Carboxylic Acids 3 Flashcards
Ester as an Carboxylic acid derivative
OH replaced by OR (where R can’t be H)
Thioester as a carboxylic acid derivative
OH replaced by SR
Acid Halide as a carboxylic acid derivative
OH replaced by a halogen
Acid anhydride as a carboxylic acid derivative
OH replaced by OCOCH3
Amide as a carboxylic acid derivative
OH replaced by NH2, NHR, NHRR’
Carboxylic acid derivatives
ester, thioester, acid halide, acid anhydride and amide. Done by nucleophilic substitution (Sn1)
Why does leaving group leave in tetrahedral intermediate of derivative formation
Most stable substituent. Donates it’s electrons to the delta negative C. This allows C to form a double bond with the negatively charged O. Resulting in the neutralisation of the molecule and the leaving of the leaving group
Nucleophilic addition of ketones and aldehydes
Nucleophile attaches to molecule breaking C to O double bond. O is negatively charged as result. Molecule is neutralised by the addition of a H
Conditions for nucleophilic addition to carboxylic acid
Leaving group must be more stable then nucleophile, nucleophile is strong enough to attach and C must be electrophilic enough
Carboxylic acid derivatives reactivity (weakest to strongest)
Amides, ester, acid anhydride and acid chloride. Can convert reactive derivatives to weaker ones (not the other way around)
Carboxylic acid leaving groups stability (least to most)
NRR’, OR, OCOR and Cl-. The better the leaving group the more reactive the carboxylic acid derivative (faster substitution occurs). The better a substance acts as a leaving group = the worse a nucleophile
Acid chloride hydrolysis
H2O reacts with acid chloride to form carboxylic acid and HCl (OH replaces halogen)
Acid Chloride Alcolysis
Alcohol reacts with acid halide to ester and HCl (OH replaces halogen)
Acid Chloride Aminolysis
Amide reacts with acid halide. Forms an amine and HCl. NH2 replaces halogen
Steps of acid halide Hydrolysis
Attachment of nucleophile (concerting C to O double bond to single), Halogen donating it’s electrons and leaving. Reforming of C to O double bond
Symmetrical Anhydrides Outline
Substituents on either end of molecule are the same. Formed by addition of 2 carboxylic acids with the elimination of water (in presence of acid catalyst and heat)
Acetic Anhydride Hydrolysis products
carboxylic acid and CH3COOH
Acetic anhydride with alcohols products
esters and CH3COOH
Acetic anhydrides with amine product
amide and CH3COOH
Aspirin Formation
Acetic anhydride and salicylic acid react to form aspirin and CH3COOH
Paracetamol Formation
p-hydroxyaniline and acetic anhydride resct to form paracetamol and CH3COOH
