1
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is -
A
20 g
B
4 g
C
80 g
D
10 g
2
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
JEE Main 2019 (Online) 12th January Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 125 English
can not be prepared by -
A
CH3CH2COCH3 + PhMgX
B
PhCOCH3 + CH3CH2MgX
C
PhCOCH2CH3 + CH3MgX
D
HCHO + PhCH(CH3)CH2MgX
3
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
In the following reaction, products A and B are –

JEE Main 2019 (Online) 12th January Morning Slot Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 173 English
A
JEE Main 2019 (Online) 12th January Morning Slot Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 173 English Option 1
B
JEE Main 2019 (Online) 12th January Morning Slot Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 173 English Option 2
C
JEE Main 2019 (Online) 12th January Morning Slot Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 173 English Option 3
D
JEE Main 2019 (Online) 12th January Morning Slot Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 173 English Option 4
4
JEE Main 2019 (Online) 12th January Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The correct order for acid strength of compounds :

CH $$ \equiv $$ CH, CH3–C $$ \equiv $$ CH and CH2 = CH2 is as follows
A
CH $$ \equiv $$ CH > CH2 = CH2 > CH3 – C $$ \equiv $$ CH
B
CH3 – C $$ \equiv $$ CH > CH2 = CH2 > HC $$ \equiv $$ CH
C
HC $$ \equiv $$ CH > CH3 – C $$ \equiv $$ CH > CH2 = CH2
D
CH3 – CH $$ \equiv $$ CH > CH $$ \equiv $$ CH > CH2 = CH2
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