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4 -- Part 9: THE STUDY OF CHEMICAL REACTIONS
A copla nar arrangement of bonds to the atoms being eliminated is required for the E2 reaction to take place.
The transition state may be syn-coplanar in rigid systems.
There is no geometry required for the slow step.
The coplanar arrangement is required for the transition state.
The carbocation intermediate is involved in the E1 reaction.
The interme diate can be rearranged by the shift of a hydride or an alkyl group.
The reaction takes place in one step.
There is no way to make a change in the E2 reaction.
Rearrangements are common.
Predict the major and minor elimination products of the reactions.
Explain if you expect the elimination to be E1 or E2.
The substitution and eliminations require the base to react with a nucleophile.
Both substitution and elimination may be possible in the same reaction because most bases are also basic.
Predicting the products of a reaction requires both the mechanism and substitution or elimination to be present.
The strength of the reagent that serves as the base determines the order of the reaction.
Bimolecular, second-order reactions are favored by strong reagents because they attack the substrate faster than the ionizes.
NaOCH3 is a strong base and a strong nucleophile.
The narrowing of the likely reaction mechanisms to SN2 and E2 does not mean which of these will dominate.
The following examples show a mixture of the two reactions.
The principles that apply to bimolecular substitution and elimination are shown in these examples.
The SN2 mechanism causes primary halides to react with strong nucleophiles.
A small amount of elimination might occur if the reagent is a strong base.
A strong base will react with a tertiary halide to give only elimination by the E2 mechanism.
It is difficult to predict secondary halides.
They react with strong bases by the E2 mechanism.
The proportions of substitution and elimination can be controlled by other factors such as temperature, solvent, and structural effects.
A few strong nucleophiles can promote substitution over elimination.
Halide ion are examples of weak bases.
Bulky bases are often used to promote E2 elimination.
Bulky bases are not allowed to approach the back side of a carbon atom to do the SN2 displacement, but they can approach a hydrogen atom on a neighboring carbon and abstract a protons.
In the absence of a strong base or strong nucleophile and in the presence of a good ionizing solvent, the reaction rate is first-order.
We don't have control over the fast second step when the reagent attacks the carbocation or removes an adjacent protons.
The carbocation intermediate can lead to more products.
Unimolecular substitution and eliminations of alkyl halides are rarely used for synthesis due to lack of control.
The substitution and elimination mechanisms are compared with those that occur under first-order conditions.
A strong reagent is likely to remove a protons from a neighboring carbon.
A weak reagent is unlikely to react.
It can form a bond to the carbocation or remove a protons from a neighboring carbon.
The competition between substitution and elimination can be influenced by temperature.
Eliminates produce more products than reactants.
The substitution or elimination of a secondary alkyl halide is possible under second-order conditions.
SN1 and E1 are possible under first-order conditions.
An increase in temperature favors elimination, while a decrease in temperature favors substitution.
When other factors are balanced, this principle applies to unimolecular and bimolecular reactions.
The strength of the nucleophile and the type of alkyl halide affect the reaction products.
In the presence of a strong nucleophile, 1deg alkyl halides give good yields of substitution products with little elimination.
The 1deg alkyl halides can be forced to ionize by AgNO3/EtOH and heat.
Note 3 says that any strongly hindered alkyl halide will not undergo a SN2 reaction at all.
Predict the major products of the reactions and indicate the mechanisms that are likely responsible for their formation.
NaOCH3 is a strong base and a strong nucleophile, suggesting that the likely reactions are SN2 and E2.
The only remaining possibility is elimination by the E2 mechanism.
One of the CH2 groups can be eliminated by the use of a ring protons on one of them.
Zaitsev's rule probably applies to NaOCH3 because it is not a bulky base.
A chair transition state is needed to show how a trans-diaxial arrangement of the H and Br allows E2 elimination to give the major product.
Methanol is heated for an extended period.
A slow ionization of this secondary alkyl halide leads to formation of a 2deg carbocation, which can lose a protons in either direction to give products of E1 elimination.
Methanol can attack the 2deg carbocation, followed by the loss of a protons, to give a product of SN1 substitution.
The 2deg carbocation can be rearranged by a hydride shift.
The carbocation can lose a protons in either direction to give products of E1 elimination, or it can react with methanol and deprotonation to give a product of SN1 substitution.
We don't know if there will be a major product or not.
Zaitsev's rule tells us that the product with the most substituted double bond will dominate.
The same product is used for both the initial and rearranged carbocation.
A substitution product could be the major product.
The rearrangement is faster than the attack by the solvent.
If the rearrangement is quicker, we can see more rearranged products.
The E1 and SN1 reactions of alkyl halides are rarely used for synthesis.
There are mechanistic steps omitted in the solution shown above.
Students sometimes feel overwhelmed.
Base memorizing is not the best way to approach this material because the answers are not absolute and there are too many factors involved.
The real world is not as clean as the paper world.
S solvate ion or react as a base.
If you can eliminate some of the possibilities, you can make some accurate predictions.
If a strong base is present, it will force second-order kinetics.
You should consider first-order reactions if there is no strong base.
Adding silver salts to the reaction can cause some problems.
The E2 reaction is sometimes seen in primary halides.
Unless the carbocation is resonancestabilized, primary halides rarely undergo first-order reactions.
SN2 substitution is usually observed with good nucleophiles.
Sometimes a no reaction E2 elimination can be observed.
The SN2 reaction cannot be performed by R tertiary halides.
Second-order kinet ics are eliminated by the E2 mechanism when a strong base forces it.
The ratio of substitution to elimination is determined by the reaction conditions.
It is difficult to predict the reactions of secondary halides.
The E2 reaction is possible with a strong base.
Both the SN1 and E1 reactions are possible, but they are slow.
There are a lot of products.
substitution or elimination is favored by some nucleophiles and bases.
The base should not attack a carbon atom in order to promote elimination.
Butoxide enhances elimination.
In most cases, higher temperatures favor elimination.
The conjugate base of a strong acid is a highly polarizable species.
Weak bases and favor substitution are what C X are.
Explain which are most likely when more than one product is possible.
We have studied the eliminations that convert alkyl halides to alkenes.
Other types of compounds are used to make alkenes.
The elimination of elements of water from alcohols is a common laboratory synthesis of alkenes.
In most cases, the equilibrium constant is not large.
The reverse reaction is a method for converting alkenes to other substances.
The products from the reaction mixture are respirated in one of the steps.
The alcohol is hydrogen-bonded and the alkene is boiling at a lower tion.
The alkene is removed while the alcohol is left in the reaction to form a mixture.
Concentrations of sulfuric acid and/or concentrated phosphoric acid can be used as conditions in the laboratory for dehydration because they act both as acidic catalysts and as dehy reaction agents.
The acids are exothermic.
Only one enantiomer of the product is formed by the use of sulfuric acid.
The loss of water from the alcohol causes a carbocation.
Alcohol dehydrations usually involve E1 elimination.
Loss of H+ is the last step.
Deprotonation to speed up the alkene.
Deprotonation to speed up the alkene.
Like other E1 reactions, alcohol dehydration follows an order of reactivity that shows carbocation stability.
In alcohol dehydrations,arrangements of the carbocation intermediates are common.
Zaitsev's rule states that the major product is usually the one with the most substituted double bond.
The first step is converting the hydroxy group to a good leaving group.
Rearrangements are common.
The second step is giving a carbocation.
The mechanism is complete with the abstraction of a protons.
Predict the products and propose a mechanism for dehydration.
A carbocation occurs after the loss of water.
A more stable carbocation can lose a protons.
The unrearranged 2deg carbocation can lose either of two protons to give two of the following products.
The rearranged 3deg carbocation can lose either of two protons to give two products.
An alkene and a shortened alkane are formed by bond cleavage.
Small alkenes can be made up to six carbon atoms with cracking.
It depends on the market for different alkenes and alkanes.
Depending on the temperature, catalyst, and concentration of hydrogen in the cracking process, the average and relative amounts of alkanes and alkenes can be controlled.
The mixture is ready to be packaged and sold.
The products are always mixed so labo ratory synthesis of alkenes is not possible.
Better methods can be used to make relatively pure alkenes from other functional groups.
The methods shown in this chapter are better than the methods shown in the summary.
Alkene is given by the dehydrogenation of an alkane.
This reaction has an unfavorable enthalpy change.
There are two products called C + H.
The enthalpy term is favored at room temperature.
Dehydrogenation is favored at high temperatures.
Dehydrogenation is similar to catalytic cracking.
Dehydrogenation and catalytic cracking have the tendency to produce a mixture of products, and neither reaction is suited for the laboratory synthesis of alkenes.
There are three major classes of reaction mechanisms that we have seen at this point.
The process can be broken down into a series of logical steps using some general principles.
If you use a systematic approach, you can come up with a mechanism that is at least possible and that explains the products.
There are more complete methods in Appendix 3A.
You must determine what kind of reaction you are dealing with before you propose a mechanism.
A free-radical initiator such as chlorine, bromine, or a peroxide suggests that a free-radical chain reaction is most likely.
Chapter 4 discussed free-radical reactions.
There are mechanisms such as the SN2 or E2 that involve attack by the strong base or nucleophile.
There are mechanisms that involve acidic intermediates, such as the SN1, E1, and alcohol dehydration, that are suggested by strong acids.
When you have decided which type of mechanism is most likely, some general principles can be used to propose the mechanism.
Chapter 4 discussed some principles for free radical reactions.
We now consider reactions that involve strong nucleophiles as intermediates.
We will apply these principles to more complex mechanisms in later chapters.
The reactive intermediates are likely to be three-bonded carbon atoms.
If you attempt to draw a line-angle formula, you will most likely misplace a hydrogen atom and show the wrong carbon atom as a radical, a cation, or an anion.
Two steps must be shown for a carbocation to occur.
To show water falling off, you must not circle the hydroxy and the proton.
The arrow going from the electrons of the hydroxy oxygen to the protons must never be from the protons to the hydroxy group.
When a strong base is present, we expect to see intermediates that are both strong bases and strong nucleophiles.
Weak acids and philes are generally found in such a reaction.
Draw strong acids such as H3O+.
They are not likely to coexist with strong bases.
A strong nucleophile can be converted to a functional group by deprotonation or reaction.
There is a mechanism for the dehydrohalogenation of 3-bromopentane.
Several general principles of proposing mechanisms would be violated by this mechanism.
The carbocation and the H+ ion are unlikely in the presence of ethoxide ion.
The mechanism doesn't explain why the strong base is needed, and the rate of ionization wouldn't be affected by ethoxide ion.
Even in an acidic reaction, H+ must be removed by a base.
The presence of ethoxide ion in the reaction suggests that the mechanism involves only strong bases and nucleophiles.
The E2 mechanism is an example of a reaction involving a strong base.
As the electron pair left behind form a pi bond, ethoxide ion expels bromide ion.
When a strong acid is present, expect to see intermediates that are also strong acids.
Cationic intermediates are common, but don't draw any species with more than one charge.
Bases and nucleophiles are weak in such a reaction.
Avoid drawing strong bases.
They are not likely to coexist with strong acids.
Functional groups can be converted to strong electrophiles by either reaction with a strong electro phile or by protonation.
The presence of sulfuric acid indicates that the reaction is acidic.
The product's carbon skeleton is different from the reactant.
Formation and rearrangement of a carbocation is likely under these acidic conditions.
The hydroxy group is a poor leaving group, it cannot ionize to give a carbocation and -OH, and we do not expect a strong base such as -OH in this acidic reaction.
The hydroxy group can become protonsated in the presence of a strong acid.
The OH group becomes a good group to leave.
OH2 is a good group to leave.
A primary carbocation would be formed by a simple ionization.
They are very unstable.
A primary carbocation is never formed when a methyl shift occurs.
It's incompatible with the acidic solution.
Two types of protons, labeled 1 and 2 in the figure, could be lost to give alkenes.
The required product is given by the loss of proton 2.
Zaitsev's rule predicts that the more stable product will be the major product.
You may be asked to explain unusual compounds that are only minor products in other problems.
Your mechanism needs to explain the products shown.
Rearrangements are very common.
A lot of the time, it gives a mixture.
Alkene was replaced by N(CH3)3.
There are examples of Alkyl shifts.
Adding to or removing from a molecule.
There is a dihedral angle of 180 degrees.
There is a dihedral angle.
A species that can abstract a protons.
A species that can give a pair of electrons to form a bond.
Unless one of the rings contains at least eight carbon atoms, a stable bridged bicyclic compound cannot have a double bond at a bridgehead position.
There are two rings.
Three links connecting the bridgehead carbons have at least one carbon atom.
There are three bridges of bonds connecting the carbon atoms.
The heating of petroleum products in the presence of a catalyst causes bond cleavage to form alkenes and alkanes.
The cis-trans arrangement on a ring or double bond can be different.
diastereomers are a subclass ofcis-trans isomers.
There are groups on the same side of a ring.
There are groups on opposite sides of a double bond.
The higher priority groups are on the same side of a double bond.
The higher priority groups are on opposite sides of the double bond.
The breaking of bonds and the formation of new bonds occur at the same time.
Two bonds are separated by one bond.
Extra stability is contributed to by the interaction between the pi bonds.
Acid-catalyzed elimination of water from a compound.
The elimination of hydrogen from a compound is done with a catalyst.
Base-promoted elimination is the elimination of a hydrogen halide from a compound.
Constitutional isomers have the same position of a double bond.
The same alkane can be given by double-bond isomers hydrogenate.
Two elements of unsaturation are a double bond and a triple bond.
The formation of a pi bond is a result of a reaction that involves the loss of two atoms or groups.
A multistep elimination where the leaving group is lost in a slow ioniza tion step, and then a protons is lost in a second step.
The Zaitsev orientation is preferred.
A transition state where the base is abstracting a protons at the same time the leaving group is leaving is a concerted elimination.
The anti-coplanar transition state is preferred.
Unless the base or leaving group is bulky, Zaitsev orientation is usually preferred.
There are examples of hydrride shifts.
There are two or more bonds.
In a simple alkene, isolated double bonds react indepen dently.
A species that can give a pair of electrons to form a bond.
A simple addition of units creates a new substance.
A change in the bonding sequence of a molecule.
In reactions such as the SN1 and E1,arrangements are common.
The energy of a compound with an equivalent number of isolated double bonds is greater than that of a conjugated system.
Being incapable of undergoing addition reactions, having only single bonds.
A substitution is when the solvent serves as the attacking reagent.
Giving rise to stereoisomers.
A stereoisomer is created by characteristic of an atom or a group of atoms.
The starting material reacts to give different stereoisomers of the product.
Adding or removing the same face of a molecule.
There is a dihedral angle.
An anti and coplanar arrangement allows E2 elimination on a cyclohexane ring.
Both substituents must be in the same place on the ring.
Multiple bonds can undergo addition reactions.
Alkenes with more double bonds are more stable.
The most stable alkene product is usually the most substituted alkene.
Zaitsev's rule doesn't always apply to a bulky base or leaving group.
The elimination gives the Zaitsev product.
Each skill is followed by problem numbers.
Draw and name alkenes.
Predict the relative stabilities of alkenes and cycloalkenes using their structure and stereochemistry.
Zaitsev's rule can be used to predict the major and minor products of dehydration reactions.
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