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27 -- Part 3: Reactions of Organic Compounds
There is a strong preference for forming the internal double bond.
The most stable product of an elimination reaction is usually the major one.
The order of stability is based on experimental data.
The transition state leading to a network is lower in energy than the alkenes.
The less highly substituted alkene.
The major components of the interactions of filled and the product mixture are based on double bonds and the internal explanations.
A word of caution is in order.
The major product in an elimination without an explanation is the most highly substituted alkene, but important exceptions exist.
The stability of the base is required for the formation of the most highly substituted alkene.
Access to an alkene increases with the hydrogen atoms on a secondary or tertiary carbon atom.
A large base will attack the hydrogen atoms with the least amount of alkene.
In the first reaction, a smaller, less hindered base is used and the major elimination product is the more highly substituted alkene.
The major elimination product in the second reaction is the less highly substituted alkene.
The H atom on the b carbon must be anti to the leaving group in an E2 reaction.
There is a restriction on the elimination products that can be formed.
Section 27B, A Closer Look at the E2 Mechanism, can be found on the MasteringChemistry site.
haloalkanes can undergo a variety of reactions.
A variety of products are possible because of the competitive nature of these reactions.
Many factors must be considered when determining whether a reaction will proceed through an E1 mechanism.
There is a tentative yes to this question.
The donor and acceptor of the electron pair are identified.
Consider the electrolyte.
Consider the donor of the electron pair.
Determine if the dominant reaction will be SN2, E2, or E1.
The first consideration is the base strength of the electron pair donor, [?]B.
steric hindrance, nucleophilicity, and solvent effects are additional considerations.
There are two possible reactions for primary alkanes.
There are three reactions for primary, secondary, and tertiary haloalkanes.
There are additional considerations in the red box.
The possibility of (E) and (Z) stereoisomers exists for an alkene with an internal C " C bond.
In polar protic solvent, a secondary haloalkane can react with a weak nucleophile to give products.
The CH3X is not included because it undergoes only one reaction.
A b carbon will have a protons removed from it.
If the electron donor is strong, elimination by E2 is the main reaction and the major product is an alkene with an internal C bond.
If a polar apro tic solvent is used, the main reaction will be SN2.
The configuration of a carbon in the haloalkane can be changed by the SN2 reaction.
Along with the SN1 products, E1 products will also form.
The possible reactions are SN1, E1, or E2.
A carbon that is too sterically hindered for backside attack is not very likely.
Predict the mechanisms by which the products are formed for each reaction.
The primary haloalkane is identified as CH3 CH2 CH2Br.
The sterically strong base is 1CH323CO-.
Although a strong base is usually a strong nucleophile, the bulkiness of this base disfavors substitution, and we expect that elimination by E2 will provide the major product.
A secondary haloalkane is the electrophile.
The weak base of the electron pair donor is negatively charged and it is a good nucleophile in a polar solvent.
The main reaction will be SN2.
The solvent's molecule serves as an electron pair donor.
The reaction involves a weak base and weak nucleophile in a polar protic solvent.
These conditions do not favor either E2 or SN2 Both SN1 and E1 occur at the same time.
In a polar protic solvent, such as water, secondary carbocations are stable.
We expect both products.
substitution will dominate over elimination because H2O is a very weak base.
The mechanism for the reaction was not shown.
We need to make sure we can show the steps involved in forming the products.
Predict the substitution and elimination products.
There is a summary given at the end of Section 27.
The carbocation in these reactions is susceptible to attack from a variety of species.
The reactions of haloalkanes are not often used to make other organic compounds.
The rearrangement of the carbocation intermediate to form a more stable intermediate is a consequence of the SN1 and E1 reactions.
When 2-bromo-3-methylbutane is heated with water, a situation arises.
The major product is 2-methylbutan-2-ol.
Two possible rearrangements, labeled (a) and (b), can occur.
The movement of a hydrogen atom is involved in both rearrangements.
38 carbocation more stable than 28 Rearrangement does not happen because it results in a 1deg carbocation, which is highly unstable.
Rearrangement converts a 2deg carbocation into a 3deg carbocation.
The major substitution product comes from the 3deg carbocation.
Reconfiguration of the carbocation is a possibility whenever a carbocation forms.
The rearrangement can involve the movement of alkyl chains.
It comes as no surprise that SN1 and E1 reactions are not used as often as other reactions for organic synthesis.
Alcohols can be easily converted into other compounds in organic synthesis.
In Chapter 26, we learned that primary alcohols could be converted to aldehydes or carboxylic acids.
Alcohols react with carboxylic acids to form esters.
The name of the chemical is pyridinium chlorochromate.
Section 27-2 can be seen.
Table 27.3 summarizes the reactions of alcohols.
OH group is replaced by a halogen atom or eliminated as H2O.
A group of alcohol can be replaced with a halogen atom or H2O.
The feasibility of the following substitution reactions can be considered using concepts we discussed in this chapter.
OH I is the leaving group in both reactions.
The OH I ion is a poor leaving group and it is a good nucleophile.
The first reaction doesn't mean that the base is weak.
It is better if it is a leaving group in the presence of strong acid.
The leaving group is H2O instead of OH- because of the weak base.
OH H2O is a weaker base than OH-, it is a better leaving group.
The conditions are just right for a carbon leaving group reaction: a strong base and poor reaction.
A SN2 reaction is followed by a reversible protonation step.
An SN2 reaction involves a backside attack at a carbon.
The structure of the transition state can be drawn.
If a tertiary alcohol, such as 1CH323COH, is used in place of the primary alco hol in reaction, the substitution occurring in the second step occurs by an SN1 reaction.
The oxygen atom of the alcohol functional group is protonsated in the first step.
The second step is an attack on a carbon.
The substitution occurs because the carbon is primary and I is a strong nucleophile.
The carbocation is stable by the alkyl groups.
In the previous sections, substitution reactions competed with elimination reactions.
Alcohols can also have elimination reactions.
An important method of synthesizing alkenes is OH2 from an alcohol.
An acid catalyst is needed for dehydration of alcohol.
The leaving group will be H2O rather than OH- if the acid catalyst is used.
N2 doesn't happen.
The alcohol is protonsated in the first step.
The major product in an elimination reaction is usually the more highly substituted alkene.
When H is attacked, the blue arrows show the movement of electrons.
When H is attacked, the red arrows show the movement of electrons.
The more highly substituted alkene is the major product.
In the last step, the H3O+ ion is regenerated.
Dehydration of secondary alcohols can occur by E1 or E2.
If appropriate, suggest a way to show the movement of electrons by using arrows.
We consider the role of alcohol in each case after we write structural formulas for the reactants.
Alcohols can be deprotonated.
They can act in a substitution reaction or be dehydrated.
To make a decision, we need to consider the carbon atoms in the reactants.
The other reactants, solvent, and reaction conditions must also be considered.
The structural formula for propan-1-ol is CH3 CH2 OH.
It's a primary alcohol.
Under the conditions specified, substitution and elimination are not possible.
We need to compare the strengths of the acids on the side of the equation.
Both bases have a negative charge on oxygen, so we expect them to be similar in strength.
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