Question:
In Carius tube, an organic compound ‘X’ is treated with sodium peroxide to form a mineral acid ‘Y’. The solution of \(BaCl_2\), is added to ‘Y’ to form a precipitate ‘Z’. ‘Z’ is used for the quantitative estimation of an extra element. ‘X’ could be
In Carius tube, an organic compound ‘X’ is treated with sodium peroxide to form a mineral acid ‘Y’. The solution of \(BaCl_2\), is added to ‘Y’ to form a precipitate ‘Z’. ‘Z’ is used for the quantitative estimation of an extra element. ‘X’ could be
Updated On: Oct 16, 2024
- A nucleotide
- Chioroxylenol
- Methionine
- Cytosine
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The correct option is (C): Methionine
Was this answer helpful?
0
1
Top Questions on Classification Of Organic Compounds
- Match List I with List II.Choose the correct answer from the options given below:
List I
(Molecule)List II
(Number and types of bond/s between two carbon atoms)A. ethane I. one σ-bond and two π-bonds B. ethene II. two π-bonds C. carbon molecule, C2 III. one σ-bonds D. ethyne IV. one σ-bond and one π-bond - NEET (UG) - 2024
- Chemistry
- Classification Of Organic Compounds
- The final product A, formed in the following reaction sequence is:
- JEE Main - 2024
- Chemistry
- Classification Of Organic Compounds
- Given below are two statements: Statement I:
IUPAC name of \(HO–CH_2–(CH_2)_5–CH_2–COCH_3\) is 7-hydroxyheptan-2-one.
Statement II: 2-oxoheptan-7-ol is the correct IUPAC name for above compound.
In the light of the above statements, choose the most appropriate answer from the options given below:- JEE Main - 2024
- Chemistry
- Classification Of Organic Compounds
- Bond line formula of HOCH(CN)$_2$ is:
- JEE Main - 2024
- Chemistry
- Classification Of Organic Compounds
- Among the given organic compounds, the total number of aromatic compounds is
- JEE Main - 2024
- Chemistry
- Classification Of Organic Compounds
View More Questions
Questions Asked in JEE Main exam
- A rectangular loop of length 2.5 m and width 2 m is placed at 60° to a magnetic field of 4 T. The loop is removed from the field in 10 sec. The average emf induced in the loop during this time is:
- JEE Main - 2024
- Electromagnetic induction
- Two coils have mutual inductance 0.002 H. The current changes in the first coil according to the relation \( i = i_0 \sin \omega t \), where \( i_0 = 5 \, \text{A} \) and \( \omega = 50\pi \, \text{rad/s} \). The maximum value of emf in the second coil is \( \frac{\pi}{\alpha} \). The value of \( \alpha \) is ______.
- JEE Main - 2024
- Electromagnetic induction
- The 20th term from the end of the progression \[ 20, 19, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots, -129 \frac{1}{4} \] is:
- JEE Main - 2024
- Arithmetic Progression
- We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentration 0.1 M, 0.01M & 0.001 M, respectively. The value of vant’ Haft factor (i) for these solutions will be in the order.
- JEE Main - 2024
- Colligative Properties
- $\Delta_{vap}H^\circ$ for water is $+40.49 \, \text{kJ mol}^{-1}$ at 1 bar and $100^\circ\text{C}$. Change in internal energy for this vaporisation under same condition is .............. kJ mol$^{-1}$. (Integer answer)
(Given R = 8.3 J K$^{-1}$ mol$^{-1}$)- JEE Main - 2024
- Enthalpy change
View More Questions