This equation is useful for two reasons. Firstly, it gives us a test for the SN2 mechanism. Let’s illustrate this with another example: the reaction between NaSMe (an ionic solid—the nucleophile will be the anion MeS−) and MeI to give Me2S, dimethyl sulfide.

To study the rate equation, fi rst, we keep the concentration of NaSMe constant and in a series of experiments vary that of MeI and see what happens to the rate. Then in another set of experiments we keep the concentration of MeI constant and vary that of MeSNa and see what happened to the rate. If the reaction is indeed SN2 we should get a linear relationship in both cases: the graphs in the margin show a typical set of results. The fi rst graph tells us that the rate is proportional to [MeI], that is, rate = ka [MeI] and the second graph that it is proportional to [MeSNa], that is, rate = kb [MeSNa]. But why are the slopes different? If you look at the rate equation for the reaction, you will see that we have incorporated a constant concentration of one of the reagents into what appears to be the rate constant for the reaction. The true rate equation is

rate = k 2[MeSNa][MeI]

If [MeSNa] is constant, the equation becomes

Rate= k_b[MeSNa], where k_b = k₂ [MeI]

If you examine the graphs you will see that the slopes are different because

slope 1 = ka=k2 [MeSNa], but slope 2 =k_b= k₂ [MeI]

We can easily measure the true rate constant k2 from these slopes because we know the constant values for [MeSNa] in the first experiment and for [MeI] in the second. The value of k₂

from both experiments should be the same. The mechanism for this reaction is indeed SN2: the nucleophile MeS− attacks as the leaving group I− leaves. The second reason that the SN2 rate equation is useful is that it confirms that the performance of an SN2 reaction depends both on the nucleophile and on the carbon electrophile. We can therefore make a reaction go better (speed it up or improve its yield) by changing either. For example, if we want to displace I⁻ from MeI using an oxygen nucleophile we might consider using any of those in the table below.

The same reasons that made hydroxide ion basic (chiefly that it is unstable as an anion and therefore reactive) make it a good nucleophile. Basicity can be viewed as nucleophilicity towards a proton, and nucleophilicity towards carbon must be related. So, if we want a fast reaction, we should use NaOH rather than, say, Na2SO4 to provide the nucleophile. Even at the same concentration, the rate constant k2 with HO− as the nucleophile is much greater than the k2 with SO4 − as the nucleophile. But that is not our only option. The reactivity and hence the structure of the carbon electrophile matter too. If we want reaction at a methyl group we can’t change the carbon skeleton, but we can change the leaving group. The table below shows what happens if we use the various methyl halides in reaction with NaOH. The best choice for a fast reaction (greatest value of k2) will be to use MeI and NaOH to give methanol.

● The rate of an SN2 reaction depends on:

• the nucleophile

 • the carbon skeleton

• the leaving group

along with the usual factors of temperature and solvent.

مواضيع ذات صلة

Mechanisms for nucleophilic substitution

Resolutions using diastereoisomeric salts

Separating enantiomers is called resolution

Chiral compounds with no stereogenic centres

Diastereoisomers can arise when structures have more than one stereogenic centre

Absolute and relative stereochemistry

Diastereoisomers can be chiral or achiral

Dia stereoisomers are stereoisomers that are not enantiomers

The rotation of plane-polarized light is known as optical activity

R and S can be used to describe the configuration of a chiral centre

Many chiral molecules are present in nature as single enantiomers

Stereogenic centres