate step 2
Rate = k2[NO3][NO]
The problem is that NO3 is contained in the proposed rate law. But because NO3 is an intermediate there s no way to vary it or measure it experimentally. For this reason you have to substitute the NO3 in terms of reactants and/or products. In the above example, the equation for the production of NO3 can be written from the first step
K1 = [NO3][NO][O2]
From this, you can solve for [NO3] and get:
[NO3] = K1[NO][O2]You can then substitute NO3 in the rate determining step found above, for what we solved here. You then get:
Rate = k2(K1[NO][O2])[NO]When you solve it out you get:
Rate = k2K1[NO]2[O2]After this you can see that the proposed rate law is consistent with the observed rate law. Therefore the proposed rate law is a candidate (poo only a candidate =( ) for the true chemical reaction mecha-nism.
i. catalysisA catalyst is a substance which may increase or decrease the rates of reactions without itself being changed or consumed
There are two types of catalyst:
1. Homogenous - reactants and catalysts in same phases
2H2O2(aq)I−(aq)−−−−→ 2H2O(l)O2(g)
2. Heterogeneous: Reactants and catalysts in different phases
2SO2(g) + O2(g)
V2O5(s)−−−−−→ +2SO3(g)
Catalysts can dramatically increase the rate of reaction. Catalysts provide a new mechanistic path-way for reaction which has a lower Ea than the uncatalyzed reaction.
[Look at diagram below]
Blue is the catalyzed reaction while the green is the uncatalyzed reaction.
Sometimes when you catalyze a reaction, you may turn the reaction into a two step reaction. i.e. when you look at the graph, there will be two hills plus a valley in between them. There s no set rule on what the catalyst does, so this is possible.
j. chain reactionsA special class of reactions called chain reaction. Self contained, and self propagated reactions.
It s a “self sustaining” reaction in which a reaction intermediate species is:
1. Consumed in an early mechanistic step, and then
2. Regenerated in a later step.
Since the regenerated reaction intermediate species are so reactive, it fuels the decomposition of many reactant molecules. For a real life example, the danger of the chlorflurocarbons (CFCs) in the upper atmosphere is that one chlorine atom can systematically remove about 1000 other atoms.
Example:
H2 + Cl2 −→ 2HCl
Mechanism:
The first step in a chain reaction is the initiation step
1. Cl2hu−−→ Cl + Cl
Notice that the Cl atom is 7 valence species, ie it is reactive. The following steps are what are called the propagation steps. The chain reaction cycles between these steps, where the product of one re-action fuels another reaction
2. Cl + H2 −→ HCl + H
3. H + Cl2 −→ HCl + Cl
Notice that the H atom is also reactive because of it non empty valence shell.
Cl atoms are the reactive intermediate. The Cl reacts with H2 giving rise to an H atom. The H atom is another reaction intermediate. In all chain reactions, you need a initiating step, and a propagation step. In this case 1. is the initiation step and 2. and 3. are the propagation steps.
Net Reaction - Steps 2 and step 3:
H2 + Cl2 −→ 2HCl
The termination step(s) are those that remove the reaction intermediate. In the case of this reaction:
4. Cl + Cl −→ Cl2
5. H + Cl −→ HCl
6. H + H −→ H − H
Reactive intermediates:
Cl atom and H atoms are reactive intermediates in the reaction mechanism. Cl atom is said to “propagate” the chain reaction. The Cl atom is consumed in step 2 and regenerated in step 3.
Applied Example:
Allow 1 mole of H2 to react with 1 mole Cl2
Result: 2 moles of HCl produced
k. little k and big KIs there a relationship between k & K?
Example:
N2O4 � 2NO2
at equilibrium the rate forward reaction = rate reverse reaction
[Note kf = kforward and kr = krevese]
1. Rate forward reaction = kf[N2O4]
2. Rate reverse reaction = kr[NO2]2
Therefore if you make all these into an equation:
kf [N2O4] = kr[NO2]2
kf
kr= [NO2]
2
[N2O4]= K
∴ K = kf
kr
Therefore the equation constant directly equal to the ratio the of rate constant
l. radionuclide datingRadionuclide dating is an application of first order kinetics, specifically first order decay reactions.
In the example with Carbon, Carbon has an unstable isotope called Carbon-14. In living organisms, the ratio of Carbon-14 to Carbon-12 (the regular carbon) is the same. However when the organism dies, the Carbon-14 starts to decay, whereas the Carbon-12 stays constant. The half life of Carbon-14 is 5,700 years. One can look at the ratio between Carbon-12 and Carbon-14 in the dead organism and the ratio of Carbon-12 and Carbon-14 in the living organism to calculate how many half lives the Carbon-14 has undergone. Using this one can calculate how old an organism is .
However because the half life of Carbon-14 is only 5,700 years, it is only reliable to calculate ages of around 60,000 years. However there are other elements one can use, such as the isotope of Potas-sium, Pottassium-13. The half life of Potassium-13 is 1.3 million years.
m. pseudo first-order kineticsSecond order reactions are rarely studied under “second order” conditions, ie [A] approximately equal to [B]
For the general reaction
A + B −→ CIf [A] >> [B] (much larger than), then [A] changes very little as [B] decreases to a large extent.
For instance:
[A] = 50 [B]. Reaction kinetics appear to be first order in B under these conditions since [A] is a considered a constant.
Rate = k [A] [B] k [B]
(Where k = k[A])
The kinetics of such pseudo first order reactions are greatly simplified. Such reactions are character-ized apparent “half lives”
n. enzyme kineticsThe general equation for an enzyme is:
Substrateenzyme−−−−−−→
(catalyst)Product
If the amount of enzyme is kept constant and the concentration of the substrate is increased, you get a plot like the y = x , ie the rate of reaction increases, but as you add more and more substrate, the rate does not increase as rapidly. It is said to approach a maximum rate (velocity)
The entire reaction consisting of the enzyme is:
E + S � ESk2−→ E + P
Where ES is the enzyme-substrate complex.
However for this reaction, you are making some assumptions, such as:
1. The enzyme exists only as E and ES
2. ES is in true equilibrium with E + S
3. Rate of the reverse reaction (P to S) is negligible in early stages
If all these assumptions are held, then the Rate becomes:
Rate = v = d[P ]dt = k2[ES]
and
[E]TOT = [E] + [ES]When [S] is high
[E] ≈ [ES]
and the maximum velocity is:
V = k2[E]TOT
At this point, the enzyme is said to be saturated. The equilibrium between E, S and ES is described by the equilibrium constant Km (the Michaelis constant). The Michaelis constant states that:
Km = [E][S][ES] and
[S] = Km( [ES][E] )
When the enzyme is “half-occupied” by S, [E] is said to equal [ES]. At that point [S] = Km
Therefore Km is directly equal to the substrate concentration at which the reaction velocity is half maximal (that is that v = 1/2 V)
when [S] << Km - Rate or velocity increases linearly with increasing [S]. ie v∝[S]
When [S] >> Km, v =V and v is constant. Therefore [S] does not affect v
a. types of equilibriaThere are two types of equilibria
1. Static
meaning that opposing forces are balanced
For instance: tug of war equilibria
2. Dynamic
opposing forces are taking place at some rate
that is that when forward reaction is directly equal to reverse reaction there is no net change.
physical and chemical equilibria exist
Example:
You put a sample of I2(s) in a closed vessel. What happens?
a) I2(s) −→ I2(g)Molecules in I2 (s) escape to form I2 (g)
b)I2(g) −→ I2(s)Molecules in I2 (g) condense to form I2 (s)
Thus at equilibrium, these two processes occur at the same rate, keeping the amount of I2 (g) and I2 (s) constant until the equilibrium is disrupted.
I2(s) � I2(g)Dynamic Physical Equilibrium involves materials undergoing changes of state at constant T (like the above example)
Dynamic Chemical Equilibrium involves different chemical species reacting with each other. At equi-librium, the concentrations of products and reactants do not change. (Given that T is constant)
Name 5 important things related to chemical equilibrium
1. K equilibrium constant
2. Temperature
3. Le Chatelier s principal
4. Concentration of reactant / products
5. G
Name some practical applications of chemical equilibria:
1. Production of Ammonia
NH3 � 3H2 + N2
2. Hemoglobin
Hb + O2 � Hb·O2
Hb·O2 + O2 � Hb·(O2)2
Hb·(O2)2 + O2 � Hb·(O2)3
Hb·(O2)3 + O2 � Hb·(O2)4
b. the equilibrium constant KLets talk about the equilibrium constant, and how it changes. We ll talk about Kc (K in terms of con-centration) and Kp (K in terms of partial pressures) and then a relationship between these two
From the law of mass action, it follows that for the general equation:
aA + bB � cC + dDwe can write an expression to describe this equilibrium mathematically, in terms of stichiometric coef-ficients and concentrations.
k = [C]c[D]d
[A]a[B]b
Remember K is constant for constant T.
A minor note, that concentrations are in mol/L
Q. Can we predict where equilibria lie? ie on he side of the products or an side of reactions?
A. K is determined by G (Gibb s Free Energy)
ΔG = −RT lnKWhen looking at this equation, we really have 3 Cases:
K ln K G Reaction
Description
K >> 1 positive negative Reaction proceeds to product side (C & D) Reaction said to be spontaneous
1 0 0 Reaction approxi-mately half way
K << 1 negative positive Reaction lies on re-actant side (A & B) Reaction is said to be non spontaneous
If K is greater than one that means that the reactants are much bigger than the product, (numerator is much greater than the denominator) and visa versa. Therefore the reaction moves the respective di-rection to reach 1.
The reaction quotient Q and the equilibrium constant K
For any chemical reaction system at a given temperature, the equilibrium state is attained no matter how the reaction is conducted.
So you can come at it from the product side or reactant side, it doesn t matter
For any general chemical reaction
aA + bB � cC + dD
we can write an expression for Reaction Quotient Q where
Q = [C]c[D]d
[A]a[B]b
As the reaction proceeds in the direction given, value of Q changes continuously until the equilibrium is attained
At equilibrium Q = K
Q: When reactants are mixed, how can you predict whether a reaction will proceed in a forward or backward direction on its way to equilibrium?
A. Calculate the reaction quotient Q by substituting initial reactant concentrations and calculate Q.
Compare Q to K
Condition Explanation
Q = K System at Equilibrium - no net change. A G = 0 sce-nario (very rare)
Q < K We re on the reactant side. This is because there must be a greater number on the denominator than the nu-merator. We re going to have to increase the numerator relative to the denominator, ie move material from A and B to C and D
Reactants will be converted to products as the net reac-tion proceeds in forward di-rection
G = negative
Q > K Numerator is larger than the denominator. Products will be converted “back” to reactants or reaction said to be moving in the reverse direction
G = positive
When you talk about G vs reaction coordinate, the Grxn = -RT ln K. Think of the following analogy. I m holding my pen and I m about to drop it, it ll reach equilibrium on the floor. You want the G from the height the pen is dropped from, to the floor.
c. equilibria involving gasesYou can express gases that involve equilibria in gases in terms of partial pressures.
aA(g) + bB(g) � cC(g) + dD(g)The total pressure of the system is the sum of all the partial pressures. We talk about Pa, the notion of partial pressure of a
PA = P.PreasureofA = (#ofmolesgasA
total#molesofgasinsystem) ∗ Ptotal
Kp =Pc
c ∗ Pdd
Paa ∗ Pb
b
usually K is not equal to Kp, however K is directly equal to Kc
Relationship between K & Kp:
n gases = product gas moles - reactant gas moles
Δngases = (c + d) − (a + b)Relationship between Concentration and Gas Law:
Pv = nRt
P = (nv )RT
and C = n
v
∴ C = PRT
Now if you take a reaction and want to write it in terms of concentration, you can actually convert everything
K = [C]c[D]d
[A]a[B]b= ( Pc
RT )c∗( PdRT )d
( PaRT )a∗( Pb
RT )b
K = (Pcc∗Pd
d
Paa∗Pb
b ) ∗ ( 1RT )
K =Pc
c ∗ Pdd
Paa ∗ Pb
b∗ ( 1
RT )c ∗ ( 1RT )d
( 1RT )a ∗ ( 1
RT )b
However you know from above that the left part of the equation just equals to Kp
K = Kp ∗ ( 1RT )(c+d)
( 1RT )(a+b)
K = Kp ∗ (1
RT)(c+d)−(a+b)
Therefore:
K = Kp ∗ (1
RT)Δngases
Which means that K = Kp only if n gases = 0. Otherwise K does not equal Kp
ΔG = −RT lnKlnK = −(
1T
)(ΔG
R)
K = e−( 1T ) + e(ΔG
R )
This shows that the relationship between K and T is non-linear.
d. homogenous and heterogeneous equilibria
There are two types of equilibria.
Homogenous Equilibria:
1) All chemical species are in the same phases
Example: all gas or all species in solution
Co(H2O)62+ + 4Cl− � CoCl4
2− + 6H2OPink Blue
All the species here are in aqueous solution, and we can write an equilibrium equation for it.
K =[CoCl4
2−][H2O]6
[Co(H2O)62+[Cl−]4
However the water is a constant, so we can just remove it. For those wondering how you can just re-move water from the equation, even if it s “used” in the equation. The reason is that the volume of wa-ter stays constant in the solution. Water dilutes molecules. When water is used in an equation, you don t realize it but the concentrations of the molecules change slightly (it has to since the volume of the water decreased). If water was in the equilibrium equation, the gain in water and the change in concentration of the other molecules actually balance out. The effect is that water stays constant, as though it had no role in the reaction. Therefore we can just leave it out.
K =[CoCl4
2−][Co(H2O)6
2+[Cl−]4
Heterogeneous equilibria:
In this case more than one phase is present
Example:
Li2O(s) + CO2(g) � LiCO3(s)We can also write the equilibrium equation for this. However note that for solids and liquids, the con-centration in mol/L is constant, and therefore do not have to be included in the equilibrium expression.
K =1
[CO2]
e. le chatelier s principalLe Chatelier s Principal describes factors that affect equilibria. His principal states that if stress is im-posed on a system at equilibrium, the system will shift its equilibrium position in a direction that re-duces stress.
The term “stress” can mean change in:
- concentration (reactant or product)
- pressure
- volume
- temperature
1. Change in concentration
Example:
Taking the equation
Co(H2O)62+ + 4Cl− � CoCl4
2− + 6H2OThe following constitutes as a change concentration, and the resultant effect
Changes in Conc. Effect on Equilibrium Position
Add NaCl (Na+ is a spectator) Equilibrium shifts to CoCl4 side
Add H2O (l) Equilibrium shifts to Co(H2O)6 due to dilution. All concentra-tions are affected equally however [Cl]- is raised to the 4th power and therefore its dilution effect is huge
Add AgNO3 (aq) Equilibrium shifts away from CoCl4 side to the Cl- side
Therefore you can conclude that:
1. If a reactant or product is added to the equilibrium system, the system shifts away from the added material.
2. If a reactant or product is removed from the system at equilibrium, the equilibrium will shift towards the removed material.
2. Changes in Pressure (or Volume)
Changes in system allow you to alter equilibrium positions.
There are really only three things you can do to change the pressure of a reaction involving gas com-ponents.
a) You can add or remove reactants or products.
b) You can change the volume of the container or apply an external pressure. For example squeeze a balloon, compress a piston.
c) Add a gas that is not involved in the reaction (possibly an inert gas)
Which of these factors will change the equilibria?
Gaseous equilibria will be affected by a) and possibly b)
Gaseous equilibria will be unaffected by c)
This can be illustrated in the following example:
H2 + I2 � 2HIYou add in Ar gas. You know that
Ptotal =∑
Px
If you write out the equilibrium expression of the reaction
Kp =(PHI)2
(PH2) ∗ (PI2)As you can see when you enter the partial pressures of the species, there is no place to enter Ar, and therefore although the total pressure changes, the partial pressures of the other species remain the same, and therefore equilibrium is unaffected.
3. Changes in Temperature
ΔG = −RT lnKFrom this equation you know that K changes with T
Illustrated Example:
We have an equation:
N2(g) + 3H2 � 2NH3(g) (Exothermic by 92kJ)
Rewrite the reaction with “energy” as a product
N2(g) + 3H2 � 2NH3(g) + 92 kJmole
The increase in T causes equilibrium to move to the left hand side of the equation , that is toward H2 and N2
Decrease in T causes equilibrium to shift to the right hand side, ie to NH3 side.
4. Addition of a catalyst:
Does not alter K and does NOT shift equilibrium position. However addition of a catalyst will increase rate of attaining equilibrium.
f. an approach to solving problemsThere s a generic approach to solving equilibrium problems, and it s pretty full proof
1. Write a balanced chemical equation. You might also want to make sure the reaction is going in the right direction.
2. Write the equilibrium equation for K
3. List the initial concentrations.
4. If needed calculate Q, compare Q to K and determine which way the reaction proceeds to attain K
5. Define changes needed to attain equilibrium and create an ICE table
6. Substitute equilibrium vales into equilibrium expression and solve the problem
Example: The formation of HF from its elements has K = 115 at given a given T. 3.0 moles of each component was placed in a 3.0 L vessel and allowed to reach equilibrium. Calculate concentrations of all gases at equilibrium
Answer:
1.H2(g) + F2(g) � 2HF (g)
2.
K =[HF ]2
[H2][F2]
3. Initial concentrations --->
C =3.0mole
3.0L= 1
mole
L
4.
Q =[HF ]2
[H2][F2]=
12
1 ∗ 1= 1
Therefore Q = 1, and we know that K = 115.
We want 1 to increase to 115. So the numerator has to be 115 times bigger. So the reaction has to shift to the HF side
5.
[H2] [F2] [HF]
a. Initial 1.0 M 1.0 M 1.0 M
b. Change - x - x + 2x
c. Final Equilibrium 1 - x 1 - x 1 + 2x
K =[1 + 2x]2
[1 − x]2= 115
You solve for x and find that
x = 0.764The final values for H2 and F2 will be 1 - x and HF will be 1 + 2x
Example 2:
Formation of HI (g) from H2 (g) & I2 (g) has K = 100. Initial pressures of HI, H2 & I2 were 0.5 atm, 0.01 atm and 0.005 atm respectively. Calculate equilibrium pressures of all species.
1.H2(g) + I2 � 2HI(g)
2.
K = 100 =[HI]2
[H2][I2]3. Initial pressures in system are given
4.
Qinitial =[HI]2
[H2][I2]=
(0.5)2
(0.01)(0.005)= 5000
Therefore Q >> K, and the reaction has to go towards the left (ie the H2 and I2 side)
5.
P[H2] P[I2] P[HI]
a. Initial 0.01 0.005 0.5
b. Change + x + x - 2x
c. Final Equilibrium 0.01 + x 0.005 + x 0.5 - 2x
6.
Solve:
K =[0.5 − 2x]2
[0.01 + x][0.005 + x]= 100
Solve for x and you get:
x = 0.0355 or -0.07119 atm. However you ignore the negative root
Therefore
PH2 = 0.01 + 0.0355 = 0.0455 atm
PI2 = 0.005 + 0.0355 = 0.0405 atm
PHI = 0.50 - 2(0.0355) = 0.429 atm
f. relative magnitudes of K and GWe know that G = -RT ln K. Because of the exponential relationship, you don t need a large change in G to have a huge change in K.
Simple Example:
A � B @ T = 25
K =(B)(A)
= 10
ΔG = −RT lnK = −(8.314J
mol∗ 298 ∗ ln 10)
= −5705J
mol
= −5.7kJ
molTherefore using this formula you can construct a table of values with K and G
K G (kJ/mol)
109 -51.3
106 -34.2
103 -17.1
102 -11.4
10 -5.7
2 -1.72
1 0
0.5 +1.72
0.1 +5.7
10-2 +11.4
10-3 +17.1
10-6 +34.2
10-9 +51.3
So what? What do these numbers mean? That s where the applications come in. Sometimes num-bers and magnitudes get confused. For example, to break a C-H bond you need 480KJ/mol, that s about 1072 in terms of K. Almost everything you heard about chemistry is in the bonds. The H-bond is typically given a value of 15KJ/mol. What s the G of a H-bond interaction?
A + B � AB“free and interacting species respectively”
Simultaneous equilibria.
You might have a series of equations where the product of one equilibrium might start another equilib-rium equation.
Depending on what you add, you may get a different effect:
or
Think of the Creb cycle. The molecules are in equilibrium with each other. There s one reaction how-ever in the cycle that is not reversible. This is because you lose a carboxyl group. The effect of this is that the equilibrium can only go in a clockwise direction.
The idea of acid and bases arose from ionization. Ionic molecules form ions in water. For instance HCl ionizes into H+ and Cl- ions. H+ is what makes a solution more acid and OH- ions is what makes a solution more basic.
However the problem with this theory was that some bases (such as NH3) have no OH- in their com-position, yet they are basic.
a. self-ionization of waterAcid/bases always cause the most grief =*(. Water reacts with itself. H2O is amphateric, meaning that it can act both as an acid and as a base. It produces H3O+ (which is the conjugate acid) and OH- which is the conjugate base.
H2O + H2O � H3O+ + OH−
K =[H3O
+][H2O]2
We know that the concentration of water is constant. and therefore
Kw = K[H2O]2 = [H3O+][OH−] = 10−14
@ 25˚C
What is the numerical value of [H2O]?
1000 gH2OL
18 gmol
= 55.5mol
LAs [H3O+] increases [OH-] decreases and visa vera.
In pure H2O the concentration of H3O+ and OH- are equal to 10-7
A long time ago someone made a notation to deal with very small numbers. The notation was called log
pH = − log10[H3O+]
pOH = − log10[OH−]
Kw = [H3O+][OH−] = 10−14
If you take the negative log of both sides of the equation you get:
pKw = pH + pOH = 14Change of 1 in log scale is a factor of 10 in concentration.
b. strong acids and bases in waterNote: What you re expected to know is the list of strong acids.
HBr + H2O −→ H3O+Br−
acid 1 base 1 acid 2 base 2
The definition of a strong acid is the dissociation of the molecule is complete. Ie there is complete (100%) dissociation.
HBr is a strong acid - it is full dissociated in H2O
Br- is a extremely weak base, ie it has no tendency to pick up a proton and go back to being an HBr. In this case no tendency is to remove H+ from H2O.
Therefore another definition of strong acid is that it s conjugate base has no tendency to pickup a pro-ton and go back.
The conjugate acid of H2O is H3O+
All these strong acids will all produce 1 Molar of H3O+
The strongest acid that can exist in H2O is H3O+
H2O is said to level the strengths of H3O+
The numerical value of K for a strong acid is infinity (very very large) because practically 100% of the reaction is complete complete.
c. weak acids and bases in waterWeak acids only undergo partial ionization in water while weak bases only undergo partial reaction in water. The reaction of weak acids and weak bases in water are equilibrium reactions.
Weak acids and bases tend to stay on the reactant side, as opposed to strong acids and bases which like to stay on the product side. The reason for this is that in the dissociation of a weak acid or base, the corresponding acid usually pushes the equilibrium to the product side. This can be seen above with OH- and H3O+.
Example:
What is the pH and % dissociation of 0.5 M HNO2 ? Where Ka (HNO2) = 4.5 x 10-4
There are 4 easy steps to solving a question like this:
1. Write the equation
HNO2 + H2O � H3O+ + NO2
−2. Write the Equilibrium Expression
Ka =[H3O
+][NO2−]
[HNO2]3. Setup an ICE table and solve
[HNO2] [NO2-] [H3O+]
1. Initial Concentrations
0.5 M 0 M 10-7 M
2. Change - x + x + x
3. Final Equilibrium Concentration
0.5 - x x x
Ka =(x)(x)
(0.5 − x)= 4.5 ∗ 10−4
Which gives:
x2 + (4.5 ∗ 10−4)x − 2.25 ∗ 10−4
The quadratic equation can now be solved and x values plugged in to determine equilibrium concen-trations. However, believe it or not there are 2 ways to solve this equation
1. Exactly, using the quadratic formula
2. By the approximation method
Approximation Method? How and when to use it:
The quadratic equation above arises because of the denominator (0.50 - x)
1. If we assume x to be much less than 0.50 (ie a very small number) then 0.50 - x 0.50.
so then
x2
(0.5 − x) becomes
x2
0.52. Solve for x (which will be much easier because we don t have to worry about quadratics)
x2
0.5= 4.5 ∗ 10−4
∴ x = 0.015
3. Check that x is actually much less that 0.5 (that is can we actually ignore x?)
If x<= 5% of the number we subtracted from, the approximation method gives us a good and accurate result
However, if x > 5% then solve using the exact method with the quadratic formula
[back to question ^^]
4. Solve the problem and check assumption
x2
0.5= 4.5 ∗ 10−4
[H3O+] = x = 1.5 ∗ 10−2
pH = − log10(1.5 ∗ 10−2)pH = 1.82
In this case, H3O from water can be ignored
Does assumption fall with the guideline (ie <5%) - How to check:
(x
0.5) ∗ 100 = (
1.5 ∗ 10−2
0.50) ∗ 100 = 3
3 < 5 and therefore the approximation method is valid
Question: How do % dissociation and [H3O+] vary with [acid]?
- for a strong acid eg. HCl
- for a weak acid eg. CH3COOH
HCl CH3COOH
[Acid] % Dis. [H3O] pH % Dis. [H3O] pH
1.0 M 100% 1.0 M 0 0.42 % 0.0042 M 2.38
0.10M 100% 10-1 1 1.34 % 0.0013 M 2.39
0.010M 100% 10-2 2 4.24 % 0.0042 M 3.38
0.0010 M 100% 10-3 3 41.5 % 0.0042 M 3.38
0.00010 M 100% 10-4 4 100% 0.0001 M 4.00
If weak acids and bases are dilute enough, you can accomplish 100% dissociation.
If you notice, from 0.010 M to 0.0010 M for weak acids, the dissociation goes from 4.25% to 41.5% That s a huge, almost 10-fold increase. But the concentration of H3O+ does not change. The reason for this is because weak acids and bases are non-linear, so you can have these weird and funny bumps and kicks and non-lineararities (don t even know if that s a real word ^^). Weak acids and bases tend to plateau off as you increase the dilution (refer to figure).
As you can see HCl is very linear, CH3COOH on the other hand is not so linear =S
Somewhere between 10-3 and 10-4, the CH3COOH hits 100% dissociation.
Conclusion:
For weak acids HA, as the concentration of HA decreases, the % dissociation of HA increases. They all reach 100% at some stage.
Polyprotic Acids
All you are responsible for in the polyprotic acid section is being able to recognize what a polyprotic acid is, (i.e. acids which have more than 1 readily ionizable Hydrogens)
For instance H3PO3 is a polyprotic acid with 2 readily ionizable H s.
Another polyprotic acid is H3PO4 with 3 readily ionizable H s
Remember, you are not responsible for any calculations involving polyprotic acids =)
Just a comment on weak bases. In the equilibrium equation with ammonia, if there s a strong base on one side, it ll drive the reaction to the opposite side.
The conjugate acids of weak bases, is itself is a weak acid.
d. salts of weak acids and bases in water: hydrolysis equilibria
Weak acids, HA , are partially ionized in water because the conjugate base A- efficiently removes a H+ from the strong acid H3O+, to give back HA.
If you look at the reactions, in the first reaction equilibrium shifts away from the side with the strong acid. and likewise for the second reaction.
Think about who has the proton, and who s getting it.
Now we re going to take a salt, Na+A-,and put it into water, what s going to happen?
A- comes from an NaA dissociation. The A- behaves as a weak base, ie a proton acceptor. It has to accept it from water, because that s all that s available. When that happens it produces HA and OH- .
This says that A- is a weak base, but it will produce an amount of HA, but the OH- will drive the equi-librium to the reactant side.
[Demo]
1. Sodium acetate adding HCl should force it to the right
2. Sodium benzoate
3. Sodium carbonate
4. Ammonium Chloride
--
Conjugate bases of weak acids (A-) are weak bases and react with H2O to afford OH- and thus basic solutions.
A− + H2O � HA + OH−
KbforA− =[HA][OH−]
[A−]
then divide the top and bottom by HA and multiply by [H3O+]/[H3O+]
=[OH−]
[A−][HA]
∗ [H3O+]
[H3O+]
and you know that [OH-][H3O+] is simply Kw and [A-][H3O]/[HA] is simply Ka
Kb =[H3O
+][OH−][H3O+][A−]
[HA]
=Kw
Ka
∴ Kb =Kw
Ka
This is important because this means that there is no need to tabulate Kb for A- species, as you can easily covert between Ka and Kb
The weaker the acid HA is, the stronger the conjugate base A- is, and the more basic the solutions of A- salts are.
For example NaA(aq). If you alter the concentration (ie dilute it more and more) you get a stronger and stronger base.
Example: What is the pH of 0.30M KCN? Ka(HCN) = 4.9 x 10-10
KCN is strong acid and therefore fully dissociated in H2O. This means that [CN-] = 0.30M
Step 1. Write reaction:
CN− + H2O � HCN + OH−Step 2. Write equation
Kb =[HCN ][OH−]
[CN−]=
Kw
Ka
Step 3. Write expressions of concentrations
[CN-] [HCN] [OH-]
1. Initial concentrations 0.30 M 0 0 (10^-7)
2. Change due to reaction - x + x +x
3. Final equilibrium 0.30 - x x x
Kb =Kw
Ka=
x2
(0.3 − x)=
10−14
(4.9 ∗ 10−10)Step 4. Solve it
Assume that x << 0.30
x2
0.30= 2.04 ∗ 10−5
x = [OH−] = 7.82 ∗ 10−4MCheck assumption that 7.82 x 10-4 is less than 5%
(7.82 ∗ 10−4M
0.30M) ∗ 100 = 0.26
0.26% < 5% and therefore the assumption is valid.
pOH = −log10(7.82 ∗ 10−4) = 3.11
pH = 14 − 3.11 = 10.89---
In HNO3 vs HNO2, the central oxygen is what determines the acidity. The acidity of HNO3 is much more. If you think about it, the extra oxygen in the HNO3 helps pull the electrons towards the oxygen side. This in effect makes the bond between the H and the N weaker, and thus it is easier to break off the H and thus more acidic.
Why is it that a weak acid is weak, and why is a strong acid strong? It all depends on ionization. HCl is a strong acid because it dissociates really easily. Part of the reason could be due to the fact that it s an ionic compound. However make sure you don t think “just because it s ionic it has to be strong”. A good example of this is with HF. HF is a weak acid. The molecules tend to stick together, and so you have very little ionization into H+ and F-.
A question you might have is, why is it that when KCN is placed in water and dissolves to make K+ and CN-, why do we ignore K+ and any reaction that it could possibly have with water?
The reason for this is is the K+ is a cation for an extremely strong base (KOH). If you think about it, if you had KOH in water, it will completely dissociate to form K+ and OH-. In other words, the reaction lies far on the K+ side, and there s is no tendency for the K+ is react with OH- to reform KOH. There-fore in essence, K+ just sits there and does nothing, and so we can leave it out the the equation
This same concept applies to anions of strong acids (ie Cl- , Br- , I-, NO3- , ClO4-, ect.)
Another way to look at this whole idea is through hydrolysis equilibria.
There are 2 simultaneous equilibria occurring in this reaction:
The first reaction we know that is taking place, is the self ionization of water. That s what s shown in
the purple box. Water and Water makes H3O+ and OH- where Kw = [H3O+][OH-].
In the orange, we can see that we dissolve KCN it in water, producing a CN- and K+. The question is do those two ions and the H3O+ and OH- interact? That s where you have to know a little about what s going on. You know that H3O+ and CN - interact producing HCN and H2O. The question then is: Does OH - and K + do anything to each other? OH- is a strong base, and therefore it full dissociates, pushes equilibrium to the other side. In this case, the OH- and the K+ interact to produce KOH, which is un-dissociated because of the OH-. Which means that the concentration of OH- doesn t decrease.
How about the equilibrium of the green box?
Ka = [H3O+][CN−][HCN ]
The common species of equilibria with the purple and the green box is the H3O+. We know from pre-vious discussion that the green box equilibrium lies way on the HCN side, because it is a weak acid reaction. So if you mix H3O+ and CN- you ll make HCN and H2O.
When the green boxes pulls a proton from the H3O, this causes the H3O concentration to decrease. However due to le chatilier s principal, more H3O, to produced to achieve equilibrium. The extra H3O is produced by the 2 reacting water molecules.
If you remember though, the other product of the reaction between Water and Water is OH-. OH-, as already explained before, is not consumed in the reaction. Therefore as the H3O is constantly being produced to maintain equilibrium, the concentration of OH- keeps increasing. This continues to hap-pen until CN- reaches its desired Ka. In the end, you have a lot of OH-, which gives the fairly high pH.
Explanation: (shorter ^^)
CN- (from KCN) reacts with H3O (from the H2O) to produce some HCN. H2O must self ionize to re-store H3O and restore Kw equilibrium. The Net result is that “extra” OH- is produced to from the deple-tion of the H3O by CN-.
Therefore [OH-] increases above 10-7 of pure H2O. Basically both the Ka and Kw equilibria work to-gether simultaneously.
The overall way of looking at this is Kb = Kw/Ka
e. reservoir picture of weak acid/base and salt equilibra
You can represent amount of dissociation diagrammatically, using reservoir pictures.
For instance, comparing the amount of dissociation of of 1.0M HF and 1.0M KF
a. 1.0M HF
We know that the equilibrium is
HF + H2O � H3O+ + F−
and you re given that ~2% is dissociated. [ Ka = 3.5 x 10-4 ]
If you have 1.0 M in each tank , how much would you have have in each tank at equilibrium?
Tank 1 = 0.98 M Tank 2 = 0.02 MTank 3 = 0.02 M
The concentration of H3O, is much greater than 10-7 and therefore the solution is acidic
Remember that HF & F must always obey:
Ka =[H3O
+][F−][HF ]
at all times.
b. 1.0M KF
F− + H2O � OH− + HFgiven that only 0.0005% reaction of F- occurs, how much is in each tank?
There s be 1.0 M and 5x10-6 M respectively. Technically the first tank does not have exactly 1.0M be-cause some dissociation did occur. However the change is so small, that it has not major effect on calculations.
e. buffers
A buffer is a solution that contains “adequate” or “reasonable” quantities of both:
1. a weak acid & it s conjugate base (eg. HF and F-)
OR
2. a weak base and it s conjugate acid (eg. NH3 and NH4+)
Note: In case you forgot conjugate acid and conjugate base, the conjugate acid is the molecule that donates a proton, and the conjugate base is the molecule that results from the loss of the proton.
For example: Consider a buffer solution composed of:
0.50M HF and 0.50M KF
KFaq −→ K+aq + F− ⇒∴ 0.50MF−
Reservoir picture of buffers:
A quick review (that is tips) on recognition of Acid-Base Reactions
1. all strong acids are fully dissociated in water to give H3O+
eg. 0.50M HNO3 produces 0.50M H3O+
2. all strong bases are fully dissociated in water to give OH-
eg. 0.22M KOH produces 0.22M OH-
3. ALL ACIDS (strong and weak) react FULLY with STRONG BASES to give salts and water
4. ALL BASES (strong and weak) react FULLY with STRONG ACIDS to give salts and water
[H3O+] = Ka ∗ [HF ]
[F−]
By taking the negative log of both sides we get
pH = pKa − log( [HF ][F−] ) = pKa + log [F−]
[HF ]
Therefore in general
pH = pKa + log [base][acid]
This is known as the Henderson-Hasselbach equation.
What happens when you place various chemicals into a buffer. What are the effects?
Consider: A buffer containing 0.5 mol HF and 0.5 mol KF, made up to 1.0L with water
1. OH- Addition
If you add 0.1 mol KOH to this buffer, what happens?
If you add 0.1 mol of OH- it ll react with 0.1 mole of HF to produce F-. So what you ve done is shifted the reservoirs slightly on either side. If you look at the ratio between the 2 concentrations, it hardly changes.
In the example pH changes from 3.46 to 3.64:
2. Addition of H3O+ (to 0.5 mole HF, 0.5 mole KF in 1.0L H2O)
We re going to add 0.1 mole of HCl (which reacts 100% with water 100% to produce H3O+)
We also know that we ve got this H3O+ and it will react with an F- or any base it can find thus
Initial situation: We have 0.1 mole of H3O+ with 0.5 mole of F- and 0.5 mole of HF
You know that 0.1 moles of the H3O+ is going to consume 0.1 moles for F- . Therefore the reaction will lead to-0.1 mole on the left side and +0.1 mole on the other side
So the net result is “0 moles” of H3O + 0.4 mole of F- and 0.6 mole of HF
To summarize:
0.1 mole added H3O+ converts 0.1 mole F- into 0.1 mole of HF leaving 0.4 mole F- remaining
What's going to happen with H3O+ ? We ll find out that the value of 3.5x10-4 M = [H3O+] before H3O+ added in
In the first example when we added OH the pH went up by 0.18 and H3O+ makes it go down by 0.18
Think: If you have pure water, and you added 1 mol HCl, the pH would be 1
General Procedure for Solving:
Example: What is the pH of a solution that is 0.1 mol in HF and 0.2 m is NaF where Ka[Hf] = 3.5x10-4
Remember that NaF(aq)---> Na+(aq) + F-(aq)
Step 1: Write reaction:
HF + H2O � H3O+ + F−
Step 2: Write equations :
Ka = [H3O+][F−][HF ]
Step 3: Set up the problem:
HF H3O+ F-
1. Initial Con. 0.1 0 (10-7) 0.2
2. Change -x + x +x
Final Equ. Con 0.1 - x x 0.2 + x
Step 4: Solve it (hehe the lazy way ^^)
Q: What pH change will occur on the addition of 0.05 mol HCl to 1.00 L of this solution?
Reaction: What happens?
The HCl dissolves in H2O & ionizes to give H3O+ . Also H3O+ reacts with F- to produce HF
Steps:
The chemistry you have to know about is that
F− + H3O+ −→ HF + H2O
[HF] [H3O+] [F-]
1. before HCL addition
0.1 mole 1.25 x 10 -4 0.2 mole
2. addition of 0.05 mol HCl
0.1 mole 0.05 mole 0.20 mole
3. Change due to reaction of HCl
+0.05 mole -0.05 mole -0.05 mole
4. Net effect of HCl addition
0.15 mole ~ 1.75 x 10-4 0.15
The above steps show how added HCl resulted in the adjustment of the HF and F- concentrations
New Equilibrium established between HF & F-
So then the net effect is:
[HF] [H3O+] [HF-]
1. NEW initial conc. 0.15 M (~ 1.75 x1--4) 0.15M
2. Change in cons. -x +x x
[HF] [H3O+] [HF-]
3. Find new equilibrium conc.
0.15 - x x 0.15
Ka = [H3O+][F−][HF ] = x(0.15+x)
(0.15−x)
assume x << 0.15
x( 0.150.15 ) = 3.5x10−4
x = [H3O+] = 3.5x10−4
∴ pH = 3.46
Direct use of Henderson Hasselbach Equation:
pH = pKa + log [F−][HF ]
pH = 3.46 + log[ 0.150.15 ]
pH = 3.46The pH “pivot point” of a buffer is the pKa for a weak acid or (pKb for a weak base)
Henderson Hasselbalch Equation says:
Depending on your range, you have a weak acid or base
g. titrations and indicators
Titration: Procedure whereby a solution of known concentration is reacted with a solution of unknown concentration. Examples acid-base titrations, redox titrations ect.
The range of actual useful pH range is only about 2. The idea of buffer is to resist addition of acid or base.
In the acid base context, we talk about the equivalence point. It s when the moles of acid equals moles of base.
Equivalence Point: (in an acid base titration) is attained when moles of acid is identical to the moles of base. Product at equivalence is a salt. This is an abstract concept,
Example:
HCN + NaOH ---------> NaCN + H2O
0.1 mole add NaOH to total 0.1 mole 0.1mole at equivlance point
The problem is finding out when you re at the equivalence point. However if you can match a colour change to the titrations, you can do it
End Point: Point in titration where indicator changes colour. The indicator is selected so that the end point (color change) happens at the same point as the equivalence point (moles acid = mole base)
The notion that you want a visual change coupled to the abstract concept change
Indicators: are weak acids or weak bases which change colour on conversion from their acid form to their base form.
The color change range compasses approximately 2 pH units ( i.e. 1/10 --> 10/1 range or acid base species); at 1/1 point the is the pKa of the weak acid indicator.
Indicators should be elected so that the pH at the end point corresponds to the pH at the equivalence point. (i.e the pH of the salt solution at the equivalence point)
Type of Titration Example: Reactant
---> salt formed
pH of salt solution -
identical pH at
equ. pt.
1. Strong acid/Strong base HCL + KOH --> KCl + H2O 7
2. Weak acid/strong base HNO2 + NaOH --> NaNO2 + H2O
> 7
3. String acid/weak base HNO3 + NH3 --> NH4NO3
or
HNO3 + NH4OH --> NH4NO3 + H2O
< 7
This means that for the first titration, you want a indicator that changes around 7
Consider the titration of 100mL of 0.1 M HF with 0.1 M NaOH (as much as you want)
HF + NaOH −→ NaF + H2O WA WA SB
Lets consider the situation at 4 points
1. 0 ml 0.1 M NaOH added
Only have 100 mL of 0.1 M HF
This is simply a weak acid problem
HF + H2O � H3O+ + F−
Ka = [H3O][F ][HF ]
2. 50.0 mL of 0.1 NaOH added “buffer problem” - This is the Half- equivalence point. Only 1/2 of acid HF has been reacted . Other 1/2 of HF is now NaF
moles HF = moles NaF = 1/2 moles original HF
pH = pKa
3. 100 mL of 0.1 M NaOH added (“salt” problem) - We have reached the equivalence point. Where moles acid = moles of base. Now we have a solution of NaF (0.05 M) solution has
F− + H2O � HF + OH−solution has pH > 7
We ve diluted the F by 2 essentially by adding equal volume.
4. 100 mL of 0.1 M NaOH added - adding excess NaOH to NaF solution.
Question. What is the pH? This is simply a strong base problem
The real issue in chemical bonding is covalent bonds and not anything else (ionic and such).
Covalent bonding between atoms involves the sharing of electron density between the atoms. The greater the sharing of the electron density between atoms, the stronger the bond between the atoms.
In fact this whole issue of the covalent bonds, how it works, how to measure are all still conjectures. (A conjecture is an opinion or conclusion formed on the basis of incomplete information). Research is still ongoing. All we have now are theories. Some theories are more correct than others, but none are actually full complete.
An example of a covalent bond is with 2 hydrogen atoms. Why do the H atoms want to come together and sometimes split apart? It all comes down to the concept of charges. What do you know about charges? Like charges repel and opposites attract.
As atoms approach other atoms 3 interactions come into play
1. Attraction between the the positive nuclei and electrons (negative charges).
2. Balancing out of the repulsion between the electrons (negative charges) in atoms
3. Repulsion between nuclei (positive charges)
With all these interactions, at a specific point (ie distance from each other) there ll be a balance. At this point, it ll reach a minimum energy, this is called the bonding state. The energy between a free state and the lowest energy state, is Hbond energy
The balance between these forces determines whether a bond will form and to the degree that the bond is strong
In this diagram of the potential en-gery vs. intermolecular distance, as the atom is far away (effec-tively known as an infinite distance away), the attraction is almost 0. However as the atoms become closer, and in this case reach an intermolecular distance of of 74 pico meters, the atoms are at it s lowest energy level. The energy from 0 to the lowest point is known as the Hbond energy
b. valence-bond approach to chemical bonding
According to the professor, this is a relatively easy and simple mined approach to bonding. VESPR doesn t really give a explanation to why a certain molecule is a certain way. It s like a “one size fits all” type approach to bonding.
An approach (which was surprisingly proposed earlier) goes a step further. Covalent bond formation that occurs through the overlap of electron density in atomic orbital (s, p, d, ect. orbitals) is called the valence-bond theory.
The valence bond method correctly predicts the structures using atomic orbitals in some cases. In other cases however, using atomic orbitals only leads to “failure” to predict correct structure.
The thing is, in many cases, we know structures of molecules. There is sometimes absolutely no question, as structure is determined from different methods (such as x-ray). The problem however, is how to account for the structure.
Taking an example of the valence - bond structure of H2S and CH4:
If we follow Hund s rule, we can put the electrons in their appropriate orbitals.
The question is that, what s available for bonding? In S, the electrons are in the so called “filled shells” and therefore can t take any more electrons. But the “partially filled orbitals” however, they still can take an electron. Formally this means that they can take in more electrons and interact with something.
From a very simple minded perspective we should be able to do something with the 3py and 3pz electrons.
The picture of the electron configuration tells us that there is an electron in each orbital, and that arbi-trarily the 3Py and 3Px shells are partially filled. Then when an H atom comes along, it can overlap with the 3Pz with the sulfur, thus forming 2 covalent bonds, due to electron density overlap. The S forms 2 S-H bonds. This also predicts a bond angle. We know the angle between the p orbitals is 90 degrees and therefore it predicts that the H-S-H bond angle is also 90 degrees. The observed bond angle is actually 92 degrees.
[Note that in the picture, only valence electrons are shown, and not 1s, 2s ect.]
We also know a molecule named methane: 1 carbon 4 H s
We can do the same thing that we did with the SH. We see that because the 2Py is empty, the result is CH2. So this is a problem. we know that CH4 exists, and that the H-C-H angle is 109 degrees.
[Note that CH2 does exist but is VERY reactive]
The question is: how can we account for bonding in CH4?
c. hybrid atomic orbitalsThe problem was that we were using s and p orbitals separately. Sometimes this works, but in the case of CH4, no way.
Answer: We use s & p atomic orbitals and mathematically “combine” them to make new hybrid atomic orbitals. The atomic orbitals that we know and love, only really apply to gas phases atoms. They re still useful, you just need some sort of context
With carbon:
We start off with the orbital layout for carbon. Then add a little energy and we can move an electron to the following orbital.
When we consider hybrid orbitals we can consider them the same in terms of energy distributions and wave functions ect.
This can give us:
1) 4 sp3 hybrid orbitals - all with the same energy 3 different orbitals arranged tetrahedrally in space:
You don t have to mix the orbitals in order to get 4 identical orbitals. You can actually have 3 identical orbitals (ie 3 sp2 orbitals) with a p orbital “left over”
Or finally you can have sp1 orbitals where you have 2 identical orbitals and 2 p orbitals “left over”
Now that we have a better understanding of the hybrid orbitals, we can apply it to molecules, and de-termine bond angles using it.
Hybrid
Orbitals
Geometric
Orientation
Example
sp Linear BeCl2
sp2 Trigonal-Planar BF3
sp3 Tetrahedral CH4
sp3d Trigonal-bipyramidal PCl5
sp3d2 Ocathedral SF6
Bonding in CH4
Bonding in NH3
d. types of covalent bondsThere are two types of covalent bonds. Sigma and pi bonds. A sigma bond is defined as electron density overlap that forms a bond that lies ALONG the internuclear axis. in other words it can be though of as in the same plane, or “head-on” bonding.
A pi bond on the other hand is when two lobes of one orbital, overlap two lobes of another orbital. In other words, in a pi bond, half the bond is half above the plane, half below the plane.
Pi bonds are usually weaker than sigma bonds because the negatively charged electron density is further from the positive charge of the atomic nucleus. In terms of quantum mechanics, because pi bonds are parallel in orientation, there is less overlap.
However, although the bond is weaker, it is found in combination in multiple bonds together with a sigma bond, creating a stronger combination than either body by itself.
Two atoms connected by a double or triple bond usually have 1 sigma bond and one or more pi bonds.
e. bonding in ethene and ethyneEthene (or the common name, ethylene) C2H4, is known to have the following structure:
1. molecule is planar - all atoms are in the same plane
2. H-C-H angle is 120 degrees. H-C-C angle is also 120 degrees
3. C2H4 bond is shorter than C2H6 bond
On the right diagram , the bond in the middle represents the sigma bond from the carbon carbon overlap. The green bonds above and below represent the pi bonds.
A double bond is equivalent to a sigma + a pi bond, and a triple bond is equivalent to a sigma + 2 pi bonds.
A difference in double and triple bonds compared to single bonds is that there is no free rotation about the carbon-carbon bond, due to the presence of the pi bonds. However, it is possible to “dis-rupt” the pi bonds temporarily, which will allow free rotation to occur. After the rotation has occurred, the pi bonds reform.
Ethyne (common name: acetylene), the the molecular formula C2H2. The molecule is known to be lin-ear
H − C ≡ C − H
C ≡ C is equivalent to a sigma bond plus 3 pi bonds
[Note that you do NOT need to know the molecular orbital approach to chemical bonding]
To give you an idea of average bond strengths:
C − C is approximately 350kJ/mol
C = C is approximately 650kJ/mol
C ≡ C is approximately 830kJ/mole
Example with organic molecules:
[Note that the notes here may be more detailed than the petrucci text book covers. =S]
Organic chemistry refers to the study of the compounds of carbon. Almost all chemical substances produced in nature are organic compounds.
Q: Why do carbons have such a vast chemistry compared to all other elements?
1. Carbon forms strong covalent bonds to itself. Long chains of carbon-carbon bonded systems pos-sible. Of all elements only, C, Si, & S can form long chains of neutral species.
2. Carbon forms strong covalent bonds to many other elements in the periodic table, metals and non-metals
3. Carbon can form strong MULTIPLE bonds with itself and other atoms such as O and N.
Below is a list of average bond energy
b. alkanesHydrocabons:
Compounds containing C and H
Alkanes:
Hydrocarbons with only C-C single bonds
Example:
If we remove an H from CH4, by putting in approximately 450J of energy, we can get a “CH3” a methyl group which has a valency of 1 for bonding
By joining 2 CH3 groups we can make a C-C bond
Each carbon has 4 bonds, and each has 3 hydrogens attached to it, and the other bond is to the other carbon. There is a sigma bond between the carbons which exhibits “free” rotation. Technically it s not totally free, but it has very little barrier.
We can keep going on. We can take any of the 6 H s of ethane and replace it with another “CH3” group
Note: He s using, dash and dark lines. The dashed lines means that “this H lies behind the plane” The dark lines means “This H lies in front of the plane” all relative to the screen.
Soon we ll find out that there s no free rotation between the pi bonds
Above you can see a cartoon representation of methane, ethane and propane. a) shows the ball-stick representation, and b) shows the space-filling (cloud model) representation. Rotation about the carbon-carbon bond is indicated by the arrows. Rotation is facile and rapid.
By replacing an “H” on either terminal (terminal means anything at the end) CH3 group of propane we can “prepare” the linear alkane C4H10. You can continue this hypothetical process and prepare linear hydrocarbons with the formulas:
[Formula] [Chemical Structure]
C5H12 CH3CH2CH2CH2CH3
C6H14 CH3CH2CH2CH2CH2CH3
C7H16 CH3CH2CH2CH2CH2CH2CH3
... ...
CnH2n+2 CH3(CH2)n-2CH3 (General Formula)
As you can see there is a general formula to produce the formula and structure for an alkane with any number of carbons
Table of boiling points and melting points of various alkanes
You can see about that there are a variance of boiling and melting points. You can also see that every time you add a CH2 group into the hydrocarbon chain you increase the boiling point by about 25 - 30 degrees. The reason for this is due to the number london forces increases.
Some general statements can be made from these observations
Alkane General Formula : CnH2n+2
Where n is an integer (n = 1, 2 , 3 ect.)
The alkanes form a homologous series, where each member of the series differers from the next member by an increment of CH2 (Meaning molecular mass increases by 14 each time)
There is a regular progression of:
1. Boiling Points: as the molecular size increases, London forces increase, and therefore the boiling points also increase. The london forces increases, because you have more electrons that undergo this instantaneous [something]
2. Melting Points: not quite as regular as boiling points. - London forces are involved, but also packing in the solid is important. Meaning some molecules pack better than others, and thus has a higher melting point than other molecules.
The chemical reactions of one member of the series is typical of all members. This is because for the most part, the reactions only involve C-C and C-H bonds
If you were to make a table of carbon-chain lengths vs. physical properties, you would get a table like the following:
Which hydrocarbons do we use? Natural gas (methane) and propane for BBQs.
c. bonding in organic moleculesBonding in organic molecules primarily involves covalent bonding. Carbon can form both single cova-lent bonds and multiple covalent bonds to itself and many other atoms (eg. O and N)
If we had the molecule ethane, we d have a structure like the following:
If we wanted to know the geometry of the bonds, we re interested in the angle between the C-H and the H-H bonds. The carbons in this molecule are sp3 hybridized and therefore all the angles are 109.5 degrees.
Ethene would look like the following:
The hybridization of this is sp2, and therefore the bond angles are approximately 120 degrees. This molecule has a double bond, and for that reason we talk about 2 kinds of bonds, so called sigma bonds, and pi bonds. Remember that you re not saying that top bond is a sigma bond and bottom is pi bond, it s just that one of them is sigma and one of them is pi.
Ethyne, more commonly known as acetelyne, looks like the following:
This molecule is sp hybridized. The molecule is therefore linear, and the angles between H-C-C-H are both 180 degrees.
Methonol, looks like the following:
The hybridization in both the carbon and the oxygen is sp3. If you look closely, all this is, is actually a fancy water molecule. However, instead of H-O-H, one of the H s is replaced by a methyl group.
Classification of sp3 hybridized carbons attached to substituents or functional groups:
If you consider the general diagram above, the carbon directly bonded to the X can be classified based on the number to carbons directly bonded to IT (i.e. 1,2,3 on the diagram). This classifies the carbon as primary, secondary or tertiary (also called 1st degree, 2nd degree or 3rd)
a. Primary Carbon (1st degree)
Only 1 Carbon directly bonded to the Carbon bearing X
The carbon bearing the OH (ie the “X”) only has one carbon attached to it (circled in green), and is therefore a primary carbon. This molecule is primary alcohol
b. Secondary Carbon (2nd degree)
2 Carbons directly bonded to the Carbon bearing X
As you can see, the Carbon bearing the Br (“X”) has two carbons attached to it (circled in green). This molecule is a secondary bromide.
c. Tertiary Carbon (3rd degree)
3 Carbons directly bonded to Carbon bearing X
This molecule is a tertiary amine
c. cycloalkanesCycloalkanes are simply alkanes which have a ring form.
The simplest cycloalkane is cyclopropane (C3H6).
Note: Rarely are cycloalkanes drawn like they are done so above. There is shorthand notation, where either/both carbon and hydrogen symbols are left out of the diagram assumed to be there
Cyclobutante (C4H8)
You can create a general formula from this, that will apply to all cycloalkanes: CnH2n
Cyclopentane (C5H10)
Looking at this diagram, you would expect the angles between the carbons to be 108 degrees each. However, analysis of cyclopentanes show that the angles are actually 109.5 degrees
Cyclohexane (C6H12)
Cyclohexane is the most ubitiquous of all cycloalkanes. Much like the case of cyclopentane, you would expect cyclohexane to have bond angles of 120 degrees, but in reality it is actually 109.5.
All the carbons in cycloalkanes are sp3 hybridized. This means that the bond angles are 109.5 de-grees. This creates a problem in cyclopropane. Cyclopropane, because of its geometric configuration, “wants” 60 degree angles. To try and accommodate this requirement, the sigma bonds actually bend. This causes a phenomenon called ring strain. The sigma bonds don t align up properly and creates a sort of “half-bond”, sometimes referred to as a banana bond. In essence it means that it is actually more difficult for the molecule to stay in cycloalkane form, than linear form. This causes cyclopropane to be highly unstable.
This same problem also applies to cyclobutane, because of the 90 degree bond angles. However, 90 degrees is much closer to 109.5 degrees, and therefore it is more stable than cyclopropane, however ring strain is still present.
To get these bended bonds, you need to apply some energy. It is approximately 120KJ/mol
Cyclohexane (C6H12)is NOT planar. A perfectly flat/linear hexagon would create a molecule with con-siderable bond strain (as the bonds would not be 109.5 degrees). To solve this problem, cyclohexane usually conforms to a 3D “chair” structure. Cyclohexane can also exist in another “boat” form, how-ever it is not very stable, and thus usually reverts back to the chair conformation.
Of all cycloalkanes, cyclohexane has the lowest angle strain, and is regarded as 0 ring strain. For this reason also, cyclohexane is the most stable cycloalkane, and produces the least amount of heat when burned.
d. structural isomersStructural isomers are compounds with the same molecular formula but different arrangements of bonded atoms
Starting with a propane, depending on where you add the the methyl group, you will get two different molecules. They both have the formula C4H10, however as you can see, they look very different. These two molecules therefore are structural isomers.
If you look at the table of different conformations of C4H10 and C5H12, you can see two general pat-terns. Firstly, as the number of carbons increase, the boiling point also increases. This makes sense because more carbons means that more bonds need to be broken, and thus requires more energy. The second pattern you can see is that the more linear the molecule is, the higher the boiling point is. This also makes sense because linear molecules have more interactions with other molecules, lon-don forces and hydrogen bonding. The effect of this is a higher boiling point.
For the molecule C5H12, there are 3 possible structural isomers:
There are:
5 possible isomers for C6H14
9 possible isomers for C7H16
4347 possible isomers for C8H18
4 x 109 possible isomers for C9H20
e. conformations of alkanesThere is “free rotation” about the C-C single bonds in alkanes. There are an infinite number of con-formations (or rotational arrangements) about any C-C bond.
If you take ethane as an example, there are two major conformations of interest: staggered and eclipsed.
Below are two different ways to show the staggered and the eclipsed conformations. The staggered conformation is energetically favoured because in the eclipsed form, the electron densities of the C-H are closer together, which causes repulsion among the electrons. The conformation of the staggered conformation is only 12 kJ/mol less is energy (ie more stable) than the eclipsed conformation. How-ever, even this slight difference means that during free rotation, the ethane molecule spends a little bit more time in the staggered conformation than the ecliptic one.
The difference between the staggered and the ecliptic conformations is only 60 degrees. Therefore if you looked at a 360 degree rotation for ethane, you would see the following:
If you were to look at the Newman projections for each of the above mentioned angles, you d get:
Sometimes you might be exposed to an angle called the “dihedral angle”. This is nothing more than the angle between two planes.
If you were to think about the carbon to the Y as one plane, and the carbon the the X as another plane, the angle between them would be the dihedral angle.
f. nomenclature of alkanesFrom previous examples, you saw that pentane had 3 different isomers. How would we distinguish between the three isomeric pentanes, without drawing any of the structures?
We need a systematic naming method that leads to unambiguous chemical structures. That naming method is referred to as the IUPAC rules of naming. The rules are as follows:
1. Find the largest continuous carbon chain in the molecule and name the compound as a derivative of that alkane.
2. Locate and name the substituent groups attached to the longest carbon chain. If a substituent ap-pears more than once, then use the prefixes di, tri, tetra (for 2, 3, 4 ect.)
3. Number the substituents, starting at one end of the longest carbon chain, as to give the lowest sum of substituent numbers.
4. Arrange the the substituent groups in alphabetical order (ignoring the prefixes or di, tri, ect.) For in-stance, ethyl will come before methyl. Triethyl will also come before methyl.
Some substituent group examples are:
g. cycloalkane isomersThere are two types of isomers, structural isomers and geometric isomers.
Consider the the molecule dimethylcyclopropane. There are two possible structural isomers of this molecule:
1,1 - dimethylcyclopropane
1,2 - dimethylcycloproane
However, because there is no free rotation about the C-C bonds, whether the substituents (methyl) are pointing upwards or downwards, is important.
When the two substituents are on the same side, we refer to them as a cis isomer. When the sub-stituents are on different sides, they are refereed to as trans isomers. These are called geometrical isomers.
Also remember that, with cis, the substituents pointing both up, or both down are equivalent:
h. alkenes: carbon-carbon doublebonds
Hydrocarbons containing carbon-carbon double bonds are called alkenes. Ethene (also commonly known as ethylene) is the simplest alkene.
According to VSPR theory, the carbon center is trigonal planar and the bond angles are approxi-mately 120 degrees.
There are 2 types of bonding electrons in a double bond. One pair of electrons along the internuclear axis (sigma bond) and the other pair of electrons located above and below the sp2 plane in the al-kene (pi-bond)
There s a corollary that comes out from this. A pi bond above and below the sigma bonds means that there is no free rotation about the C-C double bond of alkenes due to presence of the pi-bond. Rota-tion can only occur if pi bonds are “broken”. It s more like disrupted actually. If you use light or heat, it ll temporarily disrupt the bond, then it can rotate, and then re-form the bond.
If you keep adding homologous units of alkenes and cycloalkenes you can form a long series
The general formula for alkenes is CnH2n
The general formula for cycloalkenes is CnH2n-2
If we lose an H and add a CH3 (ie go from ethene to propene), we can do the same thing for the re-sulting molecule. Turns out that there are 4 different H groups (labeled a,b,c, d). The reason that they re different, is because there is no free rotation about the C-C bond and therefore the 4 different H s of propene can be replaced by a CH3 group to give 4 new four-carbon alkenes.
The normal ones are:
But you can also get:
Here is where things get interesting, we have 2 different butenes with different arrangements of the CH3 group. In these two compounds, both would be 2-butene. But they are actually different, and so we need a way to name these. The CH3 groups are on different sides of the alkene bond. The have the same structural form but different geometries and are therefore they are geometric isomers. The way we name these is not cis and trans (that was the old way to name these compounds).
There are 4 Isomers in all, with:
Z (zusammen) on the same side of alkene bond (formerly cis)
E (entygegen) on opposite sides alkene bond (formerly trans)
Therefore in essence, we have 3 structural isomers, where one structural isomer can exist as either of 2 geometrical isomers.
Why do we worry about these? The human body can tell the difference between a cis and trans ge-ometry and therefore it is important to know about.
The carbons are sp2 hybridized and therefore everything attached to it, has to be on the same plane. The carbons of the methyl groups, and the carbons of the C-C bond and the H s are on the same plane. However the methyl groups themselves are not planar.
Nomenclature of alkenes:
In alkenes there are two extra steps, where you must give the positive of alkene bond and indicate any geometrical isomerism.
1. Find the longest continuous chain of Cabons that contains a double bond and naming system is that same as an alkane except you have to replace the “-ane” of alkanes by “ene” to get alkene. For instance, a 5 Carbon alkene would be called a pentene.
2. Number chain so as to keep the numbering of the alkene bond (ie the C=C double bond) as low as possible.
3. Name and number all substituents - using E and Z as prefixes where appropriate.
Examples:
We see that there are 6 carbons on the longest chain. The double bond is at 2. In addition the 4,5,6 is on the same side as 1.
Therefore: Z-4-methyl-2-hexene
Note that the use of “cis” and “trans” for alkenes is still very common, so you should know them.
i. nomenclature of other functional groups
Functional groups, eg alkene and alcohols are always numbered as low as possible
Priority of Dominance: functional groups have an order in which they dominate the numbering in the naming of organic compounds
Random note:
If you take a look at the terminal atom:
The textbook says that the terminal atom in multiple bonds is NOT hybridized.
The year 1 instructors disagree with this statement. They believe that the terminal atoms are hybrid-ized, which predicts the sigma, pi bonds and the locations of the electron lone pairs more accurately than the text book s approach.
This is the view that the professors will follow in the tests and quizzes so remember that =).
1. Alcohols
The H on an alkane is replaced by an OH
For naming, you drop the -e in the alkane name and add an -ol
For instance, alkane would become alkanol
2. Aldehydes
There are characterized by the at the end of the hydrocarbon chain or on a ring.
For naming, you drop the -e in the alkane and add -al to make alkanal
3. Carbonyl (pronounced car-bun-eel)
Are characterized by a in the middle of the hydrocarbon chain or ring.
For naming, you drop the -e and you add -one to produce alkanone.
4. Carboxylic Acids
They are characterized by the -COOH at the end of a chain, or attached to a ring.
For naming you drop the -e and you add -oic acid to produce alkanoic acid
j. chemistry of alkanes
Preparation of Alkanes
1. Hydrogenation - the addition of H2 gas in presence of a catalyst. The barrier for the reaction is very high, and therefore a catalyst decreases the barrier. The barrier would be Platinum, Pd or N. They fa-cilitate the reaction, making the the corresponding alkane. Turns out that alkynes are just super-dooper, alkenes, and therefore they can also use the hydrogenation of alkynes.
2. Decarboxylation of Carboxylate Salts
It s actually possible to remove the CO2 group. If you heat up the molecule in the presence of sodium hydroxide, you can produce the corresponding alkane:
3. Coupling of Alkyl Halides
In this particular case, we can use the halogen, and remove it.
Reactions of alkanes: There are 2 principal reactions:
(Random note: These molecules are sometimes called paraffin. This comes from the latin root “pa-rum” meaning barely (i.e. unreactive). In chemist s terms the chemistry is really boring ^^ )
1. Combustion
If you take a hydrocarbon and heat it in the presence of oxygen, you can cause it to combust
CH4 + O2 −→ CO2 + H2O + heat
C10H10 + 15 12O2 −→ 10CO2 + 11H2O + heat
2. Halogenation
Turns out you can only do halogenation with Cl2 and B2. F2 is too reactive, and I2 just doesn t do it.
CH4 + Br2hv−−→
orΔCH3Br + HBr
This is similar to the reaction:
H2 + Br2Δ−→ CH3Br + 2HBr
Steps:
Steps 2 and 3 are the propagation steps, while 4, 5 and 6 below are the termination steps:
Note: The purple boxed species aboe are the reactive intermediates.
Polyhalogenation
The C-H bonds in alkanes & haloalkanes have similar H bond association energies. So haloalkane products can also be reactions for another halogenation step.
CH3 − Br + Br2hv−→ CH2Br2 + HBr
CH2Br2 + Br2hv−→ CH2Br3 + HBr
CHBr3 + Br2hv−→ CBr4 + HBr
Example reaction:
CH4 + 2Br2hv−→ CH4 + CH3Br + CH2Br2 + CHBr3 + CBr4
k. chemistry of alkenes & alkynesAs mentioned before the general formulas of alkenes and alkynes are:
Alkenes : CnH2n
Alkynes: CnH2n-2
Preparation:
1. Alkenes:
The elimination of H2O from alcohols produces an alkene. This is an acid catalyzed reaction
2. Alkynes
They can be produced by the reacting acetylides (an acetylene [aka ethyne] with one of the hydro-gens replaced with a metal) with alkyl halides
This shows the preparation of sodium acetylide from acetylene. The Carbon on the acetylide acts as a strong lewis base. You can then react it with an alkyl halide, where the alkyl acts as a strong lewis acid center, to get the following reaction:
Reactions:
Addition Reactions:
Alkenes and alkynes react by addition of an X2 or an X-Y species to the pi bond of the alkene/alkyne. The X becomes attached to to one carbon of the alkene/alkyne bond and the Y becomes attached to the other alkene carbon.
The sigma bond framework of the alkene/alkyne remains, and the pi-bond electrons are used to form bonds to X and Y.
Note that X2 refers to Cl2, Br2 and I2 and X-Y refers to Br-Cl , I-Cl and Br-I
A summary of the types of addition reactions possible are (will be explained in detail next):
1. Addition of X2
2. Addition of H-X
3. Addition of H2O
4. Addition of H2
5. Addition of H2 to alkynes
1. Addition of X2 to alkenes
The general reaction is:
alkene + X2solvent−−−−−→ dihaloalkane
Example:
Oxidant
You can also consider the case of the reaction with Br-Cl
In this reaction you can see that only one combination of the resultant molecule is observed. This phenomenon will be explained after addition of H-X is explained (because the same phenomenon is observed in those reactions too)
2. Addition of H-X to alkenes:
The general reaction is:
alkene + H − X −−−−−→solvent
haloalkane
Example 1:
Example 2:
Markinov s Rule:
In the addition of unsymmetrical chemical reagents to unssymetrical alkene, the H portion of the reagent (or the electropositive, ie. the Lewis Acid Por-tion) adds to the alkene Carbon with more H s
Unsymmetrical alkenes means that the number of carbons attached to alkene carbon, differs
Alkynes also behave similarly:
3. Addition of H2O to alkenes:
This addition is catalyzed by dilute H2SO4 or dilute H3PO4
The general equation is:
alkene + H − OHH2SO4−−−−→
Δalcohol
This reaction also follows Markinov s rule which is why you see an exclusive product. This reaction is reverse of the formation of an alkene bond from an alcohol.
4. How can the results observed in the reactions (ie. Markinov s rule) be explained?
This explanation will be illustrated with propene
In the reactions we are interested in, in the molecule H-X , the X is always more electronegative that the Hydrogen. This means that the bonding electrons are pulled towards the X end of the bond a bit more that the hydrogen side. The effect of this is that the hydrogen becomes slightly positively charged.
The slightly positive hydrogen is an electrophile and is attracted to the pi bond in the propene.
The next thing to determine is which side of the alkene bond the H will bond to. Lets consider both cases.
In both cases the pi bond is broken, and the electron pair forms a bond with the hydrogen. While this is happening, the electrons in the H-X bond are given to the X, forming an X- ion.
Case 1:
The electron pair forms a bond between hydrogen and the left hand carbon.
After the bonding with hydrogen has finished, the second part of the mechanism occurs: The lone pair of electrons on the X- ion forms a bond with positive carbon cation. [Just note that this reaction does not follow Markinokov s rule, and does not occur. I will explain in just a bit ^^]
Case 2:
The electron pair forms a bond between hydrogen and the right hand carbon.
The same thing happens, except that this method is the correct one =)
Why? Why is one of these cases better than the other?
The reason one mechanism works better than the other is because of the reaction intermediate formed.
In the better correct/better reaction method, you get a secondary carbocation formed
In the other case you get a primary carbocation
A secondary carbocation is energetically more stable, the H-formation is lower for a secondary car-bocation than a secondary carbocation.
You should note that a tertiary carbocation (if it is possible to form) is even MORE energetically stable (ie the activation energy is lower) and therefore it is the preferred product.
5. Addition of H2 to alkenes
The catalysts for this reaction are Ni, Pt or Pd
The general reaction is:
alkene + H2catalyst−−−−−→ alkane
Example:
6. Addition of H2 to alkynes
The general reaction is:
alkyneH2−−−−−→
catalystalkene
H2−−−−−→catalyst
alkane
Example:
Almost all catalysts reduce alkynes to alkanes rapidly and therefore the alkene intermediate is not observed.
However, if for some reason you want to halt to the alkene stage, you can use one of two methods.
a. Use of Lindler s Catalyst (Pd/BaSO4) - Z-Alkenes
b. Use of “dissolving metal” reductions - E-Alkenes
The mechanism for reducing an alkyne to an alkene:
The metal catalysts absorb the H2 to their surfaces and transfer the “H species” to one face of an al-kene or one side of an alkyne, resulting in:
l. aromatic hydrocarbonsAromatic hydrocarbons are planar, cyclic molecules containing alternating single and double bonds throughout the molecule.
This alternating pi system is said to be “conjugated”
Other examples of Aromatic Hydrocabons:
The above compounds are found in combustion emissions, e.g. wood, smoke, diesel exhaust ect.
Valence bond “picture” of bonding in benzene:
sp2 bonded C-C and C-H framework and p orbitals overlap to form a de-localized pi system. There-fore all the C-C bonds equal in bond length.
Naming Benzene derivatives:
Naming substituents - use 1,2,3, ect.
There is common naming method you can use, called ortho, meta and para
Ortho means 1,2 - distributed benzes
Meta means 1,3 - distributed benzes
Para means 1,4 - distributed benzes
Some parent benzene derivative:
Some reactions with benzene and benzene derivatives:
Reactions with benzene derivatives:
Substituent groups already on benzene system “direct” incoming substituent groups in one of two ways:
a. to produce ortho and para products predominantly over meta products
Z = NH2 , OH , OR , -R, -X
b. to produce meta substituted products primarily over ortho/para products
Z = NO2, -CN , -SO3H , -CHO, -COOR
(all of the above have X=Y multiple bonds bonded directly to the ring
Examples:
m. degrees of unsaturationAs you know, alkanes have are saturated, because they have the maximum number of possible hy-drogens bonded to it. Alkenes on the the other hand are missing hydrogens because of the presence of a double bond. The same thing applies to ring structures, and alkynes. What if you re given a for-mula and you want to quickly know whether the structure has multiple bonds or not?
The degrees of unsaturation (sometimes known as the index of hydrogen deficiency) is a useful con-cept for determining the number of rings and/or multiple bonds that are possible for a given molecular formula. The formula for calculating degrees of unsaturation is:
2C+2−H−X+N2
Where C is the number of carbon atoms, H is the number of hydrogen atoms, X is the number if hal-ide atoms and N is the number of Nitrogen atoms. Remember do not include group VI atoms (such as oxygen and sulphur, but do count the hydrogens attached to oxygen - ie hydroxyl)
So, say you calculate something and you got 3 degrees of unsaturation. What does that mean? That means that 3 x H2 , ie 6 hydrogen atoms are missing from the molecule being fully saturated.
n. alcohols, phenols and ethers
Alcohols, phenols and ethers are structurally related to water (H-O-H); Replacement of an H in water by an R group yields an R-OH (an alcohol or a phenol), whereas replacement of both H s of water by R groups yields an ether R-O-R .
Examples:
Preparation of alcohols:
1. Hydration of Alkenes
2. Substitution of X in alkyl halides by OH
3. Commercial preparation of Methanol:
Ethers (Preparation):
o. aldehydes and ketonesThese contain carbonyl groups such as methanol and formaldehyde.
Carbonyl Groups Resonance Structures:
Preparation of Aldehyde and Ketones by:
1. Oxidation of Secondary Alcohols [producing a ketone]
Oxidant Colour Change Upon Oxidation:
For balanced 1/2 reactions:
Try to work them out yourself then check table D-4 (Appendix D)
2. Oxidation of Primary Alcohols using PCC [ produces aldehydes ]
For oxidation to occur you need H s or C bearing OH. What you do is you look at the last carbon, in this case CH2, and you see if there s an O attached an a corresponding H, ie an alcohol. Oxidation is essentially the removal of H2, one from the OH and one from the last carbon atom. You know that if such conditions are present oxidation can take place
3. Tertiary Alcohols + K2Cr2O7/H+ (or KMnO4)
In the teritary alcholols there is no H on the C bearing OH and therefore there is no reaction takes place. This can therefore be a test for tertiary alcohols. If no reaction occurs, you know that the sub-stance is a tertiary alcohol.
Note: Oxidation of Primary Alcohols using excess K2Cr2O7/H+ (or KMnO4) [to produce carboxylic acids]
The Breathalyzer Test:
Ethanol in breath detector reacts and changes colour
The Silver Mirror Test:
Aldehydes are readily oxidized by mild oxidizing agents to carboxylic acids
Therefore the test of aldehydes is the silver mirror test
Reactions Of Aldehydes & Ketones
We ve talked how to make them, but now what about reactions with them.
Addition Reactions:
You know that aldehydes and ketones polarized, so you can essentially have two types of reactions, ones with a polarized molecules and ones with non-polarized molecules.
With the polarized reactions, generally you ll see that A is attracted to the negative end, and B is at-tracted to the positive end. This is what you should be look for, because most of the reactions we cover are going to be like this
The other case is when A and B are the same, for instance H2
1. The addition of H2 (g) [ production of Alcohol ]
You start off with a ketone and by using hydrogen and platinum as a catalyst, you can produce a sec-ondary alcohol. There is a redox relationship between these two.
2. “Net addition” of H2 using NaBH4
There s a number reagent which used borohydride which is a salt of Na+ and BH4- which is the lewis base adduct of BH3- and H-
The aldehyde has bond polarity, where is the NaBH4 going to go ? The H- is powerful protic base, and also a strong lewis base, strongly attracted to the positive end, which forms a CH bond. But to add an H, carbon has to remove a bond, in this case a pi bond. Then if you keep track of your electrons, you see that then the O becomes O- and then the Na+ is strongly attracted to the result O. In the end, you end up with the alcohol.
Basically you start with an aldehyde and you end up with a primary alcohol. There is also a redox re-lationship between these two.
3. Addition of Organometallic Reagents to aldehydes and Ketones
Normally we ve been reacting with fluorine chlorine ect. But you can actually join in metals
For example:
We know formally that Mg is 2+ indicated by the ++, and Br is - therefore C must be also be - because the whole molecule is neutral
In this case CH2 is a lewis base
General Reaction:
We re going see that the AB groups stay bonded in the system and we form a new carbon-carbon bond. One of the reasons for including these reactions is that it allows us to build carbon frameworks. Then you add alkoxide (the thing that s prodcued) and you add dilute H3O+ and you get this alcohol. Depending on what A and B are, you can produce primary alcohols, secondary alcohols or tertiary al-cohols at will. It s the ultimate mix and match ^^.
Specific Examples:
A. Preparation of Primary Alcohols: Reaction with Formaldehyde
The key thing to understand is what s happening in the general reaction.
We re going to see that CH2CH2 forms this adduct. So there s a new Carbon-Carbon bond (that s the hallmark there ^^) and the reason it happens is because of the electronegativity differences. After that you just need to add a proton to finish off the reaction, which makes it collapse down to CH2-OH.
Tada you just made a primary alcohol =)
[Note: Remember that reactions occur in ether]
B. Preparation of Secondary Alcohols: Reaction with an Aldehyde
If we play the same game, except replace one of the H s with a Carbon group, you get an generic al-dehyde.
C. Preparation Of Tertiary Alcohols: Reaction with Ketones
D. Preparation of Carboxylic Acids: Reaction with CO2 (s)
4. Addition of HCN to Carbonyl Group
p. carboxylic acids, esters and amides
Preparation of Carboxylic Acids:
1. Oxidation of Primary Alcohols - with excess oxidant [ produces carboxylic acids ]
You treat it with excess dichromate and you get the corresponding acid, which is butanoic acid. (The sell of rancid butter ^^)
2. Addition of Organometallic Reagents to CO2
We know that carbon dioxide is very polar-phillic, and what you end up with is a magnesium bromide salt of the carboxylate.
Then if you want to recover the pure carboxylic acid, (ie remove the metal), by adding dilute H3O+ and then the the magnesium will just dissolve away in water.
Preparation of Esters and Amides:
Really what an ester and amides are, are just acids plus an alcohol / amine. The way we bring them together is a bit different, but the following is a simple way to do it
Esters:
Alcohol + CarboxylicAcidCatalyzedbyAcid(H+)−−−−−−−−−−−−−−→ Ester + Water
We re doing a dehydration reaction. It s actually an equilibrium reaction, that s why we want to re-move the water to drive the equilibrium to our desired side. To do this we add concentrated H2SO4. Concentrated H2SO4 acts as an acid catalyst and a dehydrating agent.
Amides:
If you remember that this an acid and a base makes a salt. In this case, you can make an ammonium salt. It turns out that if you heat these resulting salts, you can actually drive off the water, and get the corresponding amide. This isn t the most efficient way to make an amide, but it works ^^.
Reactions of Esters and Amides
We talked about hydride donor molecules. This is a powerful source of H-. When you throw this in water, you instantly make hydrogen.
The beauty of the reagent is that it does some pretty powerful reductions.
It s the reverse reaction basically from an ester to a primary alcohol and an amide group will be brought back down to a primary amine.
1. Reduction with LiAlH4 (Lithium aluminum hydride)
2. Dehydration of an Amide [ production of a nitrile ]
Why is P4O10 a good dehydration agent? It s the anhydride to phosphoric acid. It loves to suck up wa-ter and become phosphoric acid.
Amide Resonance Structures:
It turns out that what the nitrogen does is that it donates electrons, and becomes an pi bond pair with carbon. We also know from previous discussion that if this was an alkene, we d say that there is no rotation about the C=C double bond. If you had the first structure with only the C-N sing bond, then in principal you should have free rotation about the NH2 quite rapidly. whereas no rotation in the second form with the double bond. So which form is it? It s actually in between. It turns out the H s actually change places, and they have very different resonance frequency, Once you start to heat things up, you see rotation. The diagram is somewhere in between these two diagrams. Just imagine it ^^.
What this really means is that all the atoms in the box don t rotate and are planar. So if you look at the polypeptide chain, they are fixed in groups that move together.
q. amines and heterocycles
Amines are carbon-substituted derivatives of NH3.
Heterocycles are carbon-based ring systems in which one or more carbons is substituted by O, N, S, ect.
Amines:
Amines behave as weak bases in water. Amine salts on the other hand:
act as weak acids in water.
Types of Amines:
Preparation of Amines
1. LiAlH4 Reduction of Amides
2. Reduction of NO2 Functional Group
There are only two ways to this as the text book describes. You can either use metals that undergo reduction (for this you need the assistance of an acid).
Also it is possible to use hydrogen as the corresponding reducing agent.
The introduction of the nitro group is just to get the N into the benzene. This is the first step for creat-ing bigger and more complex molecules.
3. Reaction of NH3 or an Amine with alkyl halides
The lewis base goes for the lewis acid center, and the result is an amine salt. The difference between the top and bottom is HBr. The bottom molecule is a weak base. The top molecule is the conjugate acid. Therefore to turn the amine salt in to the base, you need to add dilute OH-.
4. Reaction of Amines with Excess Alkyl Halide (under basic conditions)
r. interconversions of functionalgroup synthesis
The chemical transformations of organic functional groups provides numerous ways to interconvert the functional groups and to build more complex organic structures. This is the field of organic syn-thesis.
An overview scheme of the reactions discussed in chapter 26 is something in Fg 26-16
There are other reactions presented in these lecture notes which are NOT present in the Pretricci text.
s. optical isomerism - chiralityTo date we have discussed two types of isomerism - structural isomerism and (within a single struc-tural isomer it is possible to exhibit) geometrical isomerism
Some molecules can exhibit yet another type of isomerism - optical isomerism. The most common examples of optical isomerism occurr when there are 4 different groups attached to a tetrahedral car-bon atom
The way this works is, if you put a mirror plane, the molecules are mirror images or each other. Any-thing can have a mirror image, but these are special ones. The C reflects to C and a reflects to a , ect. so that they are true true mirror images.
These are mirror images, but NOT superimposable.
So what s the difference between these two molecules? They have absolutely identical physical prop-erties (b.p., m.p, refractive index) except for the different ability to rotate plane polarized light.
Every molecule has as mirror image, and for most cases the mirror image and the original molecule are identical. Thus, there is nothing unused about this mirror image relationship.
However there are examples where the mirror image and and the original molecule are not super im-poseable (much like your left and right hand). The following molecules are chiral or asymmetric.
Nomenclature of Enantomers (The R / S System):
Consider the molecule 2-butanol. It can exist as an optical isomer (an enantomer)
Consider the following general case:
The convention is to look at the molecule along the the c-d plane, and what you should see is that ra-dioactive looking sign. After this you can see that the letters a, b, c, can be clockwise or anti-clockwise. The naming convention states if the molecule letters go clockwise, it is considered an R type configuration, and if the molecule letter go anti-clockwise, it is an S type configuration. The next step therefore is to figure out how to label the individual atoms as a, b and c to figure out which direc-tion it is in.
How to assign priorities to optical isomers
Rules for assigning group priorities
Rule 1: The higher the atomic number, the higher the priority
Start at the middle carbon, and then you go directly to the next atom, and see what the highest num-ber is
Example : 1 bromo-fluoropropane
Rule 2: If two substituent atoms attached directly to the stereocentre have identical priorities, proceed along the two substituent carbon chains, until a point of difference is reached.
Example: 2-butanol
Example: 1 - chloro-3-hexanol
Rule 3: Double bonds & Triple bonds are treated as “single” bonds; however atoms are duplicated or triplicated by the particular atom at the other end of the multiple bond
Example: 2,3 dihydroxy propnal (glyceraldehyde)
Name: R-2,3-dihydroxypropanal
E, Z - Nomenclature of alkenes:
This is directly based on the priority system for the R/S nomenclature system. Groups with higher pri-orities determine E or Z.
t. substitution reactions of alkyl halides
Reactivity of Haloalkanes - C-X Bond Polarity
We already know that the C-X bond is polarized
The positive end of the C-X dipole behaves as a lewis acid (ie reacts with Lewis Bases) or nuclio-philes
eg. OH-, NH3, I-, CN-, N3-
The negative end of the dipole behaves as a lewis base and reacts with lewis acids
eg. BF3, Ag+
Two type of substitution Reaction Mechanisms
mechanisms kinetically different - different orders of reaction
You might take a primary halide, and mix it with a salt (using an appropriate solvent). In this case ace-tone is used as a solvent as it dissolves bromopropane and KCN
Mechanism of SN2 Substitution
Example:
Reaction mechanism involves attack of I- on carbon bearing Br from the “backside” of the C-Br bond
If this is the reaction mechanism, then this is a single step reaction. In which case a reaction coordi-nate diagram can be graphed. Change in enthalpy vs some sort of reaction progress. This can in-clude something like distance of C-I bond distance for example.
If we look at this reaction from the point of view of tetrahedral geometry, basically think of it as an um-brella. The Br being the handle. The I comes from the other side and the Br is removed and the um-brella is inverted
Therefore SN2 reactions are said to proceed with inversion at the tetrahedral center.
Evidence for Sterochemical Change at Reaction Center
This is going from R reactant to a S product. This is known as stereochemical change. However note that the relative priorities stay the same.
Conclusion:
Stereochemical center undergoing the SN2 reaction was “inverted” from R to S absolute sterochemis-try during the course of the reaction
Example 2:
Going from our previous example, if we know that inversion is going to take place, then you know that the I is going to become an H and the H is going to become a C. Only the cis product is observed. This is consistant with inversion of sterochemistry.
Take home message: That SN2 chemistry makes inversion
Relative rates of SN2 reactions for primary, secondary and tertiary halides
Organohalide & [X-] kept the same for all reactions Temp. also kept cosntant
Each CH3 group added to C bearing Br provides additional “bulk” or “blockage of access” to C bear-ing Br. Each additional CH3 “slows” reaction by ~ 30X.
Reaction of Alkyl Halides with Strong Protic Bases - Substitution vs. Elimination
Strong protic bases, eg. OH-, -OCH3, -OR , ect. can react as either (deepening on reaction condi-tions):
1. Nucleophile (Lewis base) in SN2 (or SN1) reactions & replaces X by OH, OR ect.
2. Strong bases, leading to loss of HX from haloalkanes & formation of alkenes
If you control the reaction conditions
1? Why do
Question: Why do “same reactants” give different products? under different conditions?
List of Substituting Nucleophiles (ie Lewis Bases)
Charged Species:
I-, Br-,Cl-, OH-, -OR, N3-, CN-, -SH
Uncharged species
H2O, NH3, amines (NR3)
Example: amine synthesis:
Example: Synthesis of acetylcholine (a neurotransmitter)
Mechanism of SN1 Substitution
Tertiary halides react via a first order kinetics mechanism - the SN1 reaction
Example:
Mechanism: (2 Steps)
1)
Simple ionization reaction. Note that is the slowest step, ie, it is the rate determining step.
2)
Now the cation. This lewis acid reacts with a lewis base N3- to form a new carbon center, which now includes N3-. This step accounts for the formation of the final product.
3)
Is the reverse of step 1, and is unlikely to occur if there is a lot of N3- present.
Carbocations:
1. Carbocations have only 6 valence electrons about the carbocation carbon.
2. Therefore there is a full blown positive charge on the center carbon, and is a very powerful Lewis acid.
3. Nucleophiles (Lewis bases) can provide electrons, or in other words, attack the carbocation from either face of the carbocation plane, the top or bottom face with equal probability.
A Carbocation is an sp2 hybridizied carbon, and therefore it s planar.
Stereochemical Fate of SN1 Reaction:
In SN1, when you start with R or S, you get 50:50 R and S. In SN2 reactions however you get inver-sion. If you start with R you get S.
Reaction Progress Diagram - SN1 Reaction:
How do you know this? It s a two step mechanism, and therefore there are 2 humps. The first hump is the higher than the second hump, because we know that it s the rate determining step. What lives in the valley? The cation, also known in this case as the reaction intermediate.
Q: Why do Tertiary halides react via SN1 mechanisms rather than SN2 mechanism?
A: There are two main reasons.
1) In the case of SN2, you have to do a backside attack. When you have a molecule that s tertiary, there are so many carbons and hydrogens in the way, that it s really, really hard for a backside attack. In short, a backside attack required for SN2 mechanism in tertiary halides, backside is too hindered by carbon-bases groups
2) A carbocation formed in the rate determining step of the SN1 mechanism (not Sn2 mechanism)
Remember that the H-formation of tertiary is less than secondary which in turn is less than primary.
Recall relative ease of carbon cation formation.
Table:
Thus within this group of very reactive carbon-cation species, the tertiary carbocations have lowest Hf and are said to be “most easily formed” or “most stable”
u. polymersPolymers are large, high-molecular-mass molecules prepared by the linkage of small low-molecular-mass molecules called monomers
There are two major classes of Polymers
1. Condensation Polymers: (or step reduction polymers)
Condensation between monomers occurs by loss of a small molecule such as H2O or an alcohol to form the polymer
Polyamides - Nylons
Means many amides. The amide bonds are formed through the loss of water.
Polyester
Forming multiple esters.
2. Chain-reaction Polymers
Monomers (usually alkenes or alkynes) are joined together to form long chains by radical cationic or anionic mechanisms
Radical Chain Polymerization:
Peroxides commonly used as a source of radicals to initiate the polymerization reaction
Physics properties of Polymers depend on:
1. Molecular mass of polymer
2. Degree of inter-winding, cross-linking chains
Chemical Properties of Polymers depend on:
1. Chemical composition of the polymer
