Question

Question:21-64How would you distinguish spectroscopically between the followingisomer pairs? Tell what differences you would expect to see.
(a) N-Methylpropanamide and N,N-dimethylacetamide
(b) 5-Hydroxypentanenitrile and cyclobutanecarboxamide
(c) 4-Chlorobutanoic acid and 3-methoxypropanoyl chloride
(d) Ethyl propanoate and propyl acetate

Short answer

The differences are

a)N-Methyl propanamide

N,N-dimethyl acetamide

b) 5-Hydroxypentanenitrile and cyclobutanecarboxamide

c)4-Chlorobutanoic acid and 3-methoxypropanoyl chloride

d)Ethyl propanoate and propyl acetate

Step-by-step solution

Step 1: Introduction to the Concept

The arrangement of chemical bonds between atoms in a molecule (in anion or radical with many atoms), particularly which atoms are chemically bound to which other atoms with what kind of chemical bond, as well as any geometric shape information needed to uniquely identify the type of molecule.

Step 2: Determination of differences between N-Methylpropanamide and N,N-dimethylacetamide

a) The bond location in amides is affected by the degree of substitution. The greater the degree of substitution in disubstituted amides, the more electron density (resonance) from nitrogen is delocalized into the carbonyl, lengthening the C= 0 bond and lowering its stretching frequency.

The stretching absorption bonds in the infrared are,

N-Methyl propanamide

Step 3: Determination of differences between 5-Hydroxypentanenitrile and cyclobutanecarboxamide

b) The C= O bond is lengthened and its stretching frequency is reduced as electron density from nitrogen (resonance) is delocalized into carbonyl.

As a result, the amide has a lower stretching frequency than nitrile.

Step 4: Determination of differences between 4-Chlorobutanoic acid and 3-methoxypropanoyl chloride

c) Both electron and drawing function groups exist in nature. The larger the removal of inductive electrons, the shorterthe C=Obond and the higher thestretching frequency.

Step 5: Determination of differences between Ethyl propanoate and propyl acetate

d) Around a% 1735 cm^-1 is absorbed by ester. Due to the stronger delocalization of electron density caused by the resonance of which increases the bond length and decreases stretching frequency, the ester A should have a lower IR than the ester B.