Predict the major organic product of the reaction between N-methylbenzamide and sodium propanoate.
b. phenyl propanoate
c. benzoic propanoic anhydride
d. propyl benzoate
e. no reaction
Answer to question 1
In the first four reactions, Alkene acts as a nucleophile and adds electrons to the sigma* orbital of H-Br bond.
First the alkene is protonated at the carbon with more no. Of Hydrogen atoms. Then the nucleophile attacks the carbocation to form the more stable saytzeff product.
In the 4th reaction, rearrangement of carbocation takes place to give a more stable tertiary carbocation.
Answer to question 2
e. no reaction
The reactivity of carboxylic acid derivatives in nucleophilic acyl substitution is as below from most reactive to least reactive:
Acid chloride > Acid anhydride > Thioester > Ester > Amide
Amide is the least reactive due to the resonance structure 2 as shown below. Nitrogen is less electronegative than oxygen, and the resonance form 2 is formed more often compared to the resonance in other carboxylic acid derivatives. Thus the C-N bond has slight double bond property and is the less reactive in nucleophilic acyl substitution than ester. Therefore, when N-methylbenzamide is mixed with sodium propanoate, instead of forming the less stable propyl benzoate, there is no reaction happen.
Answer to question 3
This is a addition type of reaction in which hydrogen halide is added to the alkenes. The intermediate for this reaction is a stable carbocation. In this reaction there is a rearrangement of 2° carbocation to 3° carbocation as 3° carbocation is more stable than 2° and 1° carbocation. The major product will obtain when the intermediate is more stable therefore in this reaction rearrangement take place. The mechanism for this reaction to predict the major product in given in a attached image.