For one reaction studied , F as the leaving group was observed to be times faster than iodine! So what could be different about nucleophilic aromatic substitution that makes the rate of reaction much less sensitive to the identity of the leaving group than the S N 1 and S N 2 reactions? Well, for one thing, this would suggest that, unlike the S N 1 and S N 2 reactions, C-F bond cleavage does not occur in the rate-determining step. This information is helpful in coming up with a mechanism for the reaction.
The position of substitution is controlled by the placement of the leaving group. For example, nucleophilic aromatic substitution of p -nitrophenyl fluoride is orders of magnitude faster than m -nitrophenyl fluoride, even though the NO2 is closer to the leaving group and should presumably exert more of an inductive effect. In the course of adding nucleophiles to various electron-poor aromatic molecules with a leaving group, intermediates have been isolated.
The intermediate is the non-aromatic addition product between the aromatic ring and the nucleophile. In the case below, the negative charge is delocalized to an oxygen on one of the nitro groups:.
Meisenheimer intermediates can be isolated and characterized. However, if heated, the compound goes on to form the final nucleophilic aromatic substitution product. The first step is attack of the nucleophile on the electron-poor ring to generate a negatively charged intermediate e. In electrophilic aromatic substitution EAS we saw that electron-rich substituents stabilized the electron-poor intermediate.
But in nucleophilic aromatic substitution NAS the tables are turned! Instead, the intermediate is electron-rich , and is stabilized by electron-withdrawing substituents, such as NO 2. This two-step mechanism where addition is the rate-determining step helps to explain our earlier puzzle of why the reaction with para- nitro is faster than the meta- isomer. Note how the anion in the para- intermediate can be delocalized to the oxygen on the nitro group, putting a negative charge on more electronegative oxygen.
In the meta- intermediate, the negative charge cannot be delocalized to the nitro group, and is stuck on less electronegative carbon. It also helps to explain why fluorine substituents increase the rate of nucleophilic aromatic substitution: the rate determining step is attack on the aromatic ring, not breaking the very strong C-F bond. The highly electronegative fluorine pulls electron density out of the ring, activating it towards attack. Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one.
In all simple elimination reactions the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons. The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases, you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors. For example, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly substitution.
However, you can influence things to some extent by changing the conditions. For a given halogenoalkane, to favour elimination rather than substitution, use:. When t -butyl bromide reacts with ethanol, small amount of elimination products obtained via E1 mechanism.
This is a slow bond-breaking step, it is also the rate-determining step for the whole reaction. Since only the bromide substrate involved in the rate-determining step, the reaction rate law is first order. That is the reaction rate only depends on the concentration of CH 3 3 Br, and has nothing to do with the concentration of base, ethanol. Solvolysis is a more general term, used when a bond in a reagent is broken by a solvent molecule: usually, the solvent in question is water or an alcohol such as methanol or ethanol.
What new functional group has been formed? Because the nucleophile is free to attack from either side of the carbocation electrophile, the reaction leads to a mixture of two stereoisomeric products.
In other words: In general nonenzymatic SN1 reaction can occur with either retention or inversion of configuration at the electrophilic carbon, leading to racemization if the carbon is chiral. The result of this nonenzymatic reaction is a racemic mixture of chiral alcohols. It is important to remember, however, that enzymatic reactions are in almost all cases very specific with regard to stereochemical outcome.
Before we go on to look at some actual biochemical nucleophilic substitution reactions, we first need to lay the intellectual groundwork by focusing more closely on the characteristics of the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group.
For the sake of simplicity, we will continue to use simple, non-biological organic molecules and reaction examples as we work through the basic concepts. This is illustrated by the reaction between chloromethane and hydroxide ion: Recall that the hydroxide ion in this reaction is acting as a nucleophile an electron-rich, nucleus-loving species , the carbon atom of chloromethane is acting as an electrophile an electron-poor species which is attracted to electrons , and the chloride ion is the leaving group where the name is self-evident.
For an example, consider the hydrolysis of S chloromethylhexane.
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