What is multistep synthesis?

Multistep synthesis is the process of taking a readily available compounds (ones you can buy) and converting them into desired products using known reactions. Multistep syntheses require more than one step, and so one or more intermediate compounds are formed along the way.

What are multistep synthesis problems

Typical multistep synthesis problems give a starting material and a product and instruct you to devise a route that takes the starting material into the product. For example, you might be asked to convert starting material W into product Z using reagents you've learned.

What's an acceptable answer to these problems?

The way to answer these problems is to show the reagents that convert the starting material into the intermediate compounds, and, finally, into the product. For example, if you were asked to convert compound W into compound Z, you might convert W into an intermediate compound (X), which could in turn be reacted to form another intermediate compound (Y), which could be reacted once more to form the product (Z). Note that no mechanisms (arrow-pushing) are shown for the individual reaction steps, just the reagents and any intermediate compounds formed along the way.

Six tips for working through multistep synthesis problems:

Multistep syntheses problems can be very challenging. So here are six tips that can aid you in solving these types of problems.

1. Know the reactions.

This is the basic requirement. No matter how smart you are, you don't stand a chance on synthesis questions unless you know the reactions. Memorize the reagents, use flash cards, use whatever techniques you find most helpful, but get the reactions down cold. Since organic chemistry is a cumulative course, you can't afford to forget any reactions that have been previously covered, so never throw your stack of old flash cards away, but rather keep adding to the pile (the deck will be thick by the end of the course). Often, textbooks have end-of-chapter reaction summaries that can be helpful in making up flash cards.

2. Compare the carbon skeletons.

Compare the carbon skeleton of the starting material to the product. Were any carbons lost or added? If so, can you identify where they were added or lost? A carbon count of the reactant and the product doesn't take long, but can help you determine what kind of reactions you are dealing with.

Take the following simple example. The red portion of the molecule identifies where the likely carbon skeleton of the reactant is found in the product. Doing this allows you to clearly see what portion needs

to be added or lost during your synthesis (it may seem trivial in this obvious example, but it can be a quite helpful to organize your thoughts in tougher problems).

3. Work backwards

Ever find it easier to get through a maze starting at the finish and working back to the start? The same thing applies to multistep synthesis (working backwards like this is a technique termed retrosynthesis). Look at your product, and think of all the reactions that you know of that could form it, ignoring your starting material.

If your product is an alkene, think potentially of alkene-forming reactions like elimination reactions or the Wittig reaction. Write all these reactions out and look what reactant would be required for each. Now look at your starting material. The reaction that most resembles your starting material is probably the best one to select as a potential candidate.

For example, if you were asked to do the following synthesis:

After completing the first two steps above, you would want to think of ways to make the alkene in the product. Ignore the starting material for the moment. Just brainstorm all the ways you can think of to make the alkene and write them down on your scratch paper. You should get something that looks like this:

Now you have three possible routes to choose from. The route to choose is the one that uses a reactant that most resembles our original starting material. If you did step 2 (accounting for the carbon skeleton), you would know that the product has one carbon more than the starting material. Only the first reaction, the Wittig reaction, accounts for this additional carbon, and since the reactant for the Wittig reaction most resembles our starting material, this would be the reaction to tentatively choose. If it turns out to be wrong, we can always go back and try another route.

Looking at our reaction scheme, we now have something that looks like this:

Now repeat the same procedure for cyclohexanone, thinking of all the different ways you could make the ketone. One pointer here is that the closer you get to completing a retrosynthesis, the more you can reference the starting material in your thinking. At this point, for example, you may want to tune your thinking from "I need to think of all the ways I can make cyclohexanone" to perhaps something more on the lines of "I need a reaction that converts an alcohol to a ketone". If you did step one, you would know several ways (different chromate reagents, KMNO4, Ag2O, etc).

If you get stuck, go back and try one of the other pathways. If the Wittig reaction in our example had let to a dead end, then we could have gone back and tried one of the elimination reactions. Choosing the correct way back is often a manner of feel, and that only comes after working a lot of problems (See tip 5).

 

4. Making your life easier: Sorting your synthetic tools

There are pretty much two types of synthetic tools available to you: (1) carbon-carbon bond forming reactions, and (2) functional group reactions that convert one functional group into another (like a reaction that converts an alcohol into a ketone, for example).

You may have noticed that many, many multistep syntheses involve making carbon-carbon bonds (which, in turn, you will have noticed by comparing the carbon skeleton of the starting material and product as suggested in a previous tip). In the toolbox of all your reactions, these carbon-carbon bond forming reactions should go right on the top where they are easy to reach. The functional group transformations are of more secondary importance and go on the bottom of the toolbox to be dusted off when needed.

Synthetic toolbox.

Step 1: Compiling a list of all the carbon-carbon forming reactions you've learned.

Carbon-carbon reactions are your primary tools to build up your molecules, and are perhaps the most important and valuable reactions to remember. Once you know which carbon-carbon reactions to use to make the product, the other reactions often seem to magically fall into place.

At first, your list of carbon-carbon forming reactions will be small. But this list will get bigger. At the end of your first semester, you may have a list that contains some of the following carbon-carbon reactions (and perhaps others, depending on what material your professor and textbook chooses to cover):

Your most important synthetic tools: Carbon-carbon forming reactions

Acetylide reactions. These reactions involve use of an acetylide (a deprotonated terminal alkyne) as a nucleophile. Typically, the acetylide is used to attack a primary halide (in an SN2 reaction) or a carbonyl group to make an alcohol.

Cyanide additions to primary halides. Cyanides can be substituted for halides in a SN2 or SN1 substitution reaction (although I'd recommend you stick with SN2 reactions rather than SN1 reactions in your multistep syntheses).

The Wittig Reaction. Makes a carbon-carbon double bond starting with a carbonyl compound phosphonium ylide

Friedel-Crafts reactions (for aromatic rings). This reaction makes an aromatic-carbon bond.

Diels-Alder reaction. This reaction takes a diene and a dienophile to make ringed and bicyclic products. This reaction makes two new carbon-carbon bonds.

Grignard reaction. This reaction adds a halomagnesium reagent (Grignard reagent) to a carbonyl to make an alcohol.

Some common carbon-carbon forming reactions in first-semester undergraduate organic chemistry

Later (second semester typically), you may add enolate and enol reactions to the list (like the aldol reaction, Claisen reaction, Michael reaction, etc).

Tip 4 continued... Adding carbon-carbon forming reactions into your retrosyntheses

All right. You've learned your reactions and organized them into (at least) two categories: functional group conversions and carbon-carbon bond forming reactions. You've got a fresh synthetic problem in front of you. So now what?

Now you want to think of which carbon-carbon bonds you have to form to take the starting material into the product.

Take an example synthesis:

In this example, you can think, "well, if I'm starting with cyclohexane, the carbon-carbon bond I'm going to have to make connects the cyclohexane ring to the chain containing the carbonyl (C=O) group."

Then you can think "which of the carbon-carbon forming reactions would be useful to make this carbon-carbon bond?" Then check off the list to see which ones might be used and which can be eliminated from consideration.

Acetylide chemistry: This reaction won't work to make this bond. Acetylide reactions work well only with primary halides. The acetylide chain would have to attack a ring carbon which would be a secondary carbon. Also, it's not clear how to selectively take the resulting alkyne to the ketone since it would be an internal alkyne.

Cyanide addition: This won't work. Cyanides add only one carbon. You need to add three carbons. Wittig reaction: This one might work. Of course, Wittig reactions form a carbon-carbon double bond

and you want a carbon-carbon single bond. Probably it's best to see if there's a better route to go before trying this one.

Friedel-Crafts: You don't have an aromatic ring in this problem, so this reaction's out of the question. Diels-Alder reaction. This won't work. The Diels-Alder reaction forms rings and bicyclic compounds.

You already have the ring in the starting material. Grignard reagents. This reaction should work. Of course, a Grignard reagent reacts with a carbonyl

compound to make an alcohol, not a ketone. Fortunately, since you've learned all of your functional group transformations, you know that it's a straightforward task to take a secondary alcohol into a ketone. This reaction looks the most promising so try this carbon-carbon bond forming reaction.

Now that you've chosen the carbon-carbon forming reaction, notice how all the other functional group conversions required to complete the synthesis seem to fall neatly into place. It's usually best to work backwards (using the retrosynthesis approach I discussed in a previous tip), so do that for this problem.

Since you know the carbon-carbon bond-forming reaction forms an alcohol, the last step must convert that alcohol to the ketone in the product. In this case, a number of oxidizing reagents could be used. PCC or Jones' reagent would work just fine here. (You could use other chromium reagents as well)

To make this alcohol you use the Grignard reagent in the carbon-carbon bond-making step that you decided upon earlier. To make a secondary alcohol, you must react the Grignard reagent with an aldehyde. I chose cyclohexyl magnesium chloride here as the starting material, but you could just as easily have gone with cyclohexyl magnesium bromide (either works fine)

To make Grignard reagents you add magnesium turnings to an alkyl halide. Since I chose in the previous reaction to make a chloride Grignard reagent, the starting material I choose in this case is chlorocyclohexane (it would be bromocyclohexane if you went with the bromo Grignard reagent).

The way to add a chlorine to an alkane is to chlorinate using free-radical chemistry in the presence of light.

 

And that's the retrosynthesis for this molecule. Notice how the steps all seemed to be logical once the carbon-carbon bond forming reaction was chosen.

5. Check your answer.

Once you have a potential synthesis, go back and make sure all of your reagents are compatible with the functional groups on your molecule. Make sure, for example, if you are proposing a Grignard reaction, that there are no alcohols or other incompatible functionalities on your reagent. Undergraduate organic professors often seem to take delight in creating challenging (read: tricky) exam questions, giving little partial credit for incorrect answers, so double check every detail of your synthesis for correctness. Which leads us to the most important tip at becoming good at multistep synthesis questions, which is:

6. Work lots of problems.

There's no way around it, no magic formula. A good textbook will have plenty of problems to practice on. Start with easy synthesis problems to get the feel of what is required, then work your way to harder problems. Get help from a tutor if you need it. If you have a solutions manual to your text, don't refer to it until after you have completed the problem. Looking at the solution manual and thinking "yeah, I could do this problem," or "yeah, that looks about right," is no substitute for actually doing it. On an exam, the question will never be "Does this look right to you, check yes or no." So you'll need experience to get the feel of how to work problems. Get lots of experience. Working in groups can help, but make sure that you do the work yourself. Don't let someone else do it for you. You're on your own when exam time comes.

 See other tutorials ::

MECHANISMS

Free radical halogenation

Click here for an animation of the free radical halogenation mechanism of methane (opens in a popup). Does not show termination steps. Requires the Flash plugin, standard with most new browsers.

Free radical halogenation is a reaction that substitutes a chlorine or a bromine for a hydrogen on an alkane. This reaction is a photochemical one. That is, it occurs only when performed in the presence of uv light (abbreviated hv).

Typically, free radical reactions are described in three steps: initiation steps, propagation steps, and termination steps (described below). Note the use of a single headed arrow when describing the movement of a single electron.

Initiation Step:

The reaction begins with an initiation step, which is the separation of the halogen (X2) into two radicals (atoms with a single unpaired electron) by the addition of uv light. This is called the initiation step because it initiates the reaction.

Propogation Steps:

The initiation step, or the formation of the chlorine radicals, is immediately followed by the propogation steps--steps directly involved in the formation of the product. As an example, isobutane (C4H10) will be used in the chlorination reaction. The first step is the abstraction of the hydrogen atom from the tertiary carbon (a tertiary carbon is a carbon that is attached to three other carbon atoms) Note that these are not protons (H+ ions) that are being abstracted, but actual hydrogen atoms since each hydrogen has one electron. This first propogation step forms the tertiary radical.

In the last step, the tertiary radical then reacts with another one of the chlorine molecules to form the product. Notice that another chlorine radical is regenerated, so this reaction can, in theory, go on forever as long as there are reagents. This is called a chain reaction.

A sidenote on free radical stabilities:

Hydrogens attached to more highly substituted carbons (ie. carbons with many carbons attached to them) are more reactive in free-radical halogenation reactions because the radical they form is stabilized by neighboring alkyl groups. These neighboring alkyl groups have the ability to donate

some of their electron density to the electron-deficient radical carbon (a radical is short one electron of filling the atom's valence octet). Thus the hydrogen on the tertiary carbon here is abstracted in preference to the 9 other hydrogen atoms attached to a primary carbon (a carbon that is attached to only one other carbon atom) because it forms a more stable radical.

Here, the tertiary radical is stabilized by electron donation from neighboring alkyl groups.

Selectivity of free-radical halogenation

A point of note about free radical processes is that the intermediates are so highly reactive and short lived that usually you obtain a mixture of products, even though there is preference for forming more highly substituted free radical intermediates. In this example with isobutane, for instance, there would certainly be some abstraction of hydrogens attached to the primary carbons, leading to a different product than the above product (can you draw it out?).

Bromine reacts exactly the same way as chlorine; however, it is far more selective. If propane (CH3CH2CH3), for example, was the substrate, 2-bromopropane would be the dominant product, and there would be only a small amount of 1-bromopropane. Free radical chlorination, though, would not be quite as selective, and there would be a greater amount of the chlorination of the primary carbon than in the bromination reaction.

Termination Steps:

Side reactions that can stop the chain reaction are called termination steps. These termination steps involve the destruction of the free-radical intermediates, typically by two of them coming together.

Other Halogens?

So why can't the other halogens such as fluorine or iodine be used? Iodine reacts endothermically (energetically uphill) and too slowly to be of much good in these free radical processes, while fluorine is at the other pole--it reacts too violently and too quickly to be selective, and can, if uncontrolled, even break carbon-carbon bonds. To understand why this is so, derive the H's for the 4 reactions and compare them (you will find that flourination is highly exothermic, while

iodonation is endothermic; chlorination and bromination, however, are right in the middle).

SN2 Mechanism

Overview:

The general form of the SN2 mechanism is as follows:

nuc: = nucleophileX = leaving group (usually halide or tosylate)

The SN2 reaction involves displacement of a leaving group (usually a halide or a tosylate), by a nucleophile. This reaction works the best with methyl and primary halides because bulky alkyl groups block the backside attack of the nucleophile, but the reaction does work with secondary halides (although it is usually accompanied by elimination), and will not react at all with tertiary halides. In the following example, the hydroxide ion is acting as the nucleophile and bromine is the leaving group:

Because of the backside attack of the nucleophile, inversion of configuration occurs.

Solvents: Protic solvents such as water and alcohols stabilize the nucleophile so much that it won't react. Therefore, a good polar aprotic solvent is required such as ethers and ketones and halogenated hydrocarbons.

Nucleophiles: A good nucleophile is required since it is involved in the rate-determining step.

Leaving groups: A good leaving group is required, such as a halide or a tosylate, since it is involved in the rate-determining step.

 

SN1 Mechanism

Overview:

The general form of the SN1 mechanism is as follows:

Because the mechanism goes through a carbocation, the leaving group must be attached to either a tertiary or secondary carbon to stabilize the intermediate. A methyl or primary leaving group will not form a carbocation. Since it goes through a carbocation intermediate, there are possibilities for alkyl and hydrogen rearrangements (HINT: In mechanism problems if you see a change in the carbon skeleton between the reactant and the product, automatically suspect a carbocation intermediate (ie, E1, Sn1) stabilized by an alkyl or hydrogen rearangement). .

An example ofthe Sn1 Mechanism

Base Strength: Base strength is unimportant, since the base is not involved in the rate determining step (the formation of the carbocation). .

Leaving groups: A good leaving group is required, such as a halide or a tosylate, since the leaving group is involved in the rate-determining step.

Notes: Be wary of rearangements that can occur with the SN1 reaction. Because it goes through a carbocation intermediate, both hydrogen shifts and alkyl shifts can occur!

 

E1 Mechanism

Overview:

The general form of the E1 mechanism is as follows:

B: = baseX = leaving group (usually halide or tosylate)

In the E1 mechanism, the the first step is the loss of the leaving group, which leaves in a very slow step, resulting in the formation of a carbocation. The base then attacks a neighboring hydrogen, forcing the electrons from the hydrogen-carbon bond to make the double bond. Since this mechanism involves the formation of a carbocation, rearangements can occur.

An example of the E1 reaction:

Base Strength: A strong base not required, since it is not involved in the rate-determining step

Leaving groups: A good leaving group is required, such as a halide or a tosylate, since it is involved in the rate-determining step.

Rearangements: Since the mechanism goes through a carbocation intermediate, rearangements can occur.

E2 Mechanism

Overview:

The general form of the E2 mechanism is as follows:

B: = baseX = leaving group (usually halide or tosylate)

In the E2 mechanism, a base abstracts a proton neighboring the leaving group, forcing the electrons down to make a double bond, and, in so doing, forcing off the leaving group. When numerous things happen simultaneously in a mechanism, such as the E2 reaction, it is called a concerted step.

An example of the E2 reaction:

Base Strength: A strong base is required since the base is involved in the rate-determining step.

Leaving groups: A good leaving group is required, such as a halide or a tosylate, since it is involved in the rate-determining step.

Stereochemistry requirements: Must occur with antiperiplanar stereochemistry.

Electrophilic Addition to Alkenes Mechanism

Overview:

Electrophilic addition to alkenes takes the following general form:

nuc: = nucleophileE+ = electrophile

Electrophilic addition to alkenes starts with the pi electrons attacking an electrophile, forming a carbocation on the most stable carbon. A nucleophile then attacks the carbocation to form the product. There are many different kinds of such addition, including:

Hydroxylation Hydrogenation Halogenation Oxidative Cleavage Hydration Epoxidation Cyclopropanation Halohydrin Formation

Clearly, there are numerous kinds of products that can be formed as a result of this mechanism.

Orientation of Addition: Electrophilic Addition adds to give the Markovnikov Product, with the nucleophile added to the more highly substituted carbon. This is because the carbocation intermediate is significantly stabilized by alkyl substituents.

Example of electophilic addition to alkenes:

First, formation of the carbocation on the most highly substituted carbon

Followed by attack of chloride on the carbocation to give the addition product

Hydroboration of alkenes

Overview:

The general form of the hydroboration of alkenes mechanism is as follows:

First step is the attack of the alkene on BH3, which then forms a four membered ring intermediate of partial bonds. It is because of this intermediate that hydroboration forms the anti-Markovnikov product. The boron atom is highly electrophilic because of its empty p orbital (ie. it wants electrons), and forms a slight bonding interaction with the pi bond. Since some electron density from the double bond is going towards bonding with the boron, the carbon opposite the boron is slightly electron deficient, left with a slightly positive charge. Positive charges are best stabilized by more highly substituted carbons, so the carbon opposite the boron tends to be the most highly substituted. Once the transition state breaks down, BH2 is attached to the least substituted carbon.

Peroxide then removes the borane and replaces it with the alcohol to form the anti-markovnikov product.

 

An example of the hydroboration mechanism:

Nucleophilic addition to carbonyl groups

Overview:

The general form of the nucleophilic addition to carbonyl group mechanism is as follows:

First step is the attack of the nucleophile on the partially positive carbon to make the tetrahedral intermediate with the full negatively charged oxygen. The oxygen then becomes protonated to yield the alcohol.

Variety of nucleophiles:

Grignard Reagents Alcohols Amines Alkyl Lithium Reagents Acetylide Ions

Example of nucleophilic addition to carbonyl groups:

In this case, acetylide anion is acting as the nucleophile

Alcohol dehydration

Overview:

The general form of alcohol dehydrations is as follows:

The first step involves the protonation of the alcohol by an acid, followed by loss of water to give a carbocation.

Elimination occurs when the acid conjugate base plucks off a hydrogen. Alcohol dehydrations generally go by the E1 mechanism.

Example of alcohol dehydration:

Fischer esterification mechanism

Overview:

The general form of Fischer esterification mechanism is as follows:

The first step involves protonation of the carbonyl oxygen, followed by the nucleophillic attack of the alcohol.

Then a loss and regain of a proton,

followed by loss of water as electrons from the alcohol oxygen kick down to form the double bond. Loss of a proton yields the ester.

Example of fischer esterification:

 

 

Williamson ether synthesis

Overview:

The Williamson ether synthesis is a reaction that converts alcohols (R-OH) into ethers (R-O-R). The first step in this reaction is forming the conjugate base of the alcohol (called an alcoxide) by reacting the alcohol with sodium metal. This reaction forms hydrogen gas (H2) as a biproduct, so if you perform this reaction take caution to keep all flame sources away during sodium addition.

The alcoxide can then be added to a suitable alkyl halide (typically a primary halide) to form the ether via an SN2 mechanism.

Example:

An example of the Williamson ether synthesis to make diethyl ether.

 

Friedel-Crafts alkylation

Overview:

The general form of the Friedel-Crafts alkylation mechanism is as follows:

Adding an alkyl halide to the Lewis acid aluminum trichloride results in the formation of an organo-metallic complex. In this complex the carbon attached to the chlorine has a great deal of positive charge character (in fact, for practical purposes when dealing with this reaction, you can think of the partially positive charge as a carbocation).

The pi electrons in a benzene ring are mildly electrophilic, and can attack the partially positive carbon to create a non-aromatic intermediate (note that this intermediate has several resonance structures, so that it is not as unstable as it might appear). Elimination of a proton re-establishes the aromaticity of the ring, and the aluminum trichloride catalyst is regenerated along with a molecule of hydrochloric acid.

A word of caution about this reaction: because the aluminum trichloride generates what can

essentially be thought of as a carbocation, rearrangments can occur to produce a more highly-substituted carbocation.

For example: Addition of 1-Chloro-2-Methylpropane to benzene with aluminum trichloride results in the rearranged product, t-butyl benzene, and not the product that you might initially expect (work out the mechanism if you cannot see how that product is attained).

An example of a Friedel-Crafts alkylation:

Claisen condensation

Overview:

The general form of a Claisen condensation is as follows:

The first step involves adding a strong base to an ester to generate an enolate at the carbon (note that the enolate has an additional resonance structure).

The enolate can then add to another ester molecule by attacking the carbonyl to make the tetrahedral intermediate. The carbonyl reforms with loss of the alcoxy group to make the -keto ester.

 

Example of a Claisen condensation:

This is an example of an intramolecular Claisen reaction, called a Dieckmann condensation.

 

Baeyer-Villiger oxidation

Overview:

Baeyer-Villiger oxidations are a really handy way to make esters from ketones. The general form of this reaction is as follows:

Under basic conditions, a peroxide can be deprotonated. This nucleophilic species can then attack a carbonyl group to form a tetrahedral intermediate. Once the tetrahedral intermediate collapses, instead of kicking the peroxide back off, the more highly substituted alkyl substituent makes a sigmatropic shift to the oxygen, kicking off the alcoxide as the leaving group, forming the ester.

Example of a Baeyer-Villiger oxidation:

Here the common MCPBA (m-chloroperoxybenzoic acid) is used as the peroxide. It is one of the most common peroxides because it is cheap and crystalline, and can be used in stoichiometric quantities.

Diels-Alder Reaction

Overview:

The Diels-Alder reaction combines a diene (a molecule with two alternating double bonds) and a dienophile (an alkene) to make rings and bicyclic compounds. The three double bonds in the two starting materials are converted into two new single bonds and one new double bond. Since this reaction forms two new carbon-carbon bonds in a single step, it is a very useful and powerful reaction (one which earned Otto Diels and Kurt Alder a Nobel prize in chemistry for discovering it).

Typically, the Diels-Alder reaction works best when either the diene is substituted with electron donating groups (like -OR, -NR2, etc) or when the dienophile is substituted with electron-withdrawing groups (like -NO2, -CN, -COR, etc).

Conformational requirements of the diene

One quirk of the Diels-Alder reaction is that the diene is required to be in the s-cis conformation in order for the Diels-Alder reaction to work. The s-cis conformation has both of the double bonds pointing on the same side of the carbon-carbon single bond that connects them. In solution, the carbon-carbon single bond in the diene that connects the two alkenes is constantly rotating, so at equilibrium there is usually some mixture of dienes in the s-trans conformation and some in the s-cis conformation. The ones that are at that moment in the s-trans conformation do not react, while the ones in the s-cis conformation can go on to react.

Because of the Diels-Alder's requirement for having the diene in a s-cis conformation, dienes in rings react particularly rapidly because they are "locked" in the s-cis conformation. Unlike dienes in open chains in which there is usually some proportion of the diene in the unreactive s-trans conformation, dienes in rings are held in the reactive conformation at all times by the constraints of the ring, making them react faster.

Stereochemistry of Diels-Alder reaction

What about the stereochemistry of the Diels-Alder reaction? If your dienophile is disubstituted (substituted twice), there is the possibility for stereochemistry in the product. In the Diels-Alder reaction, you end up with the stereochemistry that you started with. In other words, if the substituents started cis (on the same side) on the dienophile, they end up cis in the product. If they started trans (opposite sides) on the dienophile, they end up trans in the product.

Formation of bicyclo products.

When the diene is in a ring, the product of the Diels-Alder reaction is a bicyclo ring system (which can be somewhat intimidating to draw at first). A bicyclo ring system is a compound in which two rings share more than two carbons. There are two main bicyclo ring systems that you typically have to deal with, and these are the ones that come from the diene being in a five-membered ring system and the diene being in a six-membered ring system. (The nomenclature of bicyclic alkanes can be found elsewhere).

When you make bicyclic products (that is, when the diene is in a ring), and you have a dienophile that is substituted, there are two possible products that you can form from the Diels-Alder reaction--the endo product, in which the substituent points down from the top of the bicyclic molecule, and the exo product, where the substituent points towards the top of the bicyclic molecule. In general, you form the endo product preferentially over the exo product in the Diels-Alder reaction.

Example of the Diels-Alder reaction

In this example, since the diene is in a six-membered ring, you make a bicyclo product. Since the dienophile is cis disubstituted, you get the endo stereochemistry with the two cyano (CN) substituents pointing away from the top of the bicyclo compound.

PRACTICE TESTS

Introduction to organic chemistry testNote that you must have a Javascript capable browser to take testHints cost 5 points each

Difficulty Rating: = easy

= medium

= hard

= extremely hard

1. What is the hybridization of the carbon atom in Urea?

Sp

Sp2

Sp3

Sp4

2. Which of the following compounds are not consistent with

valence rules?

     

     

     

     

3. Label the strongest Bronsted-Lowry acid in the following

equation: (see here for explanation)

A

B

C

D

4. Assuming equal concentrations of the acid and base, the equilibrium for the

reaction from question 3: (see here for explanation)

Favors the Products

Is approximately 1

Favors the Reactants

Cannot be determined

5. Which of the following acids will be almost completely deprotonated

by NaOH?

phenol pKa = 10.0

Ethanol CH3-CH2-OH pKa = 16.0

Water HOH pKa = 15.7

Both Ethanol and Water

6. This compound has the following functional groups:

       

ketone, alkene, carboxylic acid, ester

alkyne, ester, carboxylic acid, aldehyde

carboxylic acid, alkene, ketone, ester

ester, aldehyde, carboxylic acid, alkene

7. What is the structure of trans-decalin

8. How many quaternary carbons are on the following molecule:

1

2

3

4

9. What is the IUPAC name of the following molecule: (see here for explanation)

Bicyclo [2,2,1] heptane

Bicyclo [2,2,2] heptane

Bicyclo [1,2,1] heptane

Bicyclo [2,2,1] pentane

10. The following compound is:

A ketone

An amine

A ketone and an amine

A lactam

  

Alkyl halides test

Note that you must have a Javascript capable browser to take quiz. Hints cost 5 points each.

Difficulty Rating:

= easy

= medium

= hard

= extremely hard

1. Which of the following alkyl bromides will undergo the Sn2

reaction the fastest?(see here for explanation)

2. Rank the following carbocations in order of increasing

stability (least stable to most stable)

1 2 3

1, 3, 2

1, 2, 3

3, 2, 1

2, 1, 3

2, 3, 1

3. Of the carbocations in question 2, which ones are prone to

rearangement?

1 and 2

1 only

2 only

2 and 3

None of them will rearange

4. What is a product of the following reaction:

5. What is the major product of the following reaction?

None of the Above

6. Which is the major product of the following reaction?

7. How many potential E2 elimination products can form from the addition of NaOH to the following:

(free hint: trans-decalin cannot undergo a ring flip)

1

2

3

No Elimination Products are Possible

8. Which of the following alkyl halides will react more quickly

in the E1 reaction

(see here for more)

9. Free radical chlorination (Cl2, hv) is not very selective. Which of the following compounds will give

the most selective mono-chlorination (ie. only one product)?

10. What compound could not be a product of the following reaction?

All can be products  

Stereochemistry test

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1. Optically active molecules which rotate plane-polarized light in a counterclockwise direction are

said to be:

levorotary

of R configuration

dextrorotary

of S configuration

2. The specific rotation of a compound is denoted by the symbol:

R

S

3. Assign R or S configuration to the chiral carbon marked with an asterisk

R

S

4. Assign the correct term describing the relationship to the following two isomers:

enantiomers

diastereomers

identical

5. How many stereoisomers of S-Adenosylmethionone are possible in principle (Free Hint: there are 5

stereogenic centers):

8

16

32

None of the Above

6. (R)-(+)-Glyceraldehyde:

is levorotary

rotates plane-polarized light in a counterclockwise direction

rotates plane-polarized light in a clockwise direction

is racemic

7. What is the stereochemistry of the following compound:

R

S

Meso

Not Chiral

8. Would the following compound have an enantiomer:(See explanation)

 

Yes

No

9. Is the following compound meso:

Yes

No

10. Does the following compound rotate plane-polarized light:

Yes

No

Alkenes test

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1. Calculate the degree of unsaturation in Testosterone,

C19H26O2 (See explanation)

5

6

7

8

None of these

2.

Both B and C

None of the Above

3.

4. What is the product of the addition of MCPBA to styrene?

None of the Above

5. Choose the best reagent from the list below to do the

following conversion:

1. O3 2. Zn, H3O+

KMnO4, Acid

1. OsO4  2. NaHSO3, H20

CH2I2, Zn(Cu)

6. Rank the following substituents by the Cahn-Ingold-Prelog sequence rules and put them in

decreasing order (from the highest priority substituent to the lowest priority substituent)

1, 3, 4, 2

4, 1, 2, 3

1, 3, 2, 4

3, 1, 2, 4

None of the Above

7. 8. What would probably not be a product of the following reaction:

8. One of the following cycloboranes undergoes elimination

much faster than the other. Which one is it?

A

B

9. What is the most stable conformation of cis-

1,4 di t-butylcyclohexane?

None of these are the most stable

10. What is the major product of the following reaction:

No reaction

Alkynes Test

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= medium

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None of the Above

2. What is the common name of ethyne?

Acetone

Acetylene

Angstrom

Agriculture

3. Which of the following bases is the weakest base?

4. What is the product of the following reaction:

None of the above

5. What is the product of the following reaction:

None of the Above

6. Which of the following bases are strong enough to deprotonate acetylene?

Concentrated H2SO4

NaNH2

CH3-CH2:- Li+

Both B and C

7. Which of the following reagents will best convert methyl acetylene to acetone?

H2SO4, H20, HgSO4

1. BH3, THF 2. HOOH

H2, Lindlar

1. HCl 2. NaNH2, NH3

8. What is the major product(s) of the following reaction:

Both the first and third choice (A and C)

No Reaction

9. What is the major product of the following reaction:

None of these

10. What is the major product of the following reaction :

None of the Above

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Alcohols testNote that you must have a Javascript capable browser to take quizHints cost 5 points each

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1. Rank the following alcohols from most acidic to least acidic in

ascending order (least acidic to most acidic):

A, B, C,

C, B, A

B, C, A

A, C, B

2. For questions 2 and 3, consider the following transformation:

What reagents from the list below would best accomplish the

above transformation?

1. OsO4, pyridine 2. NaHSO3, H2O

1. Hg(OAc)2, H20 2. NaBH4

1. RCO3H, CHCL2 2 H30+

1. BH3, THF 2. H2O2, OH-

3. For the reaction in question 2, the alcohol product is classified as a:

Primary Alcohol

Secondary Alcohol

Tertiary Alcohol

Quaternary Alcohol

4. What is the product of the following reaction:

5. What is the major product of the following reaction:

None of the Above

6. Choose the best reagents to carry out the following reaction:

D2, Pt

1. Mg, Ether 2. D20

D2SO4, D20,

D3O+, Acetone,

7. What is the major product of the following reaction:

No reaction

8. Disregarding elimination products, what is the major product

of the following reaction:

None of the Above

9. What reagent would undergo the following reaction the

quickest:

10. The product of the following reaction immediately undergoes adimerization at room temp (Diels-Alder reaction). To regain its normal monomer state, it must be heated. What is the structure of the dimer?

**Note: Grader will only explain incorrect answers**

Grading ScaleA 80-100 OutstandingB 70-79 Very Good C 60-69 Average

D 50-59 Below AverageF Below 50 Failing

Conjugated systems and aromatics testNote that you must have a Javascript capable browser to take quizHints cost 5 points each

Difficulty Rating:

= easy

= medium

= hard

= extremely hard For Questions 1 and 2, refer to the following reaction:

1. The kinetically controlled product in the above reaction

is:

3-Chloro-1-Butene

1-Chloro-2-Butene

2. For the reaction in Question 1, which one is the result

of 1,4-addition?

3-Chloro-1-Butene

1-Chloro-2-Butene

3. What is the major product of the following reaction:

None of the Above

4. Which of the following compounds have no conjugated

portions?

5. For a diene to undergo Diels-Alder reaction it must:

be substituted with electron-withdawing groups.

be able to adopt an s-trans conformation

be substituted with electron-donating groups

be able to adopt the s-cis conformation

6. Which of the following would have the longest

wavelength absorption in its UV spectrum.

7. Which of the following are aromatic:

Both A and D8. Indicate which spectral technique would best be used

to distinguish between the following compounds:

and

1H NMR

IR Spectroscopy

Mass Spectrometry

UV Spectroscopy

9. Which of the following are antiaromatic:

Both A and B

Both C and D

10. What is the major product of the following

reaction:

No Reaction

Ketones & aldehydes testNote that you must have a Javascript capable browser to take quizHints cost 5 points each

Difficulty Rating:

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1. What is the IUPAC name of the following compound:

Diphenyl Ketone

Benzophenone

Dibenzyl Ketone

1,1-Diphenylmethanone

2. What is the correct structure for 2-Hydroxyacetophenone:

3. What is the product of the following reaction:

None of the Above

4. What is the structure of Benzaldehyde (Oil of Bitter

Almonds)?

None of the Above

5. Choose the best reagent(s) from the list to do the following

conversion:

H2/Ni (R)

Hg(OAc)2, H30+

1. BH3, THF 2. H2O2, NaOH

NaBH4, THF

6. What is the product of the following reaction:

None of the Above

7. One of the following dioxanes is potentially explosive.

Which one is it?

8. What is the major product of the following reaction:

None of the Above

9. What is the major product of the following reaction:

No Reaction

10. Which of the following would give a positive Tollens

Test:

CH3CH2COCH3

CH3CH=CH-CH=CH-OH

CH3CH2CHO

Both B and C

Amines testNote that you must have a Javascript capable browser to take quizHints cost 5 points each

Difficulty Rating:

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1. Classify the following amine:

Primary Amine

Secondary Amine

Tertiary Amine

Quaternary Ammonium Salt

2. What is the name of the following amine:

Pyrrole

Pyridine

Pyrimadine

Piperdine

3. Which of the following amines is more basic:

A

B

4. What is the major product of the following reaction:

None of the Above

5. Which of the following amines can be resolved into

enantiomers:

ethyl amine

N, N-Dimethylaniline

1-Methylpiperdine

6. Tell whether the products or reactants are favored in

the following acid-base reaction:

Reactants

Products

7. What would not be a product of the following

reaction:

Br-

8. What is the major product of the following reaction:

No Reaction

9. What is the major product of the following reaction:

None of the Above

10. What is the major product of the following

reaction:

It forms a tosyl amide

No Reaction

  

**Note: Grader will only explain incorrect answers**

Grading ScaleA 80-100 OutstandingB 70-79 Very Good C 60-69 AverageD 50-59 Below AverageF Below 50 Failing