1
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
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The structure of the major product formed in the following reaction is :

JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English

A
JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English Option 1
B
JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English Option 2
C
JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English Option 3
D
JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Haloalkanes and Haloarenes Question 10 English Option 4
2
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
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The elemental composition of a compound is $54.2 \% \mathrm{C}, 9.2 \% \mathrm{H}$ and $36.6 \% \mathrm{O}$. If the molar mass of the compound is $132 \mathrm{~g} \mathrm{~mol}^{-1}$, the molecular formula of the compound is : [Given : The relative atomic mass of $\mathrm{C}: \mathrm{H}: \mathrm{O}=12: 1: 16$ ]

A
$\mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_2$
B
$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$
C
$\mathrm{C}_4 \mathrm{H}_9 \mathrm{O}_3$
D
$\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_3$
3
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
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Based on the data given below :

$$\begin{array}{ll} \mathrm{E}_{\mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}}^{\circ}=1.33 \mathrm{~V} & \mathrm{E}_{\mathrm{Cl}_2 / \mathrm{Cl}^{(-)}}^{\circ}=1.36 \mathrm{~V} \\ \mathrm{E}_{\mathrm{MnO}_4^{-} / \mathrm{Mn}^{2+}}^0=1.51 \mathrm{~V} & \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} \end{array}$$

the strongest reducing agent is :

A
$\mathrm{Cl}^{-}$
B
$\mathrm{MnO}_4^{-}$
C
$\mathrm{Cr}$
D
$\mathrm{Mn}^{2+}$
4
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
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For reaction

JEE Main 2025 (Online) 24th January Evening Shift Chemistry - Compounds Containing Nitrogen Question 10 English

The correct order of set of reagents for the above conversion is :

A
$\mathrm{Ac}_2 \mathrm{O}, \mathrm{Br}_2, \mathrm{H}_2 \mathrm{O}(\Delta), \mathrm{NaOH}$
B
$\mathrm{H}_2 \mathrm{SO}_4, \mathrm{Ac}_2 \mathrm{O}, \mathrm{Br}_2, \mathrm{H}_2 \mathrm{O}(\Delta), \mathrm{NaOH}$
C
$\mathrm{Br}_2 \mid \mathrm{FeBr}_3, \mathrm{H}_2 \mathrm{O}(\Delta), \mathrm{NaOH}$
D
$\mathrm{Ac}_2 \mathrm{O}, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{Br}_2, \mathrm{NaOH}$
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