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8.4: Nucleophilic Substitution - 1st Order

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    321475
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    Let’s now consider what is happening in an SN1 reaction:

    Step 1: Many organic reactions are acid-base reactions. In this reaction, we have an acid (HCl, pKa = –8). Quick question: do we have a base that we can protonate easily? In other words, is there a pair of electrons that can grab a proton and move the reaction forward?

    Screen Shot 2021-05-20 at 10.30.14 AM.png

    In the first step, the lone pairs on the alcohol pick up a proton from HCl. The HOMO in this step is nO and the LUMO is \(σ^{*}_{H-Cl}\). The right hand side of the equation is favored because the pKa of the protonated alcohol is approximately –1.7.

    Step 2: At this point, we must consider where we are going – we need to form a C-Cl bond. Is there a way to do this? Certainly, and there are actually two ways:

    (a)

    Screen Shot 2021-05-20 at 10.30.22 AM.png

    (b)

    Screen Shot 2021-05-20 at 10.30.26 AM.png

    Which one is correct? If I told you that step 2 was the rate determining step of an SN1 reaction, which one must it be? Well, if the rate is 1st order, then the rate determining step must involve only one molecule. So, it must be (b).

    Step 3: At the end of step 2, we will have generated an intermediate that is highly reactive (a carbocation). What happens next is that the nucleophilic chloride (Cl) reacts with the electrophilic carbocation and forms a new bond. The HOMO is the nCl and the LUMO is the atomic p orbital on carbon.

    Screen Shot 2021-05-20 at 10.30.31 AM.png

    Now we have the product! Let’s put everything together into one mechanistic diagram:

    Screen Shot 2021-05-20 at 10.30.37 AM.png

    What would the potential energy diagram of an SN1 reaction look like?

    Screen Shot 2021-05-20 at 10.30.43 AM.png

    The rate of the reaction should be dependent on only one molecule (remember, 1st order), and indeed TS #2 is the highest in energy. It is dependent on all of the steps that come before it. What this also means is that the rate is independent of the concentration of Cl (increasing Cl has no effect on rate).

    So, what does effect the rate? The rate of the reaction depends on the formation of the carbocation. There are two factors that contribute to this. The first is that using a stronger acid can accelerate the protonation step. This depends on the pKa of the inorganic acid. For example, hydroiodic acid (pKa = –10.0) is stronger than hydrobromic acid (pKa = –9.0), so the protonation step should be faster. The conjugate base, in this case iodide (I), is more stable because it is more polarizable than bromide and better able to stabilize the negative charge.

    The second factor that contributes to the rate of formation of the carbocation is less intuitive, but involves the structure of the starting material itself. The rate of formation of the carbocation matches the stability of the carbocation, but remember that stability and reactivity are two different phenomena. Stability is a thermodynamic phenomenon, and we have said before that more highly substituted carbocations are more stable because of s-donation. Reactivity is a kinetic phenomenon, which states that carbocations that are more highly substituted are formed faster. The reason for this also has to do with s-donation, but this time it occurs in the transition state (TS #2 from above). Remember that the transition state for an SN1 reaction has a bond-breaking event that involves one species. By increasing the substitution on this molecule/ion, the energy needed to overcome this transition state is lowered (it’s an easier hill to go over). Tertiary alcohols form tertiary carbocations faster than secondary alcohols form secondary carbocations (assuming they’ve been protonated). Increased substitution lowers this transition state maximum because the increased \(σ\)-donation is weakening the \(σ_{C-O}\) bond. Why? Because there is delocalization from \(σ_{C-C}\)/\(σ_{C-H}\) into the \(σ^{*}_{C-O}\). Since more \(σ\-donation will populate the \(σ^{*}_{C-O}\) to a greater extent, the bond will lengthen/weaken, lowering the energy needed to break the bond entirely.

    Screen Shot 2021-05-20 at 10.34.51 AM.png

    Screen Shot 2021-05-20 at 10.34.57 AM.png

    So, to review, the rate of an SN1 reaction is faster when a stronger acid is used, and also faster when the starting material is more highly substituted. In general, SN1 reactions occur only for tertiary and secondary alcohols, which can generate stable carbocations. Primary alcohols do not undergo SN1 reactions, with the exception of allylic/benzylic alcohols (which form resonance-stabilized primary carbocations – quite stable!). And finally, the solvent for SN1 reactions is normally a polar, protic solvent like H2O, MeOH, or EtOH. The reason for this is that polar, protic solvents stabilize the charged carbocation formed during the course of the reaction.

    Screen Shot 2021-05-20 at 10.36.56 AM.png

    There are other types of SN1 reactions that do not involve substitution of alcohols. The mechanism is quite similar, but eliminates the protonation step entirely. In order for this to happen, the starting material must be tertiary or secondary (or benzylic/allylic) and have a good leaving group. A leaving group is a functional group that “leaves” during the rate-determining step to generate the carbocation. For example, in tert-butyl bromide, we can generate the tert-butyl carbocation if the \(σ_{C-Br}\) bond breaks to give a carbocation and Br. We say that the leaving group is “good” because Br is stabilized (polarizable). The more \(σ\)-donation there is, the more the \(σ^{*}_{C-Br}\) is populated, and the faster the leaving group leaves. Remember that this is a bond-breaking event which involves only one molecule, so it is still 1st order.

    Screen Shot 2021-05-20 at 10.37.01 AM.png

    The leaving group ability of a functional group will depend on the pKa of its conjugate acid. The stronger the conjugate acid, the better leaving group ability of that particular group. For substitution reactions, the leaving group ability cut-off point is a conjugate acid pKa of 4. Thus, OH is never a good leaving group because the pKa of water is 15.7. This is why we protonate alcohols in the first SN1 mechanism you learned – protonation turned a good leaving group (OH, pKaH = 15.7) into a much better leaving group (H2O, pKaH = –1). Several common leaving groups are listed below:

     

                Let’s do a few examples. What is the mechanism of these reactions?

    1.

    Screen Shot 2021-05-20 at 10.38.08 AM.png

    2.

    Screen Shot 2021-05-20 at 10.38.14 AM.png

    Draw the starting material, reagents, or products of the following reactions:

    1.

    Screen Shot 2021-05-20 at 10.38.20 AM.png

    2.

    Screen Shot 2021-05-20 at 10.38.24 AM.png

    3.

    Screen Shot 2021-05-20 at 10.38.28 AM.png

    So, to summarize, substituted alkanes (alkyl halides, sulfonates, etc.) that have good leaving group can participate in SN1 reactions. The solvent for these reactions is normally polar and aprotic (DMF or THF). Alcoholic solvents are not suitable because they can themselves act as nucleophiles (unless one is performing a solvolysis reaction, in which you WANT an alcoholic solvent which acts as your nucleophile). DMF and THF are used because the inorganic salts that are often the source of the nucleophile need to be dissolved in something polar because they are charged.


    8.4: Nucleophilic Substitution - 1st Order is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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