An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen. The molecular mass of the compound is 86. It does not reduce Tollen's reagent but forms an addition compound with sodium hydrogen sulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible sturcture of the compound.

Open in App

Solution

SetP I: To determine the molecular formula of the compound.
$\begin{matrix} E l e m e n t & P e r c e n t a g e & A t o m i c m a s s & N o . o f m o l e s & S i m p l e s t m o l a r r a t i o C & 69.77 & 12 & \frac{69.77}{12} = 5.81 & \frac{5.81}{1.16} = 5 H & 11.63 & 1 & \frac{11.63}{1} = 11.63 & \frac{11.63}{1.16} = 10 O & (100 - 69.77 - 11.63) & 16 & \frac{18.60}{16} = 1.16 & \frac{1.16}{1.16} = 1 = 18.60 \end{matrix}$
Empirical formula of the given organic compound = $C_{5} H_{1} 0 O$ .
$M o l e c u l a r f o r m u l a = n \times (e m p i r i c a l f o r m u l a) w h e r e, n = \frac{M o l e c u l a r m a s s o f t h e c o m p o u n d}{E m p i r i c a l f o r m u l a m a s s o f t h e c o m p o u n d}$
Given, molecular mass = 86
$E m p i r i c a l f o r m u l a m a s s o f C_{5} H_{10} O = (12 \times 5) + (10 \times 1) + (16) = 60 + 10 + 16 = 86 ∴ n = \frac{86}{86} = 1 M o l e c u l a r f o r m u l a = 1 \times C_{5} H_{10} O = C_{5} H_{10} O$
Step II: Predicting the structure of the compound.
Criteria:

Formation of addition compound with $N a H S O_{3}$ depicts the presence of aldehyde or ketone.

The given compound does not reduce Tollen's reagent but gives positive iodoform test, so it is a methyl ketone.

On oxidation, the compound forms a mixture of ethanoic and propanoic acid, so it is

Step III: Chemical equations:

Suggest Corrections

Similar questions

69.77

11.63

86

(X)

C_{9} H_{10} O

(X)

C_{8} H_{8} O

2, 4 - D N P

1, 2

H C N

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program