Paracetamol is produced by the reaction between 4-aminophenol and acetic anhydri

Question

Paracetamol is produced by the reaction between 4-aminophenol and acetic anhydride at room temperature. What class of compounds does paracetamol belong to? Alcohol Amine Anhydride Amide Ester Aspirin is produced by the reaction between 2-hydroxybenzoic acid and acetic (ethanoic) anhydride with an acidic catalyst under heat. What class of compounds does aspirin belong to? Phenol Carboxylic acid Anhydride Amide Amine Which of the following combination of reagents will produce a polyamide upon heating at the 1:1 molar ratio? Hexanedioic acid and dimethylamine Hexanoic acid and l, 6-diaminohexane Hexanedioic acid and 1, 4-diaminobutane Hexanedioic acid 1, 2-etbanediol The resonance stabilization of which functional group is shown on this scheme? Carboxy-group Carboxylate (conjugate base of a carboxy-group) Ester Alcohol Enol What is the chemical consequence of the concept, presented in question 8? pKa of the group E is lower than pKa of all other groups pKa of the group E is higher than pKa of all other groups pKa of the group D is lower than pKa of all other groups pKa of the group D is higher than PKa of all other groups pKa of the group A is lower than pKa of all other groups How did the concept from question 8 help you answer question 9? It helped me evaluate stability of the conjugate base It helped me evaluate stability of the conjugate acid It helped me evaluate reaction rate of deprotonation It helped me identify the rate-limiting step of protonation It helped me identify the rate-limiting step of deprotonation Which of the following compounds will not form a carboxylic acid upon acidic hydrolysis? Ethyl acetate Diethylacetamide Diethylmethylamine Acetic anhydride

Explanation / Answer

Q5

paracetamol --> it has NH2 group and C=O grup, so it must be an AMIDE

Q6

Asipirin is a very common acid (i.e. Acetyl Salycilic Acid)

so..

choose Carboxilic acid

Q7

Polyamide upon heating:

Hexanedioic acid + 1-amiinohexane

amide --> C=O and NH2 groups

Q8

This is an enol ,since double bond is changing, i.e. the lone pairs/pi electrons resonance

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