1
JEE Main 2022 (Online) 27th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Match List I with List II.

List I List II
(A) Benzenesulphonyl chloride (I) Test for primary amines
(B) Hoffmann bromamide reaction (II) Anti Saytzeff
(C) Carbylamine reaction (III) Hinsberg reagent
(D) Hoffmann orientation (IV) Known reaction of Isocyanates.

Choose the correct answer from the options given below:

A
A-IV, B-III, C-II, D-I
B
A-IV, B-II, C-I, D-III
C
A-III, B-IV, C-I, D-II
D
A-IV, B-III, C-I, D-II
2
JEE Main 2022 (Online) 26th July Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The correct sequential order of the reagents for the given reaction is

JEE Main 2022 (Online) 26th July Evening Shift Chemistry - Compounds Containing Nitrogen Question 63 English

A
$$\mathrm{HNO}_{2}, \mathrm{Fe} / \mathrm{H}^{+}, \mathrm{HNO}_{2}, \mathrm{KI}, \mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}$$
B
$$\mathrm{HNO}_{2}, \mathrm{KI}, \mathrm{Fe} / \mathrm{H}^{+}, \mathrm{HNO}_{2}, \mathrm{H}_{2} \mathrm{O} /$$ warm
C
$$\mathrm{HNO}_{2}, \mathrm{KI}, \mathrm{HNO}_{2}, \mathrm{Fe} / \mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O} / \mathrm{H}^{+}$$
D
$$\mathrm{HNO}_{2}, \mathrm{Fe} / \mathrm{H}^{+}, \mathrm{KI}, \mathrm{HNO}_{2}, \mathrm{H}_{2} \mathrm{O} /$$ warm
3
JEE Main 2022 (Online) 26th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The correct stability order of the following diazonium salt is

JEE Main 2022 (Online) 26th July Morning Shift Chemistry - Compounds Containing Nitrogen Question 64 English

A
$$(\mathrm{A})>(\mathrm{B})>(\mathrm{C})>(\mathrm{D})$$
B
$$(\mathrm{A})>(\mathrm{C})>(\mathrm{D})>(\mathrm{B})$$
C
$$(\mathrm{C})>(\mathrm{A})>(\mathrm{D})>(\mathrm{B})$$
D
$$(\mathrm{C})>(\mathrm{D})>(\mathrm{B})>(\mathrm{A})$$
4
JEE Main 2022 (Online) 26th July Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : Experimental reaction of $$\mathrm{CH}_{3} \mathrm{Cl}$$ with aniline and anhydrous $$\mathrm{AlCl}_{3}$$ does not give $$o$$ and $$p$$-methylaniline.

Reason (R): The $$-\mathrm{NH}_{2}$$ group of aniline becomes deactivating because of salt formation with anhydrous $$\mathrm{AlCl}_{3}$$ and hence yields $$m$$-methyl aniline as the product.

In the light of the above statements, choose the most appropriate answer from the options given below :

A
Both A and R are true and (R) is the correct explanation of (A).
B
Both A and R are true but (R) is not the correct explanation of (A).
C
(A) is true, but (R) is false.
D
(A) is false, but (R) is true.
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