COMEDK 2024 Evening Shift | Haloalkanes and Haloarenes Question 13 | Chemistry

No.	Name of reaction	No.	Equations representing reactions
(A)	Sandmeyer	(P)	$$ 2 \mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+\mathrm{Na} \text { (dry ether) } \rightarrow \text { Diphenyl. } $$
(B)	Fittig	(Q)	$$ \mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CH}_3+\mathrm{AgF} \rightarrow \mathrm{CH}_3-\mathrm{CHF}-\mathrm{CH}_3+\mathrm{AgBr} $$
(C)	Swarts	(R)	$$ \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Br}+\mathrm{Nal} \text { ( Acetone) } \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{I}+\mathrm{NaBr} $$
(D)	Finkelstein	(S)	$$ \mathrm{C}_6 \mathrm{H}_5 \mathrm{~N}_2+\mathrm{Cl}-+\mathrm{Cu}_2 \mathrm{Br}_2 / \mathrm{HBr} \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{Br}+\mathrm{N}_2 $$

$$ \mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{R} $$

$$ \mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{P} $$

$$ \mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P} $$

$$ \mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{P} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{Q} $$

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

Arrange the compounds $$\mathrm{A}, \mathrm{B}, \mathrm{C}$$ and $$\mathrm{D}$$ in the increasing order of their reactivity towards $$\mathrm{S_N}1$$ reaction.

$$ \begin{aligned} & \mathrm{A}=\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl} \\ & \mathrm{B}=\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl} \\ & \mathrm{C}=\mathrm{CH}_2=\mathrm{CH}-\mathrm{Cl} \\ & \mathrm{D}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Cl} \end{aligned}$$

$$\mathrm{D}<\mathrm{B}<\mathrm{A}<\mathrm{C}$$

$$\mathrm{D}<\mathrm{C}<\mathrm{B}<\mathrm{A}$$

$$\mathrm{B}<\mathrm{D}<\mathrm{C}<\mathrm{A}$$

$$\mathrm{C}<\mathrm{A}<\mathrm{D}<\mathrm{B}$$

COMEDK 2024 Evening Shift

MCQ (Single Correct Answer)

-0

Identify the products C, D and F formed in the following sets of reactions.

COMEDK 2024 Evening Shift Chemistry - Haloalkanes and Haloarenes Question 12 English

$$ \mathrm{C}=\mathrm{p} \text {-nitrobromobenzene } \mathrm{D}=\mathrm{m} \text {-nitrobromobenzene } \mathrm{F}=\text { o-nitrobromobenzene } $$

$$ \mathrm{C}=\text { o-nitrobromobenzene } \mathrm{D}=\mathrm{m} \text {-nitrobromobenzene } \quad \mathrm{F}=\mathrm{p} \text {-nitrobromobenzene } $$

$$ \mathrm{C}=\mathrm{o} \text {-nitrobromobenzene } \mathrm{D}=\mathrm{p} \text {-nitrobromobenzene } \mathrm{F}=\mathrm{m} \text {-nitrobromobenzene } $$

$$ \mathrm{C}=\mathrm{m} \text {-nitrobromobenzene } \mathrm{D}=\mathrm{p} \text {-nitrobromobenzene } \mathrm{F}=\text { o-nitrobromobenzene } $$

COMEDK 2024 Afternoon Shift

MCQ (Single Correct Answer)

-0

Arrange the following compounds in the decreasing order of reactivity towards electrophilic substitution reaction.

(I) Chlorobenzene

(II) Nitrobenzene

(III) Benzene

(IV) Isopropylbenzene

$$ \text { IV }>\text { III }>\text { I }>\text { II } $$

$$ \mathrm{I}>\mathrm{II}>\mathrm{IV}>\mathrm{III} $$

$$ \mathrm{IV}>\mathrm{I}>\mathrm{II}>\mathrm{III} $$

$$ \mathrm{III}>\mathrm{II}>\mathrm{IV}>\mathrm{I} $$

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