datavorous/entrance-exam-dataset huggingface dataset:
In the following reaction, $A$ and $B$ respectively are
$A \stackrel{\mathrm{HBr}}{\longrightarrow} \mathrm{C}_2 \mathrm{H}_5 \mathrm{Br} \stackrel{B}{\longrightarrow} A$

Options:
A $\mathrm{C}_2 \mathrm{H}_4$ and alcoholic $\mathrm{KOH} / \Delta$
B $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}$ and aqueous $\mathrm{KOH} / \Delta$
C $\mathrm{C}_2 \mathrm{H}_2$ and $\mathrm{PBr}_3$
D None of these

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_4$ and alcoholic $\mathrm{KOH} / \Delta$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the following reaction A is

Options:
A
B
C
D

The correct answer is:

Solution:
The above reaction sequence is given below. In the first step, halogenation reaction is occurred then elimination reaction gives cyclohexene and reductive ozonolysis gives the dialdehyde and then aldol condensation takes place and final desired product is formed.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2020 (09 Jan Shift 2)
In the following reaction, ‘A’ is

Options:
A
B
C
D

The correct answer is:

Solution:
Given reactant undergoes SN2 reaction with diethyl carbonate.

Initially lone pair electron of –NH2 attack on electrophilic carbon, after then lone pair electron of oxygen attacks leading to formation of cyclic compound.

Topic: Chemistry, Amines, JEE Main, JEE Main 2023 (01 Feb Shift 1)
In the following reaction ' $A$ ' is

Options:
A $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CHO}$
B $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$
C $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$
D $\mathrm{C}_2 \mathrm{H}_5 \mathrm{CHO}$

The correct answer is:
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{OH}$

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the following reaction
Aldehyde+Alcohol→HClAcetal

Aldehyde	Alcohol

HCHO	t-BuOH

CH3CHO	MeOH

The best combination is:

Options:
A HCHO and MeOH
B CH3CHO and t-BuOH
C CH3CHO and MeOH
D HCHO and t-BuOH

The correct answer is:
HCHO and MeOH

Solution:
Formation of an acetal occurs when the hydroxyl group of a hemiacetal becomes protonated and is lost as water. The carbocation that is produced is then rapidly attacked by a molecule of alcohol. Loss of the proton from the attached alcohol gives the acetal.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2019 (12 Jan Shift 1)
In the following reaction, ‘B’ is

Options:
A

B

C

D

The correct answer is:

Solution:
The given reaction is an example of dehydration of alcohol. In this reaction the first step is protonation of alcohol which leads to the formation of carbocation. The carbocation formed undergo rearrangement to get stability. The tertiary carbocation formed in this reaction involved in ring closure, which loses proton to from alkene. The steps are shown below.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (06 Apr Shift 2)
In the following reaction,
$$
\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br} \frac{\text { (i) } \mathrm{Mg}, \text { Ether }}{\text { (ii) } \mathrm{H}_{3} \mathrm{O}^{+}} \mathrm{X}
$$
the product ' $\mathrm{X}^{\prime}$ is

Options:
A $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{C}_{6} \mathrm{H}_{5}$
B $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}$
C $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}$,
D $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{5}$

The correct answer is:
$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{3}$,

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the following reaction,

CH3COCl â BaSO4 Pd/H2X

Identify X out of the following

Options:
A Acetaldehyde
B Propionaldehyde
C Acetone
D Acetic anhydride

The correct answer is:
Acetaldehyde

Solution:
CH 3 −COCl acetyl chloride  +  H 2 → Pd-BaSO 4 CH 3 CHO acetaldehyde  + HCl
This reaction is called Rosenmund's reaction.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the following reaction
carbonyl compound +MeOH⇌HClacetal
Rate of the reaction is the highest for:

Options:
A Acetone as substrate and methanol in stoichiometric amount
B Propanal as substrate and methanol in excess
C Propanal as substrate and methanol in stoichiometric amount
D

Acetone as substrate and methanol in excess

The correct answer is:
Propanal as substrate and methanol in excess

Solution:
Acetal is form from the reaction of aldehyde with excess of MeOH

Formation of an acetal occurs when the hydroxyl group of a hemiacetal becomes protonated and is lost as water. The carbocation that is produced is then rapidly attacked by a molecule of alcohol. Loss of the proton from the attached alcohol gives the acetal.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2019 (09 Apr Shift 2)
In the following reaction, find the number of products formed which contain carbon atom(s):
Sodium bicarbonate \( \stackrel{\Delta}{\longrightarrow} \) Products

Options:

The correct answer is:
2

Solution:
$2 \mathrm{NaHCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}$
Products containing C-atom: $\mathrm{Na}_{2} \mathrm{CO}_{3}$ and $\mathrm{CO}_{2}$

Topic: Chemistry, s Block Elements, JEE Main
$$
\text { In the following reaction }
$$
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Br} \frac{1 . \mathrm{Mg} \text {, Ether }}{2 . \mathrm{H}_3 \mathrm{O}^{+}} \mathrm{X}$
the product ' $X$ ' is

Options:
A $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OCH}_2 \mathrm{C}_6 \mathrm{H}_5$
B $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{OH}$
C $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3$
D $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{C}_6 \mathrm{H}_5$

The correct answer is:
$\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_3$

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2010 (Mains)
In the following reaction product ' $\mathrm{P}$ ' is:

Options:
A $\mathrm{RCH}_2 \mathrm{OH}$
B $\mathrm{RCOOH}$
C $\mathrm{RCHO}$
D $\mathrm{RCH}_3$

The correct answer is:
$\mathrm{RCHO}$

Solution:
The given reaction represents Rosenmund's reaction.
Mechanism of Rosenmund's reaction

Topic: Chemistry, General Organic Chemistry, JEE Main
In the following reaction sequence, [C] is :

Options:
A
B
C
D

The correct answer is:

Solution:
To form product [C] multiple name reactions are taking place.

To form product [A] when aromatic amine react with nitrous acid it forms diazonium salt which when treated with copper salts as reagent forms aryl halides and reaction is called as Sendmeyer reaction.

Now product [A] that is when treated with Cl2 in presence of sunlight, free radical substitution takes place and [B] is formed.

Now when [B] is treated with sodium metal in presence of dry ether forms product [B] by Wurtz reaction.

Overall reaction takes place as:

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (04 Sep Shift 2)
In the following reaction sequence:

Compound I is

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2017 (09 Apr Online)
In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are
S2O32-→Ag+Xclear solution →Ag+Ywhite precipitate →with timeZblack precipitate

Options:
A [Ag(S2O3)2]3-,Ag2S2O3,Ag2S
B [Ag(S2O3)3]5-,Ag2SO3,Ag2S
C [Ag(SO3)2]3-,Ag2S2O3,Ag
D [Ag(SO3)3]3-,Ag2SO4,Ag

The correct answer is:
[Ag(S2O3)2]3-,Ag2S2O3,Ag2S

Solution:
S2O32-→Ag+[Ag(S2O3)2]3-X→Ag+Ag2S2O3Y→Ag2SZ black ppt

Topic: Chemistry, d and f Block Elements, JEE Main
In the following reaction sequence, structures of A and B are, respectively

Options:
A
B
C
D

The correct answer is:

Solution:
Initially electrophilic addition reaction takes place and form alcohol and alkyl bromide. Then Wurtz reaction takes place.

The Wurtz reaction is a very useful reaction in the fields of organic chemistry and organometallic chemistry for the formation of alkanes. In this reaction, two different alkyl halides are coupled to yield a longer alkane chain with the help of sodium and dry ether solution.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2020 (07 Jan Shift 2)
In the following reaction sequence, structures of P and Q, are respectively

Options:
A
B
C
D

The correct answer is:

Solution:
Here the reaction is as follows

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is ____.

(Atomic weights in g mol-1:H=1,C=12,N=14,O=16,Br=80. The yield (%) corresponding to the product in each step is given in the parenthesis)

Options:

The correct answer is:
495

Solution:
molecular weight of= 330
Weight of 1 mol of 2, 4, 6 tribromoaniline = 330 gm
â´ Weight of 1.5 mol of 2, 4, 6 tribromoaniline = 330 Ã 1.5 = 495 gm

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the following reaction sequence, the correct structures of $\mathrm{E}, \mathrm{F}$ and $\mathrm{G}$ are

$\stackrel{\text { Heat }}{\longrightarrow}[\mathrm{E}] \frac{\mathrm{I}_{2}}{\mathrm{NaOH}}[\mathrm{F}]+[\mathrm{G}]$
$\left[*\right.$ implies ${ }^{13} \mathrm{C}$ labelled carbon)

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the following reaction sequence,
the major product B is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Amines, JEE Main, JEE Main 2020 (07 Jan Shift 2)
In the following reaction sequence the major product B is

Options:
A
B
C
D

The correct answer is:

Solution:
Here the major product is formed as follows

Topic: Chemistry, Amines, JEE Main
In the following reaction, the final product can be prepared by two paths (I) and (II) .
Which of the following statements is correct?

Options:
A Path (I) is feasible
B Path (II) is feasible
C Both paths are feasible
D Neither of the two paths is feasible

The correct answer is:
Path (I) is feasible

Solution:
Since the reactant is a 3o alkyl halide, so in the presence of NaCN, it will follow E2 path rather than S N 2, so path II is not feasible.
The possible product by path II is

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
In the following reaction, the major product is

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the following reaction, the major products are

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\text { Thus, option (2) is correct. }
$$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the following reaction,
The organic product X has the structure

Options:
A
B
C
D

The correct answer is:

Solution:
It is a case of elimination reaction

Topic: Chemistry, Amines, JEE Main
In the following reaction, the reason why meta-nitro product also formed is:

Options:
A Formation of anilinium ion
B -NH2 group is highly meta-directive
C -NO2 substitution always takes place at meta-position
D low temperature

The correct answer is:
Formation of anilinium ion

Solution:
Aniline on protonation gives anilinium ion which is meta directing. So considerable amount of meta product is formed.

Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2021 (24 Feb Shift 1)
In the following reaction, the total number of oxygen atoms in X and Y is______

Na2O+H2O→2X

Cl2O7+H2O→2Y

Options:

The correct answer is:
5

Solution:
Let's first balance the two given chemical equations:

Na2O+H2O⟶2NaOHX

In this equation, there are two NaOH molecules, so the total number of oxygen atoms in the products is 1.

Cl2O7+H2O⟶2HClO4Y

In this equation, there are two HClO4 molecules, so the total number of oxygen atoms in the products is 4.

So X has one O and Y has four O.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (10 Apr Shift 1)
In the following reaction

​​​​​​​

The % yield for reaction I is 60% and that of reaction II is 50%. The overall yield of the complete reaction is%

Options:

The correct answer is:
30

Solution:
Let initial moles of reactant taken =n
Total moles obtained for benzene sulphonic acid (with % yield =60%) =0.6 n

Moles of benzene sulphonic acid before reaction II=0.6 n

Moles obtained for phenol (with % yield = 50%) =0.6×0.5 n=0.3 n

So over all % yield of complete reaction =0.3nn×100=30

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (27 Jul Shift 1)
In the following reaction, three products a, b, c are obtained.

The approximate experimental yields of the three compounds were 64%, 33% and 3%. Which of the following is the correct with respect to yield and the corresponding product?

Options:
A (a) 33% ; (b) 64% ; (c) 3%
B (a)â3%â;â(b)â64%â;â(c)â33%
C (a) 3% ; (b) 33% ; (c) 64%
D (a) 64% ; (b) 3% ; (c) 33%

The correct answer is:
(a) 3% ; (b) 64% ; (c) 33%

Solution:
% yield ∝ stability of Alkene
Decreasing order of stability

Topic: Chemistry, Hydrocarbons, JEE Main
In the following reaction, which of the following steps is wrong?

Options:
A Step 1
B Step 2
C Step 3
D None

The correct answer is:
Step 2

Solution:
Step 2 is wrong because Friedel-Crafts reaction will not take place in the presence of e- -withdrawing, m-directing -NO2 group.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the following reaction; xA→yB

log10⁡-dAdt=log10⁡-dBdt+0.3010

‘A’ and ‘B’ respectively can be:

Options:
A C2H4 and C4H8
B C2H2 and C6H6
C n-Butane and Iso-butane
D N2O4 and NO2

The correct answer is:
C2H4 and C4H8

Solution:
From the rate law:

log-dAdt=logdBdt+0.3010=log⁡-dAdt=log⁡dBdt+log⁡2

-dAdt=2×dBdt

2A→B and we know 2C2H4→C4H8

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (12 Apr Shift 1)
In the following reaction ‘X’ is

CH3CH24CH3→HCl△Anhy.AlCl3Xmajor product

Options:
A

B

C CH3CH24CH2Cl
D Cl-CH2-CH24-CH2-Cl

The correct answer is:

Solution:
When n-Hexane is heated with anhydrous AlCl3 and HCl gas at 573K under a pressure of about 30-35atm it Isomerises to give a branched chain alkanes.

CH3CH24CH3→HCl△Anhy.AlCl3CH32CHCH2CH2CH32-methylpentane+3-methylpentane

Major product in the above reaction is 2-methylpentane.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2023 (13 Apr Shift 1)
In the following reaction

Options:
A

B

C

D

The correct answer is:

Solution:
Aldehydes and ketones can react with Grignard reagents to form alcohols. Grignard reagents are organometallic compounds containing a carbon-magnesium bond (R-Mg-X).

The reaction between ketone and a Grignard reagent involves the nucleophilic addition of the carbon atom of the Grignard reagent to the carbon-oxygen double bond of the ketone. This leads to the formation of a metal alkoxide intermediate. Subsequently, the metal alkoxide reacts with water (or a protic solvent) to yield the alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (12 Apr Shift 1)
In the following reactions, identify $P, Q$ and $R$, respectively
$\text { I. } 3 \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \longrightarrow 2 \mathrm{P}+\mathrm{CO}_2 \uparrow$
II. $\mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{CO} \longrightarrow 3 Q+4 \mathrm{CO}_2 \uparrow$
III. $\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \longrightarrow 2 R+\mathrm{CO}_2 \uparrow$

Options:
A $\begin{array}{lll}\mathbf{P} & \mathbf{Q} & \mathbf{R} \\ \mathrm{Fe} & \mathrm{Fe}_3 \mathrm{O}_4 & \mathrm{FeO}\end{array}$
B $\begin{array}{lll}\mathbf{P} & \mathbf{Q} & \mathbf{R} \\ \mathrm{Fe}_3 \mathrm{O}_4 & \mathrm{FeO} & \mathrm{Fe}\end{array}$
C $\begin{array}{lll}\mathbf{P} & \mathbf{Q} & \mathbf{R} \\ \mathrm{Fe}_3 \mathrm{O}_4 & \mathrm{Fe} & \mathrm{FeO}\end{array}$
D $\begin{array}{lll}\mathbf{P} & \mathbf{Q} & \mathbf{R} \\ \mathrm{FeO} & \mathrm{Fe}_3 \mathrm{O}_4 & \mathrm{Fe}\end{array}$

The correct answer is:
$\begin{array}{lll}\mathbf{P} & \mathbf{Q} & \mathbf{R} \\ \mathrm{Fe}_3 \mathrm{O}_4 & \mathrm{Fe} & \mathrm{FeO}\end{array}$

Solution:
Among the given equations, option (c) will give the required answer, i.e.,
$(P)=\mathrm{Fe}_3 \mathrm{O}_4,(Q)=\mathrm{Fe},(R)=\mathrm{FeO} \text {, i.e. }$
I. $3 \mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \longrightarrow 2 \mathrm{Fe}_3 \mathrm{O}_4+\mathrm{CO}_2$
II. $\mathrm{Fe}_3 \mathrm{O}_4+4 \mathrm{CO} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{CO}_2$
III. $\mathrm{Fe}_2 \mathrm{O}_3+\mathrm{CO} \longrightarrow 2 \mathrm{FeO}+\mathrm{CO}_2$

Topic: Chemistry, d and f Block Elements, JEE Main
In the following reactions, products A and B , respectively, are:
NaOH+Cl2→A+side products
hot and conc.

CaOH2+Cl2→B+side products
dry

Options:
A NaClO3 and CaOCl2
B NaClO3 and CaClO32
C NaOCl and CaOCl2
D NaOCl and CaClO32

The correct answer is:
NaClO3 and CaOCl2

Solution:
Here,

NaOH gives NaClO3 on reaction with chlorine gas.

CaOH2 gives CaOCl2 on reaction with chlorine gas.

6NaOH+3Cl2hot and conc→NaClO3A+5NaCl+3H2O2CaOH2dry+2Cl2→CaOCl2B+CaCl2+2H2O

CaOCl2 is known as bleaching powder.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2020 (07 Jan Shift 2)
In the following reactions, products (X) and (Y) respectively are

NaOH hot and conc. â+â Cl 2 â(X)â+ side products

Ca(OH) 2 dry + Cl 2 â(Y)+ side products

Options:
A NaOCl and Ca(ClO3)2
B NaClO3 and Ca (OCl)2
C NaOCl and Ca(OCl)2
D NaClO3 and Ca(ClO3)2

The correct answer is:
NaClO3 and Ca (OCl)2

Solution:
Here the products formed (X) and (Y) are NaClO3 and Ca(OCl)2 respectively
6 NaOH+3  Cl 2 →5 NaCl+ NaClO 3 +3  H 2 O
2  Ca(OH) 2 + Cl 2 → Ca(OCl) 2 + CaCl 2 + H 2 O

Topic: Chemistry, s Block Elements, JEE Main
In the following reactions:
the major product $X$ is :

Options:
A
B
C
D

The correct answer is:

Solution:
Ozonolysis and intramolecular aldol

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the following reactions
$\mathrm{X}$ and $\mathrm{Y}$ are

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
In the following reactions
$\mathrm{X}$ and $\mathrm{Y}$, respectively, are

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the following reactions, $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are

Options:
A $\mathrm{X}=\mathrm{CH}_{3} \mathrm{Cl} ; \mathrm{Y}=$ anhydrous $\mathrm{AlCl}_{3} ; \mathrm{Z}=\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}$
B $\mathrm{X}=\mathrm{CH}_{3} \mathrm{COCl} ; \mathrm{Y}=$ anhydrous $\mathrm{AlCl}_{3} ; \mathrm{Z}=\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}$
C $\mathrm{X}=\mathrm{CH}_{3} \mathrm{Cl} ; \mathrm{Y}=$ conc. $\mathrm{H}_{2} \mathrm{SO}_{4} ; \mathrm{Z}=\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}$
D $\mathrm{X}=\mathrm{CH}_{3} \mathrm{Cl} ; \mathrm{Y}=$ dil. $\mathrm{H}_{2} \mathrm{SO}_{4} ; \mathrm{Z}=\mathrm{HNO}_{3}$

The correct answer is:
$\mathrm{X}=\mathrm{CH}_{3} \mathrm{Cl} ; \mathrm{Y}=$ anhydrous $\mathrm{AlCl}_{3} ; \mathrm{Z}=\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4}$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
In the following reactions, ZnO is respectively acting as a/an,
(i) ZnO+Na2O→Na2ZnO2
(ii) ZnO+CO2→ZnCO3

Options:
A base and base.
B acid and acid.
C acid and base.
D base and acid.

The correct answer is:
acid and base.

Solution:
Amphoteric means it behaves as both acid and base. Na2O is basic oxide and ZnO will act as acidic oxide in reacting with this oxide. CO2 is acidic oxide and ZnO will act as basic oxide in reacting with this oxide. So, with Na2O, it acts as an acid and with CO2, it acts as a base.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2017 (02 Apr)
In the following reversible reaction
$2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3+Q_{\mathrm{Cal}}$
Most suitable condition for the higher production of $\mathrm{SO}_3$ is

Options:
A High temperature and high pressure
B High temperature and low pressure
C Low temperature and high pressure
D Low temperature and low pressure

The correct answer is:
Low temperature and high pressure

Solution:
Reaction is exothermic and volume is decreasing from left to right so for higher production of there should be low temperature and high pressure.

Topic: Chemistry, Chemical Equilibrium, JEE Main
In the following sequence of reaction, the major products \(B\) and \(C\) respectively are :

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2024 (09 Apr Shift 1)
In the following sequence of reactions a compound A, (molecular formula C6H12O2 ) with a straight chain structure gives a C4 carboxylic acid. A is :

ALiAlH4H3O+B⟶ Oxidation C4- carboxylic acid

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2021 (01 Sep Shift 2)
In the following sequence of reactions, the alkene affords the compound ' $B$ ' $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}_3 \stackrel{\mathrm{O}_3}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{H}_2 \mathrm{O}}{\mathrm{Zn}} \mathrm{B}$.
The compound $\mathrm{B}$ is

Options:
A
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$

B
$\mathrm{CH}_3 \mathrm{COCH}_3$

C
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COCH}_3$

D
$\mathrm{CH}_3 \mathrm{CHO}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CHO}$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2008
In the following sequence of reactions
The compound $C$ formed will be

Options:
A butanol-1
B butanol-2
C 2-methyl propanol-1
D 1,1 -dimethylethanol

The correct answer is:
1,1 -dimethylethanol

Solution:

Topic: Chemistry, Amines, JEE Main
In the following sequence of reactions,
the compound ' $\mathrm{D}$ ' is

Options:
A
butanal

B
$n-$butyl alcohol

C
$n-$propyl alcohol

D
propanal

The correct answer is:
$n-$propyl alcohol

Solution:
$\therefore$ the compound $\mathrm{D}$ is $n-$propyl alcohol.
Hence, (C) is correct option.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the following sequence of reactions,
the compound ' $\mathrm{D}$ ' is

Options:
A
butanal

B
$n-$butyl alcohol

C
$n-$propyl alcohol

D
propanal

The correct answer is:
$n-$propyl alcohol

Solution:
$\therefore$ the compound $\mathrm{D}$ is $n-$propyl alcohol.
Hence, (C) is correct option.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2007
In the following sequence of reactions ;
The compound \( \mathrm{A} \) is

Options:
A Propane nitrile
B Ethane nitrile
C \( \mathrm{CH}_{3} \mathrm{NO}_{2} \)
D \( \mathrm{CH}_{3} \mathrm{NC} \)

The correct answer is:
Ethane nitrile

Solution:
The reactions involved are,
Hence, the compound $\mathrm{A}$ is ethane nitrite.

Topic: Chemistry, Amines, JEE Main
In the following sequence of reactions, the final product D is:

Options:
A

B
C
D

The correct answer is:

Solution:
Terminal alkyne reacts sodamide to produce sodium salt of alkyne by elimination of ammonia molecule. Sodium salt of alkyne reacts with alkyl halide by nucleophilic substitution reaction mechanism. Secondary alcohol is oxidised by CrO3 to produce ketone.

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (27 Aug Shift 1)
In the following sequence of reactions the maximum number of atoms present in molecule ‘ C ’ in one plane is_______
A→Cu tubeRed  hotB→Anhydrous AlCl3CH3Cl1.eqC
(A is a lowest molecular weight alkyne)

Options:

The correct answer is:
13

Solution:
Number of atoms in one plane = 13

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2020 (08 Jan Shift 2)
In the following sequence of reactions the maximum number of atoms present in molecule C in sp 3 hybridised state is
A → Red hot Cu tube B → AlCl 3 CH 3 Cl (1 eq.) C
( A is a lowest molecular weight alkyne)

Options:

The correct answer is:
1

Solution:
Here only one C-atom of CH3 group is sp3 hybridised rest all C- atoms of benzene ring are sp2 hybridised.

Topic: Chemistry, Hydrocarbons, JEE Main