The order of reactivity is different from the order of carbocation stability.
The stability of the carbocation is related to the rate-determining step in the SN1 reaction.
The rate of the SN1 reaction will increase if reagents or reaction conditions favor the formation of a carbocation of bromoalkanes.
It is important to point out that the vantage point of a larly stable is not part of the picture.
The a mediates are attacked by conjugates with short lifetimes.
The positive charge on the carbon decreases when alkyl groups are stable.
The explanation of how alkyl groups are stable is a topic of debate.
A s bond is formed with a hydrogen atom.
A greater degree of stabilization is the result of this type.
FIGURE 27-7 shows that because no alkyl groups are present in a methyl carbocation, they are least stable.
Other explanations for the stabilization of alkyl groups carbocation through hyperconjugation are also used.
As the number of alkyl groups bonding to a car increases, the stability of the carbocation also increases.
Take a look at the following combinations of reactants.
Predict whether a substitution reaction will occur.
Pick out the products and suggest the likely mechanism.
The first thing we need to determine is whether the reaction will take place.
The order of the primary and secondary schools is different.
The leaving group is the ion, the chloropropane, and the nucleophile.
The equilibrium constant for the reac tion should be large because the ion is a stronger base.
The chloropropane is a primary haloalkane, so the likely mechanism is SN2 and the product will be CH3 CH2CN.
The potential leaving group is the ion, and it's called the br- ion.
There will be no reaction.
The leaving group is the ion.
We need to know the relative basicities of CH3OH and Cl Cl to make a decision on which direction the reaction will go.
We expect an equilibrium to be established if we assume that the basicities of methanol and chloride ion are the same.
The equilibrium is shifted by the fact that we are using a large amount of methanol.
The likely mechanism for 1CH323COCH3 is SN1, since it is a tertiary haloalkane.
In some of the reactions we will see that other products, not just substitution products, are possible.
Write a mechanism for the reaction.
A sample of 2-iodobutane is dissolved.
The solution of 2-methoxybutane is not active.
A negatively charged ion or a polar molecule is the nucleus in a substitution reaction.
polar solvent is generally used to dissolution the starting materials.
A reaction will follow an mechanism.
The examples are shown in the margin.
There are two examples of nonpolar solvents.
The formation of a carbocation is a N an S 1 reaction.
An S 1 reaction will not occur because Dimethylsulfoxide lizes the ion formed.
H2O bonds are formed through the stabilization of anions.
CH3 is an attacking nucleophile.
Computer simulations show that the F- ion shown in green is more strongly solvated by water than by DMSO.
When the solvent is water, the spheres enclosing the F- ion emphasize that the solvent is close to the F- ion.
When solvated by water.
Nucleophilicity is a measure of how quickly a nucleophile attacks a carbon atom.
It is tempting to think that there is a simple relationship between basicity and nucleophilicity.
There is no simple relationship because basicity and nucleophilicity are fundamentally different properties.
Basicity is a property related to the stability of a base.
Trends in basicity and nucleophilicity are correlated in many texts.
The correlations are of little use without an understanding of the underlying reasons for them.
The explanation of trends in nucleophilicity is based on factors we use for explaining trends in basicities.
The effect of electron-withdrawing or electron-donating groups and the size of the atom are some of the factors that affect basicity.
Trends in nucleophilicity depend on other factors, such as interactions between the nucleophile and solvent molecule, steric effects, and the nature of the electrophile.
Trends in basicity do not always follow trends in nucleophilicity.
We can use a few principles to understand trends in nucleophilicity.
Our point of reference is the nucleophilic atom, which has the lone pair that is used to form a bond.
We will see similarities to explanations given before as we work our way through these principles.
An un charged nucleophile will react faster than a negatively charged one.
HO- is a stronger nucleophile than CH3O H2O.
A negatively charged atom is more attracted to anphilic center than a partially negative charge atom.
The nucleophilic atoms are from the second period.
A lone pair on F is less available for bonding than is the lone pair on H3C-, which is why increasing nucleophilicity is not as good as H3C F.
Predicting the effect of other factors, such as charge delocalization or the presence of electron-withdrawing or electron-donating groups, is sometimes possible.
The atoms are from the second period.
Increasing nucleophilicity H2N- is the strongest nucleophilic because N is less negative than O, and thus the lone pair on N is more available for bonding than is a lone pair on O.
Negatively charged nucleophiles are more reactive in polar protic solvent than they are in polar protic solvent.
In a polar protic solvent, anions are less reactive.
The importance of considering how the solvent affects nucleophilicity cannot be overstated.
Changing the solvent can reverse the trend in reactivities of a series of nucleophiles.
The larger the nucleophilic atom, the easier it is for the charge cloud to be distorted towards the carbon atom.
The rate of reaction is increased because of the distortion of the charge cloud towards the carbon atom.
The reactivity of a nucleophile is reduced by groups adjacent to the nucleophilic atom.
The spherical charge distribution is represented by the polarizability of the nucleophile.
The ability of the nucleophile to bond with the carbon atom is affected by solvent-nucleophile interactions.
1CH323CO- is a weak nucleophile.
There is no substitution reaction at a hybridized carbon atom.
Table 27.2 summarizes some key ideas from this section and 100 means that the nucleophile reacts identify the combinations of reactants and reaction conditions that are 100 times faster than water.
The symbols 1deg, 2deg, and 3deg stand for primary, secondary, and tertiary, respectively.
Use arrows to show the movement of electrons as you write the steps of the mechanism.
There is a haloalkane.
Secondary haloalkanes undergo substitution reactions depending on the solvent and the nucleophile.
The haloalkane and Solve Methanol are the same substance.
The polarizability of the O is the most important factor in determining the nucleophilicity.
The O atom is small and not very polarizable.
The solvent will help to fix a carbocation.
The substitution reaction is expected to occur with a weak nucleophile and a polar protic solvent.
An ether is obtained when a CH3OH molecule is transferred from the solvent to the protons.
To name the products, you may want to review the rules given in Chapter 26.
The reaction considered in this example is called a solvolysis reaction because the solvent acts as the nucleophile.
A haloalkane can undergo a substitution reaction in which the halogen atom is replaced by another group.
The halogen atom and hydrogen atom can be removed from the molecule in an elimination reaction.
There are different mechanisms that can cause elimination reactions of haloalkanes.
The competition that occurs between elimination and substitution reactions is discussed in Reactions of Organic Compounds.
The electron pair donor reacted as either a base or a nucleophile in all of the reactions we have examined so far.
The situation is not always easy.
Two different products are obtained when 2-bromo-2-methylpropane is dissolved in methanol.
The expected substitution product is the major product.
The minor product is 2-methylprop-1-ene.
The formation of a carbocation is the first step.
The first step in the mechanism is the same as this one.
In the second step, a methanol molecule reacts as a base and removes a protons from the carbocation, yielding an alkene.
The positively charged carbon atom of the carbocation is adjacent to the positively charged carbon atom of the b carbon.
The formation of a carbocation is the first step.
The bond is not completely broken.
In the second step, CH3OH acts as a base and removes a protons from the b carbon.
A new p bond is forming between two C atoms in the transition state for the second step.
A bond is forming.
After the formation of the carbocation, methanol can act like a nucleophile and attack the carbon atom, or it can act as a base and remove a protons.
The role of the electron pair donor in the second step is different between the two mechanisms.
The electron pair donor forms a s bond with the carbon atom of the carbocation.
The electron pair donor acts as a base in the E1 mechanism.
The nucleophile was CH3OH.
The CH3OH molecule is a neutral molecule with a small nucleophilic atom and a weak base.
The CH3CH2O- ion is a stronger base than CH3OH, so let's consider what would happen in the reaction of 1CH323CBr and CH3CH2ONa.
The elimination reaction cannot be the E1 mechanism because of the use of a stronger base.
The rate-determining step must be bimolecular because the elimination reaction is second order.
The E2 mechanism consists of a single step that proceeds through a single transition state where three changes are occurring simultaneously: (1) removal of a protons from the b carbon; (2) departure of the leaving group; and (3) formation of a p bond between the a and b
A product in the reaction of 1CH323CBr and Na CH3CH2O is alkene.
Only one elimination product has been studied so far.
More than one elimination product can be produced.
The product that is formed the fastest will be richer.