Identifying Where Substitution and Elimination Reactions Happen

Identifying Carbons Where Substitution and Elimination Reactions Can Take Place

• Substitution and elimination reactions need a leaving group in order for them to occur.
• Look for a good leaving group on the substrate undergoing the substitution or elimination reaction.
• If there is no good leaving group, then there won’t be a substitution or elimination reaction. It’s that simple.
• What makes a good leaving group? Review here (See Article: What Makes a Good Leaving Group), but the bottom line is look for halogens and other species that are weak bases.
• Hydroxyl groups (HO) are poor leaving groups unless acid is present.
• Furthermore, SN1/SN2 and E1/E2 generally only happen on sp³-hybridized carbons.
• sp²-hybridized carbons such as alkenyl, aryl and alkynyl halides will not undergo SN1/SN2/E1/E2

Table of Contents

1. Look For A Good Leaving Group On The Substrate
2. The Leaving Group Must Be On An sp³ Hybridized Carbon
3. Identify The Type of Carbon As Primary, Secondary, or Tertiary
4. Acid Makes Alcohols (and Ethers) Into Better Leaving Groups
5. Multiple Functional Groups
6. Epoxides
7. Summary
8. Notes
9. Quiz Yourself!
10. (Advanced) References and Further Reading

1. Look For A Good Leaving Group On The Substrate

This article presumes you are familiar with the S_N1, S_N2, E1, and E2 reactions. If you need a review on any of these reactions, please follow these links and then come back.

• The SN1 Reaction
• The SN2 Reaction
• The E1 Reaction
• The E2 Reaction

In nucleophilic substitution reactions, the substrate or electrophile (an electron-pair acceptor, e.g. an alkyl halide) undergoes attack by a nucleophile (the electron-pair donor).  A new carbon-nucleophile bond forms, and a carbon-leaving group bond breaks off from the substrate.

In elimination reactions, a base breaks a C-H bond on the substrate, and a new C-C pi bond forms, with the breaking of a carbon-leaving group bond from the substrate.

Both of these reactions require the loss of a leaving group from the substrate (i.e. the molecule which is accepting a lone pair from the nucleophile, or is being deprotonated by a base).

If there’s no good leaving group on the substrate… then no substitution or elimination reaction can happen! It’s that simple.

Common examples of good leaving groups are species that can be weak bases once they accept a lone pair of electrons:

• Halogens (Cl, Br, I, but not F, as the C-F bond is very strong)
• Sulfonates [OTs or OMs]
• Acyloxy groups (O-C(O)R)
• Positively charged groups containing oxygen [OH₂ (+),  OH(R)(+)], nitrogen [NR₃(+)], or even sulfur [SR₂(+)]

(There is an excellent correlation  between low pK_a and leaving group ability of the conjugate base. For more on this, see article: What Makes A Good Leaving Group?)

Examples of poor leaving groups include strong bases such as hydroxide (HO^–), alkoxides, amides (NH₂^–)hydride (H^–) and carbanions (R^–). The fluoride ion (F^–) also tends to be a poor leaving group in substitution and elimination reactions since it forms extremely strong bonds to carbon. [Note 1] The cyano group (CN) is a weak base but does not tend to act as a leaving group in substitution or elimination reactions. [Note 2]

So the first order of business in determining the product of any of these reactions is to identify the site(s) on the substrate where there is a good leaving group. 

In other words, look for groups that will be weak bases when they accept a lone pair of electrons.

Not all molecules are capable of substitution or elimination!

For example, hydrocarbons do not act as substrates in S_N1/S_N2 or elimination reactions because the leaving groups would have to be the extremely strong bases H(-) or alkyl anions (e.g. CH₃(-) ).

Some hydrocarbons can act as nucleophiles in substitution and elimination reactions, but that’s different than being the substrate. (See post: Acetylides from Alkynes And Their Use In Substitution Reactions)

These exercises give you the challenge of identifying sites on a molecule where substitution or elimination can occur.

If you identify the leaving group, then you are identifying at least one of the bonds that will break. You are halfway there!

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Here’s another set of examples.

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As these quizzes should hopefully have driven home, you want to look for halogens, sulfonates, or other species that will be weak bases once they depart from the substrate.

Some groups that are poor leaving groups can be made into better leaving groups through the addition of acids. More on that below.

2. The Good Leaving Group Should Be Attached To An sp³ Hybridized Carbon

In substitution and elimination reactions, the transition state generally involves the formation of partial positive charge at the carbon bearing the leaving group.

Carbocations are most stable when they are formed through the loss of a leaving group on an sp³-hybridized carbon. Removing a leaving group from an sp² or sp-hybridized carbon is very difficult, since the greater s-character means the electrons are held closer to the nucleus. [See article: Carbocation Stability] [Note 3]

So another key to identifying where a nucleophilic substitution or elimination reaction will occur is to examine the carbon attached to the good leaving group. 

• If the leaving group is attached to an sp³ hybridized carbon, then this is a good candidate for the site of the reaction.
• If it is attached to an sp² or sp hybridized carbon, then it is unlikely to undergo nucleophilic substitution or elimination.

In this exercise, identify the carbon atoms which are capable of undergoing substitution or elimination.

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Here is another set of examples.

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It’s important to know the few exceptions. [Note 4] Elimination to give alkynes occurs on sp² hybridized carbon using the extremely strong base NaNH₂. [See article: Alkynes via Elimination Reactions]. Another exception, more common in advanced courses, is a reaction related to the S_N2 called the S_N2-prime (S_N2′) where formation of C-Nu and breakage of C-LG is accompanied by the shifting over of a C-C pi bond. [Note 5]

3. Identify The Carbon Attached To The Leaving Group As Primary, Secondary or Tertiary

Once a plausible site for substitution or elimination has been found, the next step is to be able to classify that carbon as primary, secondary or tertiary.

Why is this important?  Because it will greatly help in determining whether a reaction proceeds via SN1, SN2, E1 or E2 pathway.

I have more to say on this in the subsequent article (See article: SN1/SN2/E1/E2 – The Substrate)

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Here is another quiz:

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4. Alcohols And Ethers Can Participate In Substitution and Elimination Reactions If Acid Is Present

If you have made it this far, and have not covered alcohols or epoxides, then you are probably ready to proceed to the next article.

If you are covering the substitution and elimination reactions of alcohols and ethers, however, you’ll need to know another important twist.

Hydroxyl groups (OH) are poor leaving groups in substitution and elimination reactions. Unless strong acid is present, in which case they can be protonated to give their conjugate acids. (See article: The Conjugate Acid Is A Better Leaving Group)

How do you know if the alcohol or ether might be protonated? Look for strong acids such as HCl, HBr, HI, H₂SO₄ or TsOH.

The restriction that substitution and elimination reactions only happen on sp³ hybridized carbons still applies, however.

See if you can identify the carbons in each case which can undergo substitution or elimination reactions.

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As a side note, positively-charged sulfur, nitrogen and even phosphorus can also be good leaving groups. The leaving group in these cases are neutral amines (NR₃) sulfides (SR₂) or phosphines (PR₃).

5. Situations With Two Leaving Groups

There are times where you will be presented with cases where multiple good leaving groups are present, and your challenge will be to find the one site on the molecule where substitution or elimination is most likely to occur.

As we saw before, look for a good leaving group attached to an sp³ hybridized carbon.

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You might see other groups like OH, NH₂, or SH in addition to the good leaving group. Those are likely nucleophiles that will react with the substrate via an intramolecular substitution reaction. (See article: Intramolecular SN2 Reactions)

6. Epoxides

Epoxides are a special case of ethers that will undergo substitution reactions without the addition of acid. That is due to the considerable ring strain of the three-membered ring (See article: Epoxide Ring-Opening With Base).

7. Summary

In summary, the process for identifying where S_N1/S_N2/E1/E2 reactions happen looks like this.

• Identify good leaving groups (such as halides, but also OTs and OMs, among others)
• … that are attached to sp³ (alkyl) hybridized carbons.
• Additionally, alcohols and ethers can undergo S_N1/S_N2/E1/E2 if strong acid is present.

The next step is to identify the carbon bearing the leaving group as primary, secondary, or tertiary. That will help to narrow down whether the reaction will participate in an S_N1, S_N2, E1 or E2 reaction. More on that in the next post.

Next Post: Deciding SN1/SN2/E1/E2 – The Substrate

Notes

Note 1 – The bond dissociation energy of fluorine in CH₃F is about 109 kcal/mol, even larger than the C-H bond dissociation energy (104 kcal/mol).

Note 2 – Despite being a weak base (the pK_a of HCN is 9.2) the cyanide ion does not act as a good leaving group in S_N2 reactions for reasons similar to that for fluoride ion; the C–CN bond is just too strong. (One estimate is that the reaction is unfavorable by about 15 kcal/mol, which makes displacement of CN(-) very unlikely)

In fact, methyl cyanide, more commonly known as acetonitrile, is a commonly used polar aprotic solvent for S_N2 reactions.

Note 3 – Why do SN1/SN2/E1/E2 reactions happen only on alkyl halides and not on alkenyl or alkynyl halides?

Here’s one way to look at it.

Remember how s-orbitals are held closer to the nucleus, which is why acetylide ions are more stable than alkyl anions? You can think of alkenyl and alkynl carbons as having a great effective electronegativity than alkyl carbons, which makes them “hold on” to electrons more tightly and making it more difficult for an electron pair to leave.

So in other words, alkenyl and alkynyl carbons form stronger bonds.

Another, related explanation has to do with the relative abilities of carbon to stabilize positive charge as the amount of s-character increases.

The E1 and S_N1 reactions go through carbocations, whereas the S_N2 reaction requires a partially-positive charged carbon in the transition state.

So each of these reactions involves a transition state where the central carbon bears a partial positive charge.

If you recall that the ability to stabilize a lone pair of electrons increases as s-character increases (sp³ < sp² < sp ) then think of the trend in the opposite direction; the ability of a carbon to stabilize a positive charge decreases as s-character increases, as the empty orbital is held closer to the positively charged nucleus.

For more on the stability of carbocations, see this post on Carbocation Stability. 

Note 4 -One exception to the “no S_N1/S_N2/E1/E2 on sp² hybridized carbons” is formation of alkynes from alkenyl halides via use of the extremely strong base NaNH₂. See article – Synthesis of Alkynes via Elimination of Alkenyl Halides

Note 5. Another exception to the “no S_N1/S_N2/E1/E2 on sp2 hybridized carbons” is found in a reaction known as the S_N2-prime (S_N2′) reaction. It’s similar to the S_N2 in that it is concerted and bimolecular, but slightly different in that the attack does not occur directly on the sp2-hybridized alkyl halide.

It’s not a reaction that comes up too much in introductory organic, but a good exam question.

Quiz Yourself!

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(Advanced) References and Further Reading

In Streitweiser’s Solvolytic Displacement Reactions, page 31, :

[According to calculations] the reaction of methyl bromide with hydroxide ion is exothermic by 16 kcal/mol. The equivalent increase in the activation energy of the reverse reaction corresponds to a factor of 10^-12 in rate; i.e. the rate is far too slow to be observed. Since the reactions of alkyl halides with ethoxide, phenoxide, acetate, hydrosulfide, cyanide, thiosulfate and sulfite ions are all exothermic by 10-20 kcal, the reverse reactions will be slower by factors of 10^-7 to 10^-14 and will, therefore, not be observed except under the most exceptional conditions.”

1. Mechanism of ionic reactions. The heat of ionic substitution reactions
   R. A. Ogg
   Trans. Faraday Soc., 1935,31, 1385-1392
   DOI: 10.1039/TF9353101385
   Measurement of the heats of reaction for some simple nucleophilic substitution reactions of alkyl halides with various nucleophiles.