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17 Reactions of Aromatic Compounds -- Part 17
The E2 mechanism requires a strong base to eliminate the quaternary ammonium salt.
The base is provided by converting the quaternary ammonium iodide to the hydroxide salt.
The elimination is a one-step, concerted E2 reaction using an amine as the leaving group.
A mixture of but-1-ene and but-2-ene can be formed when butan-2-amine is converted to hydroxide salt and heated.
The eliminations of alkyl halides usually follow Zaitsev's rule.
The rule applies because the most stable alkene is the most substituted.
The sheer bulk of the leaving group is one of the reasons why the Hofmann elimination prefers the least-substituted alkene.
The E2 mechanism requires an anti-coplanar arrangement of the leaving group and the protons.
The large trialkylamine leaving group in the Hofmann elimination can interfere with the coplanar arrangement.
The stereochemistry of the elimination of butan-2-amine is shown in Figure 19-11.
The salt eliminates by losing trimethylamine and a proton.
An anti-coplanar arrangement between a C3 proton and the leaving group requires an unfavorable gauche interaction between the C4 methyl group and the bulky trimethylammonium group.
The leaving group and one of the protons have an anti relationship.
The elimination of butan-2-amine was done by Hofmann.
There is an anti relationship between C3 bond and the leaving group.
There is an anti relationship between the leaving group and a protons.
The Hofmann product is given by the abstraction of a protons from C1.
The elimination of Hofmann can be used to convert complex amines to simpler amines.
The least-substituted alkene is usually the direction of elimination.
Predict the major product when the following amine is treated with excess iodomethane and then heated with silver oxide.
Finding every possible elimination of the salt is required to solve this problem.
There are three possible elimination routes shown by the green, blue, and red arrows.
There are many ways the compound can be eliminated.
The first alkene has a double bond.
The red alkene has an unsubstituted double bond and the second alkene gives the least substituted alkene.
Predict the major products formed when the following amines undergo elimination are best methylation, treatment with Ag2O, and heating.
Oxidation is a side reaction in amine syntheses.
During storage, amine oxidizes in contact with the air.
One of the reasons for converting amine to their salts is to prevent air oxidation.
The states are more oxidation as you go from left to right.
Common oxidants include H2O2, permanganate, and peroxyacids.
As the amine becomes more oxidation from left to right, the following sequence shows increasingly oxidized products of a primary Primary amine.
The symbol is used for monoamine oxidase.
Side products are often sants, but they are rarely formed, and the yields may be low.
The mechanisms of amine oxidations are due to many side not well characterized.
This oxidation can be done with either H2O2 or a peroxyacid.
A full positive charge on nitrogen and a negative charge on oxygen must be drawn from an amine oxide.
O bond of amine oxide formed by donation of electrons on nitrogen is often written as an arrow in older literature.
A strong base is not needed because the amine oxide acts as its own base.
The least-substituted alkene is given the same orientation as Hofmann elimination.
The base and leaving group are eliminated using an amine oxide.
The elimination of Cope requires stereochemistry.
Amine elimination occurs under milder conditions than Hofmann elimination.
It's useful when a sensitive or reactive alkene needs to be synthesised.
syn stereochemistry is involved in the elimination of Cope.
When the following compound is treated with H2O2 and heated, predict the products expected.
The tertiary amine is converted to an amine oxide.
It can give either of two alkenes.
The less-hindered elimination is expected to be favored.
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