Question

Compound A (CH2O) gives off oxygen on treatment with sodium metal and also decolonizes Br in CCL to give organic compound B. Compound A on treatment with b in NaOH gives iodoform and a salt C which after acidification gives a white solid D (CHO). Using knowledge of organic chemistry, identify structures A, B, C, and D.

a) A: Formaldehyde, B: Ethylene, C: Sodium chloride, D: Formic acid
b) A: Methanol, B: Chloroform, C: Sodium bromide, D: Formaldehyde
c) A: Glucose, B: Bromine, C: Sodium iodide, D: Acetic acid
d) A: Acetaldehyde, B: Ethanol, C: Sodium acetate, D: Formic acid

Answer 1

Compound A is formaldehyde, which when treated with sodium metal releases oxygen and gives ethanol (B). Compound A treated with I2 in NaOH gives iodoform and sodium acetate (C), which yields formic acid (D) upon acidification.

Step-by-step explanation:

The compound A (CH2O) is likely formaldehyde since it releases oxygen when treated with sodium metal and decolorizes bromine in CCl4 to give an organic compound B. When A is treated with I2 in NaOH, it gives iodoform, which suggests the methyl ketone characteristic of acetaldehyde, and a salt C, which would be sodium acetate. Once acidified, this salt yields a white solid D, which is formic acid (CHO). Considering all the given information and reactions:

• A is formaldehyde (CH2O)
• B is ethanol (C2H5OH), which upon oxidation can lead to acetaldehyde
• C is sodium acetate (CH3COONa)
• D is formic acid (HCOOH)

Thus, the correct identification of the structures is: A: Formaldehyde, B: Ethanol, C: Sodium acetate, D: Formic acid, which corresponds to option (d).