Tips for tomorrow

Hello All,

I was going through my notes and I thought these few tips would be very useful for tomorrow's test:

-Methoxy and hydroxy substituents on benzene rings are strong activators and so does not need the Lewis acid FeBr3. If you use the FeBr3, you may get a tribromo substituted benzene.

-Meta directors on benzene rings does not undergo Friedel-Craft reactions.

-Also, Aniline and N-substituted anilines do not under Friedel-Craft reactions. The lone pair of electron on the nitrogen forms a bond with the lewis acid.

-A tertiary alkyl group attached to a benzene cannot form the benzenecarboxylic acid because there is no hydrogen attached to it.

-If you want to form a trisubstituted benzene and you have two groups already attached to it, the stronger activator determines the position of the third group.

-Electron withdrawing groups on a benzene increases its acidity. Electron-donating groups decreases its acidity.

-Amides are the strongest of the Class I carboxylic acids because it forms strog dipole-dipole interactions as well as intermolecular hydrogen bonds and therefore have the highest boiling points.

-Cl- is the weakest base in the class I carbonyl groups but it is the most reactive. Amide is the least reactive. Remember, weaker bases are better leaving groups.

-The reaction of an anhydride with water always yields two carboxylic acids.

-Esters do not react with Cl-

- In acidic conditions (hydrolysis of esters) you do not want to have negative charges. Only positive or neutral charges.

-The acid makes the carbonyl more reactive and the carbonyl a better leaving group.

-In basic conditions you do not want to have positive charges; only negative or neutral charges.

-Amides react only with water in acidic solutions to form carboxylic acids and NH4+.


I hope these few points will help!