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Lecture Comments (10)

2 answers

Last reply by: Kamawak Anota
Mon Jan 20, 2014 10:32 AM

Post by Kamawak Anota on January 8, 2014

Is the energy released during bond formation the same as Bond dissociation energy? Because you used the values from the same chart. Right?

1 answer

Last reply by: Professor Starkey
Fri Sep 20, 2013 1:45 PM

Post by Vinit Shanbhag on September 20, 2013

ATP hydrolysis gives energy which can be coupled to other endergonic reactions, so how is bond breaking requires energy? Thanks.

1 answer

Last reply by: Professor Starkey
Thu Oct 20, 2011 11:31 PM

Post by Darcy Kuczajda on October 11, 2011

Can you tell me where to find more information on identifying nucleophiles and electrophiles??? For example, chloromethane. The carbon will have a partial positive charge and be electrophilic, and the chlorine will have a partial negative charge making it nucleophilic, so then chloromethane can act as both a nucleophilic and an electrophilic molecule. Is this right?

1 answer

Last reply by: Professor Starkey
Thu Oct 20, 2011 11:16 PM

Post by Erik Hord on October 11, 2011

Dr. Starkey, You said at about (15:40) that sigma bonds are stronger than pi bonds and I'm assuming you mean "per bond"? Looking at the chart it looks like the C-C sigma bond energy is 83 and the C=C B.E. is 146 so I'm reading that as the overall energy in the pi bond is higher but the energy per bond is lower (73), is that right? Thanks, Erik

Chemical Reactions

Give the IUPAC name for the following compounds:
  • Compound A is an ether. It has a larger group of 6 C's cyclohexane and a substituent methoxy group.
  • Compound B is an alcohol with a 5 carbons chain (pentanol)
  • Compound C is an alkene
A. methoxycyclohexane
B. 3,3-dimethyl-1-pentanol
C. 1-methylcyclopentene
Give the IUPAC name for the following compounds:
  • Compound A is an aldehyde:
  • Compound B is a ketone
  • Compound C is an ester:
A. 3,3,4,4-tetramethylpentanal
B. 2-methyl-3-pentanone
C. ethyl acetate
Give the IUPAC name for the following compound:
Give the IUPAC name for the following compound:
  • Find the longest chain containing COCl:

    hexanoyl chloride
  • Number and name the substituents:
2,4-dimethylhexanoyl chloride
Give the IUPAC name for the following compound:
Give the IUPAC name for the following compound:

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Chemical Reactions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Chemical Reactions 0:06
    • Reactants and Products
    • Thermodynamics
    • Equilibrium Constant
    • Equation
    • Organic Reaction
  • Energy vs. Progress of Rxn Diagrams 3:48
    • Exothermic Reaction
    • Endothermic Reaction
  • Estimating ΔH rxn 9:15
    • Bond Breaking
    • Bond Formation
    • Bond Strength
    • Homolytic Cleavage
    • Bond Dissociation Energy (BDE) Table
    • BDE for Multiple Bonds
    • Examples
  • Kinetics 20:35
    • Kinetics
    • Examples
  • Reaction Rate Variables 23:15
    • Reaction Rate Variables
    • Increasing Temperature, Increasing Rate
    • Increasing Concentration, Increasing Rate
    • Decreasing Energy of Activation, Increasing Rate
  • Two-Step Mechanisms 30:06
    • E vs. POR Diagram (2-step Mechanism)
  • Reactive Intermediates 33:03
    • Reactive Intermediates
    • Example: A Carbocation
  • Carbocation Stability 37:24
    • Relative Stability of Carbocation
    • Alkyl groups and Hyperconjugation
  • Carbocation Stability 41:57
    • Carbocation Stabilized by Resonance: Allylic
    • Carbocation Stabilized by Resonance: Benzylic
    • Overall Carbocation Stability
  • Free Radicals 45:05
    • Definition and Examples of Free Radicals
  • Radical Mechanisms 49:40
    • Example: Regular Arrow
    • Example: Fish-Hook Arrow

Transcription: Chemical Reactions

Back to Educator.0000

Next we are going to talk about studying chemical reactions.0002

If we are going to consider a chemical reaction, let's say we have reactants in equilibrium with some products.0005

Sometimes I will refer to these reactants as starting materials; I might abbreviate that SM--short for reactants.0018

If we take a look at a chemical reaction such as this, what are some of the things that we can investigate about that?--what are some questions we can ask and things we can study?0029

One area of interest is the area of chemistry called thermodynamics.0037

That is where we compare the energy or the stability of our reactants or starting materials and compare those to the energy of our products.0042

Using thermodynamics is going to help us be able to predict the direction of this equilibrium.0051

Is the forward reaction going to be favored?--in which case we would built up our concentration of products.0056

Or is the reverse reaction going to be favored?--in which case we build up the concentration of our reactants.0061

To describe the direction of this equilibrium, we use an equilibrium constant; and that is called Keq.0067

Keq is always defined as the concentration of the products divided by the concentration of the reactants.0075

If you have Keq equal to 1, that means you have exactly the same amount of reactants and products; and they are converting the forward reaction and the reverse reaction.0083

Of course, in any equilibrium, they are occurring at the same time; but you never have a buildup of one or the other.0096

But if you have a Keq that is greater than 1; if we take a look at this ratio.0103

A Keq greater than 1 means that our concentration of products are going to be greater than our concentration of reactants.0107

We would describe that as the forward reaction being favored; the forward reaction is favored because we have a buildup of products.0115

If the Keq is less than 1--is a very tiny number, that means our denominator here is the larger value.0128

The concentration of reactants is greater; and so that means that the reverse reaction is favored.0134

If we are asking in which direction an equilibrium lies, we can either ask--is it forward or reverse?0144

Or perhaps we can ask it in terms of Keqs; is Keq greater than 1 or less than 1?0150

Another equation that is going to be important to us in thermodynamics is the following: ΔG=ΔH-TΔS.0156

Here ΔH is our Gibbs free energy; this describes the spontaneity of a reaction.0166

The ΔH is our change in enthalpy or the heat of reaction; and the ΔS is our change in entropy.0174

For most organic reactions, this value is usually very small; that might still have an impact depending on the temperature of a reaction.0186

But since ΔS is usually a very small number, this value is typically or very often insignificant; so for organic reactions, we find that ΔG is very often nearly equal to ΔH.0197

A lot of times we are going to see reactions or descriptions where ΔH and ΔG are used somewhat interchangeably; in the context of organic reactions, that is often the case.0214

If we take a look at an energy versus progress of reaction diagram, the way it is going to look is going to vary depending on whether we have an exothermic reaction or an endodermic reaction.0230

Let's look at each of those one at a time; in an exothermic reaction, let's assume we have just a single step mechanism.0241

Our starting materials are going to be at some initial energy; and our products are going to be at a lower energy.0248

This difference in energy between the starting material energy and the product energy is defined as our ΔH--the change in the enthalpy of the reaction.0260

In this case, because we started at a higher number and ended at a lower number, our ΔH is less than 0; that is how we define an exothermic reaction.0272

The pathway to get from the starting material to the product is not just simply a downhill run; we need to go through a high energy state called a transition state; TS stands for transition state.0282

This is where we have a high energy complex; we have partial bonds--bonds that are starting to break; bonds that are starting to form; partial charges--charges that are developing or dissipating.0298

That is what a transition state looks like; and in order to go from starting material to product, you need to pass through this transition state in order to get down to product.0310

The energy required to get up to that transition state is defined as EA; that is the energy of activation.0319

Sometimes this is defined as ΔG‡ you might see that sometimes; the double dagger is something that often refers to a transition state species.0328

What are some things that we can observe about an exothermic reaction??we said that the ΔH is less than 0; it is a negative change.0341

That means that our ΔG is also going to be less than 0 for a typical organic reaction where the ΔS term is very small.0353

That tells us that the products are favored; the products are favored because the products are the lower energy species; they are more stable.0363

If the product is favored, what do you expect for the Keq?--how would you describe the Keq?0371

That is going to be greater than 1 because the products are in the numerator so it is going to be a large number when we look at that Keq ratio.0377

Products are favored; Keq is greater than 1; and we describe the forward reaction as being spontaneous.0386

The reaction where we are going from starting materials to products, again, is our general reaction that we are looking at.0393

An exothermic reaction--we would typically translate that to being a spontaneous reaction for organic chemistry because we are going in a direction of lower energy products.0402

An endothermic reaction is one in which we start at a relatively low energy for our starting materials, and our products end up in a higher energy.0416

I seem to have exaggerated that a little bit more; but still, in order to get from the starting materials to the products, we are going to have to go through some kind of transition state.0429

This is what an energy diagram might look like for an endothermic reaction.0442

Again, we can identify our ΔH; it is always the difference between the starting energy and the final energy; that is the overall change in the heat of the reaction.0446

How would you describe this change--if we start out at a lower number and we end up at a higher number?--that is a positive change; it is going to be ΔH is greater than 0.0458

Just like if you start out with $10 in your wallet, and you go to the bank and take out money, and you now have $100 in your wallet.0468

That is a positive change; you have increased the amount of money you have; that would be a ΔH of greater than 0.0479

That must mean that our ΔG is also going to be a positive number greater than 0.0489

ΔH greater than 0 tells us it is an endothermic reaction; energy is consumed.0493

ΔG greater than 0 tells us that this reaction is not a spontaneous one; because the reactants are lower energy, the reactants are the ones that are more stable so the reactants are favored.0499

How would you describe the Keq if the reactants, the starting materials, are going to be more stable?0516

The reverse reaction is going to be favored; it is going to be the spontaneous one so our Keq in this case is going to be less than 1.0522

We would say that the reverse reaction is the spontaneous one or we would describe the forward reaction as the nonspontaneous one.0530

We can also point out here, again, our energy of activation is always the energy required from the starting material to achieve the transition state.0538

Those are a few things we can label on our energy versus progress of reaction diagram.0548

We can estimate the ΔH of a reaction--on whether it is going to be an exothermic or endothermic reaction.0558

By taking a look at the types of bonds that are being formed in the reaction and the bonds that are being broken.0565

Bond breaking and bond forming--one of those processes requires energy; one of those processes produces energy or releases energy.0573

It is very common for students to have a misconception on which is which; or not a firm grasp on which is which.0584

Let's imagine having a stick and trying to break the stick; what do you have to do to break that stick?0589

You have to put energy into the system; you have to put force into the system in order to snap the stick into two.0596

Guess what?--chemical bonds are the exact same; bond breaking is a process which requires energy.0601

In other words, the ΔH of such a reaction--if all we do is we broke a bond, that would be a positive ΔH; it would be an endothermic reaction.0611

That must mean that the opposite reaction, any reaction which forms a bond, is something that releases energy.0626

Such a reaction would have a ΔH that is less than 0; it would be an exothermic reaction.0639

This is a very important chemical fact that you should have solid; and that is that bond breaking is something that is difficult to do; it requires energy; it costs energy.0646

Bond formation is a very favorable process; it releases energy; it is an exothermic process.0659

Knowing that this is true, what we can do is we can use a table of average bond strengths to estimate the heat of the reaction.0667

We can look at what is known as a bond dissociation energy to decide the relative strengths of the bonds that are being formed and the bonds that are breaking.0677

Let's take a look at a sample of a table that might show some bond dissociation energies.0687

A bond strength is defined or described by the standard bond dissociation energy; you can call that BDE for short or DH0.0695

That is defined as the energy required to break a bond in a homolytic fashion in standard conditions or standard states.0705

We are going to take this hydrogen-hydrogen bond, for example, and break it so that one electron goes to the left and one electron goes to the right; so our two products are two hydrogen atoms.0719

We describe that as a homolytic cleavage because they each get one electron; they get the same thing.0729

The heat of this reaction is +104kcals per mole; does that make sense that it would be a positive number?0734

Yes, because in order to break a hydrogen-hydrogen bond, you must have to put energy into it; so this is an endothermic process.0743

You can find in any textbook a table of such energies; these are average energies so you can read the table as such.0750

A hydrogen-hydrogen bond is worth about 104kcals per mole; a hydrogen-carbon-carbon bond is 99; and so on.0757

A few things that are maybe interesting to point out about this table is we can see a trend here.0767

When we go down the periodic table, we go from fluorine to chlorine to bromine to iodine; on hydrogen, you see that it becomes easier and easier to break that bond.0775

The trend we have here is that our bonds are getting longer; and longer bond are going to be weaker; that is something that we can observe in this.0786

Another interesting thing to see is something like... let's say a carbon-hydrogen bond or a carbon-carbon bond is worth about 100kcals.0801

Take a look at an oxygen-oxygen bond; that is a very low number; that is a very weak bond; as is a nitrogen-oxygen bond and all these.0809

When we have something like an oxygen-oxygen or any two very electronegative atoms attached to one another.0819

This is also a bond that is very weak and very easy to break as we can see by these average bond dissociation energies.0826

We also have a chlorine-chlorine bond, for example; or a bromine-bromine bond is very weak bonds that are easy to break.0839

I also see a number that jumps out here; silicon-oxygen is an extremely strong bond; silicon loves to bond to oxygen; we will see some examples of that way down the line.0854

Another thing that is interesting to note is to compare the strengths of single bonds?sigma(σ) bonds to double bonds; so down here we have some examples of multiple bonds.0865

For example, we could take a look at a carbon-carbon double bond.0877

If we were to totally break this carbon-carbon double bond, it would cost us 146kcals per mole or we would get 146 kcals per mole if we were to form a carbon-carbon double bond.0881

A carbon-carbon single bond is worth about 100kcals per mole; so what does that tell you about the strength of a pi(π) bond compared to that of a σ bond?0894

In order to break the σ bond, you have to put in 100kcals, but it only takes another extra 50kcals or so to break that π bond.0906

This is demonstrating to us that π bonds are weaker than σ bonds; it is easier to break a π bond; and it is going to be more difficult to break a σ bond.0914

But let's take a look at an interesting example; how about a carbonyl?0932

This is called a carbonyl; a carbonyl--when we have a C-O double bond is a very important arrangement of atoms called a functional group.0936

It is a very important functional group because if we compare the energy of a C-O single bond, a C-O single bond is right here; it is worth 86kcals.0947

If we were to have a double bond, what would we expect that energy to be?--well, 86 times 2 is 172kcals; so if it was worth twice as much as a σ bond, we would expect it to be 172.0956

We expect that a π bond is weaker so we would expect a carbonyl to be somewhere less than 170kcals per mole.0976

But look at the number; it is actually more; it is actually greater; all these C-O double bonds have actually more energy in them than two separate C-O σ bonds would have.0984

That tells us something very special about the carbonyl is that it is very stable; it is very strong; it is because it has resonance stabilization.0999

Again, carbonyls are going to be a very important functional group to us; we will see any reaction that forms a carbonyl would be a very favorable reaction; we will see lots of those.1013

You can see that there are slightly different numbers here depending on the structure in which the carbonyl exists.1022

This is for carbon dioxide; and this is if we have an aldehyde--that means that the carbonyl has a hydrogen attached; this is a for a ketone--that means there are two carbons attached to this.1028

Again, these are just average values so you can see that there is some slight differences.1039

What we can do is we can use a table like this to estimate for a given reaction whether or not it is going to be exothermic or endothermic.1044

Let's take a look at an example here; here are our starting materials; they are combining and reorganizing to give this product.1055

To estimate the ΔH of a reaction, it doesn't matter what the path is to go from the starting materials to the product.1065

Because remember, in thermodynamics, all we are doing is comparing the energies of our starting materials to the energy of our products; the path taken to get from here to there is not relevant.1071

What we can do is we can inspect this reaction and see what bonds have been broken and what bonds have been formed.1084

Here we have a carbonyl on our starting material and that carbonyl is gone; so we see that we are breaking this C-O double bond.1091

We are also breaking this C-H bond; it is no longer here in the product; these two C-H bonds are still there; so the bonds that we are breaking are a carbonyl and a C-H.1101

The bonds that we are forming are going to be--this O-H bond is new; there is no OH group in the starting material.1115

This C-O bond is new because the way that we do this calculation is we assume that we are completely breaking this C-O double bond and then reforming the C-O single bond.1124

That is not really what is happening in the mechanism; but again, the math works out the same.1134

Then this carbon-carbon bond is a new bond that is going to be formed in the reaction.1138

What we can do is we can look up these values on our table.1144

The aldehyde carbonyl was worth 176kcals; this C-H was an average of 99; the O-H was 111; the C-O is 86; and the C-C is an 83--the carbon-carbon is 83.1148

What we can do then is we can add up how much energy did we have to put into the system to break those bonds; so we add 176 plus 99; that is the energy consumed from the bonds broken.1167

Then we subtract from that the energy released from the bonds that are formed; these are the bonds formed.1185

When we do this math, we see that we come up with an overall -5kcals per mole as our overall estimated ΔH.1199

Does that tell us our reaction is exothermic or endothermic?--our ΔH is less than 0; this is an exothermic reaction.1208

Based on thermodynamics, is this a favorable reaction?--yes, this is a favorable reaction.1219

We would expect it to be spontaneous--something that the forward reaction is going to be favored; based on thermodynamics.1224

Another thing that we can study for a chemical reaction is the kinetics of the reaction; and this is where we take a look at the rate of the reaction or the speed of the reaction.1237

Maybe it is a favorable reaction but it is very very slow or it is very very fast; what can we study when we look at kinetics?1247

Let's consider a reaction such as this; let's say we have A and B as our starting materials and C and D as our products.1255

This reaction can be described by a rate expression as shown where the rate is equal to some kind of rate constant that is defined at a given temperature.1264

Rate studies are done at a given temperature to come up with this rate expression.1277

The rate will be dependent on the concentration of your starting materials; [A] to some exponent--little a; and the concentration of [B] to some exponent b.1282

The way that this rate expression is determined experimentally; this is experimentally determined.1291

The concentrations are varied and the rate is measured and that is how the exponents are set.1303

Here is an example--in this reaction, we have chloromethane reacting with hydroxide to give methanol and chloride.1310

When this reaction was studied, the following observations were made.1317

If the concentration of the hydroxide starting material was doubled, then the rate doubled; in other words, the reaction was twice as fast to go to completion.1320

If the concentration of the other starting material was doubled, then the rate also doubled.1330

Based on this experimental evidence, the rate for this reaction is described as k times the concentration of hydroxide times the concentration of the chloromethane.1336

Each of these is just to the first power; because if you double this number, you double the rate; if you double this number, you double the rate; so they have a 1:1 correspondence.1348

The way we describe this reaction given this rate expression is we say that the reaction is first-order with respect to each of the reactant; that simply describes the exponent which is 1 in each case.1361

We describe the reaction as being second order overall because we add up the exponents to describe the order of the overall rate; so we are adding a 1 and a 1; so it is a second-order reaction.1374

We are going to be using those terms later on; so it is good to review some of these kinetics principles.1388

What are some things we can do to speed up a reaction or to slow down a reaction?--we call these reaction rate variables.1397

Remember, to do our reaction, we need to achieve our transition state in order to go from starting materials to product.1403

Let's review that we have our energy versus progress of reaction diagram.1412

Whether a reaction is endothermic or exothermic, either way you still need to go through some kind of transition state to have that reaction occur.1419

What we need is a collision; if we have more than one starting material, we need a collision of those molecules to occur.1432

With enough energy to reach up to this activated complex where we are starting to form bonds and break bonds as transition state.1439

What are some things we can do?--well, one thing we can is we can increase the temperature; if you increase the temperature, we increase the rate.1449

This is true; whether you have an exothermic reaction or an endothermic reaction, this is always going to be true; we speed up a reaction by heating it up.1466

What is happening there is if you heat up a reaction, you increase the kinetic energy of the molecules; in other words, they are moving faster.1474

If you imagine a solution with our starting materials stirring around in the solution and you heat it up and now they are all stirring around; they are moving around faster.1483

What happens is when they collide, that is now going to be a high energy impact; and what you will end up with is more collisions that have the necessary energy to achieve the transition state.1493

So our reactions are going to go faster, the hotter the condition is.1513

This is why certain reactions like food spoiling, that is going to happen faster at room temperature or faster in hot foods than in cold foods; that is why we store our foods in the refrigerator.1517

That is why maybe you store your batteries in the refrigerator or your medicines in the refrigerator; the colder you keep them, the slower all the reactions that might happen for degradation.1529

Another possibility is we can increase the concentration and we can increase the rate.1540

We just saw on the previous slide how the rate expression shows the rate being proportional to the concentration of the starting materials.1545

If we increase that concentration, you increase the rate; how is that happening?1555

What we are doing here is we are going to increase the probability of having collisions.1560

If you have more molecules in a smaller area, it is more likely they are going to collide, and therefore, the reaction proceeds at a faster rate.1573

One thing I could point out for these first two to describe it is if you think of our molecules that are colliding as if they are bumper cars.1585

Let's say we have two bumper cars that need to collide at a certain speed in order for a reaction to occur.1593

Increasing our temperature is like having our bumper cars racing around a lot faster; and if they are racing around, then when they do collide, it is going to be a very high energy collision.1602

If they are going very very slowly, if we have nice cold cold temperatures, it slows down our molecules; and so when those bumper cars do collide, they are going to do so very low energy.1613

That is not going to cause a reaction; so there is most definitely a relationship between temperature and rate.1623

Increasing the concentration; if we think about our bumper cars, it is like we have ten bumper cars and we spread them out all over the United States; and they are wandering around the United States.1629

How likely is it they are going to collide?--very unlikely; and therefore, it is going to be a very very slow reaction if that reaction requires a collision.1644

However, if we take those ten bumper cars and put them in the room with us in our classroom and have them racing around.1652

Of course, they are more likely to collide, and therefore, a reaction is going to happen faster; so increasing the concentration is another way to increase our rate.1660

The last way to increase our rate is if somehow we could decrease our energy of activation; that would make it a faster reaction.1670

Where was our energy of activation?--it was right here; it was the energy required for our starting materials to reach this transition state energy.1681

If we could decrease the height of the hill that we have to climb, we could do that by somehow stabilizing--bringing the energy of the transition state down by stabilizing the transition state.1693

If somehow we could stabilize the transition state, that would be a faster reaction; how could we do this?--well, there is a few different things that we might find.1718

It can be done with solvent effects; certain solvents might be better at stabilizing a transition state than others.1728

You could have steric effects when you have crowding; that is going to be something that maybe raises a transition state energy.1743

Or reduced crowding will lower it; that will make one reaction faster or another.1752

What is another way?--have you ever seen a reaction diagram that looks something like this?--have you ever seen that picture where suddenly a different path is taken that is lower in energy?1757

What did that?--perhaps if we added a catalyst; that would be a way that we could lower our transition state energy.1766

That would be the role of a catalyst; and of course, catalysts are there to speed up a reaction.1773

So all sorts of things can be used to decrease our transition state energy; and the key is that if we have a lower transition state energy, we have a faster reaction.1779

We are going to see lots of examples of this as we study organic reactions.1793

This is a concept that is really important reaction to keep in mind because it will explain a lot of the observations that we make in the future.1798

What if we were to have a two-step mechanism rather than a single-step mechanism?--how might that look when we look at our energy diagram for example?1808

A two-step mechanism means that instead of going from a starting material directly to a product, we are going to first go to some kind of intermediate structure.1821

Then that intermediate structure is going to continue on to product.1831

If we had a two-step mechanism, then we have two transition states; one for each step.1836

Let's say that we have our starting material energy somewhere here; and at the very end, we have our products; let's assume that we have an exothermic reaction.1843

Because that is going to be true; most of the reactions we are going to be studying that are of interest to us are interesting because they are spontaneous; because they are favorable and exothermic.1856

If it was a two-step reaction, then instead of going straight from starting material to product, we are going to stop somewhere in between at an intermediate.1866

Step one is to go from your starting material to the intermediate; and that is going to require going through some kind of transition state; we will call that transition state 1.1879

Then after, at the intermediate, it is going to have to go up again to some other transition state before it can get to product.1889

It doesn't matter too much who is higher in energy; of course, that is going to be depend on the exact reaction we are looking at; this is just the overall picture we are going to have.1896

When we have a two-step reaction, we can describe one of those steps as the rate-determining step; and the rate-determining step is going to be the step with the highest energy of activation.1907

The step with the highest energy of activation is going to be our slowest step; so the rate at which that step occurs is going to define the rate of the overall reaction.1919

Let's take a look at this diagram to see where we have our rate determining step.1930

The energy of activation for our first step is here; we go from our starting material up to this transition state energy of activation 1.1936

The energy of activation for the second step is going from this intermediate up to this second transition state; this is energy of activation 2.1947

In this example, where would our rate determining step be?--well, this is a much higher hill to climb; this is going to be a much slower reaction; this will be our rate determining step.1957

It is whichever step is the slowest step; and the step with the highest energy of activation, we call a rate determining step.1974

Another thing to consider if we have a two-step reaction or multistep reaction is taking a closer look at these reactive intermediates.1984

These are the species that are involved in the reaction; they are going to be high energy species, meaning they are unstable.1991

The relationship between stability and energy is another understanding that we have to have in order to fully understand organic reactions or any chemical reactions really.2001

The good news is that chemicals and chemical species are a lot like humans in this respect.2015

If you are a very low energy stable person, you are calm; you are cool; you are relaxed; you could be described as stable; and you could be described as having very low energy.2023

Molecules would be the same; a very stable molecule would have a very low energy.2039

However, if you are a very high energy person, you are very unstable; you are always on the verge of exploding or getting into a fight or something like that.2044

Again, molecules would be the same; if the molecule is very unstable, we would describe it as being a high energy species.2052

When we say that a reaction intermediate is something that is high energy, that is the same thing as saying it is an unstable species.2060

It is defined as an intermediate because it is produced in the reaction and then it is consumed; it does not appear in the net reaction.2068

Remember our net reaction is simply the starting materials that we added at the beginning and the products that we recover at the end.2074

An intermediate is a fleeting species that is created and then continues reacting; it is not stable enough to stay on its own as either a starting material or a product.2082

It is different from our transition state; again, this will be more apparent when we actually look at some specific reactions and see the intermediates and the transition states.2094

But the transition state is something that will have partial bonds and partial charges; remember the transition state is a very high energy maximum.2102

A reactive intermediate has full bonds and full charges so it is going to look better than a transition state; but it still clearly going to be an unstable species.2113

For example, one reactive intermediate we are going to be encountering in organic chemistry is called a carbocation; a carbocation is a carbon with three bonds and therefore it has a + charge.2121

If we look at this species a little more closely, how would you describe its hybridization?--what is the hybridization of the carbon of a carbocation?2134

It has one, two, three groups around it; any carbon with three groups around it will be an sp2 hybridized carbon.2145

What kind of geometry do you expect for sp2?--we expect trigonal planar.2154

If we were to draw a sketch of a carbocation, we could draw those three R groups as defining a plane.2165

What else does an sp2 hybridized atom have besides these three sp2 hybrid orbitals?--it also has an unhybridized p orbital that is perpendicular to that plane.2177

Here is a question: what if we ask what is in the p orbital of a carbocation?--are there any electrons in it?--well, there can't be.2191

This is a positively charged carbon so it has just one, two, three electrons around it for its formal charge count which means that is going to be an empty p orbital.2203

Here are the two lobes of the p orbital; so what is in the p orbital?--nothing.2220

A carbocation has three bonds; it is a trigonal planar species; and it contains an empty p orbital.2225

This is going to be a very important species we are going to be seeing involved in a number of mechanisms.2232

Let's look a little more closely at the carbocation and what are some things that we should know about it moving forward.2239

One thing that we should have an understanding of is the relative stability of different carbocations.2246

Clearly, no carbocation is stable because it is missing an octet; it has got a positive charge and only three bonds; we know that carbon wants to have four bonds to be stable.2252

But not all carbocations are created equal; we could describe a carbocation by saying how many carbon groups are attached to the carbocation carbon.2261

Here we have a CH3 so this is just described as a methyl carbocation; this carbocation is described as a primary carbocation because it has just one carbon group attached.2273

This carbon has two carbon groups attached so we call it a secondary carbocation; and finally, we have a carbocation with three carbon groups attached--we call that a tertiary carbocation.2289

It turns out that the tertiary is the most stable of the carbocations shown; so that is more stable than a secondary which is more stable than a primary which is more stable than a methyl.2302

A methyl is the least stable carbocation; we really don't ever want to draw a methyl carbocation if we can help it.2316

Why is that?--what is there that changes this stability?--well, the alkyl groups that are attached are described as being polarizable.2326

That means that we can look at the cloud of electrons in that alkyl group and those electrons can be drawn toward the electron deficient carbocation.2339

It would be polarizable which means you can stretch out your electron density.2347

We can draw that as... of course, a methyl has no alkyl group.2353

But a primary has this alkyl group, and it can share some of its electron density with the carbocation; so we could draw this arrow showing it is sharing its electron density.2358

We would describe it then as being electron donating groups; alkyl groups are electron donating groups inductively.2371

We are just talking about the effects of this σ bond and sharing the electrons to the electron deficient carbon.2378

A primary carbocation has one carbon group that can do that; a secondary has two carbon groups that can do that; and tertiary has three carbon groups that can do that.2386

Is that a good thing?--is that something that would be good for a carbocation?2397

Yes, this is a stabilizing feature because a carbocation is electron deficient; so anything that can help add electron density to that center is going to help make it less electron deficient.2402

Obviously, tertiary has three groups that can do that; so this is going to be the best carbocation we can find.2416

Our take home message is that as you increase the number of alkyl groups on a carbocation, you increase the stability.2422

What is the maximum number of carbon groups you can have on carbocation?--just three; because a carbocation always has three groups attached; so a tertiary would be the best here.2435

This effect of donating electron density inductively is called hyperconjugation.2445

There is a way we can maybe draw a picture illustrating this electron donation and what is going on.2450

If we take a look at the orbital picture of this example of a tertiary carbocation, so here we have a carbocation; it has got an empty p orbital; it has got three methyl groups.2456

I've drawn out one of these methyl groups to show the C-H bonds because there are electrons in those σ bonds; it has the sp3 hybrid orbitals overlapping with the hydrogens.2466

What happens in this hyperconjugation is these alkyl groups can share the electrons and their σ bonds and donate some of them to the empty p orbital.2478

You get partial overlap between the empty p orbital and the σ bond here; and what is going to happen is it is going to share its electrons.2492

This is a drawing that you can have and you can look at; typically the way we just represented those with this arrow showing the inductive donation; that would be plenty.2508

One other way that we can stabilize a carbocation is by having resonance; so there is a few other carbocations that are exceptionally stable.2519

One is described as an allylic carbocation; the word allylic means that it is next to a π bond.2528

In other words, this carbocation is not on the carbon-carbon double bond, it is next to the carbon-carbon double bond.2538

Every time you have this allylic relationship, there will be a resonance form we could draw for this where this π bond shifts over to fill that vacancy; and that moves the positive charge to the other side.2545

Any time you have resonance, that is going to be something that delocalizes the charge; and delocalization of a charge is always a good thing; that is always a stabilizing thing.2559

Resonance is a great way to stabilize any molecule, but especially an unstable carbocation.2573

Another example of a resonance stabilized carbocation is a benzylic one; the word benzylic means that it is next to a benzene ring.2579

A benzene ring is a special group that has a six-membered ring with three π bonds; that is called benzene.2591

If you have it next to a benzene ring, you can have that same sort of resonance where this π bond can move over; that is going to move the positive charge onto one of the carbons within the ring.2599

Because benzene has three π bonds to delocalize, it can actually continue; this now is allylic to this π bond; so we can have additional resonance.2616

Actually a benzylic carbocation has multiple resonance forms, et cetera; you could continue moving this π bond over; there are quite a few resonance forms you could draw for carbocations.2628

If you ever find an allylic or benzylic carbocation, those are exceptionally stable.2640

If we look at overall carbocation stability, the best possible ones we can have are those that are benzylic or allylic, meaning they have resonance stabilization.2646

Or maybe tertiary because it has three alkyl groups donating electron density.2656

Those are going to be great carbocations; the most stable carbocations; very easy to form.2663

Those are all going to be secondary, and those are going to be much much better than either primary or methyl.2671

Primary or methyl are going to be really awful carbocations; we are going to avoid making them.2676

We typically even go to the extreme of saying that these are not formed under normal circumstances; if we needed to invoke a mechanism that involved a primary or methyl carbocation.2685

We would probably say that it is going to be so slow and so difficult to make that very very high energy carbocation, that the reaction would be described as no reaction.2696

Another type of reactive intermediate that we are going to be encountering in organic chemistry is a free radical.2707

Here are some examples of free radicals; this would be described as a chlorine atom; and this is a carbon radical.2715

In general, we could just describe or represent a radical as some R· because what defines a free radical is something that has an unpaired electron.2727

Rather than having an electron as either a σ bond or a π bond or a lone pair of electrons, it is a single electron on its own that is not being shared or paired up.2736

What is the charge of these free radicals?--these are interesting species.2748

If you go to calculate the formal charge of this chlorine atom, you would add up one, two, three, four, five, six, seven.2752

When you check with the periodic table, you would see that chlorine requires seven electrons; so this is actually a neutral species; there is no charge.2760

Same thing with this carbon; we would count that as one, two, three, four electrons, and carbon's valence has four electrons; so this is also neutral.2767

Most radicals that we are going to be dealing with in our organic mechanisms are going to be neutral species.2775

Still they are reactive intermediates; they are high energy and unstable even though they are neutral; and that is because they are missing an octet, right?2782

We always know we want to have this magic number of eight electrons; so if you have an odd electron anywhere then you can't ever be at eight.2790

The geometry of a carbon radical is approximately planar; it is either sp2 hybridized; it could be described as sp2.2798

Or it is a rapidly inverting pyramidal so you could either draw it as like we did the carbocation with a single electron in the p orbital that could be above or below the plane.2810

Or you could show it as an sp3 hybridized which would make it a tetrahedral arrangement; we describe it as pyramidal because it has just three bonds.2825

But if it is pyramidal, it is rapidly inverting so you would have inversion of stereochemistry; it is not possible to have a chiral radical because it is essentially planar.2841

Either one of these drawings is good; a lot of times, we just draw it in the p orbital here.2852

If you take a look at a carbon radical, you might wonder--is it electron rich or is it electron deficient?--because you do see that it has the electrons here.2858

But in fact, because it is missing an octet, it is an example of an electron deficient species; and so like carbocations are electron deficient, they have the same relative stabilities.2868

Just like we saw for carbocations, the best radicals are going to be benzylic or allylic.2879

Let's review what allylic means--remember allylic means that you are next to a π bond; that would be an example of a allylic radical.2889

A benzylic radical--what does a benzylic radical look like?--it doesn't mean you have a benzene with a radical on it.2899

That would not be a benzylic radical because benzylic means that it is next to a benzene ring; so you are not on the double bond; you are one carbon away.2907

You have an extra carbon; now it is benzylic; now you have that allylic relationship and you can have resonance; so remember both of these are resonance stabilized.2917

Tertiary is good because we have these three alkyl groups donating electron density; so just like a carbocation would like that electron donation, a radical would like that as well.2925

These are all the most stable; those are better than secondary; those are better than primary.2935

Here is an example of a primary radical; it has just one carbon group attached; those would be better than methyl; this is the least stable.2946

But it is still possible to make a methyl radical; remember it is not so electron deficient like a carbocation.2954

A carbocation is so electron deficient, it is positively charged; so this is a little less unstable; a little more stable.2960

We will see methyl and primary and secondary radicals; these are all OK.2968

But if given a choice, we are going to want to try and form a tertiary radical or maybe a resonance stabilized radical; those would be our best.2971

Finally, when we do encounter radical mechanisms, we need to think about how we would use curved arrows to show electrons moving.2983

Let's think about this resonance of this allylic carbocation; this is actually called the allyl carbocation; this three-carbon system.2994

If we want to convert from one resonance form to the other, what we do is we pick up that π bond and we shift it over to the other side.3000

That is what we have been using up to now; we could just describe it as a regular arrow; every time we use a regular arrow, it moves two electrons at a time.3008

How do we convert from this resonance form of the radical to this resonance form of the radical?--how have these electrons been redistributed?3018

You could imagine breaking this π bond and moving just one electron over to pair with the radical; and the other electron stayed behind and that is the new radical.3027

What we are going to be using for our radical mechanisms is called a fishhook arrow; it has got a barb on this one side; and that is what we use to move just a single electron.3042

This will come in handy when it comes time to do either resonance for a radical or if we want to do any radical mechanisms.3051

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