datavorous/entrance-exam-dataset huggingface dataset:
The reaction in which the hybridisation of the underlined atom is affected is

Options:
A H3P¯O2→Disproportionation
B H2S¯O4+NaCl→420 K
C N¯H3→H+
D X¯eF4+SbF5→

The correct answer is:
X¯eF4+SbF5→

Solution:
(1) H2Ssp3O4+2NaCl→Na2Ssp3O4+2HCl

(2)

(3) H3Psp3O2→H3Psp3O4+PH3

(4) XeF4+SbF5⟶XeF3+SbF6-sp3d2 sp3d

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (04 Sep Shift 2)
The reaction
is called

Options:
A Etard reaction
B Perkin's reaction
C Claisen condensation
D Claisen Schmidt reaction

The correct answer is:
Claisen condensation

Solution:
This is a claisen condensation reaction.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reaction
is fastest when X is

Options:
A
Cl

B
$\mathrm{NH}_2$

C
$\mathrm{OC}_2 \mathrm{H}_5$

D
OCOR

The correct answer is:
Cl

Solution:
Conjugated acid of $\mathrm{Cl}^{-}$is a stronger acid i.e. $\mathrm{HCl}$.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2005
The reaction is:

Options:
A $\mathrm{CH}_3 \mathrm{CHOHCH}_3$
B $\mathrm{CH}_3 \mathrm{COCH}_3$
C $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
D $\mathrm{CH}_3 \mathrm{COOH}$

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

Solution:
The reaction shows a two-step process:
- In the first step, potassium dichromate ($\mathrm{K} 2 \mathrm{Cr} 2 \mathrm{O}_7)$) oxidizes the alcohol (reactant A) to a ketone or aldehyde in the presence of dilute sulfuric acid (H2SO4).
- In the second step, the ketone or aldehyde is reduced to a primary alcohol by methylmagnesium iodide (CH3Mgl) in water ($\mathrm{H} 2 \mathrm{O})$.
The reactant $\mathrm{A}, \mathrm{C} 2 \mathrm{H} 5 \mathrm{OH}$, is ethanol, a primary alcohol. The product of the first step is acetaldehyde, a ketone. The second step reduces the ketone to propanol, another primary alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The reaction \( \mathrm{A}+\mathrm{H}_{2} \mathrm{O} \stackrel{\mathrm{H}^{+}}{\rightarrow} \mathrm{B}+\mathrm{C} \) is called:

Options:
A acid catalysed reaction
B mutarotation if \( \mathrm{A} \) is sucrose.
C inversion of cane sugar if \( \mathrm{A} \) is sucrose
D Both a \& c

The correct answer is:
Both a \& c

Solution:
1.

3.

Topic: Chemistry, Biomolecules, JEE Main
The reaction, \( \mathrm{RCl}+\mathrm{Nal} \stackrel{\text { Acetone }}{\longrightarrow} \mathrm{R}-\mathrm{I}+\mathrm{NaCl} \), is known as:

Options:
A Wurtz's reaction.
B Fittig reaction.
C Frankland's reaction.
D Finkelstein's reaction.

The correct answer is:
Finkelstein's reaction.

Solution:
The given reaction is a nucleophilic substitution reaction, in which chloride ion is acting as the leaving group and iodide ion is the attacking nucleophile. This reaction is commonly known as halogen exchange reaction or Finkelstein's reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The reaction

$\mathrm{MgO}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Mg}(\mathrm{s})+\mathrm{CO}(\mathrm{g}),$ for which

$\Delta \mathrm{H}^{\circ}=+491.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}^{\circ}=198.0 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$

is not feasible at $298 \mathrm{~K}$. Temperature above which reaction will be feasible is

Options:
A $2040.5 \mathrm{~K}$
B $1890.0 \mathrm{~K}$
C $2480.3 \mathrm{~K}$
D $2380.5 \mathrm{~K}$

The correct answer is:
$2480.3 \mathrm{~K}$

Solution:
$\mathrm{MgO}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Mg}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$

For a reaction to be spontaneous $\Delta \mathrm{G} < 0$

$\Delta \mathrm{H}^{\circ}-\mathrm{T} \Delta \mathrm{S}^{\circ} < 0$

$\Rightarrow \mathrm{T}>\frac{\Delta \mathrm{H}^{\circ}}{\Delta \mathrm{S}^{\circ}}$

$\mathrm{T}>\frac{491.1 \times 1000}{198}$

$\mathrm{T}>2480.3 \mathrm{~K}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (11 Jan Shift 2)
The reaction, $\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$ is exothermic and reversible. A mixture of $\mathrm{N}_2(g), \mathrm{H}_2(g)$ and $\mathrm{NH}_3(g)$ is at equilibrium in a closed container. When a certain quantity of extra $\mathrm{H}_2(g)$ is introduced into the container, while keeping the volume constant, then which statement among the following is true?

Options:
A The pressure inside the container will not change.
B Equilibrium condition will not change.
C The temperature will increase.
D The temperature will decrease.

The correct answer is:
The temperature will increase.

Solution:
Since, the reaction is exothermic and volume is kept constant. So, the temperature will increase.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The reaction
$\mathrm{N}_2 \mathrm{O}_5$ (in $\mathrm{CCl}_4$ solution) $\rightarrow 2 \mathrm{NO}_2$ (solution) $+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$
is of first order in $\mathrm{N}_2 \mathrm{O}_5$ with rate constant $6.2 \times 10^{-1} \mathrm{~s}^{-1}$. What is the value of rate of reaction when $\left[\mathrm{N}_2 \mathrm{O}_5\right]=1.25 \mathrm{molel}^{-1}$

Options:
A $7.75 \times 10^{-1} \mathrm{molel}^{-1} \mathrm{~s}^{-1}$
B $6.35 \times 10^{-3} \mathrm{molel}^{-1} \mathrm{~s}^{-1}$
C $5.15 \times 10^{-5} \mathrm{molel}^{-1} \mathrm{~s}^{-1}$
D $3.85 \times 10^{-1} \mathrm{molel}^{-1} \mathrm{~s}^{-1}$

The correct answer is:
$7.75 \times 10^{-1} \mathrm{molel}^{-1} \mathrm{~s}^{-1}$

Solution:
Rate $=K\left(N_2 O_5\right)=6.2 \times 10^{-1} \times 1.25$
$=7.75 \times 10^{-1} \mathrm{~mol} \Gamma^{-1} \mathrm{~s}^{-1}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The reaction occurs in which of the given galvanic cell?

12H2( g)+AgCl(s)⇌H+(aq)+Cl-(aq)+Ag(s)

Options:
A Pt|H2(g)|HCl(soln )|AgCl(s)|Ag
B Ag|AgCl(s)|KCl(soln )|AgNO3|Ag
C Pt|H2(g)|HCl(soln )|AgNO3(soln )|Ag
D Pt|H2(g)|KCl(soln )|AgCl(s)|Ag

The correct answer is:
Pt|H2(g)|HCl(soln )|AgCl(s)|Ag

Solution:
Oxidation is loss of electron and in a galvanic cell it occurs at anode. Reduction is gain of electron and in a galvanic cell it occurs at cathode.

At anode :12H2( g)⟶H+(aq)+e-

At cathode AgCl(S)+e¯⟶Ag+Cl- (aq)

Overall reaction

12H2( g)+AgCl(s)⟶H*+Cl-+Ag(s)

Cell representation :

Anode / Anode electrolyte || Cathodic electrolyte / Cathode.

The cell representation for the above cell will be

PtH2( g)HCl soln n|AgCl(s)|Ag

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The reaction of 1-bromo-3-chlorocyclobutane with metallic sodium in dioxane under reflux conditions gives

Options:
A
B
C
D

The correct answer is:

Solution:
This is Wurtz reaction. Bromides have high reactivity than chlorides in Wurtz reaction therefore, reaction occurs from Br atoms,

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The reaction of acetaldehyde and HCN, followed by complete acid hydrolysis gives

Options:
A CH2=CH-COOH
B CH3-CHOHCONH2
C CH3CHOHCOOH
D CH2=CHCONH2

The correct answer is:
CH3CHOHCOOH

Solution:
CH3C|H=O+HCN→CH3CH| CN-OH→H3O+CH3-CH-OH| CONH2 →H3O+CH3-CH-OH| COOH +NH4+

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reaction of an ester $R C O O R^{\prime}$ with an alcohol $R^{\prime} O H$ in the presence of an acid gives

Options:
A $\mathrm{RCOOH}$
B $R^{\prime} \mathrm{COOH}$
C $R^{\prime} C O O R$
D $R C O O R^{\prime}$

The correct answer is:
$R C O O R^{\prime}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reaction of aniline with chloroform under alkaline conditions leads to the formation of

Options:
A phenylcyanide
B phenylisonitrile
C phenylcyanate
D phenylisocyanate

The correct answer is:
phenylisonitrile

Solution:
Aniline is a primary amine (contains - $\mathrm{NH}_{2}$ group). When it is treated with chloroform under alkaline conditions, it results in the formation of a bad smelling compound, known by the name isonitrile or carbylamine. This reaction is known as carbylamine reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The reaction of aqueous KMnO4 with H2O2 in acidic conditions gives:

Options:
A Mn4+ and O2
B Mn2+ and O2
C Mn2+ and O3
D Mn4+ and MnO2

The correct answer is:
Mn2+ and O2

Solution:
3H2SO4+2KMnO4+5H2O2→5O2+2MnSO4+8H2O+K2SO4

Topic: Chemistry, d and f Block Elements, JEE Main
The reaction of Benzenediazonium chloride with aniline yields yellow dye. The name of the
yellow dye is

Options:
A p-Hydroxyazobenzene
B p-Aminoazobenzene
C p-Nitroazobenzene
D o-Nitroazobenzene

The correct answer is:
p-Aminoazobenzene

Solution:
The reaction involved is
\( p \)-Amino benzene (yellow dye)

Topic: Chemistry, Amines, JEE Main
The reaction of calcium cyanamide with water yields

Options:
A CaOH2 and N2
B CaC2 and N2H4
C CaHCO32 and NH3
D CaCO3 and NH4OH

The correct answer is:
CaCO3 and NH4OH

Solution:
It is good fertilizer and is prepared by heating CaC2 and N2 at 1370 K.
CaC2+N2→CaCN2+C⎵Nitrolim
CaCN2+5H2O→CaCO3+2NH4OH

Topic: Chemistry, s Block Elements, JEE Main
The reaction of $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}$ with $\mathrm{HBr}$ gives

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The reaction of chloroform with alcoholic $\mathrm{KOH}$ and p-toluidine forms

Options:
A
< smiles>Cc1ccc(Cl)cc1 < /smiles>
B
< smiles>Cc1ccc(NC(Cl)Cl)cc1 < /smiles>
C
< smiles>Cc1ccc(C)cc1 < /smiles>
D
< smiles>Cc1ccc(C#N)cc1 < /smiles>

The correct answer is:
< smiles>Cc1ccc(C)cc1 < /smiles>

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2003
The reaction of cyanamide, NH2CNs with oxygen was run in a bomb calorimeter and ΔU was found to be -742.24 kJ mol-1 . The magnitude of ΔH298 for the reaction NH2CNs+32O2g→N2g+O2g+H2Ol is kJ. (Rounded off to the nearest integer) [Assume ideal gases and R=8.314 J mol-1 K-1]

Options:

The correct answer is:
741

Solution:
ΔH=ΔU+ΔngRT

=-742.24+12×8.3141000×298

=-741 kJ/mol

Hence answer is 741

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (25 Feb Shift 1)
The reaction of H2O2 with potassium permanganate in acidic medium leads to the formation of mainly

Options:
A Mn2+
B Mn4+
C Mn3+
D Mn6+

The correct answer is:
Mn2+

Solution:
When hydrogen peroxide is treated with acidified potassium permanganate, the reduction of permanganate ion to Mn2+ion takes place by the following reaction,

2KMnO4+3H2SO4+5H2O2→K2SO4+2MnSO4+8H2O+5O2

In the above reaction, potassium permanganate acts an oxidising agent.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The reaction of $\mathrm{H}_2 \mathrm{~S}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{S}+2 \mathrm{H}_2 \mathrm{O}$ manifests

Options:
A Acidic nature of $\mathrm{H}_2 \mathrm{O}_2$
B Alkaline nature of $\mathrm{H}_2 \mathrm{O}_2$
C Oxidizing nature of $\mathrm{H}_2 \mathrm{O}_2$
D Reducing action of $\mathrm{H}_2 \mathrm{O}_2$

The correct answer is:
Oxidizing nature of $\mathrm{H}_2 \mathrm{O}_2$

Solution:
$\underset{-2}{\mathrm{H}_2 \mathrm{~S}}+\mathrm{H}_2 \mathrm{O}_2 \rightarrow \underset{0}{\mathrm{~S}}+2 \mathrm{H}_2 \mathrm{O}$
In this reaction $\mathrm{H}_2 \mathrm{O}_2$ shows oxidizing nature.

Topic: Chemistry, Hydrogen, JEE Main
The reaction of H3N3B3Cl3A with LiBH4 in tetrahydrofuran gives inorganic benzene B . Further, the reaction of A with C leads to H3N3B3Me3 . Compounds B and C respectively, are:

Options:
A Bogging god MeBr
B Diborane and MeMgBr
C Boron nitride and MeBr
D Borazine and MeMgBr

The correct answer is:
Borazine and MeMgBr

Solution:
The reaction of H3N3B3Cl3A with LiBH4 in tetrahydrofuran gives inorganic benzene B. The reaction is given below. Further, the reaction of compound A reacts with C leads to H3N3B3Me3 .

The reaction is given beow:

B3N3H3CI3A+LiBH4⟶B3N3H6B+LiCI+BCI3
B3N3H3CI3A+3MeMgBrC⟶B3N3H3CH33+3MgBrCI

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main, JEE Main 2020 (09 Jan Shift 2)
The reaction of $\mathrm{HBr}$ with in the presence of peroxide will give

Options:
A

B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}$
C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The reaction of hydrogen and iodine monochloride is given as:
$\mathrm{H}_{2(g)}+2 \mathrm{ICl}_{(g)} \rightarrow 2 \mathrm{HCl}_{(g)}+\mathrm{I}_{2(g)}$
This reaction is of first order with respect to $\mathrm{H}_{2(g)}$ and $\mathrm{lCl}_{(g)+}$ following mechanisms were proposed
Mechanism A:
$\mathrm{H}_{2(g)}+2 \mathrm{ICl}_{(g)} \rightarrow 2 \mathrm{HCl}_{(g)}+\mathrm{H}_{2(g)}$
Mechanism B:
$\begin{aligned}
& \mathrm{H}_{2(g)}+\mathrm{ICl}_{(g)} \rightarrow \mathrm{HCl}_{(g)}+\mathrm{HI}_{(g)} \text { :slow } \\
& \mathrm{Hl}_{(g)}+\mathrm{lCl}_{(g)} \rightarrow \mathrm{HCl}_{(g)}+\mathrm{l}_{2(g)} \text {; fast }
\end{aligned}$
Which of the above mechanism(s) can be consistent with the given information about the reaction?

Options:
A A and B both
B neither A nor B
C A only
D B only

The correct answer is:
A and B both

Solution:
The slowest step is the rate determining step thus the mechanism $\mathrm{B}$ will be more consistent because it involve one molecule of $\mathrm{H}_2$ and one molecule of ICl. It can be expressed as
$r=k\left[\mathrm{H}_2\right][\mathrm{ICl}]$
Related Theory
The rate determining step in a reaction mechanism is characterised by high activation energy. Usually, the first step is the slow step since it has the highest activation energy.

Topic: Chemistry, Chemical Kinetics, JEE Main
The reaction of \( \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) with \( \mathrm{NaCl} \) and concentrated \( \mathrm{H}_{2} \mathrm{SO}_{4} \) gives

Options:
A \( \mathrm{CrCl}_{3} \)
B \( \mathrm{CrOCl}_{2} \)
C \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \)
D \( \mathrm{Cr}_{2} \mathrm{O}_{3} \)

The correct answer is:
\( \mathrm{CrO}_{2} \mathrm{Cl}_{2} \)

Solution:
On heating a mixture of NaCl, K2Cr2O7 and concentrated H2SO4, red vapour chromyl chloride CrO2Cl2 is obtained.

i K2Cr2O7+H2SO4→2KHSO4+2CrO3+H2O

ii NaCl+H2SO4→NaHSO4+HCl

iii CrO3Fromi+2HClFromii→CrO2Cl2Chromy chlorideRed vopour

iv CrO2Cl2+2NaOH→Na2CrO4+2HClYellow solution

Topic: Chemistry, d and f Block Elements, JEE Main
The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below:

O3g+Clâ¢âO2g+ClOâ¢g ...(i)

ki=5.2Ã109 L mol-1 s-1

ClOâ¢g+Oâ¢gâO2g+Clâ¢g ...(ii)

kii=2.6Ã1010 L mol-1 s-1

The closest rate constant for the overall reaction

O3g+Oâ¢gâ2O2g is:

Options:
A 1.4Ã1020 L mol-1 s-1
B 3.1Ã1010Â LÂ mol-1Â s-1
C 5.2Ã109Â LÂ mol-1Â s-1
D 2.6Ã1010Â LÂ mol-1Â s-1

The correct answer is:
1.4Ã1020 L mol-1 s-1

Solution:
O3g+Clg•→O2g+ClOs•KiClOg•+Og•→O2g+Clg•Kii

_________________________________

O3g+Og•→202gKRate=Ki×Kii
Overall Rate constant = Ki × Kii
= 5.2 × 109 × 2.6 × 1010
≈ 1.4 × 1020 mol-1 L s-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2016 (09 Apr Online)
The reaction of ozone with oxygen atoms in the presence of chlorine atoms can occur by a two step process shown below:

O3g+Clâ¢âO2g+ClOâ¢g ...(i)

ki=5.2Ã109 L mol-1 s-1

ClOâ¢g+Oâ¢gâO2g+Clâ¢g ...(ii)

kii=2.6Ã1010 L mol-1 s-1

The closest rate constant for the overall reaction

O3g+Oâ¢gâ2O2g is:

Options:
A 1.4Ã1020 L mol-1 s-1
B 3.1Ã1010Â LÂ mol-1Â s-1
C 5.2Ã109Â LÂ mol-1Â s-1
D 2.6Ã1010Â LÂ mol-1Â s-1

The correct answer is:
1.4Ã1020 L mol-1 s-1

Solution:
O3g+Clgâ¢âO2g+ClOsâ¢KiClOgâ¢+Ogâ¢âO2g+Clgâ¢Kii

_________________________________

O3g+Ogâ¢â202gKRate=KiÃKii
Overall Rate constant = Ki Ã Kii
= 5.2 Ã 109 Ã 2.6 Ã 1010
â 1.4 Ã 1020 mol-1 L s-1

Topic: Chemistry, Chemical Kinetics, JEE Main
The reaction of $P_2$ with $X$ leads selectively to $\mathrm{P}_4 \mathrm{O}_6$. The $\mathrm{X}$ is

Options:
A
dry $\mathrm{O}_2$

B
a mixture of $\mathrm{O}_2$ and $\mathrm{N}_2$

C
moist $\mathrm{O}_2$

D
$\mathrm{O}_2$ in the presence of aq. $\mathrm{NaOH}$

The correct answer is:
a mixture of $\mathrm{O}_2$ and $\mathrm{N}_2$

Solution:
White phosphorus on reaction with limited supply of oxygen gives lower oxide $P_4 \mathrm{O}_6$. Therefore, air $\left(\mathrm{O}_2+\mathrm{N}_2\right)$ is a good source for controlled supply of oxygen and the best choice for controlled oxidation of white phosphorus into lower oxide $\mathrm{P}_4 \mathrm{O}_6$.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The reaction of p-HOC6H4COOH with excess Br2 forms

Options:
A
B
C
D

The correct answer is:

Solution:
Here 2, 4, 6 tribromophenol is formed as follows

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The reaction of phenol with benzoyl chloride to give phenyl benzoate is known as :

Options:
A
Claisen reaction

B
Schotten-Baumann reaction

C
Reimer-Tiemann reaction

D
Gatterman-Koch reaction

The correct answer is:
Schotten-Baumann reaction

Solution:
"Schotten-Baumann reaction".

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (23 Apr Online)
The reaction of propene with HOCl Cl2+H2O proceeds through the intermediate:

Options:
A CH3-CHOH-CH2+
B CH3-CHCl-CH2+
C CH3-CH+-CH2-OH
D CH3-CH+-CH2-Cl

The correct answer is:
CH3-CH+-CH2-Cl

Solution:
The reaction of propene with hypochlorous acid which is also the mixture of chlorine gas and water is an electrophilic addition reaction. It proceeds through the carbocation intermediate.

The whole reaction sequence is given below:

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2016 (03 Apr)
The reaction of propene with HOCl Cl2+H2O proceeds through the intermediate

Options:
A CH3-CHCl-CH2+
B CH3-CH+-CH2-OH
C CH3-CH+-CH2-Cl
D CH3-CHOH-CH2+

The correct answer is:
CH3-CH+-CH2-Cl

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
The reaction of RCONH2 with a mixture of Br2 and aqueous KOH gives RNH2 as the main product. The intermediate not involved in this reaction is (are):

Options:
A RCONHBr
B RCON-BrK+
C R-N=C=O
D RCONBr2

The correct answer is:
RCONBr2

Solution:
This is Hofmann bromamide reaction and RCONBr2 does not form in this reaction. See mechanism for it:
RCO-NH2+Br2+4KOH→R-NH2+K2CO3+2KBr+2H2O

Topic: Chemistry, Amines, JEE Main
The reaction of solid XeF2 with AsF5 in 1:1 ratio affords

Options:
A XeF4 and AsF3
B XeF6 and AsF3
C XeF+AsF6-
D [Xe2F3]+[AsF6]-

The correct answer is:
XeF+AsF6-

Solution:
XeF2 (Xenon difluoride) acts as a fluoride donor and thus, forms complex when mixed with covalent pentafluorides like AsF5 .
XeF21+AsF51→XeF+AsF6-

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The reaction of sulphur in alkaline medium is given below:

S8s+a OHaq−→b Saq2−+c S2O3aq2−+d H2Ol

The values of 'a' is______ (Integer answer)

Options:

The correct answer is:
12

Solution:
For balancing in basic medium add equal number of OH⊖. That of H+

2 S8+12H2O+24OH⊖⟶8 S2-+4 S2O82-+24H2O

2 S8+24OH⊖→8 S2-+4 S2O82-+12H2O

S8+12OH⊖→4 S2-+2 S2O82-+6H2O

a=12

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The reaction of t-butyl chloride and sodium ethoxide gives mainly

Options:
A t-butyl ethyl ether
B 2,2-dimethylbutane
C 2-methylprop-1-ene
D isopropyl n-propyl ether

The correct answer is:
2-methylprop-1-ene

Solution:
Sodium ethoxide is a strong base. The nucleophile OC2H5 cannot attack at the carbon of bond due to steric hindrance. Hence, elimination reaction occurs to form alkene.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
The reaction of the type $2 X_2+S \rightarrow S X_4$ is shown by sulphur when $X$ is

Options:
A Fluorine or chlorine
B Chlorine only
C Chlorine and bromine only
D $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ all

The correct answer is:
Fluorine or chlorine

Solution:
Fluorine and chlorine are more electronegative than sulphur

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The reaction of toluene with $\mathrm{Cl}_2$ in presence of $\mathrm{FeCl}_3$ gives predominantly

Options:
A
benzoyl chloride

B
benzyl chloride

C
$o-$and $p-$chlorotoluene

D
$m-$chlorotoluene

The correct answer is:
$o-$and $p-$chlorotoluene

Solution:
Due to $\mathrm{o}$ - and $\mathrm{p}$ - directing nature of $\mathrm{CH}_3$ group.
Hence, (C) is correct answer.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2007
The reaction of water with sodium and potassium is

Options:
A Exothermic
B Endothermic
C Reversible
D Irreversible and endothermic

The correct answer is:
Exothermic

Solution:
The reaction of water with sodium and potassium is highly exothermic in nature.
$2 \mathrm{Na}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_2 \uparrow$
$2 \mathrm{~K}+2 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{KOH}+\mathrm{H}_2 \uparrow$

Topic: Chemistry, s Block Elements, JEE Main
The reaction of white phosphorus on boiling with alkali in inert atmosphere resulted in the formation of product A. The reaction 1 mol of A with excess of AgNO3 in aqueous medium gives ________ mole(s) of Ag. (Round off to the Nearest Integer).

Options:

The correct answer is:
4

Solution:
P4+3OH-+3H2O→PH3+3H2PO2-A

H2PO2-1 mole +4Ag+ excess +2H2O→H3PO4+ 4Ag +3H+4 mole

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (17 Mar Shift 1)
The reaction of with bromine and KOH gives RNH2 as the end product. Which one of the following is the intermediate product formed in this reaction?

Options:
A
B R-NH-Br
C R-N=C=O
D

The correct answer is:
R-N=C=O

Solution:
The given reaction follows Hoffman bromamide degradation mechanism for the preparation of primary amines. In this reaction alkyl or aryl isocynate is formed as an intermediate. The formation of intermediate is shown below.

Topic: Chemistry, Amines, JEE Main, JEE Main 2022 (25 Jun Shift 1)
The reaction of zinc with dilute and concentrated nitric acid, respectively, produces:

Options:
A NOÂ andÂ N2O
B NO2Â andÂ N2O
C N2O and NO2
D NO2Â andÂ NO

The correct answer is:
N2O and NO2

Solution:
Zn on reaction with HNO3
4Zn+10HNO3→4ZnNO32+N2O+5H2O
(Dil.)
Zn+4HNO3→ZnNO32+2NO2+2H2O
(Conc.)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2016 (03 Apr)
The reaction of zinc with dilute and concentrated nitric acid, respectively produces

Options:
A NO2 and N2O
B N2O and NO2
C NO2 and NO
D NO and N2O

The correct answer is:
N2O and NO2

Solution:
The reaction of zinc with dilute and concentrated nitric acid are respectively as follows
4Zn+ 10 HNO 3 dil. →4Zn ( NO 3 ) 2 + N 2 O+5 H 2 O
Zn+ 4 HNO 3 conc. →Zn ( NO 3 ) 2 +2 NO 2 +2 H 2 O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives

Options:
A ZnOH2
B ZnO
C ZnOH42-
D ZnO22-

The correct answer is:
ZnO22-

Solution:
Zn+2NaOH→Na2ZnO2+H2↑

In the above reaction, Zinc reacts with Sodium Hydroxide, gives Sodium Zincate with evolving of hydrogen. Here Zn does not react as an acid in the reaction above since the reaction is not an acid-base one, but rather oxidation-reduction

ZnO22- is anhydrous form of ZnOH42-

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2022 (29 Jul Shift 1)
The reaction representing the Mond process for metal refining is ______

Options:
A Ni+4CO⟶ΔNi(CO)4
B 2KAu(CN)2+Zn⟶ΔK2Zn(CN)4+2Au
C Zr+2I2⟶ΔZrI4
D ZnO+C⟶ΔZn+CO

The correct answer is:
Ni+4CO⟶ΔNi(CO)4

Solution:
Mond process for metal refining: In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl:

Ni + 4CO →330 - 350K Ni(CO)4

The carbonyl is subjected to a higher temperature so that it is decomposed giving the pure metal:

Ni(CO)4 →450 - 470K Ni + 4CO

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2023 (29 Jan Shift 1)
The reaction sequence

C6H5-CH=CH-CHO→[X]C6H5-CH2-CH2-CH2OH

[X] will be

Options:
A LiAlH4
B NaBH4
C Aluminium isopropoxide
D All of these

The correct answer is:
LiAlH4

Solution:
Here X is LiAlH4 and the reaction is as follows
C6H5-CH=CH-CHO→LiAlH4C6H5-CH2-CH2-CH2OH
LiAlH4 can reduce double bond. If it is in conjugation with phenyl ring.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reaction sequence given below gives product $R$
The structure of the product $R$ is

Options:
A
B
C
D

The correct answer is:

Solution:
The reaction occurs as follows :
The above reaction is an example of Borodine Hunsdiecker reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main