The movement of electrons is not shown by the curved arrows.
Both the unrearranged carbocation and the rearranged carbocation can be affected by the solvolysis of 2-bromo-3-methylbutane.
The products from the rearranged carbocation are shown in the mechanisms.
Show which carbocation they come from and if they are the products of E1 or SN1 reactions.
There are four ways in which a carbocation can react to become more stable.
Rearrange to a more stable carbocation.
The order of stability is resonance-stabilized.
Give the substitution and elimination products.
Many compounds can be used to give a mixture of alkenes.
In a lot of cases, we can predict which elimination product will dominate.
The carbocation can lose a protons on either of the carbon atoms.
The general formula is R2C.
The order of preference is the same as the order of stability.
There are three elimina when 3-iodo-2, 2-dimethylbutane is treated with silver nitrate.
Predict which ones are formed in positive charge on a carbon atom larger amounts by giving their structures.
Solid silver iodide and a cation can be given by silver nitrate.
In either position, the tertiary cation can lose a protons.
A substitution product can be given by each of the two carbocations in Solved Problem 7-4.
The structures of the two substitution products should be given.
Two ethers and three alkenes are produced when (1-bromoethyl)cyclohexane is heated for an extended period of time.
Which of the three alkenes is the elimination product?
Under second-order conditions, eliminations can take place.
Butyl bromide and methoxide ion are in the same substance.
A strong base and a strong nucleophile can be found in methoxide ion.
It attacks the alkyl halide quicker than the halide can ionize.
The tertiary alkyl halide impedes the SN2 mechanism.
The observed product is formed of a double bond after the elimination of HBr.
The rate of elimination is proportional to the concentrations of both the alkyl halide and the base.
A strong base>nucleophile is often used with a poor 2deg alkyl halide to eliminate a strong nucleophile.
The back side of the hindered carbon is attacked by methoxide.
This reaction takes place in a single step, with bromide leaving the base.
The general mechanism of the E2 reaction involves a strong base and a carbon atom.
A double bond forms and the leaving group leaves.
In a single step, the E2 reaction takes place.
The leaving group leaves a strong base and a protons on a carbon.
The product is alkene.
There is an order of reactivity for alkyl halides.
The greater stability of double bonds is reflected in this reactivity order.
A more substituted alkene is given by the elimination of a secondary halide.
The substitution of alkene is more pronounced with the elimination of a tertiary halide.
The stabilities of the alkene products are reflected in the transition states, giving lower activation energies and higher rates for elimination of alkyl halides that lead to highly substituted alkenes.
The E2 reaction requires a protons to be on a carbon atom.
A mixture of products may result if there are two or more possibilities.
Zaitsev's rule predicts which of the possible products will be the major product.
A mixture of two products, but-1-ene and but-2-ene, can be created by the E2 reaction of 2-bromobutane.
The major product is the disubstituted isomer but-2-ene.
The mixture of a disubstituted alkene and a trisubstituted alkene was created by the reaction of 1-bromo-1-methylcyclohexane with sodium ethoxide.
The major product is the trisubstituted alkene.
The amount of substitution can be minimized by using a bulky base.
The approach to attacking a carbon atom is hampered by large alkyl groups on a bulky base.
The use of a bulky base for elimination is shown in the dehydrohalogenation of bromocyclohexane.
Both substitution and elimination can be done with bromidecyclohexane.
A bulky base such as diisopropylamine is favored over substitution for elimination.
Diisopropylamine acts as a strong base to abstract a protons.
The Zaitsev rule does not apply to bulky bases.
A bulky base can't abstract the proton that leads to the most highly substituted alkene.
Predict the products you expect and decide whether substitution or elimination is possible for each reaction.
The bonds being broken need to overlap with the bonds being formed in order for the electrons to flow smoothly from one bond to another.
Only one stereoisomer was produced.
The leaving group must overlap the H bond.
The leaving group is vacating and the groups are with it.
The hydrogen anti-coplanar elimination can be accomplished if 3 orbitals are parallel.
The hydrogen and the halogen are no longer present.
You can use the model to follow along with the discussion.
aconitic acid is seen in E2 reactions.
The anti-coplanar arrangement has a transition state.
This is usually removed in most cases.
COOH transition state is lower in energy than syn-coplanar elimination.
There is a transition state for syn-coplanar elimination.
The leaving group interfered with the attacking base.
aconitic acid is not removed.
The leaving group is bulky and negatively charged, and the repulsion between the base and the leaving group raises the energy of the syn-coplanar transition state.
The E2 Base has concerted transition states.
A hydrogen atom and a leaving group are held in a syn-coplanar arrangement.
Such compounds are likely to be eliminated by a syn-coplanar mechanism.
The hydrogen with mass number 2 is used in the following reaction to show which atom is in the base.
The bromine atom is held in a syn-coplanar position with the hydrogen atom.
Anti-coplanar eliminations are more common than syn-coplanar eliminations.
The E2 goes through an anti-coplanar transition state.
Different diastereomers of starting materials give different alkenes.
The trans isomer of the alkene product is only given by the E2 elimination of one diastereomer.
If we look at the reaction from the left end of the molecule, we can see the anticoplanar arrangement.
An anti-coplanar transition state is where most E2 reactions go.
If we look at the reaction from the hydrogen and leaving group's point of view, we can see the geometry.
You can use your models for the ones that involve stereochemistry.
Predict the products formed when the compounds react with strong bases by making models of them.
Most cyclohexanes are stable in chairs.
In the chair, all the carbon-carbon bonds are staggered, and any two adjacent carbon atoms have anti-coplanar bonds, ideally oriented for the E2 reaction.
One of the carbon atoms has its bond pointing up and the other has it pointing down.
If the leaving group can get into a trans-diaxial arrangement, an E2 elimination can take place on this chair.
The molecule must flip into the chair before it can be eliminated.
The group be trans.
The hydrogen atom and the leaving group must have a trans-diaxial relationship in an E2 elimination.
There is only one hydrogen atom in this compound.
The bromine atom is divided into two parts, one of which is a trans-diaxial arrangement.
Predict the elimination products of the reactions.
The two groups are anti and coplanar.
Look for a hydrogen trans in the chair.
Two bromodecalin stereoisomers are shown.
One isomer undergoes elimination with less time than the other.
Explain the large difference in the ease of elimination by predicting the products of these eliminations.
Explain why the following compounds will be eliminated with KOH faster.
The major product will be formed.
For each reaction, give the expected product.
The major points to remember about the E1 and E2 reactions are the factors that help us predict which of these mechanisms will operate under a given set of experimental conditions.
The nature of the base is the most important factor in deter mining.
The rate of the bimolecular reaction will be greater if a strong base is present.
A good solvent makes a unimolecular ionization likely if there is no strong base.
The elimination is caused by the loss of a protons to a weak base.
The E1 reaction usually accompanies the SN1 under these conditions.
Base strength is usually weak.
Strong bases are required.
Two ion formation is the slow step of the E1 reaction.
The E1 reaction depends on water and alcohols.
The transition state spreads the negative charge of the base over the entire molecule in the E2 reaction.
There is no need for solvation in the transition state.
The E2 is less sensitive to the solvent than some of the reagents.
A good ionizing solvent is required.
The importance of solvent polarity is not important.
The rate-limiting step in the E1 reaction is formation of a carbocation.
More substituted halides form more stable alkenes in the E2 reaction.
The rate of the E1 reaction is determined by the concentration of the alkyl halide.
The first-order rate equation is followed by it.
The rate of the E2 reaction is determined by the concentrations of the alkyl halide and the strong base.
The second-order rate equation is followed by it.
In most E1 and E2 eliminations, the product with the most substituted double bond dominates.