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JEE Main 2024 (Online) 30th January Evening Shift
MCQ (Single Correct Answer)
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-1
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m-chlorobenzaldehyde on treatment with 50% KOH solution yields :

A
JEE Main 2024 (Online) 30th January Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 19 English Option 1
B
JEE Main 2024 (Online) 30th January Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 19 English Option 2
C
JEE Main 2024 (Online) 30th January Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 19 English Option 3
D
JEE Main 2024 (Online) 30th January Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 19 English Option 4
2
JEE Main 2024 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
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JEE Main 2024 (Online) 30th January Morning Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 18 English

This reduction reaction is known as:

A
Wolff-Kishner reduction
B
Etard reduction
C
Stephen reduction
D
Rosenmund reduction
3
JEE Main 2024 (Online) 30th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
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Structure of 4-Methylpent-2-enal is :

A
JEE Main 2024 (Online) 30th January Morning Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 17 English Option 1
B
JEE Main 2024 (Online) 30th January Morning Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 17 English Option 2
C
JEE Main 2024 (Online) 30th January Morning Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 17 English Option 3
D
JEE Main 2024 (Online) 30th January Morning Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 17 English Option 4
4
JEE Main 2024 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
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Identify the reagents used for the following conversion

JEE Main 2024 (Online) 29th January Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 22 English

A
$$\mathrm{A}=\mathrm{LiAlH}_4, \mathrm{~B}=\mathrm{NaOH}_{(\mathrm{aq})}, \mathrm{C}=\mathrm{NH}_2-\mathrm{NH}_2 / \mathrm{KOH} \text {, ethylene glycol }$$
B
$$\mathrm{A}=\mathrm{DIBAL}-\mathrm{H}, \mathrm{B}=\mathrm{NaOH}_{(\mathrm{aq})}, \mathrm{C}=\mathrm{NH}_2-\mathrm{NH}_2 / \mathrm{KOH} \text {, ethylene glycol }$$
C
$$\mathrm{A}=\mathrm{LiAlH}_4, \mathrm{~B}=\mathrm{NaOH}_{(\mathrm{alc})}, \mathrm{C}=\mathrm{Zn} / \mathrm{HCl}$$
D
$$\mathrm{A}=\mathrm{DIBAL}-\mathrm{H}, \mathrm{B}=\mathrm{NaOH}_{(\mathrm{alc})}, \mathrm{C}=\mathrm{Zn} / \mathrm{HCl}$$
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