How to deduce the structure of an organic compound X - Aldehydes & Ketones

Question- An organic compound X contains 70% Carbon, 11.33% Hydrogen and 18.67% Oxygen. The molecular weight of the compound is 86. It gives negative Tollens test but forms an addition product with sodium hydrogen sulfite. It also gives a positive iodoform test. On vigorous oxidation with KMnO4, the compound X decomposes to ethanoic acid and propanoic acid. What is the structure of compound X?

Answer-  

Step 1: Obtain the molecular formula

On converting %age to mass, we can say that a 100gm sample would contain 70 gm of Carbon, 11.33 gm of Hydrogen and 18.67 gm of Oxygen.

To obtain the molecular formula, divide each mass with their respective atomic weight.

Carbon (C) = 70/ 12 = 5.83

Hydrogen (H) = 11.33/ 1= 11.33

Oxygen (O) = 18.67/ 16 = 1.16

Divide the obtained values with the lowest number that is, 1.16.

Therefore,

Carbon = 5.83/1.16 = 5

Hydrogen = 11.33/1.16 = 10

Oxygen = 1.16/1.16 = 1

There are five atoms of Carbon, ten atoms of Hydrogen and one atom of Oxygen, so the molecular formula is C5H10O.

To confirm, multiply each with its atomic weight and the result should be 86, the samples molecular mass.

C5H10O = 5 x 12 + 10 + 16 = 86. Confirmed!!

 

Step 2: Find out the degree of unsaturation

Degree of unsaturation = (2C + 2 + N – X – H) / 2= (10 + 2 – 10) / 2 = 1

Remember, Oxygen has no effect.

The molecule has 1 degree of unsaturation.

 

Step 3: Deduce the functional group

Negative Tollens test – not an aldehyde functionality

Positive Bisulfite addition product – a ketone functionality

Positive Iodoform test – a methyl ketone (CH3- CO-)

Therefore, the structure of the compound with molecular formula C5H10O and one degree of unsaturation would be,

CH3-CO-CH2CH2CH3

Step 4: Structure confirmation

On vigorous oxidation, the compound X decomposes to give ethanoic acid and propanoic acid.

CH3-CO­­-/-CH2CH2CH3

 

If the molecule cleaves next to carbonyl functionality, it will give two molecules, ethanoic acid (CH3COOH) and propanoic acid (CH3CH2COOH).

Therefore, the compound X is pentan-2-one

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