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4 -- Part 1: THE STUDY OF CHEMICAL REACTIONS
Simple reactions such as free-radical halogenation can be explained.
Draw a reaction-energy diagram and use it to identify the factors controlling the reactions.
Predict which of several possible products is the major product using the mechanism, thermodynamics, and kinetics of a reaction.
Explain the properties of reactive intermediates.
The picture shows a hole in the ozone layer caused by the release of chlorofluorocarbons in aerosol spray cans.
Free radicals in the atmosphere can generate chlorine radicals, which can lead to the breakdown of the ozone layer.
The worst chlorofluorocarbons are no longer being used.
The study of making and breaking bonds is the most interesting part of organic chemistry.
Organic chemists study reactions to learn how to make useful products.
We want to understand how reactions occur so that we can predict useful reactions.
We can organize the reactions into logical groups based on how the reactions take place and what intermediates are involved.
Alkanes are so unreactive that they are rarely used as starting materials.
The structure and properties of alkanes have already been studied, so we start with them.
We will apply alkanes to a variety of synthetic reactions once we have introduced the tools for studying reactions.
The first step in our study of a reaction is writing the overall equation with the reactants on the left and the products on the right.
The equilibrium amount of reactants and products depends on their relative stabilities.
The reaction may not take place at a useful rate even though the equilibrium may favor the formation of a product.
We can propose reaction mechanisms that are consistent with the behavior we observe by understanding the reaction's kinetics.
The chlorination of methane is an important industrial reaction that shows many of the important principles of a reaction.
The composition of the mixture of chlorinated products depends on the amount of chlorine added and the reaction conditions.
Light or heat is needed for the reaction to take place.
There are questions about the chlorination of methane.
The mechanisms, nature, and kinetics are not the most important parts of the reaction.
In what order to give the observed covered in Chapters 6 and 7 are products, the ionic reactions, bonds break and which bonds form.
We start our study with physical changes.
It allows us to compare the stability of reactants radical reactions because they are and products and predict which compounds are favored by the equilibrium.
If we change the common ionic reactions, the rate will change.
The chlorination of methane will be used to show how to study a reaction.
Before we can propose a detailed mechanism for the chlorination, we need to learn everything we can about how the reaction works and what factors affect the reaction rate and the product distribution.
When light falls on the mixture or when it is heated, the reaction begins.
The chlorine molecule is activated by light so that it starts the reaction with methane.
The product is formed for every photon of light that is absorbed.
Hundreds of individual reactions of methane with chlorine result from the absorption of a single photon by a single molecule of chlorine.
Blue light is absorbed by chlorine but not by methane.
The blue light has an energy of about 250 kJ.
One bond causes diseases and the other causes aging.
Two highly oxygen species are encountered in chlorine atoms after the splitting of a Cl2 molecule.
One of the seven short-lived hydroxyl radicals is unpaired.
Radicals don't have an octet.
The odd can combine with an electron in another atom to complete an octet and cause diseases and premature aging.
Radicals are represented by a structure with a single dot.
Lewis structures can be drawn for the free radicals.
When a chlorine radical collides with a methane molecule, it removes the regeneration of the hydrogen atom from methane.
H bond is on intermediates.
The molecule of HCl is only one of the final products.
chloromethane must be formed later.
The first propagation step begins with a chlorine atom and a methyl radical.
A propagation step of a chain reaction is the regeneration of a free radical.
The reaction can continue if another intermediate is produced.
The molecule of chlo rine reacts with the methyl radical to form chloromethane in the second propagation step.
The chlorine atom has an odd electron.
Another chlorine radical is produced by the second propagation step.
The chlorine radical can react with another molecule of methane, giving HCl and a methyl radical, which reacts with Cl2 to give chloromethane and regenerate yet another chlorine radical.
The chain reaction continues until the supply of reactants is exhausted or the other reaction consumes the radical intermediates.
Each photon of light that is absorbed creates hundreds of molecules of methyl chloride and HCl.
The reaction mechanism can be summarized.
Free-radical halogenation is a chain reaction.
A chain reaction usually requires one or more initiation steps to form radicals, followed by propagation steps that produce products and regenerate radicals.
Radicals are formed.
Radicals are formed.
Light is used to split a chlorine molecule.
Light is used to split a chlorine molecule.
A radical reacts to another radical.
A chlorine radical creates an alkyl radical.
To generate the product and chlorine radical.
The chlorine radical that was generated in step 2 goes on to react in step 1 of the chain.
Write the propagation steps that lead to the formation of dichloromethane.
Explain what free-radical halogenation does.
The chain reaction will stop if some of the free-radical intermediates are consumed without generating new ones.
The number of free radicals in the air is reduced by the combination of any two free radicals.
Other steps involve reactions of free radicals.
Although the first of these ter research is directed toward determin mination steps, it consumes the free radicals if they are necessary for the reaction to continue, thus breaking the chain.
The amount of product obtained from the reaction is small compared to the damage caused by free radicals.
The concentration of radicals is very low while a chain reaction is in progress.
The probability that two radicals will combine in a termination step is lower than the probability that each will encounter a molecule of reactant and give a propagation step.
The end of the reaction is important because there are relatively few reactants available.
The free radicals are less likely to encounter a molecule of reactant than they are to encounter each other.
The chain reaction stops.
The number of free radicals can be decreased by taking steps.
It is possible to convert cyclohexane to chlorocyclohexane in good yield.
We can consider the energetics of the individual steps now that we have a mechanism for the chlorination of methane.
We need to review some of the principles needed for this discussion.
Let's look at how energy and entropy variables describe an equilibrium.
The chlorination of methane has a large equilibrium constant.
The remaining reactants are close to zero because of the large equilibrium constant for chlorination.
The symbol deg designates a reaction involving reactants and products in their standard states.
Our intuition shows that reactions higher free energy.
The absolute temperatures in kelvins are correctly given without a degree sign.
The final concentrations of all four species should be calculated with a 1 M solution of CH3Br and H2S.
Under base-catalyzed conditions, acetone can condense to form alcohol.
5% of acetone is converted to alcohol at room temperature.
The relative strength of bonding in products and reactants is measured by the enthalpy change.
Reactions tend to favor products with the lowest enthalpy.
If bonds break and absorb energy.
The main driving force is the decrease in enthalpy.
Reactions tend to favor products.
The formation of strong bonds is the most important component in the driving force for a reaction.
The driving force for chlorination is the enthalpy change.
The enthalpy term is often larger than the entropy term in relation to it.
Some reactions have relatively small changes in enthalpy and larger changes in entropy, so we must be careful in making this approximation.
A single chlorine molecule has less freedom of motion than two isolated chlorine atoms.
The positive enthalpy term dominates the stable nature of the chlorine molecule.
Two smaller molecules add to the double bond to form ethane.
The reaction can be completed at room temperature.
We can use a hot wire to start the reaction when we put methane and chlorine into a bomb calorimeter.
105 kJ (25 kcal) of heat is evolved for each mole of methane converted to chloromethane.
We want to predict whether a reaction will be endothermic or exothermic without actually measuring the heat of the reaction.
Adding and subtracting the energies involved in breaking and forming bonds can be used to calculate an approximate heat of reaction.
We need to know the energies of the bonds to do this calculation.
Each atom retains one of the bond's two electrons.
The ability of the solvent to solvate the ion that results is very important.
The values of bond-dissociation enthalpies don't vary much with different solvent or no solvent.
A curved arrow is used to show the movement of the electron pair in an ionic cleavage, while a curved half-arrow is used to show the separation of individual electrons.
When bonds are formed, energy is released and consumed.
The bond-dissociation enthalpies are always positive.
The sum of the dissociation enthalpies of the bonds broken and the bonds formed is the overall enthalpy change for a reaction.
By studying the heat of reaction for many different reactions, chemists have developed reliable tables of bond-dissociation enthalpies.
The bonddissociation enthalpies are given in Table 4-2.
The heat of reaction for the chlorination of methane can be predicted using values from Table 4-2.
It is not necessary for each molecule of product breaking bonds to have a Bond-dissociation enthalpy for the overall enthalpy change.
The second propagation step includes the bond.
The mechanism we have used might not be the only one that explains the reaction of methane with chlorine.
The alternative mechanism is endothermic by 89 kJ>mol.
Reactions follow the lowest-energy pathway and the previous mechanism provides a lower-energy alternative.
The heat of reaction for each step in the free-radical bromination of methane can be calculated using bond-dissociation enthalpies.
The heat of the reaction should be calculated.
The position of its equilibrium is just as important as how fast a reaction goes.
A mixture of gasoline and oxygen does not react without a spark or catalyst.
If it is kept cold and dark, a mixture of methane and chlorine won't react.
The rate can be determined by measuring the increase in the concentrations of the products with time or the decrease in the concentrations of the reactants with time.
The reaction rates are dependent on the concentrations of the reactants.
We can't guess or calculate the rate equation from the reaction.
The rate equation is dependent on the mechanism of the reaction and the rates of individual steps.
Experiments show that doubling the concentration of methyl bromide will increase the rate of reaction.
The rate is doubled by doubling the concentration of hydroxide ion.
Reactions of the same type don't necessarily have the same form of equation.
The rate of this particular reaction is unaffected by the doubling of the concentration of hydroxide ion.
Zeroth order in hydroxide ion is proportional to the zeroth power.
It is the first order.
The form of the rate equation can't be predicted.
We use the rate equation to propose consistent mechanisms.
The rate increases when the concentration of chloromethane is doubled.
The rate is observed to triple when the concentration is tripled.
The rate doubles when it is doubled, which is 2 to the first power.
The first order of the reaction is chloromethane.
The reaction rate triples when [-CN] is tripled.
The first order of the reaction is cyanide ion.
The reaction rate is unaffected by the concentration of hydrogen.
Adding more ethene has no effect.
The rate equation should be written for this reaction.
Give an explanation of the rate equation and suggest a way to accelerate it.
The majority of the time, only a small fraction of the collisions occur between the same molecule with the same orientation.
The molecule bounce off each other if there is no proper orientation or enough kinetic energy.
The black curved line shows the energy distribution at room temperature, while the dashed lines show the energy needed to overcome barriers.
The curve to the right of each bar rier shows the fraction of molecule with enough energy to overcome the barrier.
The energy distribution is shifted at 100 degrees.
The 80 kJ>mol barrier is one of the energy barriers that can be overcome by more than one molecule.
For reactions with typical activation energies of 40 to 60 kJ>mol (10 to 15 kcal>mol), the reaction rate doubles when the temperature is raised by 10 degrees.
We try to find a temperature that allows the desired reaction to go at a reasonable rate without producing unacceptable rates of side reactions.
The barrier that must be overcome is the activation energy.
A transition state is unstable and cannot be late.
The transition state is a constant on the path from one intermediate to another.
In speaking of the activation energy, ++ is often used.
Transition states have high energy because bonds must break before other bonds can form.
The reaction of a chlorine radical with methane is shown in the equation.
The bond was partially formed.
The brackets emphasize the nature of transition states.
The double dagger next to the brackets means that this is a transition state, and the dot means that it has an unpaired electron.
It is easier to understand the concepts of transition state and activation energy graphically.
The total potential energy of all the species is represented by the vertical axis of the energy diagram.
The coordinate shows the progress of the reaction from the reactants on the left to the products on the right.
The C + D highest point on the graph is the transition state, and the activation energy is the difference between the reactants and reaction coordinate.
The highest point on the graph is the transition state, and the activation energy is the difference between the reactants and the transition state.
They increase the rate of reactions.
The heat of reaction and the equilibrium constant would not be affected by the addition of a catalyst.
The rate is +7 kJ>mol.
Draw a diagram of the reaction.
The barrier is 17 kJ higher than the reactants.
The equation for the reverse reaction is given.
Several steps and intermediates are involved in many reactions.
The reaction of methane with chlorine goes through two propagation steps.
The propagation steps are shown along with their reactions.
The rate of the initiation step is controlled by the amount of light or heat available to split chlorine.
Unlike transition states, reactive intermediates are stable if they don't collide with other atoms.
As free radicals, #CH3 and #cl# are quite reactive.
The unstable transition states are called the energy maxima, and the low points are called the energy minima.
Most of the important information about the reaction can be found in this complete energy profile.
Each step has its own characteristic rate in a multistep reaction.
The transition state for the rate-limiting step is the highest point in the energy diagram.
The highest point in the energy diagram of the chlorination of methane is the transition state for the reaction of methane with a chlorine radical.
This step should be rate limiting.
The rate for the overall reaction will be calculated if we calculate a rate for this slow step.
The slow step's products will be consumed by the second step as fast as they are formed.
The chlorination of methane can be accomplished with a combined reaction-energy diagram.
What we know about rates is applied to the reaction of methane with halogens.
The rate-limiting step for chlorination is the endothermic reaction of the chlorine atom with methane to form a methyl radical and a molecule of HCl.
17 kJ>mol is the activation energy for this step.
The rate is fast but controllable.
In a free-radical chain reaction, every propagation step must occur quickly, otherwise the free radicals will participate in termination steps.
The relative rates show how quickly methane reacts with different radicals.
The rate of the reaction with fluorine is very high.
If the temperature rises much, it will be difficult to control the reaction of chlorine at room temperature.
The reaction with bromine is very slow.
Iodination is very slow, even at 500 degK, so it's probably out of the question.
Our predictions are correct.
A mixture of bromine and methane must be heated to react.
Draw a complete diagram of the reaction.
The rate-limiting step should be labeled.
Each transition state has a structure.
We have limited our discussions to the halogenation of methane.
The study began with a simple compound that allowed us to focus on the reactions.
The "higher" alkanes are those with higher molecular weight.
A substitution is where a hydrogen is replaced by a halogen atom.
All four hydrogen atoms are the same in methane.
Replacing hydrogen atoms may lead to different products.
Two monochlorinated products can be used in the chlorination of propane.
One has a chlorine atom on a primary carbon atom, while the other has a chlorine atom on a secondary carbon atom.
Replacement of hydrogen atoms by chlorine is not random according to the product ratio.
Propane has six primary hydrogens and only two secondary hydrogens, yet the major product results from substitution of a secondary hydrogen.
By dividing the amount of product observed by the number of hydrogens that can be replaced, we can calculate how reactive each kind of hydrogen is.
The definition of primary, secondary, and tertiary hydrogens is shown in Figure 4-5.
Replacing either of the two secondary hydrogens accounts for 60 percent of the product, and replacing any of the six primary hydrogens accounts for 40 percent.
Each secondary hydrogen is 4.5 times more reactive than the primary hydrogen.
The 2deg hydrogens are 4.5 times more reactive than the 1deg hydrogens.
The propagation step is called a primary radical or a secondary radical.
A hydrogen atom can give either a primary or a secondary radical when it reacts with propane.
The structure of the radical formed in this step determines the structure of the observed product.
The secondary radical is formed preferentially.
The stability of the secondary free radical leads to the preference for reaction at the secondary position.
The preference is explained in more detail in the next section.
The greatest energy for a carbon is for a methyl carbon, and the lowest is for a primary carbon, a secondary carbon, and a tertiary carbon.
The less energy required to form the free radical, the more highly substituted the carbon atom.
Increasing order of stability is where the free radicals are listed.
13 kJ>mol of the secondary hydrogen is more methane than the primary hydrogen.
The primary radical is formed by 1deg ++ energies.
There is a diagram for the first propagation step.
The secondary radical is formed faster than the primary radical because the activation energy is slightly lower.
Primary hydrogen atoms react about the same as tertiary ones.
There are nine primary hydrogens and one tertiary hydrogen.
The primary product is the major product even though the primary hydrogens are less reactive.
The product ratio will be between 1.5 and 6.0.
The two possible first propagation steps in the chlorination of isobutane can be calculated using the bond-dissociation enthalpies.
This information can be used to draw a reaction-energy diagram, comparing the isobutane activation energies for formation of the two radicals.
Predict the ratios of products that are chlorinated.
The process of burning heptane in a gasoline engine takes too long.
The process of burning CH3 takes place in a slower, more controlled way.
O # radicals are more stable than alkyl radicals.
Butyl alcohol is an antiknock Additive for gasoline.
To explain why toluene has a very high octane rating, use the information in Table 4-2.
An equation can be written to show how toluene reacts with an alkyl free radical.
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