JEE Chemistry 2023 Question with Solution

Given: In Carius tube, compound 'X' is treated with sodium peroxide to form mineral acid 'Y'. Then solution of $\text{BaCl}_2$ is added to form precipitate 'Z'.

Find: Identify compound 'X'.

The Carius method is a technique used for the quantitative analysis of sulfur. Sulfur is oxidised to sulfuric acid, and then it reacts with barium chloride to form barium sulfate, which appears as a white precipitate.

Among the given options, methionine is the only compound containing sulfur that can give sulfuric acid in this method. Chloroxyleneol does not contain sulfur. Nucleotides and cytosine do not contain sulfur in a form detectable by this method.

Therefore, the correct compound 'X' is Methionine, so the correct option is C.

Discrepancy noted: The extracted option text lists Methionine as option B, but the solution states option C. Based on the solution conclusion, the answer is recorded as C.

Given: A precipitate formed after treatment with sodium peroxide and then $\text{BaCl}_2$ is used for quantitative estimation of an extra element.

Find: Which option contains that extra element.

In Carius method, sulfur present in an organic compound is converted into sulfate. Addition of $\text{Ba}^{2+}$ gives insoluble barium sulfate.

So the extra element being estimated is sulfur. Hence 'X' must be a sulfur-containing organic compound. From the given options, Methionine is the sulfur-containing compound mentioned in the solution.

Therefore, the compound is Methionine.