Question

Repeat Question 19-21 for the case when the protons are not decoupled. Note that $13C-{}^{1}H$coupling constants are generally in the range of $100$to $200$Hz.

Short answer

(a)

(b)

(c)

Step-by-step solution

(a) Methyl formate :

Methanol and formic acid combine to make methyl formate, which is the simplest ester. Between $100$and $32$degrees Celsius, it is liquid at normal pressure.

(a) Explanation :

Methyl formate structure is -

Three NMR active protons are linked to methyl C. As a result, the proton signal will split into quadrants. Since carbonyl carbon has one hydrogen linked to it, the signal will be divided into a doublet due to the coupling effect.

(b) Acetaldehyde : 

When ethanol is oxidised, it produces a colourless volatile liquid aldehyde.

(b) Explanation : 

Acetaldehyde structure is -

The carbonyl carbon in acetaldehyde has one hydrogen linked to it. As a result, the carbonyl carbon signal separates into two. Three hydrogens are connected to methyl carbon. As a result, the methyl carbon signal will be separated into four parts.

(c) Acetone : 

A colourless volatile liquid ketone that is produced by oxidising isopropanol and is utilised as an organic solvent and synthesis reagent.

(c) Explanation :

Acetone structure is -

Carbonyl carbon has no hydrogen linked to it in acetone. As a result, it will show as a singlet in C-NMR. Chemically, the other two methyl carbons are equal. However, each of them is attached to three hydrogen atoms. As a result, the methyl carbon signal divides into a quartet.