10.4 Other Pericyclic Reactions

One category of pericyclic reactions is called cyclo-addition reactions.

There are two major categories of pericyclic reactions.

The first category was introduced in this chapter.

Some similarities can be found between the other two categories and cycloaddition.
In each case, the curved arrows are moving around in a ring, and the entire process occurs in one concerted step.
These reactions are placed under the larger umbrella of pericyclic reactions.
Some important differences will be seen if we inspect the reactions more carefully.
A key difference between the categories will be seen if we count the number of bonds being broken and formed in each category.
Two p bonds are replaced with two new s bonds in a cycloaddition process.
One p bond is replaced with a new s bond.
The number of p bonds and s bonds is the same in sigmatropic rearrangements.

You should consult your textbook and/or lecture notes to see if you are responsible for any examples of sigmatropic rearrangements.

1-methoxy-3-nitrobenzene is a disubstituted benzene.

The lower number is assigned to the methoxy group.

The compound can be called a monosubstituted deriva tive of anisole.

The compound can be called a trisubstituted derivative of the methoxy group.
This means that the methoxy group is benzoic acid.
The namebenzoic acid is connected to C1 by definition.
We must assign numbers starting with the carbon atom bers in a clockwise fashion so that the carboxylic acid group is not connected to the nitro group.
The car is more important than C5.
The compound is connected to C1 by definition.

The compound can be named 2,4,6-trichlorobenzoic acid.

2-methyl-1,3-dinitrobenzene is a trisubstituted benzene.

The numbers are assigned to give the lowest possible numbers to the substitutents.
"m" precedes "n" in the name, so the methyl group appears first.

The compound can be named as a disubstituted derivative.

Toluene is assigned to the chlorine substituent.
The lower number is used because it comes first alphabetically.

The methyl group is a derivatives of phenol.

By definition, we are using a common name.
In this case, the numbers can be nol) as the parent, we must assign numbers starting with the carbon assigned either clockwise or counterclockwise.
There will be two nitro groups at C2 and C6.
The group is connected to C1.
The compound can be named 2,6-dinitrotoluene.

There are 3 hybridized carbon atoms.

There are 3 hybridized carbon atoms.

The first criterion for aromaticity is satisfied.

This cation is aromatic.

There are 6 p electrons and this pair is not available to function criterion.

This is a base and cation is aromatic.

This structure is very unstable because it is antiaromatic.

The sulfur atom has two pairs.

It is not possible for this anion to be stable.

This anion is classified as antiaromatic because it is very unstable.

There are 4 p electrons in this case.

The signal with the highest wavenumber is created by hydrogen and the carbonyl group "a" below is part of it.

The other bonds that are highlighted are part of a ketone and produce a signal of 1720 cm-1.

Single bonds are part of the carbonyl group.

Oxygen has a smaller mass than chlorine, so a C--O bond is better than a saturated one.

The signal with the lowest wavenumber will be produced by the C--O bond.

The triple bond is stronger than the double bond and will produce a signal with a higher wavenumber.
The highlighted bonds are ranked in order of decreasing wavenumber.

An additional resonance structure that gives additional single-bond character to the C-O bond gives the signal at a lower wavenumber than the saturated ketone.

The signal at a lower wavenumber is caused by the decrease in the strength of the bond.

There are no double or triple bonds in this compound.

The strongest signal will be produced by the bond that doesn't have any functional groups and is classified as an alkane.

A terminal alkyne is a compound.

The compound is a disubstituted alkene.

We expect there to be a signal above the dipole moment for a relatively strong 3000 cm-1, likely near 3100 cm-1.

The other two compounds have the same bond with no dipole moment because they occupy the same electronic environments.

We draw all resonance structures and consider the location of the partial positive charge in the last resonance structure.

There are two signals at 3350 cm-1 and 3450 cm-1.

The structure of a primary amine is consistent with the presence of these signals.

The IR spectrum has a strong signal above 1700 cm-1.

Some of the character of the last resonance structure will be explained in the solution to Problem 2.21).

The IR spectrum has a weak signal at 3400 cm-1.

This signal is consistent with the structure of a secondary amine.

There is a weak signal at around 1650 cm-1 in Spectrum A.

There is a signal just above 3000 cm-1, suggesting that the pres the C --H bond occupy very different electronic environments.

The following structure, which has an unsymmetrical C===C bond result, will produce a relatively strong signal in with at least one hydrogen atom: an IR spectrum.

The IR spectrum has a broad signal.

The structure of an alcohol above 1700 cm-1 is not consistent with the IR spectrum's signals above 2200 cm-1 and 3600 cm-1.

There is a signal at 1600 cm-1 that suggests the presence of C bonds.

A strong signal just above 1700 cm-1 is consistent with the structure of the IR spectrum, which is between 2200 cm-1 and 3600 cm-1.
The structure of a carboxylic acid is consistent with the signals in the IR spectrum.

The IR spectrum has a broad signal.

This signal is consistent with the structure of alcohol.

There is a strong signal just above 1700 cm-1 in the IR spectrum.

There is a strong signal above 1700 cm-1 in the IR spectrum Spectrum C. The is similar to the structure of a carboxylic acid.

The IR spectrum does not have any signals above that should be ignored.

It is not consistent with the structure of an alcohol or a carboxylic acid.

The IR spectrum has a broad signal.

This signal is consistent with the structure of alcohol.

The IR spectrum has an extremely broad signal between 2200 cm-1 and 3600 cm-1, as well as a strong signal just Spectrum D, which has a pair of signals at 3350 cm-1 and above 1700 cm-1.

The structure of a carboxylic acid is consistent with the signals in the IR spectrum.

The spectrum F has a broad signal between 3200 cm-1 and 3600 cm-1.

This signal is consistent with the structure of alcohol.

The protons of the five methyl groups are not all equivalent to each other, because they are via symmetry.

The left and right methyl groups represent different types of protons, while the left and right methyl groups represent the same type of protons.

The three groups on the left side of the structure are equivalent to the three groups on the right side of the structure.

In the previous problem there was additional symmetry that wasn't present in this one.

The groups on the left side are the same as the groups on the right side.
This compound will have a single signal from the protons in the spectrum.

The oxygen atoms have two signals interchanged with the protons.

The spectrum of this compound will have three signals.

The two groups occupy different electronic environments and are not the same.

On the left side of their group is closer to the oxygen atom than on the right side.

We have counted four signals so far.

The methylene (CH ) group is the same as the CH protons.

This compound will have six signals in the NMR spectrum.

The five aromatic protons give rise to three distinct signals, which represent two different types of protons, not just one responding to the highlighted protons shown here, because they do not all occupy the same electronic environments.

The spectrum of this compound will have four signals.

The structures below have the same formula for C H and all 18 protons are the same.

In the first structure, all nine CH2 groups are equivalent, giving rise to just one signal.

The compound will have four signals in the NMR spectrum.

The oxygen atom of an ether is alpha to the methyl protons on the left side of the structure.

The predicted chemical shift is based on the alpha of the protons on the right side of the structure.

Two other signals are related to methylene groups.

The methylene group is on the left.

For the remaining methylene group, the benchmark value adds +3, so the predicted chemical shift is 0.9 + 3.

This methylene group is alpha to a carbonyl group, but the left side of the structure has a mark and a mark to the oxygen atom of an ether.

This compound has two alpha to a carbonyl group, which adds +1.0, so the predicted CHEM signals are related to the groups of protons highlighted below.

The groups of protons highlighted below are related to the three signals in the protons NMR spectrum of this compound.

One signal is produced by the methyl groups, which are similar to each other.

The first methylene group should give a signal around the ring if you start at the top of the ring and go counterclockwise to two ether groups.

One signal is produced by the methylene groups, which are equivalent to each other, and the methylene group is alpha to a carbonyl group.

The next methylene group should give a signal of alpha to one ether group and alpha to another ether group because a methylene group has a benchmark value of 1.2 and +0.5.

The compound has four oxygen atoms in the NMR spectrum.

The last methylene group should give a signal near 4.2 parts per million because the benchmark value of the group is 1.2 parts per million.

The groups of protons highlighted below will correspond to the four signals in the NMR spectrum of this compound.

The first two groups that produce one signal are counterclockwise around the ring.

The methylene group should give a signal near 1.9 ppm because it is a carbonyl group with a benchmark value of 1.2 ppm and it is also abeta to the oxygen of an ether.

A methylene group should give a signal near 2.2 parts per million because it has a benchmark value of 1.2 parts per million and is alpha to a carbonyl group.

A methylene group has a benchmark value of 9.1 and we divide it by 1.2 to find the smallest integration value.

The formula shows that there is alpha to a carbonyl group.
The numbers 5:2:2:1 are not representative of the total number of protons.

The compound has 2 protons, 2 protons, and 1 protons.

The smallest integration value is identified below.

The ratio is 1:6 because we divide the integration values by that number.

The values 1 and 6 are just relative numbers because the formula indicates there are 14 protons.
In order for the total number of protons to be 14, they represent 2 protons and 12 protons.

The values 1, 1, and 1 are just relative numbers because of the fact that there are a total of 6 protons.

In order for the total number of protons to be 6 we have to see 2 protons and 2 protons in order.

There are three shifts in the spectrum of the compound's protons.

The predicted chemical shift is 1.2 + 2.5.

The predicted chemical shift is 1.2 + 2.5, because the methylene group is alpha to the oxygen atom of an ether.

Two of the signals should be very close to each other, if not overlap.

The predicted chemical shift is 1.7 + 0.5 and the methine proton is abeta to the oxygen atom of ether.

The problem statement will be preceded by three ately by the proton NMR spectrum of this compound.

The methylene groups don't split because they are both protons.

They collectively give rise to a singlet.

The groups of protons highlighted below will correspond to the four signals in the NMR spectrum of this compound.

The groups of protons highlighted below will correspond to the four signals in the NMR spectrum of this compound.

The groups of protons are highlighted below.

The groups of protons highlighted below will be reflected in the spectrum of the compound.

We doublet with an integration of 6 to calculate degrees of unsaturation.

We can ignore the presence of an isopropyl group if the two signals are characteristic.

To be fully saturated, a compound with 8 carbon atoms would need this case, 13.9), and we divide all of the integration values by 18 hydrogen atoms.

The ratio is 1 : 1.5 and the compound has only 14 hydrogen.
There is no such thing as a hydrogen atom.
To give this compound 2 degrees of unsaturation, we have to add these numbers by 2.

We can ignore the presence of 3 protons if we want to calculate degrees of unsaturation.

Now that we know the integration values, we can see that this has the same HDI as a compound with the formula C H.

A compound with 5 carbon atoms would need an integration of 3 to be fully saturated.

There are two signals that are characteristic of having 12 hydrogen atoms.
The compound has 12 hydro ethyl group.

This compound has 0 degrees of unsaturation.

The ratio is 1:6 for calculating degrees of unsaturation.

We can see that the spectrum has a septet with an atom for each nitrogen atom, if we subtract one hydrogen tion values.

A total of 14 pro this compound has 0 degrees of unsaturation according to the formula.

We can ignore the presence of protons to be 14 if we represent 2 protons and 12 protons in order for the total number of inspect the formula.

The smallest integration value is the same as a compound with the formula C H.

A compound with 8 carbon atoms needs a number to be fully saturated.

The formula indicates the number of hydrogen atoms.
The compound has 10 hydrogen, but there are only 10 protons and it is missing 8 hydrogen atoms.
The compound sent both relative values and exact values, which has 4 degrees of unsaturation.

The calculation of the HDI is the first step.

The formula of C H O has 1 degree of unsaturation now that we know the integration values.

This structure does, but not both.

The gration value of each signal is used to calculate degrees of unsaturation.

The spectrum has four signals, and we can look at the formula and determine the relative integration of each signal by dividing oxygen atoms.

A compound with 6 carbon atoms needs 1.5 to have 14 hydrogen atoms to be fully saturated.

The compound has only 10 hydrogen atoms.

These have 2 degrees of unsaturation because there is no such thing as half of a protons.

The molecu atom for each nitrogen atom is the total of the integration values.

The lar formula shows the presence of 10 protons.

The values are relative and represent the formula C H.

The analysis of each signal is the next step.

The easiest sig Gen atoms are missing.
The compound has 3 degrees of nal to analyze, and the singlet is just above 2 ppm.
The signal has an unsaturation.

The signal has been shifted because all of the given integration values by the smallest value have a benchmark value.

The process gives a ratio of downfield by a small amount.

The double bond accounts for the 1 degree of mula and indicates the presence of 10 protons in the structure.

The analysis of each signal is the next step.

The singlet just above 2 ppm is the easiest sig more than 1 ppm.

A methyl group group has a benchmark value of 0.9 parts per million, but this signal has been shifted downfield by a little more than 1 part per million.

Let's look at the last two signals.

A signal with an integration of 3 represents a group.
The double bond accounts for the 1 degree of unsatu other signal and the integration of 2 is a CH group.

A signal with an integration of 3 represents a group.

The other signal has an integration of 2.

There is a signal associated with the CH2 group that has neighbors on either side.

This gives a sextet, which is exactly what we see.

There are many ways to connect these fragments.

The calculation of the HDI is the first step.
This compound has 1 degree of unsaturation.

The structure must have either a ring or a double bond.

The number of signals and the inte gration value of each signal are the next steps.

This accounts for one of the 2 degrees of unsaturation, but we still need one more degree of unsaturation, and we have already accounted for all atoms in the formula.

There is a signal associated with the CH2 group that has neighbors on either side.

If we consider the chemical shifts of the signals associated with as if this CH group has five equivalent neighboring protons, they are all consistent with expected predictions.

The calculation of the HDI is the first step.

The calculation of the HDI is the first step.

The next step is to consider the number of signals.

The gration value of each signal is given by this process.
The spectrum has three signals and three integration values.

The molec and the integration indicate that the ring is monosubstituted.

The signals are associated with a CH2 group.

Refer to the multiplicities of these three signals.

The two fragments account for all of the atoms in the formula.

The calculation of the HDI is the first step.
This compound has 1 degree of unsaturation.

The structure must have either a ring or a double bond.

The number of signals and the inte gration value of each signal are the next steps.

The spectrum has four signals and integration values of 1H, 2H, 6H, and 3H.

We are ready to analyze each signal, one at a time, but it might be more efficient to look for patterns of signals.

The calculation of the HDI is the first step.

The presence of an aromatic ring is strongly suggestive of the two signals.

The number of signals and the inte k gration value of each signal are the next steps.

The aromatic ring is typified by the signal just above 7 ppm, and the integration indicates that the ring is monosubstituted.

The CH signal is sig nificantly downfield from the benchmark value for a CH group.

There is a quartet with an integration of 2H and a triplet.

Most of the atoms in the formula are accounted for by the two fragments, with the exception of one oxygen atom and one Notice.

The CH group is connected to the spectrum.

The carbonyl group is expected to produce signals between 150 and 220 parts per million, while the hybridized carbon atoms are expected to produce signals between 100 and 150 parts per million.

The oxygen atom connected to one of them is expected to appear between 50 and 100 parts per million.
3 hybridized carbon atoms are expected to produce a signal.

There are six carbon atoms in this compound, but two of them are the same as the other two, so we expect only five signals.

The compound has 10 carbon atoms, but also has symmetry, so we expect fewer than 10 signals.
There are six different types of carbon atoms in this compound, so we expect a total of six signals in the 13C NMR spectrum.
The highlighted carbon atoms were hybridized.
The expected location of each signal atom is expected to produce signals in the range of 0-50 parts per million.

There is no symmetry in this compound, so we expect five signals.

Two of them are connected to the k oxygen atom.

As a result of the electron-withdrawing effect from the oxygen atom, the two carbon atoms will each produce a signal between 50 and 100 parts per million, while the other three carbon atoms will produce signals between 0 and 50 parts per million.

The compound has six carbon atoms, but there is no symmetry, so we expect only three signals.

3 hybridized carbon atoms are expected to produce signals.

The signal associated with the benzylic position is expected to be shifted downfield by its proximity to the aromatic ring.

We expect one signal for each carbon atom for a total of nine signals.

One of them is connected to the oxygen atom.

The electron-withdrawing effect from the oxygen atom will cause the other two carbon atoms to produce signals between 0 and 50 ppm.

We expect five signals.

Two of the signals are connected directly to the oxygen atom.

As a result of the electron-withdrawing effect from the oxygen atom, the two carbon atoms will each produce a signal between 50 and 100 parts per million, while the other three carbon atoms will produce signals between 0 and 50 parts per million.

There are only two steps in this mechanism.

In the second step of the mechanism, water functions as a base and removes a protons from the sigma complex to restore aromaticity.

There are only two steps in this mechanism.

A chlorine atom can be installed on the ring by treating ben case, a tertiary carbocation is involved, and it cannot rearrange to zene with Cl in the presence of a Lewis acid.

The transformation requires the installation of a butyl group.

Friedel-Crafts a mixture of products can be used to achieve this alkylation.
The desired alkylation process, without the concern of rearrangements, can be achieved via a Friedel-Crafts acylation.

The installation of a primary k alkyl group is required.

A secondary carbocation is involved in the Friedel-Crafts alkylation process, and there is no way for it to rearrange to become tertiary.

A primary alkylchloride is used in the Friedel-Crafts alkylation process.

The primary carbocation is not likely to be formed.

The complex has the character of a primary carbocation.

The exact product mixture is difficult to predict, but this product is likely the major product.

The acylium ion can be used in an aromatic substitution reaction.

The desired transformation involves removing a sulfonate fonate group on the aromatic ring, which can be achieved with group from the aromatic ring, which can be achieved with a concentrated form of sulfuric acid.

The desired transformation involves removing a sulfonate group from the aromatic ring.

A mixture of nitric acid and sulfuric acid can be used to transform the aromatic ring.

The desired transformation involves the installation of a sulfonate group on the aromatic ring.

The desired transformation involves the installation of a chlorine atom on the aromatic ring, which can be achieved with a Lewis acid.

The desired transformation involves the installation of a carbocation rearrangements), and then reduce the resulting ketone methyl group on the aromatic ring, which can be achieved with a Clemmensen reduction to give the desired product.

Installation of a primary alkyl group on the aromatic ring is the desired transformation.

The Friedel-Crafts alkylation would cause a mixture of products.
The mechanism has three steps.

The mechanism can be drawn in two steps.

The aromatic ring is deprotonated by water in the first step to give a sigma complexample.

The ring has a halogen substituent.

It is not an exception that this ring has a hydroxyl group.

The existing substituent director must be considered.

The acylation case is expected to be a deactivator.

It is not an exception that this substituent is not giving two products.
The substituent does contain three chlorine atoms, but they are not connected to the aromatic ring, so there are no resonance effects.
The three chlorine atoms add together to give a powerful electron-drawing effect.

The problem statement shows that the substituent in this case is an activator.

The nitra director is indicated by the reagents (nitric acid and sulfuric acid).

The nitra aromatic ring is indicated by the reagents.

In order to draw the product, we must consider the tion reaction, which means that a nitro group will be installed.
The aromatic ring must be considered.
In order to draw the product, we must consider existing substituent.
The nitro group will be installed.
The existing substituent is a deactivator.
We have the existing substituent and its effects.

We need to consider where the bromine atom will be installed in order to draw the product.

The Friedel-Crafts alkylation reaction is indicated by the reagents.
A group is installed on the aromatic ring.
The existing substituent is a deactivator.
We need to consider the existing substituent and its effects.

We need to consider where the chlorine atom will be installed in order to draw the product.

We need to consider the existing substituent and its effects.
The Friedel-Crafts acylation reaction is indicated by the reagents.
An acyl group is installed on the aromatic ring.

The ring has two groups, one strong and one weak.

The directing effects will be controlled by the activator when they compete with each other.

The ring has two groups, a strong and a weak one.

The directing effects are controlled by the strongest activator.

The aromatic ring has two groups.

The strongest activator is expected to control the directing effects in the following locations.

The ring has two groups, a strong deactivator and a strong activator.

There is no competition between the weak and strong groups of this aromatic ring.

The aromatic ring is an alkyl group connected to the vator.

This substituent is a moderate aromatic substitution.

The aro k matic ring has a halogen connected to it.

There is a group connected to the tive toward aromatic substitution.

The substituent has a single pair connected to the aromatic ring.

The ring is electron-rich because there is a bond connected to it.

The effect is similar to an aromatic ring.

The bromination reaction is indicated by the reagents.

The aromatic ring is monosubstituted, and the alkyl group is a weak activator.

A methyl group will be installed on the ring after the Friedel-Crafts alkylation reaction is indicated by the reagents.

This substituent is a moderate deactivator because there is a C=O bond connected to the k aromatic ring.

A bromine atom will be installed on the aromatic ring because the reagents indicate a bromination reaction.

The ring will have only one pair of nitrogen an acyl group on it.

There are two substituents in this ring.

The more powerful activator controls the directing effects when a moderate activator competes with a weak one.

There are two substituents in this ring.

There are two substituents in this ring.

There are two substituents in this ring.

Next, we perform the desired reaction, and then we remove the blocking group to give the desired product.

Next, we perform a Friedel-Crafts acylation reaction and remove the blocking group to give the desired product.

The reagents show that this is a nitration reaction.

The reagents show that occur there quickly.

A nitration reaction can happen at either of the other two.
The aromatic ring has two products.

The aromatic ring is going to be installed.

The aromatic ring has two substituents.
Both products are expected.

The position on the left is more accessible than the position on the right.

The sulfonation reaction is indicated by the reagent.

The groups are at the same three locations.

The lower position is more accessible because of the sterically hindered upper position.

The aromatic ring needs to be installed with an isopropyl group and chlorine atom.
The aromatic ring can be treated with Cl and the chlorine atom can be installed via a Friedel-Crafts alkylation.

We must consider the order of events now that we know how to install each substituent individually.

Consider the fact that chlorine is a deactivator while an isopropyl group is an activator.
To maximize the efficiency of our synthesis, it makes more sense to install the isopropyl group first, so that we can take advantage of the enhanced reactivity of isopropyl benzene.

The aromatic ring needs a bromine atom and a nitro group installed.

The bromine atom can be installed with the help of a catalyst.

We must consider the order of events now that we know how to install each substituent individually.

The groups can be installed via Friedel-Crafts alkylation.

We must consider the order of events now that we know how to install each substituent individually.

Consider steric effects.
Any reaction we perform after installing the methyl group will produce a mixture of products.

We know how to install each substituent individually.

The order of events must be considered when installing a nitro group.
A mixture of products would be produced if we tried to install the nitro.

A Friedel-Crafts is not the correct location.

The ring is aromatic.

There is no need for a blocking group because the bromine atom can be installed in one step.
The syn is treating an aromatic ring.

We perform a Friedel-Crafts acylation, but can't install it directly via a Friedel-Crafts alkylation.

We must first perform a Friedel-Crafts acylation and then reduce the ketone with a Clemmensen reduction.

We must consider the order of events now that we know how to install each ally.

Directing effects of the carbonyl group before we reduce it.

We must consider the order of events now that we know how to install each substituent individually.

A blocking group is not needed because this is the desired product.

The synthesis of butyl group and bromine atom is summarized.
The Friedel-Crafts alky is performed on the aromatic ring.
The butyl group can be installed via a lation, followed by Friedel-Crafts alkylation and a bromine atom.

We must consider the order of events now that we know how to install each ally.

The bromine is a deactivator.
The butylbenzene is related to benzene.

We remove the blocking group with acylation and then reduce the resulting ketone with a Clemmensen dilute sulfuric acid to give the desired product: reduction.

We must consider the order of events now that we know how to install each ally.

The propyl group and chlorine are deactivators.
propylbenzene is used to install Cl.

There is a bromine atom on the aromatic ring.

We must consider the order of events now that we know how to install each ally.

There is a chance that this compound can serve as an electrolyte in a nucleophilic aro fore.

The compound cannot function as a leaving group because a negative charge on a NOT is very unstable.

This compound will not cause a stitution reaction.

This group does not have a leaving group.

A negative charge on a carbon ophilic aromatic substitution reaction will not allow alkyl groups to serve as a suitable electrophile in a nucle function.

This compound isn't a good choice for an aromatic substitution reaction.

The Meisenheimer complex has resonance structures that should be drawn.

This could be a complete mechanism.

The product is deprotonated under basic conditions.

The Meisenheimer complex has resonance structures that should be drawn.

This could be a complete mechanism.

The product is deprotonated under basic conditions.

The Meisenheimer complex has resonance structures that should be drawn.

This could be a complete mechanism.

The product is deprotonated under basic conditions.

This compound has a leaving group, but it lacks a powerful drawing group, so the reaction does not proceed via an S Ar mechanism.

The two products shown are due to the reaction involving a benzyne intermediate which can be attacked by hydroxide in either of two locations.

Proceed through an elimination-addition process.

This compound has a leaving group, but it lacks a powerful drawing group, so the reaction does not proceed via an S Ar mechanism.

The reaction is expected to involve a benzyne intermediate.

The two products shown are the result of the attack on the benzyne intermediate.

The compound lacks a powerful electron-drawing group.

The reaction doesn't proceed via an S Ar mechanism.

The reaction is expected to involve one of two benzyne intermediates, which can be attacked by hydroxide to give one of the three products shown.

The three criteria for S Ar have been met.

The solution to Problem 5.11 is for the mechanism.

We can rule out an aromatic substitution reaction because the reagent is strong.

The starting compound doesn't have all the criteria for an S Ar mechanism.
The compound has a leaving group, but it lacks a drawing group, so the reaction is not an S Ar mechanism.

The elimination-addition process is indicated by the high temperature.

This could be a complete mechanism.

The product is deprotonated under basic conditions.

The reagent in the first step is a strong nucleophile, and the starting compound meets all three criteria for an S Ar N mechanism: an electron-drawing group and a leaving group that are para to each other.

The Meisenheimer complex has resonance structures that should be drawn.

This could be a complete mechanism.

The product is deprotonated under basic conditions.

We can rule out an aromatic substitution reaction because the reagent is strong.

The reaction must be an elimination-addition process.

This could be a complete mechanism.

A primary alcohol is converted into an aldehyde after being cleaved.

Two bonds of this compound are cleaved, as shown.

A primary alcohol can be converted into an aldehyde after treatment with a mild oxidizing agent.

A carboxylic acid would be produced by treating this alcohol with a stronger reagent.

A primary alcohol can be converted into a carboxylic acid via a chromic acid oxidation using either sodium dichromate and sulfuric acid or the Jones reagent.

When an alkene undergoes ozonolysis, the C==-C bond is cleaved and replaced by two C==-O bonds.

A secondary alcohol can be converted into a ketone with an oxidizer such as sodium dichromate and sulfuric acid or the Jones reagent.

Two equivalents of a ketone are given by ozonolysis of the alkene.

Here are the curved arrows for protonation of the other oxy with a water source.

Each of the oxygen atoms is protonsated when water is introduced into the reaction flask.
It doesn't matter which one is shown first if you draw each step separately.
The mechanism has seven steps.
The first three steps produce a hemiacetal, while the last four steps convert it to an acetal.

The carbonyl group is even more electrophilic because the ketone is protonsated.

The carbonyl is attacked by the alcohol.

This intermediate is deprotonated to give a hemiacetal.

An excellent leaving group is generated by the hemiacetal.

The carbonyl group is regenerated by expelling the leaving group.

The oxonium ion is given by the carbonyl group.

The mechanism has seven steps.

The first three steps produce a hemiacetal, while the last four steps convert it to an acetal.

The carbonyl group is even more electrophilic because the ketone is protonsated.

The carbonyl is then attacked by a liquid.

This intermediate is deprotonated to give a hemiacetal.

An excellent leaving group is generated by the hemiacetal.

The carbonyl group is regenerated by expelling the leaving group.

The oxonium ion is given to the carbonyl group.

The mechanism has seven steps.

The first three steps produce a hemiacetal, while the last four steps convert it to an acetal.

The carbonyl group is even more electrophilic because the ketone is protonsated.

The OH group attacks the carbonyl in an intermolecular fashion.

This intermediate is deprotonated to give a hemiacetal.

An excellent leaving group is generated by the hemiacetal.

The carbonyl group is regenerated by expelling the leaving group.

The oxonium ion is given by the carbonyl group.

The process has seven steps.

The reverse process would expect the order of the intermediates to be in the opposite direction.

The acetal is turned into a better leaving group that is consistent with acidic conditions by converting the OEt group into a better leaving group.

The leaving group is expelled to create a bond.

The oxonium ion is given by water after the C=O bond is attacked.

The oxonium ion is deprotonated.

The OEt group is converted into a better leaving group with the help of the hemiacetal.

The leaving group is given a ketone.

Under acidic conditions and with removal of water via a Dean-Stark trap, a ketone can be converted into a cyclic acetal.

The bond of the ester is not reacting under these conditions.

The desired transformation involves the reduction of the carbonyl group of a ketone.

There is a reduction in the amount of a ketone.

The carbonyl group of an aldehyde is used to give a methyl group in the desired transformation.

Raney nickel can be used to reduce the cyclic thioacetal by converting the aldehyde into it.

The same transformation can be achieved with a reduction.

Under acidic conditions, the carbonyl group is reduced to give a methylene group.

The reduction of the carbonyl group of an aldehyde is needed to give an alkane.

Raney nickel can be used to reduce the cyclic thioacetal by converting the aldehyde into it.
The same transformation can be achieved with a reduction.

The same transformation can be achieved with a reduction.

The desired transformation involves the reduction of the carbonyl group of a ketone.

The same transformation can be achieved with a reduction.

A Wolff-Kishner reduction is indicated by the starting material and the reagents.

The reduction of the carbonyl group of an aldehyde is needed to give an alkane.

The reagent is hydroxylamine and the starting material is a aldehyde.

The starting material is a symmetrical ketone.

The reagent is a secondary amine.

A ketone is converted into an enamine after being treated with a secondary amine.
We expect a mixture of enamines and less-substituted enamine reagent with the secondary amine.

The reagent is a primary amine.

A ketone is converted into an imine after being treated with a primary amine.

The starting material has a primary amine and a ketone tethered together in the same molecule.

The starting material is a symmetrical ketone.
The removal reagent is hydrazine.

When a ketone is treated with a Grignard reagent, followed by water, it is reduced to an alcohol with the installation of an R group from the Grignard reagent.

When a ketone is treated with a reagent, the reaction follows.
Under acidic conditions, a secondary amine will react with a ketone to give the installation of an R group from the Grignard reagent.

A ketone will react with a Wittig reagent.

The aldehyde is reduced to an alcohol with the installation of an R group from the Grignard reagent.

A Wittig reagent will give an alkene.

LiAlH is a reducing agent.

A ketone will react with a Wittig reagent.

The Wittig reagent does not have any R groups.

The starting atom has a negative charge and is connected to two hydrogen ketone.
A methylene group is installed in the place where the oxygen atom will be inserted.

A peroxy acid is being used to treat a ketone.

We don't have to decide where the oxygen atom inserts itself because the starting ketone is symmetrical.

We can see that the starting ketone is notsymmetrical.

Predicting where the oxygen atom will be inserted is important.
We look at both sides of the group.
The carbon atom to the left of the carbonyl group is connected to two alkyl groups, while the carbon atom to the right of the carbonyl group is connected to one alkyl group.

A peroxy acid is being used to treat an aldehyde.

With the installation of a CH group, a ketone can be converted into an alkene.

A symmetrical ketone can be converted into an ester through a Baeyer-villiger oxidation.

The product is an alcohol and the starting material is a ketone.
The carbonyl group is reduced with the installation of a methyl group.

In the presence of an acid catalyst, a ketone can be reduced to give a secondary alcohol.

There is a carbon atom in the starting material.

The starting material is an unsymmetrical ketone.

An oxygen atom needs to be inserted between the aromatic ring and the carbonyl group to make the difference between starting material and product.
The oxidation reaction can be used to accomplish this.

Wolff-Kishner reduction is used to give the desired product.

The starting material is aldehyde, and the desired product is alcohol.

We must install two groups.

We can use a Grignard reaction to convert an aldehyde to alcohol, or we can oxidize it to give a carboxylic aldehyde, and then install one methyl acid upon treatment with a peroxy acid, via a group.

The alcohol is similar in structure to the Baeyer-villiger oxidation, but we are missing a group.

The starting acetal can be used to make this ketone.

A two-step synthesis has been developed for the desired product.

The starting material is alcohol.

The starting material is an alkene.
An alkane is the last step of synthesis.
We haven't seen a direct Baeyer-villiger oxidation process, in which the product can remove a methylene group.

The starting material can be used to make this ketone.

A two-step synthesis has been developed for the desired product.

The starting alkene can be used to make this ketone.

The following synthesis was developed for the desired product.

The starting material is an acetal.

There are not many ways to make epoxides.

In this chapter, we have seen that a sulfur ylide can be used to treat a ketone.

The starting material is alkene, and the desired product is alcohol.

There are no direct ways to achieve this transformation.

The starting alkene can be used to make this ketone.

The following synthesis was developed for the desired product.
The starting material is an alkene, and the product is a car to convert it into a ketone.
The product has a reduced carbon atom to give the desired product.
The starting material can be reduced.

The desired product is an alkene, and the starting material is aldehyde.

There are no direct ways to make uct is an ester.
We haven't seen many ways to make esters.
In this chapter, we have seen that an alkene can be made from a ketone via the Wittig reac Baeyer-villiger oxidation process.

A Grignard reaction and oxida two-step process can be used to make this ketone.

The following synthesis has been developed.
The following synthesis was developed for the desired product.
The starting aldehyde is treated with a desired product.
The starting aldehyde is treated with ethyl magne sium bromide, followed by water, to give an alcohol, which is then sium bromide, followed by water, to give an alcohol, which is then oxidation to give a ketone.
The ketone is treated with an oxidizer.

We need to make a C-C bond to introduce a cyclohexyl group into the compound.

The starting material is a acetal, and the desired Grignard reagent and a carbonyl group, so we oxidize the alco uct.
We haven't seen many ways to make esters, hol with PCC to give an aldehyde, and then treat that aldehyde with water, but we have seen that an ester can be made from a ketone via a cyclohexyl magnesium bromide, followed by water.
The oxidation process.
The last step of our synthesis alcohol can be oxidation to give the ketone above, in which the product converted into the desired product as shown above.

The starting acetal can be used to make this ketone.

A two-step synthesis has been developed for the desired product.
An excess of Grignard reagent is being used to treat an acid halide.
A Grignard reagent attacks the carbonyl group of the acid, giving a tetrahedral intermediate, which then re-forms the carbonyl group by expelling chloride as a leaving group.
The Grignard reagent is used to attack the ketone and give it an alkoxide ion.

An acid halide is being treated with alcohol and pyridine.

The alcohol attacks the carbonyl group of the acid halide.
The carbonyl group is re- formed by expelling chloride.

An acid halide is being treated.

The amine attacks the carbonyl group of the acid halide.
The carbonyl group is re- formed by expelling chloride.

After treatment with thionyl chloride, an acid halide is converted into a carboxylic acid.

The carboxylic acid works as a nucleophile and attacks the S O bond.
After expelling a chloride ion, the bond is re- formed.
The first three steps resulted in the conversion of an OH group into a better leaving group.
The carbonyl group can be re- formed by expelling SO gas and a chloride ion.

A hydride reducing agent is being used to treat an acid halide.

This is the end of the reaction because the alkoxide ion cannot re-form the carbonyl group.

The acid is converted into an acid halide.

When the acid is treated with an excess of ethyl magnesium bromide, the acid reacts with two equivalents of the halide, but only one equivalent of the nucleophile.
There are two ethyl groups.

The alcohol must be made with one ethyl and one methyl group.

We need to install two different alkyl groups in order to achieve this.
The desired transformation can be achieved by converting the carboxylic acid into an acid halide and then treating the acid with a cuprate.
An excess of a hydride results in the installation of only one alkyl group to give a ketone.

The Grignard reagent can be used to treat this ketone.

After treatment with an alcohol and pyridine, an acid halide is converted into an ester.

We used Me CuLi in the synthesis.

The product is alcohol and the starting material is carboxylic acid.

We need one ethyl to make alcohol.

This can be accomplished by converting the carboxylic acid into an acid halide and then treating it with a cuprate.

Only one alkyl group is installed to give a ketone.

Reducing agent can be used to give the desired alcohol.

The desired ketone can be made by treating an acid with a cuprate.

The starting carboxylic acid can be used to make acid halide.
After treatment with a carboxylic acid in the presence of pyridine, an acid halide can be converted into an acid anhydride.
The carboxylic acid is a nucleophile that attacks the carbonyl group of the acid halide, and pyridine is a base to remove the acidic protons generated during this process.

The mechanism for this process should have six steps.

The two core steps are the attack the carbonyl group and the re-form the carbonyl group.

The carbonyl group is subjected to attack by a weak nucleophile in the first step of the mechanism.
In this case, the alcohol is tethered to the carbonyl group and allows for an attack.
The resulting oxonium ion is then deprotonated by water and hydronium ion to give an intermediate that can re-form the carbonyl group.

An alcohol and carboxylic acid can be given by an ester under acidic conditions.

The two core steps are the attack the carbonyl group and the re-form the carbonyl group.

The carbonyl group is subjected to attack by a weak nucleophile in the first step of the mechanism.
Water attacks the carbonyl group.

The alcohol and carboxylic acid are tethered to each other in one molecule after the leaving group leaves.

An alcohol and carboxylic acid can be given by an ester under acidic conditions.

The mechanism for this process should have six steps.
The two core steps are the attack the carbonyl group and the re-form the carbonyl group.
The carbonyl group is subjected to attack by a weak nucleophile in the first step of the mechanism.
Water attacks the carbonyl group.

A carboxylic acid and alcohol are given to the patient after treatment with H O+).

A carboxylic acid and an alcohol are given to the patient after treatment with hydroxide and a protons source.

A carboxylic acid is converted to an ester after treatment with an alcohol.

A carboxylic acid and an alcohol are given to the patient after treatment with hydroxide and a protons source.

In the first step of the mechanism, hydroxide attacks the cyano group.
The resonance-stabilized anion is given by a deprotonation step after the anion is protonsated by water.
A amide is made under acidic conditions.

The carbonyl group is subjected to attack by a weak nucleophile in the first step of the mechanism.

This intermediate is deprotonated by water and then protonsated by a hydronium ion to give an intermediate that can re-form the carbonyl group.

In the first step of the mechanism, hydroxide attacks the cyano group.

The resonance-stabilized anion is given by a deprotonation step after the anion is protonsated by water.

The carbonyl group of an amide can be attacked by Hydroxide to give a tetrahedral intermediate.

The carbonyl group is first protonsated under acidic conditions.

Water attacks the carbonyl group.

The carbonyl group is re- formed when a C-N bond is broken.

The nitrogen atom needs to be protonsated in order to break the C-N bond.
Deprotonation gives the products.
The final step can be drawn as either deprotonation to make the carboxylate ion or expulsion of the amine and carbon dioxide.

We do not have acid.

esters are less reactive than acid halides, so we expect the amide to undergo acid-catalyzed hydrolysis in one step.
We can first convert the ester into a carboxylic acid and then convert the carboxylic acid into the desired acid.

An acid chloride is more reactive than an ester, so it can be converted into an ester.

An alcohol and pyridine can be used to achieve this transformation.

There is a starting material and a reagent.

We don't have a way to convert an amide into an acid halide in one step.
We can first convert the amide into a carboxylic acid and then convert the carboxylic acid into the desired acid.

The carboxylic acid cuprate can be converted into the ketone after treatment with lithium diethyl ester.

The acid is converted into the desired acetal.
The desired acid anhydride can be converted into a treatment with the help of an acid catalyst.
The product is an imine, which can be made from the corre learned a way to do this directly in one step.
We can first convert the amide into a carboxylic acid and then convert the carboxylic acid into the desired methyl ester.

If we can convert the starting material into the aldehyde above, this will be the last step of our synthesis.

This aldehyde can be made in two steps.

If we can convert the starting material into the ketone above, this will be the last step of our synthesis.

The starting carboxylic acid can be used to make this ketone.

If we can convert the starting material into the carboxylic acid above, this will be the last step of our synthesis.

If we can convert ketone, this can be the last step of our synthesis.

The starting material is converted into the ketone above.
A few steps from the beginning of the amide can be taken to make the ketone.
The carboxylic acid is given by the hydrolyzed with aqueous acid.
After treatment with the carboxylic acid, amide is converted into the desired acid halide.

If we can convert the aldehyde into the carboxylic acid, this can be the last step of our synthesis.

If we can convert the starting material into the aldehyde above, this will be the last step of the synthesis.

This aldehyde can be made in two steps.

An oxidizer can be used to oxidize the starting material.

The carboxylic acid can be converted in two steps.

Two steps can be taken to accomplish this.

The carboxylic acid an ester is first converted into an acid and then into a product called ketone, which is then converted into a liquid with the help of a cuprate.

The product can be made using a Wittig reaction.

If we can convert the ester to a carboxylic acid, this will be the last step in our synthesis.

The starting carboxylic acid can be used to make this aldehyde.

After converting the carboxylic acid into an acid halide, it is reduced to an alcohol with excess LiAlH.

If we can convert the alcohol into acid halide, this will be the last step of our synthesis.

The first step of acid halide can be made from alcohol in just two steps.

After the alcohol is converted into a carboxylic acid, it is converted into an acid halide with the help of thionyl chloride.

There are two alpha carbon atoms in this compound.

The alpha of the carbonyl group, and that carbon atom is connected to one carbon atom to the right of the carbonyl group.
The carbonyl group's alpha carbon atom is not an alpha carbon because it is not connected to the left three alpha protons.

There are two alpha carbon atoms in this compound.

There are two alpha carbon atoms in this compound.
The carbon atom to the right of the carbonyl group is connected to the two alpha protons on the other side of the atom.
The alpha carbon atom to the left of the carbonyl group protons is connected to the alpha carbon atom to the right.

The compound does not have any carbon atoms because the carbonyl group is not connected to any carbon atoms.

This compound does not have any protons.

This is a tautomerization in which a ketone is converted into an enol.

Under basic conditions, we first show deprotonation to give a resonance-stabilized anion, and then we show protonsation to give the product.

This is a tautomerization in which a ketone is converted into an enol.

This is a tautomerization in which an enol is converted into a ketone and there are two transfer steps.

Under basic conditions, we first show deprotonation to give a resonance-stabilized enolate anion, and then we show protonsation to give the product.

This is a tautomerization in which an enol is converted into a ketone and there are two transfer steps.

This is a tautomerization in which an enol is converted into a ketone and there are two transfer steps.

This enolate is very stable.

A ketone will be delocalized over two oxygen atoms.

This ketone has two alpha positions, but only one of them has protons, so the reaction occurs at this location.

A Hell-Volhard-Zelinsky reaction is indicated by the starting material being a carboxylic acid.

This enolate is very stable.

The starting material is a carboxylic acid.

The starting material is a ketone.

This ketone has two alpha positions, but only one of them has protons, so the reaction occurs at this location.

An enolate ion is created when one of the alpha protons is replaced with another alpha position.

The enolate ion is caused by the starting material and reagents being in an alpha position.

A haloform reaction is indicated in this case.
The starting material is a ketone, and the Enolate ion are stable by resonance.
The alpha position will be deprotonated by Hydroxide, and it will give an enolate ion.
The bromine atom is installed at the alpha position when this enolate ion functions as a nucleophile.

There is only one alpha proton in the starting material.

There is an extra group introduced at an alpha position if we look at the difference between the starting material and the product.

At an alpha position, the starting material is an unsymmetrical ketone.

The retrosynthetic analysis shows the bond that is made during an aldol condensation.

The first thing we need to do is draw two molecules of the ketone, so that the C is in the product.

Two alpha tons of the other ketone must have originally been located at the alpha pro alpha position because the oxygen atom of one ketone is pointing there.
We push the fragments together, connecting them with a double bond, after we erased the two alpha protons and protons.
The method is useful for drawing the product of an aldol condensation.

The first thing we do is draw two molecule of the ketone, so that we can see the bond made during an aldol condensation.

We erased the alpha and the C bond from the product.

The alpha position must have been the location of two addi connecting them with a double bond because the oxygen atom was highlighted.
This isn't a mechanism, but a useful method for drawing the product of an aldol condensa the location of a carbonyl group.

The first thing we do is draw two molecules of the aldehyde.

Below is a retrosynthetic analysis for the desired product, tons and the oxygen atom, and we push the fragments showing the C==C bond that is made during an aldol condensation.

To draw the starting materials for this aldol condensation, consider stereoisomeric products obtained, and the first one shown is the C+C bond in the product.

The starting materials will react with each other to give the desired product after treatment with NaOH.

Upon treatment with NaOH, these starting materials will react showing the C+C bond that is made during an aldol condensation.

Product is the starting materials for this aldol condensation.

In the first step, hydroxide is used as a base and removed from the ketone.

The carbonyl group of formaldehyde is attacked by the enolate ion.

The b-hydroxy ketone can be eliminated to give an a,b-unsaturated ketone.

The elimination process occurs in two steps.

The alpha protons are removed from the ketone bearing alpha protons in the first step.

The carbonyl group of the other ketone is attacked by the enolate ion.
The b-hydroxy ketone is given by the alkoxide ion being protonsated by H O.

The elimination process occurs in two steps.

In the first step, hydroxide is used as a base and removed from the ketone.

The carbonyl group of another molecule of the same ketone is attacked by the enolate ion.
The b-hydroxy ketone is given by the alkoxide ion being protonsated by H O.

The elimination process occurs in two steps.

The alpha protons are removed from the ketone bearing alpha protons in the first step.

The b-hydroxy ketone is given by the alkoxide ion being protonsated by H O.

The elimination process occurs in two steps.

To draw the product of this Claisen condensation, we need to draw two molecules of the ester, so that the alkoxy of one ester is pointing at the alpha protons of the other.

We are pointing at the alpha protons of the other ester.
There is a retrosynthetic analysis for the desired product, the ester is pointing at an alpha proton of the other ester.
We push the tion when we show the C-C bond made during a Claisen condensa.
Consider the C-C bond in the product to draw the starting materials for this Claisen condensation.

To draw the product of this Claisen condensation, we need to draw two molecules of the ester, so that the alkoxy of one is pointing at the alpha protons of the other.

The two esters above can be used to make a product that erases the alkoxy group and one of the alpha protons.

The C-C bond is shown in the retrosynthetic analysis for the desired product.
The C-C bond in the product can be used to draw the starting materials.

The C-C bond in the product can be used to draw the starting materials.

The Claisen condensation reaction can be used to make the desired product.

The C-C bond is shown in the retrosynthetic analysis for the desired product.

The C-C bond in the product can be used to draw the starting materials.
The location of the alpha position must have been the same as the location of the two esters.
The desired product can be made via a Claisen condensation connected to an alkoxy group, as shown below.

The Claisen condensation reaction can be used to make the desired product.

Ethoxide functions as a base and removes a protons from the beginning of the ester.
The enolate ion attacks another molecule of the ester that has not been deprotonated to give a tetrahedral intermediate.

The carbonyl group is re- formed by expelling an ethoxide ion.

Our desired product is deprotonated under these basic conditions to give a stable enolate ion.
The formation of a particularly stable anion is a driving force for the reaction.

An enolate ion is created when a protons is removed from the alpha position of the starting ester.

The enolate ion attacks another molecule of the ester that has not been deprotonated to give a tetrahedral intermediate.
The carbonyl group is re- formed by expelling a methoxide ion.
Our desired product is deprotonated under these basic conditions to give a stable enolate ion.
The formation of a particularly stable anion is a driving force for the reaction.

In the first step of the mechanism, an alkoxide ion functions as a base and removes a proton from one of the alpha positions of the starting diester.

The other carbonyl group is attacked by the enolate ion and it gives a tetrahedral intermediate.
The carbonyl group is re- formed by expelling an alkoxide ion.
Our desired product is deprotonated under these basic conditions to give a stable enolate ion.
The formation of a particularly stable anion is a driving force for the reaction.
When the reaction is complete, we need to introduce a source of protons into the flask to make a regenerated product.

The forward process can now be drawn.

The forward process can now be drawn.

The isobutyl group was installed at the alpha position.

The desired b-ketoacid is heated to give a decarboxylation reaction, and then the desired product is given.

The forward process is shown.

We can draw the forward process after this compound is treated again with sodium ethoxide.

In the first step, ethyl lowed by the other alkyl halide, the second alkyl group acetoacetate is treated with sodium ethoxide followed by benzyl (note that the two alkyl groups can be installed in either order).

The desired b-ketoacid is heated to give a decarboxylation reaction, and then the desired product is given.

The process wouldn't work because it would have to perform an S 2 reaction with phenyl bromide.

The desired product is a derivative of acetone containing two R groups, highlighted below, and can be made via an ace.

The process is difficult to use because it is a gas at room temperature.

It is best to use a liquid called methyl iodide when using a methyl bromide.

The forward process is shown.

In the first step, ethyl acetoacetate is treated with sodium ethoxide and followed by a methyl group at the alpha position.

The two alkyl groups can be installed in either order.

The b-ketoacid is heated to give a decarboxylation reaction after the ester group is hydrolyzed.

The forward process can now be drawn.

The same strategy can be used to achieve the desired transformation.

The desired product is a derivative of acetic acid.

You might wonder why one bromide and the other iodide are both bromides.

There is an explanation in the solution to Problem 8.71.

The forward process is shown.

In the first step, diethyl mal onate is treated with sodium ethoxide and followed by a methyl group at the alpha position.
The two alkyl groups can be installed in either order.
The diacid is heated to give a decarboxylation reaction after the ester groups are hydrolyzed.

The starting material can be used as a Michael acceptor.

The reagent can be used as a Michael donor.

The desired product is a derivative of acetic acid.

We don't expect a clean Michael reaction because a Grignard reagent isn't a good Michael donor.
A 1,2-addition is expected.

The starting material can be used as a Michael acceptor.

The forward process is shown.

In the first step, diethyl mal onate is treated with sodium ethoxide followed by ethyl iodide, which will install an ethyl group at the alpha position.

R groups can be installed in either order.

The groups function as acceptors.
When we compare the starting mate and the resulting diacid, we see that we have to install a group at the reaction to make the desired product.

This is not stable enough to function as a Michael donor.

Because an enamine can function as a Michael donor, we will need to synthesise a Stork enamine, in which a ketone is first converted into an enamine.
The starting material can be used as a Michael acceptor.

The starting material can be used as a Michael acceptor.

This is not stable enough to function as a Michael donor.

This is not stable enough to function as a Michael donor.

A,b-unsaturated aldehyde can be used as a Michael acceptor.

This is stable to func tion as a Michael donor.

We don't need to use a Stork enamine synthesis in this case.

The bromide is called benzyl bromide.

The com alkyl halide should be used.
This is the only way to make a Gabriel synthesis in which benzyl bromide is used.

Gabriel synthesis requires the identification of the alkyl halide that should be used.

When planning a Gabriel synthesis, this is the only choice we need to make.

Gabriel synthesis requires the identification of the alkyl halide that should be used.

When planning a Gabriel synthesis, this is the only choice we need to make.

Gabriel synthesis cannot make the desired amine.

A alkyl bromide can be used as a substrate in an S 2 reaction.

Gabriel synthesis requires the identification of the alkyl halide that should be used.

When planning a Gabriel synthesis, this is the only choice we need to make.

The alkyl bromide is required for an S 2 reaction.

This problem involves the conversion of an aldehyde into a secondary amine, which should alert us to the possibility of a reductive amination.

The Gabriel synthesis cannot be used to make the desired amine.

This problem involves the conversion of a ketone into a reduced form to give a secondary amine, which should alert us to the possibility of a reductive amination.

The forward synthesis is shown.

First the alkene is reduced to give the desired secondary amine, which is verted into a primary alcohol via hydroboration-oxidation to give anti-Markovnikov addition of H O.

The forward synthesis is shown.

The forward synthesis is shown.

The acetal is treated with acid to make a ketone and then converted into an imine.

The forward synthesis is shown.

The imine is reduced to give the acylation after Friedel-Crafts is converted into an acid halide.

This problem can be solved by first converting the amino group into an amide group.

The forward synthesis is shown.

The ring was less activated after the acid chloride.

A disubstituted aromatic ring is the starting material.

The bromine atom must be replaced with this group.
The Sandmeyer reaction can be used to convert the starting material into a diazonium salt.

One of the substituents is an amine, and the other is an amine-based reagent.

The group was replaced with a cyano group.
A Sandmeyer reaction can be used to convert a primary amine into a diazonium salt.

A trisubstituted aromatic ring is the starting material.

A chlorine atom must be used to replace this group.
A Sandmeyer reaction is used to convert a primary amine into a diazonium salt.

One of the substituents of the aromatic compound is an amino group.

This group must be replaced with a different one.
The Sandmeyer reaction can be used to convert the starting material into a diazonium salt.

A trisubstituted aromatic ring is the starting material.

The bromine atom must be replaced with this group.
The Sandmeyer reaction can be used to convert the starting material into a diazonium salt.

The Diels-Alder reaction is between 1,3-butadiene and a center.

The starting materials are achiral, so the product is a monosubstituted ethylene.

Disubstituted erythritol.

This is a Diels-Alder reaction between 1,3-butadiene and a monosubstituted ethylene.

Disubstituted erythritol.

Disubstituted erythritol.

Disubstituted erythritol.

We expect the product to have a bicyclic skeleton because of the Diels-Alder reaction.

The Diels-Alder reaction between 1,3-butadiene and a is a monosubstituted ethylene, so we expect the following pair of disubstituted alkyne.
There are no enantiomers in the product.

We expect the product to have a bicyclic skeleton because of the Diels-Alder reaction.

We expect the product to have a bicyclic skeleton because of the Diels-Alder reaction.
We expect the product to have a bicyclic skeleton because of the Diels-Alder reaction.

We expect the product to have a bicyclic skeleton because of the Diels-Alder reaction.

The product is expected to have a bicyclic skeleton.
The mixture of enantiomers will be obtained if A is a monosubstituted ethylene.

The ebook EULA can be found at www.wiley.com/go/eula.

Document Outline

Cover

Title Page

Contents

Chapter 1 Aromaticity 1.1 Introduction to Aromatic Compounds 1.2 Nomenclature of Aromatic Compounds 1.3 Criteria for Aromaticity 1.4 Lone Pairs

Chapter 2 IR Spectroscopy 2.1 Vibrational Excitation 2.2 IR Spectra 2.3 Wavenumber 2.4 Signal Intensity 2.5 Signal Shape 2.6 Analyzing an IR Spectrum

Chapter 3 NMR Spectroscopy 3.1 Chemical Equivalence 3.2 Chemical Shift (Benchmark Values) 3.3 Integration 3.4 Multiplicity 3.5 Pattern Recognition 3.6 Complex Splitting 3.7 No Splitting 3.8 Hydrogen Deficiency Index (Degrees of Unsaturation) 3.9 Analyzing a Proton NMR Spectrum 3.10 13C NMR Spectroscopy

Chapter 4 Electrophilic Aromatic Substitution 4.1 Halogenation and the Role of Lewis Acids 4.2 Nitration 4.3 Friedel-Crafts Alkylation and Acylation 4.4 Sulfonation 4.5 Activation and Deactivation 4.6 Directing Effects 4.7 Identifying Activators and Deactivators 4.8 Predicting and Exploiting Steric Effects 4.9 Synthesis Strategies

Chapter 5 Nucleophilic Aromatic Substitution 5.1 Criteria for Nucleophilic Aromatic Substitution 5.2 SNAr Mechanism 5.3 Elimination-Addition 5.4 Mechanism Strategies

Chapter 6 Ketones and Aldehydes 6.1 Preparation of Ketones and Aldehydes 6.2 Stability and Reactivity of C===O Bonds 6.3 H-Nucleophiles 6.4 O-Nucleophiles 6.5 S-Nucleophiles 6.6 N-Nucleophiles 6.7 C-Nucleophiles 6.8 Exceptions to the Rule 6.9 How to Approach Synthesis Problems

Chapter 7 Carboxylic Acid Derivatives 7.1 Reactivity of Carboxylic Acid Derivatives 7.2 General Rules 7.3 Acid Halides 7.4 Acid Anhydrides 7.5 Esters 7.6 Amides and Nitriles 7.7 Synthesis Problems

Chapter 8 Enols and Enolates 8.1 Alpha Protons 8.2 Keto-Enol Tautomerism 8.3 Reactions Involving Enols 8.4 Making Enolates 8.5 Haloform Reaction 8.6 Alkylation of Enolates 8.7 Aldol Reactions 8.8 Claisen Condensation 8.9 Decarboxylation 8.10 Michael Reactions

Chapter 9 Amines 9.1 Nucleophilicity and Basicity of Amines 9.2 Preparation of Amines Through SN2 Reactions 9.3 Preparation of Amines Through Reductive Amination 9.4 Acylation of Amines 9.5 Reactions of Amines with Nitrous Acid 9.6 Aromatic Diazonium Salts

Chapter 10 Diels-Alder Reactions 10.1 Introduction and Mechanism 10.2 The Dienophile 10.3 The Diene 10.4 Other Pericyclic Reactions

Detailed Solutions

Index

EULA