Organic Chemistry

In this practice problem, you will need to determine the major organic product and the mechanism of each reaction. This covers the competition between SN1, SN2 nucleophilic substitution and E1/E2 elimination reactions. To correctly answer these questions, you need to review the main principles of substitution reaction and elimination reactions, as well the regio- and stereochemistry of the above-mentioned reactions depending on whether the given reaction goes via a unimolecular or bimolecular mechanism which dictates, for example, the possibility of rearrangement reactions. You can also go over the stereoselectivity and stereospecificity of the E2 and E1 reactions. The reactivity of the substrate (alkyl halide), the effect of the solvent and temperature should also be taken into consideration.

Practice

1.

Predict the mechanism as SN1, SN2, E1 or E2 and draw the major organic product formed in each reaction. Consider any regioselectivity and stereoselectivity where applicable:

a)
Check your answers

Primary alkyl halide and strong, non-bulky base/nucleophile

1-Iodopropane is a primary substrate and the ethoxide is a strong base/nucleophile. Do not worry about the ethanol because it is the solvent and even though, in general, it can work as a reactant, the ethoxide is much stronger, both as a base and a nucleophile, and the reactivity of the ethanol is suppressed.

The presence of a strong base or a nucleophile rules out the possibility of the unimolecular E1 and SN1 reactions.

If it is strong, it is going to attack and react instead of waiting for the loss of the leaving group to happen first which is the case in E1 and SN1 mechanisms.

To choose between E2 and SN2, you need to remember that the more substituted the substrate the more it prefers the E2 mechanism (3o > 2o > 1o) and vice versa, less substituted substrates prefer the SN2 route (1> 2o> 3o).

Finally, remember that these two are in competition and in most cases, it is not going to be 100% predominance but rather a mixture with a major and minor component. In this case, the SN2 product is the major.

So, as a take-home message from this exercise, remember for 1o alkyl halides (or any other substrate) SN2 > E2 and, no E1 and SN1 especially when reacted with a strong nucleophile or a base.

b)
Check your answers

Secondary alkyl halide and strong, non-bulky base/nucleophile

In this example, we have a secondary substrate reacting with a strong base and a good nucleophile. This combination will largely result in equal amounts of the E2 and SN2 products even though more often you will see E2 shown as the major product and that is why I marked it so as well, to help you correlate with your lecture notes.

Again, no SN1 or E1 when strong nucleophile or a base is present.

Considerations for the E2 mechanism: Stereochemistry and Regiochemistry

Remember, E2 reactions are stereoselective and regioselective. That means out of all the possible stereoisomers and regioisomers (constitutional isomers), the more stable ones will be formed.

The regioselectivity is dictated by the Zaitsev rule which states that the more substituted alkene will be formed and stereoselectivity implicates formation of the more stable E isomer vs Z isomer which is sterically hindered.

Considerations for the SN2 mechanism: Stereochemistry

One key factor to remember in SN2 reactions is the inversion of the absolute configuration (R to S and S to R) of the chiral center that participates in the reaction. Make sure to indicate this in your structures during the test.

Also, for drawing the curved-arrow mechanism, remember that in SN2 mechanism everything (nucleophilic attack and the loss of the leaving group is happening at the same time). You can check this here.

c)
Check your answers

Primary alkyl halide and strong, bulky base

Here again, like for the for the first example, we have a primary substrate which reacts with, this time, tert-butoxide. The key difference here is that the tBuOK is sterically hindered strong base and importantly, it is a weak nucleophile because of the bulkiness. That is why the major product of this reaction is going to be the alkene formed via an E2 mechanism with very little (~10%) of the ether which is still formed by SN2 mechanism because the substrate is primary and very open for the SN2 attack. If it was a secondary and especially tertiary substrate, no product of SN2 mechanism would have been observed. Again, no E1 and hell no SN1 in this combination of a primary substrate with a strong-bulky base.

d)
Check your answers

Primary alkyl halide and strong, non-bulky base/nucleophile

A combination of a primary alkyl halide with a strong base and a good nucleophile will result in a great excess of the SN2 product over the E2 product present in the mixture.

Again, no SN1 or E1 when strong nucleophile or a base is present.

e)
Check your answers

Tertiary alkyl halide and strong, non-bulky base/nucleophile

A tertiary substrate treated with a strong base/nucleophile. The presence of the strong reactant excludes the SN1 and E1 mechanisms. And here again, you need to remember the order of reactivity for the SN2 (1> 2> 3o) and the E2 reactions (3o > 2o > 1o). This indicates that our 3o substrate will mostly undergo E2 elimination.

Like for problem (b), you need to remember that NaOH is a sterically non-hindered base and the Zaitsev product will be the major regioisomer of the forming alkenes.

Because the alkene has two methyl groups on the same carbon, it is not stereoisomeric and no the stereoselectivity of the E2 mechanism is not considered in this reaction.

f)
Check your answers

Tertiary alkyl halide and weak base/nucleophile

If the same tertiary substrate is treated with a weak base/nucleophile, this encourages the unimolecular SN1 and E1 mechanisms. So there are two factors that contribute here. One is the substrate: it is tertiary and the resulting carbocation after the loss of the leaving group (Cl) is very stable. And the second factor is the weakness of the reactant. If it is weak, it is not going to attack the alkyl halide unless the loss of the leaving group happens first because it is not so reactive, it is fine staying by itself.

The result is going to be a mixture of SN1 and E1 with a slight preference of the SN1 unless it is mentioned on your test that heat is applied, you need to indicate elimination reaction as the major route. Remember, heat encourages elimination over substitution regardless of the mechanism.

g)
Check your answers

Secondary chiral alkyl halide and weak base/nucleophile

Alcohols, like water, are weak nucleophiles and bases and this indicates the likelihood of a unimolecular mechanism SN1 or E1 and excludes the possibility of the E2 or SN2 mechanism. Remember, most of the time, when we say excludes, it does not literally mean that no molecule will react by those mechanisms. We are talking about the major products and the main mechanism. The first step of this reaction is the formation of the secondary carbocation which then can be attacked by the nucleophile from both faces. This results in formation of two enantiomers in approximately the same amounts which, remember, is called a racemic mixture. This is followed by the deprotonation of the oxygen to yield the final product.

The SN1 mechanism is covered in great detail in these practice problems.

A few points to add here. If heat was mentioned, that would indicate encouraging elimination.

The product of an SN1 reaction can be shown with plain lines without showing any stereochemistry indicating formation of a racemic mixture.

h)
Check your answers

Primary alkyl halide and strong, non-bulky base/nucleophile

Carbanions are extremely strong bases and nucleophiles so when you see them cross out the SN1 and E1 right away. It can only be SN2 or E2 which only depends on the substrate.

If it is a 1o substrate, then SN2 is the main mechanism. If it is 2o or especially 3o, then E2 will be the major reaction taking place.

In this case, since 1-bromopropane is a primary alkyl halide, we will mostly have a substitution though SN2 mechanism.

i)
Check your answers

A bulky primary alkyl halide and strong, non-bulky base/nucleophile

How come in this reaction that looks almost identical to the previous one where a primary substrate is reacted with a strong reactant I show the E2 as the major product, right?

Well, remember, primary can still mean bulky if the carbon next to the one with the leaving group is crowded. So, here we have primary, sterically hindered substrate and E2 is going to predominate.

Another classical example of a bulky primary substrate is the neopentyl group.

j)
Check your answers

Tertiary alcohol and a hydrogen halide

The hydroxyl is a very poor leaving group and must be converted into a good leaving group to participate in substitution and elimination reactions. Reacting an alcohol with a strong acid like HBr converts the OH into H2O which is an excellent leaving group. This is one of the ways to convert it into a good leaving group and you can read more about this in the “Alcohols in Substitution Reactions” article.

Tertiary alcohols undergo an SN1 substitution reaction with hydrogen halides (HX acids) and the chirality center is racemized because the sp2 carbocation intermediate is attacked from both sides by the nucleophile. This will very often produce a pair of enantiomers. However, because our substrate has two chirality centers but only participates in the reaction, a pair of diastereomers is produced.

You can draw the mechanism of this reaction and check it in this practice problem.

k)
Check your answers

Secondary alcohol and a strong acid

 

In this reaction as well, we have an alcohol reacted with a strong acid. This time it is concentrated sulfuric acid and you need to remember that unlike HX acids, the concentrated sulfuric, phosphoric acid p-toluenesulfonic acid (abbreviated as TsOH) dehydrate of the alcohol to an alkene.

The reaction goes by E1 mechanism for 2o and 3o alcohols where a carbocation is formed as the intermediate, and by E2 mechanism for 1o alcohols since formation of a primary carbocation is not possible.

 

Remember also that E1 reactions are regio and stereoselective. That is why the more substituted alkene is the major product.

The chirality of the alcohol does not matter. It is lost once the carbocation is formed.

 

You can check draw and check the mechanism in this practice problem.

 

l)
Check your answers

Secondary alcohol and a hydrogen halide

Like in problem (j), here the alcohol is treated with HBr which indicates a substitution reaction by SN1 or SN2 mechanism. Again all the details in this post.

The key difference and the reason I put this example is to remind you about the possibility of rearrangements when the secondary carbocation is formed since it can transform into a tertiary carbocation (or any other that is more stable e.g. a resonance stabilized secondary carbocation).

In this case, specifically, it is a ring-expansion rearrangement. The detailed mechanism is shown in problem 2 under rearrangement reactions. Try it first before checking the answer!

m)
Check your answers

Secondary alkyl halide and a weak base/nucleophile

The reactions of secondary alkyl halides are the most difficult to predict since they can undergo unimolecular (SN1, E1) and bimolecular (SN2, E2) reactions. That is why whenever you need to determine the mechanism of a secondary substrate, pay a closer attention to the nucleophile/base. If it is a weak base/nucleophile, then the reaction is going to be SN1 or E1. The elimination is encouraged at higher temperatures, so in this case the major product (it is actually a mixture of two diastereomers) is formed through the SN1 mechanism. Formation of the carbocation intermediate results in racemization of the chirality center and because of the second chirality center that is not involved in the reaction, a pair of diastereomers is formed.

Draw the mechanism and check it in 3.5 here.

n)
Check your answers

A benzylic substrate and a strong base/nucleophile

OTs is a great leaving group prepared by reacting alcohols with p-Toluenesulfonyl chloride (TsCl). This is one way of converting the poor leaving group OH into a good leaving group. You can read more about this in the “Alcohols in Substitution Reactions” article.

So, don’t get confused just by seeing the OTs! It is just a good leaving group (approximately same as the iodide ion).

Now, the mechanism. We have a primary benzylic substrate with a good leaving group reacted with methoxide which is also a strong base and a nucleophile.

A combination of a primary substrate with a good nucleophile (and a strong base) gives SN2!

It is true that benzylic alkyl halides (and other substrates) can undergo both unimolecular and bimolecular reactions because of the excellent stability of the benzylic cation, in this case, however, it will only be SN2 because of the strong nucleophile.

No elimination is possible since no beta hydrogens are present in the molecule.

o)
Check your answers

A benzylic substrate and a weak base/nucleophile

When the same primary substrate (see problem (n)) is treated with an alcohol which is a weak nucleophile, the reaction will predominantly go through an SN1 mechanism.  The first step is the loss of the leaving group (OTs) followed by the nucleophilic attack of the methanol which then is deprotonated and the final product is obtained. See problem 3.1 in the SN1 mechanism.

p)
Check your answers

A chiral secondary alkyl halide and strong, non-bulky base/nucleophile

Like for the OTs (see problems n and o), OMs is also a good leaving group. So we have a secondary substrate reacted with a strong nucleophile which indicates an SN2 mechanism since the nitrile is a non-basic nucleophile and no elimination reaction is possible.

Additionally, the polar aprotic solvent (acetone) also implies an SN2 reaction.

Remember to change the chirality since the SN2 mechanism goes via inversion of the absolute configuration.

q)
Check your answers

A secondary alkyl halide with two chirality centers and strong, non-bulky base/nucleophile

This problem is similar to the previous one as we still have a secondary substrate reacting with a strong (and non-basic) nucleophile in a polar aprotic solvent DMSO.

This all plays in favor of the SN2 mechanism.

The difference and the key point here is to make sure you do not change the configuration of a chirality center that does not participate in the reaction (the wedge methyl in this case). Just leave it as it is.

r)
Check your answers

A chiral secondary substrate and a weak base/nucleophile

What if a chiral secondary substrate is reacted with a weak nucleophile/base?

There are two things here: One is that if it is a weak nucleophile/base, then the mechanism is going to be either SN1 or E1. And Two, which comes as a result of number one, is that SN1 and E1 mechanism bring the possibility of rearrangement reactions.

In this reaction, despite starting with two chirality centers, the product is a racemic mixture of two rearrangement products. The secondary carbocation is transformed into a more stable tertiary carbocation by a hydride shift and the resulting carbocation is attacked from both sides which leads to the racemization of the chirality center.

s)
Check your answers

A chiral secondary substrate and a strong non-bulky base/nucleophile

On the other hand (see problem (s)), if this secondary substrate with two chirality centers is treated with a strong base methoxide, the E2 mechanism will predominate. The E2 mechanism is rgioselective the more stable regioisomer is the major product based on the Zaitsev rule. It is also stereospecific. Stereospecific E2 occurs when only one beta hydrogen is available and the geometry of the product is dictated by the requirement of the leaving group and the beta hydrogen being in anti-periplanar orientation. This can result in Z alkene being the major product even though it is less stable than the E alkene. In this case, the E alkene is the only possible stereoisomer that can be formed.

You can find more practice problems on the  stereospecificity of the E2 reactions here.

t)
Check your answers

A chiral secondary substrate and a strong bulky base/nucleophile

Try to find the main difference between this reaction and the one in problem (s).

There are two differences; the leaving group (changed from Cl to OTs) and the base (from CH3O to DBU).

Both the chloride and the OTs are good leaving groups so the only key difference here is the CH3O to DBU change. Methoxide is a strong, non-bulky base, which favors formation of the Zaitsev product. DBU is strong, bulky base, which favors formation of the less substituted alkene (Hoffman product). Notice that the other stereogenic center is intact in this case since the double bond is formed on the less substituted side of the molecule.

u)
Check your answers

E2 elimination of substituted cyclohexanes

This is a combination of a secondary substrate with a strong base/nucleophile. The ring is a sterically hindered unit, so for these types of secondary substrates, the E2 mechanism is more common than the SN2 mechanism.

There are beta hydrogens on both sides of the leaving, so the more substituted alkene will be preferred based on the Zaitsev rule since the base is sterically unhindered.

v)
Check your answers

E2 elimination of substituted cyclohexanes

This problem is similar to the previous (see problem (u)) but with one important difference which is the orientation of the beta hydrogens next to the Cl leaving group. It is required in the E2 elimination reactions that the leaving group and the hydrogen are anti periplanar which is only accomplished when the two are at axial positions and anti to one-another. That is why the double bond can only be formed on the left side of the Cl in this example.

You can check the elimination reactions of cyclohexanes here.

w)
Check your answers

SN2 of an achiral cyclohexane with cis and trans isomerism

Acetate is a strong nucleophile and a weak base. This, combined with a secondary substrate in a polar aprotic solvent, will yield a product through the SN2 mechanism.

What might be tricky here is the fact that the carbon with the Br and the one with the methyl are not chirality centers but you still know that SN2 reactions go via inversion of configuration. There is no absolute configuration in this molecule since there are no chirality centers but the relative orientation of the Br and the methyl are important. They are trans but in the product the nucleophile and the methyl will be cis because the nucleophile had to attack the Br from the opposite side.

x)
Check your answers

OH converted into a good leaving group

The only part that is different from what we have discussed so far and lead to a confusion is the presence of two oxygens that you may want to use as the leaving group.

So once again, you need to remember that the OH and alkoxide (OR) are poor leaving groups. That is why only the OTs can be the leaving group in this reaction. Check the “Alcohols in Substitution Reactions” article.

Now lets’ determine the mechanism. We have a primary substrate mixed with N3 which is the azide ion and it is strong, non-basic nucleophile.  In addition, the reaction is carried out in acetonitrile, which is a polar aprotic solvent. There is only possibility here and is the SN2 mechanism.

y)
Check your answers

Intramolecular and Intermolecular substitution and elimination of cyclohexanes

Sodium hydride is a strong base which deprotonates the alcohol in the first step. This converts the alcohol into an alkoxide which a strong base and a good nucleophile. A mixture of a strong base/nucleophile and a secondary substrate which we normally expect to give a mixture of SN2 and E2 reactions.

Pay attention to the stereochemistry; the nucleophile attacks from the opposite side of the leaving group leading to an inversion of configuration.

z)
Check your answers

Intramolecular and Intermolecular substitution and elimination of cyclohexanes

Similar to the previous problem, the alkoxide ion will be formed in the first step. However, because the Br and the oxygen are anti which is a suitable orientation for an SN2 reaction and they are close proximity, the reaction goes mainly through an intramolecular SN2 mechanism. It is intramolecular because the nucleophile and the leaving group are on the same molecule.

Leave a Comment

×

Buy the Chemistry Study Guides and get 30% off for signing up to Chemistry Steps Prime! Check this out!