RELATIONSHIP BETWEEN Sn1 and E1 REACTIONS

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RELATIONSHIP BETWEEN Sn1 and E1 REACTIONS SYNTHETIC DRAWBACKS OF Sn1 REACTIONS In terms of synthetic value, any reactions whose mechanism involves carbocation formation suffer from some drawbacks. Once formed, carbocations can undergo several process that may result in formation of undesired side products. In the context of Sn1 reactions, some of the things that carbocations can do are: 1) They can go on to form the expected Sn1 products. 2) They can rearrange to form products whose connectivities are changed relative to that of the original substrate. 3) They can undergo elimination (E!) reactions to form alkenes. Examples: This limits the synthetic usefulness of such reactions, for one has to deal with mixtures of products and the separation of the desired ones. In fact, Sn1 and E1 reactions typically go hand in hand and are difficult to disassociate, because they share similar characteristics, and the conditions that favor one also favor the other. We’ve already learned the characteristics of Sn1 reactions and the factors that favor them. We can extend that to E1 reactions as well: Characteristics of the Sn1 and E1 mechanisms: a) They are multistep processes b) They occur with formation of carbocation intermediates in the rate determining step c) They follow first order (unimolecular) kinetics. That is, rate=k[substrate] Sn1 and E1 mechanisms are favored by using: a) Sterically hindered substrates b) Weak (neutral), small nucleophiles and heating c) Moderate to high polarity solvents The Sn1 mechanism leads to substitution products, and the E1 mechanism leads to formation of alkenes. Br CH 3 OH OCH 3 + OCH 3 + + Sn1 product Rearramged Sn1 product various alkenes CH 3 Br CH 3 OH CH 3 OCH 3 + H 3 C OCH 3 + CH 3 + CH 3 etc.

Transcript of RELATIONSHIP BETWEEN Sn1 and E1 REACTIONS

Page 1: RELATIONSHIP BETWEEN Sn1 and E1 REACTIONS

RELATIONSHIP BETWEEN Sn1 and E1 REACTIONS

SYNTHETIC DRAWBACKS OF Sn1 REACTIONS

In terms of synthetic value, any reactions whose mechanism involves carbocation formation suffer fromsome drawbacks. Once formed, carbocations can undergo several process that may result in formation ofundesired side products. In the context of Sn1 reactions, some of the things that carbocations can do are:

1) They can go on to form the expected Sn1 products.2) They can rearrange to form products whose connectivities are changed relative to that of the original substrate.3) They can undergo elimination (E!) reactions to form alkenes.

Examples:

This limits the synthetic usefulness of such reactions, for one has to deal with mixtures of products and theseparation of the desired ones.

In fact, Sn1 and E1 reactions typically go hand in hand and are difficult to disassociate, because they sharesimilar characteristics, and the conditions that favor one also favor the other. We’ve already learned thecharacteristics of Sn1 reactions and the factors that favor them. We can extend that to E1 reactions as well:

Characteristics of the Sn1 and E1 mechanisms:

a) They are multistep processesb) They occur with formation of carbocation intermediates in the rate determining stepc) They follow first order (unimolecular) kinetics. That is, rate=k[substrate]

Sn1 and E1 mechanisms are favored by using:

a) Sterically hindered substratesb) Weak (neutral), small nucleophiles and heatingc) Moderate to high polarity solvents

The Sn1 mechanism leads to substitution products, and the E1 mechanism leads to formation of alkenes.

Br

CH3OH

∆OCH3

+

OCH3

+ +

Sn1product

RearramgedSn1 product

various alkenes

CH3

Br CH3OH

CH3

OCH3+

H3C OCH3

+

CH3

+

CH3

etc.

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The same substrates that are prone to undergo Sn1 reactions also undergo E1 reactions. They are of twomajor types:

a) Secondary and tertiary alky halidesb) Secondary and tertiary alcohols.

ALKYL HALIDES AS Sn1 and E1 SUBSTRATES

As mentioned before, conditions that favor Sn1 also favor E1 reactions. The first and rate-determining stepin the process is departure of the leaving group to form a carbocation. Let’s look at one of the examples fromthe previous page, the reaction between 2-bromo-3-methylbutane and methanol. As has been mentionedbefore, commonly used solvents in Sn1 reactions are water and alcohols. They frequently also double asnucleophiles. In E1 reactions these same substances would act as bases to capture a proton in the eliminationstep.

Once formed, the carbocation can:

a) Go on to form the Sn1 product. Remember that when the nucleophile is a neutral protic solvent, itsconjugate base replaces the leaving group. This implies that a proton (H+) gets released in the process.

OCH3

CH3OH

Sn1 product

+ H

The bromide ion released in the first step and the proton released in the second step can then get togetherto form HBr, which is an inorganic product of the reaction. However, we will not focus on inorganic products.As a matter of fact, inorganic products are frequently left out when writing organic reactions and mechanismsto avoid clutter and keep the focus on the organic products.

b) Rearrange to a more stable cation. We already mentioned that this can happen by an alkyl or a hydrideshift, depending on which process yields a more stable cation. This cation can in turn form another Sn1product. This Sn1 product is a structural isomer of the one formed in (a) because its connectivity haschanged.

Br

CH3OH

∆+ Br

H shift

CH3OH

OCH3

Sn1 product

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c) Eliminate a proton to form an alkene. Remember that when there is a positive charge on carbon, theneighboring protons become highly acidic. The solvent –acting as a weak base– can capture one of theseprotons and cause an electron shift towards the positive charge that results in formation of a new pi bond,or alkene product. When the protons surrounding the positive charge are nonequivalent, several alkenesare possible, as illustrated below.

H3CCH3

CH3

H

The protons shown in red are now acidic due tothe presence of the positive charge on carbon

Secondary cation before rearrangement

The rearranged cation can undergo similar elimination processes to yield two possible alkenes, but one ofthem is the same as one of the alkenes formed above, with the pi bond located between carbons 2 and 3.As an exercise, see if you can identify the protons which are acidic, how they are eliminated, and howelectron movement takes place to form the alkene structures shown below.

E1

E1

same as( )

(same as formed above)

H3C C

CH3

H

H

H

H

H

OCH3

The solvent can nowact as a base

H3C C

CH3

HH

H

alkene product

H3C C

CH3

H

H

H

HH3C CH3

CH3

alkene product

H3C O

H

1

2

1

2

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Thus we have the net reaction we introduced on page 1, showing the formation of Sn1 and E1 products:

Br

CH3OH

∆OCH3

+

OCH3

+ +

Sn1product

RearramgedSn1 product

various alkenes

etc.

Can you now explain (mechanistically) how the various products form in the second example given on p.1?

CH3

Br CH3OH

CH3 H3C OCH3CH3 CH3 CH2

+ + + +OCH3

ALCOHOLS AS Sn1 and E1 SUBSTRATES

The classic textbook example of E1 elimination reactions is the acid-catalyzed alcohol dehydration. Strongacid catalysis is needed to protonate the hydroxyl group of alcohols and turn it into a good leaving group.For alcohols which are soluble in water, an aqueous solution of strong acid is usually used. Such solutioncontains a high concentration of hydronium ion (the conjugate acid of water), which acts as the protondonor. This first protonation step is an acid-base reaction, and as such it takes place very rapidly. Theprotonation of 4-methylcyclohexanol illustrates this step.

OH

CH3

H3O+

OH2

CH3

+ H2O

Once protonated, the hydroxyl group can leave as water, leaving behind a positively charged carbon. Thisis the rate determining step in the sequence.

OH2

CH3 CH3

+ H2O

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The elimination step takes place after formation of the carbocation, with water acting as a base.

CH3

+ H2O

H

CH3

H3O++

Notice that each step in this mechanism is reversible. This means that the acid-catalyzed alcohol dehydrationis an equilibrium process. As such, equilibrium must be manipulated to shift the outcome towards the desiredproduct. This reaction enables us to make alkenes from alcohols, or alcohols from alkenes. Typically, formationof alkenes is favored by use of concentrated acid, whereas formation of alcohols is favored by use of diluteacid. The following sequence shows the steps in reverse, starting with an alcohol, and arriving at an alkene.

OH2

CH3CH3

+ H2O

OH

CH3

H3O+

CH3

NET REACTION:

CH3

+ H2O

H

CH3

+H

O

H

H

alkene protonation

12

OH

CH3

H3O+

O

CH3

+ H2O +

HH 1

2

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ELIMINATIONS INVOLVING ASYMMETRICAL SUBSTRATES

In some eliminations the products include several possible alkenes. The next question is then, which onesform preferentially? Is there a rule for predicting which alkene will predominate? Let’s look at the followingexamples given before, but this time let’s focus only on the alkene products.

Br

CH3OH

∆+ +

Saytzeff’s rule enables us to make a prediction. According to this rule, in elimination reactions, the mosthighly substituted alkene usually predominates. For details, refer to the Wade textbook, section 6-19 (5thed.) or 6-18 (6th ed.).

The most highly substituted alkene is the one with the most alkyl groups directly attached to the carbon-carbon double bond. This of course does not include hydrogens attached to the carbon-carbon double bond.In the above example, we have mono, di, and trisubstituted products. According to Saytzeff’s rule, thetrisubstituted product will predominate because it is the most stable.

H

H

H

monosubstitutedalkene

H

H

disubstitutedalkene

trisubstitutedalkene

H

Why is the trisubstituted alkene more stable than the other two? The relative stabilities of alkenes is measuredby their heat of hydrogenation (much like the relative stabilities of cycloalkanes is measured by their heatsof combustion, see ch. 3). The relationship is inverse: the higher the heat of hydrogenation, the lower thestability, because the higher the heat of hydrogenation, the higher the potential energy of the alkene.

For a full discussion of this trend with tables, refer to the Wade textbook, section 7-7 (A-D) in both the 5thand the 6th eds. Please note that in the case of cis and trans isomers, the higher stability of the trans isomeris due to a steric effect. The alkyl substituents in the cis isomer are closer to each other than in the transisomer, leading to increased steric crowding.

H3C

CH3H3C CH3 cis trans

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CH3

Br CH3OH

CH3 CH3 CH2

+ +

Let’s look at the last example. In this case we have di and trisubstituted products. Once again, the trisubstitutedproduct will be predominant because it is the most stable.

CH3

HC

HH

disubstituted

CH3

H trisubstituted(most stable)

H