datavorous/entrance-exam-dataset huggingface dataset:
The reaction, $\mathrm{SO}_{2}+\mathrm{Cl}_{2} \longrightarrow \mathrm{SO}_{2} \mathrm{Cl}_{2}$ is
exothermic and reversible. A mixture of $\mathrm{SO}_{2}(\mathrm{~g})$, $\mathrm{Cl}_{2}(\mathrm{~g})$ and $\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{l})$ is at equilibrium in a closed container. Now a certain quantity of extra $\mathrm{SO}_{2}$ is introduced into the container, the volume remaining the same. Which of the following is/are true?

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

The correct answer is:
The temperature will increase.

Solution:
Byaddition of $\mathrm{SO}_{2}$, equilibrium will shift to RHS which is exothermic. Hence temp, will increase.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The reaction that does NOT define calcination is:

Options:
A $\mathrm{Fe}_{2} \mathrm{O}_{3} \cdot \mathrm{xH}_{2} \mathrm{O} \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{xH}_{2} \mathrm{O}$
B $2 \mathrm{Cu}_{2} \mathrm{~S}+3 \mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Cu}_{2} \mathrm{O}+2 \mathrm{SO}_{2}$
C $\mathrm{ZnCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{ZnO}+\mathrm{CO}_{2}$
D $\mathrm{CaCO}_{3} \cdot \mathrm{MgCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{MgO}+2 \mathrm{CO}_{2}$

The correct answer is:
$2 \mathrm{Cu}_{2} \mathrm{~S}+3 \mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Cu}_{2} \mathrm{O}+2 \mathrm{SO}_{2}$

Solution:
Calcination involves heating of carbonates or hydrated oxides in absence of air.

$$

\therefore 2 \mathrm{Cu}_{2} \mathrm{~S}+3 \mathrm{O}_{2} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{Cu}_{2} \mathrm{O}+2 \mathrm{SO}_{2} ; \text { is roasting }

$$

(heating in a regular supply of air).

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The reaction that does not define calcination is

Options:
A ZnCO3→ΔZnO+CO2
B Fe2O3.XH2O→ΔFe2O3+XH2O
C CaCO3.MgCO3→ΔCaO+MgO+2CO2
D 2Cu2S+3O2→Δ2Cu2O+2SO2

The correct answer is:
2Cu2S+3O2→Δ2Cu2O+2SO2

Solution:
Calcination is a process in which ore is heated mainly in the absence of air whereas roasting is a process in which ore is heated in the presence of air.
2Cu2S+3O2→Δ2Cu2O+2SO2(Roasting)

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET
The reaction that does not give benzoic acid as the major product is

Options:
A

B

C

D

The correct answer is:

Solution:
Primary aromatic alcohols on reaction with pyridinium chlorochromate (PCC) which is a mild oxidising agent forms aromatic aldehydes.
In the remaining options benzoic acid is formed as follows:
$$
\text { Thus, option (3) is correct. }
$$

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2019 (Odisha)
The reaction that does not produce nitrogen is

Options:
A heating NH42Cr2O7
B NH3 with excess of Cl2
C heating of NaN3
D heating of NH4NO3

The correct answer is:
NH3 with excess of Cl2

Solution:
Normal heating; NH4NO3âÎN2O+2H2O
but Explosion NH 4 NO 3 → Explosion Δ N 2  +  1 2 O 2  +  2H 2 O
NH3+3Cl2excessâNCl3+3HCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The reaction that does NOT takes place in blast furnace between 900K and 1500K temperature range during the extraction of iron is:

Options:
A CaO+SiO2→CaSiO3
B Fe2O3+CO→2FeO+CO2
C FeO+CO→Fe+CO2
D C+CO2→2CO

The correct answer is:
Fe2O3+CO→2FeO+CO2

Solution:
Except Option B remaining all reactions will takes place in blast furnace during extraction of iron.

Fe2O3+CO→2FeO+CO2

This reaction takes place at 500 - 800 K.

At this temperature, the iron ore is in a solid state and the carbon monoxide is a gas. The coke burns in the lower part of the furnace to produce carbon monoxide, which rises through the furnace and reacts with the iron ore in the upper part of the furnace.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, NEET, NEET 2023 (All India)
The reaction that is not involved in the ozone layer depletion mechanism in the stratosphere is

Options:
A \( \mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \)
B \( \mathrm{Cl} \stackrel{\circ}{\mathrm{O}}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \rightarrow \dot{\mathrm{Cl}}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \)
C \( \mathrm{CF}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \stackrel{\mathrm{uv}}{\longrightarrow} \dot{\mathrm{Cl}}(\mathrm{g})+\dot{\mathrm{CF}}_{2} \mathrm{Cl}(\mathrm{g}) \)
D \( \mathrm{HOCl}(\mathrm{g}) \stackrel{\mathrm{h}(\mathrm{nu})}{\longrightarrow} \dot{\mathrm{O} \mathrm{H}}(\mathrm{g})+\dot{\mathrm{Cl}}(\mathrm{g}) \)

The correct answer is:
\( \mathrm{CH}_{4}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \)

Solution:
Methane oxidation is responsible for the majority of ozone formation in troposphere. Hence, the process is not involved in the ozone layer depletion mechanism in the stratosphere.

Topic: Chemistry, Environmental Chemistry, JEE Main
The reaction that is not involved in the ozone layer depletion mechanism in the stratosphere is

Options:
A CH4+2O2→CO2+2H2O
B ClO ∙g+O g→C∙l g +O2 g
C CF2Cl2 g→ uv C∙l g+C∙F2Cl g
D HOCl g→ h(nu) O∙H g+Cl∙ g

The correct answer is:
CH4+2O2→CO2+2H2O

Solution:
Methane oxidation is responsible for the majority of ozone formation in troposphere. Hence, the process is not involved in the ozone layer depletion mechanism in the stratosphere.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The reaction that is not involved in the ozone layer depletion mechanism in the stratosphere is

Options:
A ClO⋅(g)+(O)g→C⋅l(g)+O2(g)
B HOCl(g)→O⋅H(g)+C⋅l(g)
C CH4+2O3→3CH2=O+2H2O
D CF2Cl2(g)→ hv C⋅l(g)+C⋅F2Cl(g)

The correct answer is:
CH4+2O3→3CH2=O+2H2O

Solution:
As CH4 is not present in stratosphere moreover it cannot diffuse like CFCs into stratosphere from atmosphere.

Topic: Chemistry, Environmental Chemistry, JEE Main
The reaction that occurs in a breath analyser, a device used to determine the alcohol level in a person's blood stream is

2 K2Cr2O7+8H2SO4+3C2H6O→2Cr2SO43+3C2H4O2+2 K2SO4+11H2O

If the rate of appearance of Cr2SO43 is 2.67 mol min-1 at a particular time, the rate of disappearance of C2H6O at the same time is mol min-1. (Nearest integer)

Options:

The correct answer is:
4

Solution:
13-dC2H6Odt=12dCr2SO43dt

13 (Rate of disappearance of C2H6O=12 (Rate of disappearance of Cr2SO43

Rate of disappearance of C2H6O=2.67 mol/min×32=4.005 mol/min

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The reaction used for preparation of soap from fat is :

Options:
A An addition reaction
B Reduction reaction
C Alkaline hydrolysis reaction
D An oxidation reaction

The correct answer is:
Alkaline hydrolysis reaction

Solution:
The reaction used for preparation of soap from fat is not alkaline hydrolysis, but rather saponification. In saponification, a triglyceride (fat or oil) is reacted with a strong base, such as sodium hydroxide (or potassium hydroxide, in the presence of heat to produce soap molecules and glycerol as a byproduct. The reaction involves the hydrolysis of the ester bond in the triglyceride by the base to form carboxylate ions (the soap molecules) and glycerol.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The reaction $\mathrm{X} \rightarrow \mathrm{Y}$ is an exothermic reaction. Activation energy of the reaction for $\mathrm{X}$ into $\mathrm{Y}$ is $150 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Enthalpy of reaction is $135 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$. The activation energy for the reverse reaction, $\mathrm{Y} \rightarrow \mathrm{X}$ will be :

Options:
A
$280 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$285 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$270 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$15 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$285 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$X \longrightarrow Y ; \Delta \mathrm{H}=-135 \mathrm{~kJ} / \mathrm{mol}$, $\mathrm{E}_a=150 \mathrm{~kJ} / \mathrm{mol}$ For an exothermic reaction
$$
\begin{aligned}
& \mathrm{E}_{a(\mathrm{~F} . \mathrm{R} .)}=\Delta \mathrm{H}+\mathrm{E}_{a(\mathrm{~B} . \mathrm{R} .)}^{\prime} \\
& 150=-135+\mathrm{E}_{a a \text { (B.R.) }}^{\prime} \\
& \mathrm{E}_{a(\mathrm{~B} . \mathrm{R})}^{\prime}=285 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2013 (22 Apr Online)
The reaction X + YâZ is first order with respect to X and second order with respect to Y , initial rate of formation of Zâ=âRâmolâ dm 3 â sec â1 when [X] and [Y] are 0.40âmolâ dm â3 and 0.30âmolâ dm â3 respectively. If [X] is halved and [Y] is doubled, the value of the initial rate would become

Options:
A 4R
B R4
C R
D 2R

The correct answer is:
2R

Solution:
X+Y→Z
Rate1=KXY2
Rate2=KX22Y2
Rate1Rate2=12
Rate2=2Rate1=2R

Topic: Chemistry, Chemical Kinetics, JEE Main
The reaction,
$\mathrm{Zn}(\mathrm{s})+\mathrm{CuSO}_{4}(\mathrm{aq}) \rightarrow \mathrm{ZnSO}_{4}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$ is an example of a

Options:
A spontaneous process
B isobaric process
C non-spontaneous process
D reversible process

The correct answer is:
spontaneous process

Solution:
It is spontaneous process because zinc is $\mathrm{~ i s ~ i s ~}$ more reactive than copper, hence can easily repace $\mathrm{Cu}$ from $\mathrm{CuSO}_{4}$.

Topic: Chemistry, d and f Block Elements, JEE Main
The reading in the ideal voltmeter V shown in the given circuit diagram is:

Options:
A 5 V
B 10 V
C 0 V
D 3 V

The correct answer is:
0 V

Solution:
As the voltmeter is idea, no current will flow through it. Therefore, considering left loop, we can write

i=Eeqreq=8×58×0.2

⇒I=25 A

Now, potential drop across the cell which is connected in parallel with Voltmeter,

V=E-ir

=5-0.2×25

=0 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (01 Feb Shift 1)
The reading of pressure metre attached with a closed pipe is 4.5×104 N m-2. On opening the valve, water starts flowing and the reading of pressure metre falls to 2.0×104 N m-2. The velocity of water is found to be V m s-1. The value of V is _________.

Options:

The correct answer is:
50

Solution:
The formula to calculate the change in pressure can be written as

P1-P2=12ρv2 ...1

From equation (1), it follows that

4.5×104-2.0×104=12×103×v2⇒2.5×104=12×103×v2⇒v2=50⇒v=50

Hence, V=50.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (27 Jan Shift 2)
The reading of the ammeter and voltmeter are (Both the instruments are ac meters and measures rms value) -

Options:
A 2 A, 110 V
B 2 A, 0 V
C 2 A, 55 V
D 1 A, 0 V

The correct answer is:
2 A, 0 V

Solution:
The circuit is RLC resonant circuit.
â´ Reading of voltmeter=VL- VC=0
Reading of ammeter=ErmsZ=ErmsR=11055=2A

Topic: Physics, Alternating Current, JEE Main
The reading of the ammeter for a silicon diode in the given circuit is:

Options:
A 13.5 mA
B 0
C 15 mA
D 11.5 mA

The correct answer is:
11.5 mA

Solution:
Given, R=200 Ω and V=3 V. We have to calculate the current. We know that, a potential barrier across silicon diode =0.7 V.

V=IR⇒I=3-0.7200=0.0115 A=11.5 mA

So, the current is 11.5 mA.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2018 (08 Apr)
$$
\text { The reading of the ammeter in the following figure will be }
$$

Options:
A $0.8 \mathrm{~A}$
B $0.6 \mathrm{~A}$
C $0.4 \mathrm{~A}$
D $0.2 \mathrm{~A}$

The correct answer is:
$0.4 \mathrm{~A}$

Solution:
Hints: $\frac{1}{R}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1 \Omega$
$$
\begin{aligned}
& \mathrm{R}_{\text {eq }}=1+4=5 \Omega \\
& \mathrm{I}=\frac{2}{5}=0.4 \mathrm{~A}
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
The reading of the ideal ammeter will be (Resistance of ideal ammeters is zero)

Options:
A 56 A
B 65 A
C 32 A
D 23 A

The correct answer is:
56 A

Solution:
Req=3∥6+2=2+2=4 Ω

Current from Battery, I=VReq=104=2.5 A
I1=2.5×33+6=56 A

Topic: Physics, Current Electricity, JEE Main
The reagent, from the following, which converts benzoic acid to benzaldehyde in one step is

Options:
A LiAlH4
B KMnO4
C MnO
D NaBH4

The correct answer is:
MnO

Solution:
When benzoic acid is heated with formic acid in presence of Mno, benzaldehyde is obtained.

The reaction is given below, along with benzaldehyde, carbondioxide and water are also obtained.

C6H5COOH+HCOOH→HeatMnOC6H5CHO+CO2+H2O

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The reagent in Friedel Craft's reaction is

Options:
A Pyridine
B $\mathrm{RCOCl}$
C $\mathrm{RCOOH}$
D $\mathrm{HCl}$

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

Solution:
The reaction of an alkyl halide or aryl halide with benzene in the presence of a Lewis acid, generally $\mathrm{AlCl}_3$ is known as Friedel Craft's reaction.

Topic: Chemistry, General Organic Chemistry, JEE Main
The reagent needed for converting

Options:
A
Cat. Hydrogenation

B
$\mathrm{H}_2$ /Lindlar Cat.

C
$\mathrm{Li} / \mathrm{NH}_3$

D
$\mathrm{LiAlH}_4$

The correct answer is:
$\mathrm{Li} / \mathrm{NH}_3$

Solution:
With Lithium in liquid ammonia, trans-alkene is almost an exclusive product.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2014 (11 Apr Online)
The reagent (s) which can be used to distinguish acetophenone from benzophenone is (are)

Options:
A 2,4 - dinitrophenylhydrazine
B aqueous solution of $\mathrm{NaHSO}_{3}$
C benedict reagent
D $\mathrm{I}_{2}$ and $\mathrm{NaOH}$

The correct answer is:
$\mathrm{I}_{2}$ and $\mathrm{NaOH}$

Solution:
$\mathrm{I}_{2}$ and $\mathrm{NaOH}$ react with acetophenone $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3}\right)$ to give yellow ppt. of $\mathrm{CHI}_{3}$
but benzophenone $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COC}_{6} \mathrm{H}_{5}\right)$ does not
give haloform test.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reagent used for converting acetylene to oxalic acid is

Options:
A $\mathrm{HgSO}_4$ /aqueous $\mathrm{H}_2 \mathrm{SO}_4$
B $\mathrm{HgSO} / \mathrm{CH}_{\mathrm{O}} \mathrm{COOH}$
C $\mathrm{KMnO}_4 / \mathrm{KOH}, 25^{\circ} \mathrm{C}$
D $\mathrm{Cr}_2 \mathrm{O}_3 / \mathrm{H}_2 \mathrm{SO}_4$

The correct answer is:
$\mathrm{KMnO}_4 / \mathrm{KOH}, 25^{\circ} \mathrm{C}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reagent used for the preparation of higher ether from halogenated ethers is

Options:
A conc. $\mathrm{H}_2 \mathrm{SO}_4$
B Sodium alkoxide
C Dry silver oxide
D Grignard reagent

The correct answer is:
Grignard reagent

Solution:
\(\underset{\substack{\text { Methoxymethane } \\ \text{Lower ether}}}{\mathrm{CH}_3 \mathrm{OCH}_3} \xrightarrow{\mathrm{Cl}_2 / \mathrm{hv}} \underset{\alpha-\text { Chlorodimethyl ether }}{\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{Cl}} \xrightarrow[-\mathrm{MgBrCl}]{\mathrm{CH}_3 \mathrm{MgBr}} \underset{\text { Methoxyethane Higher ether }}{\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CH}_3}\)

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The reagent used in Clemmensen's reduction is

Options:
A conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$
B $\mathrm{Zn}-\mathrm{Hg}$ conc. $\mathrm{HCl}$
C $a q \mathrm{KOH}$
D alc. $\mathrm{KOH}$

The correct answer is:
$\mathrm{Zn}-\mathrm{Hg}$ conc. $\mathrm{HCl}$

Solution:
In Clemmensen's reduction
$\mathrm{Zn}-\mathrm{Hg} /$ conc. $\mathrm{HCl}$ is used.
$>\mathrm{C}=\mathrm{O}+4 \mathrm{H} \stackrel{\mathrm{Zn}-\mathrm{Hg}+\text { conc. } \mathrm{HCl}}{\longrightarrow} \mathrm{CH}_{2}+\mathrm{H}_{2} \mathrm{O}$
This method is used to convert carbonyl compound into alkane.

Topic: Chemistry, Hydrocarbons, JEE Main
The reagent used in Gatterman Koch aldehyde synthesis is

Options:
A $\mathrm{Pb} / \mathrm{BaSO}_4$
B alkaline $\mathrm{KMnO}_4$
C acidic $\mathrm{KMnO}_4$
D $\mathrm{CO}+\mathrm{HCl}$

The correct answer is:
$\mathrm{CO}+\mathrm{HCl}$

Solution:
This is known as Gatterman-Koch reaction

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reagent used in the preparation of aspirin from salicylic acid is

Options:
A $\mathrm{SOCl}_2 /$ pyridine
B $\mathrm{CH}_3 \mathrm{COOH} / \mathrm{HCl}$
C $\left(\mathrm{CH}_2 \mathrm{CO}\right)_2 \mathrm{O} / \mathrm{Conc} . \mathrm{H}_2 \mathrm{SO}_4$
D $\mathrm{CH}_3 \mathrm{Cl} / \mathrm{AlCl}_3$

The correct answer is:
$\left(\mathrm{CH}_2 \mathrm{CO}\right)_2 \mathrm{O} / \mathrm{Conc} . \mathrm{H}_2 \mathrm{SO}_4$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reagent used in the Wolff-Kishner reduction is

Options:
A $\mathrm{H}_2 \mathrm{~N}-\mathrm{NH}_2 / \mathrm{KOH}$
B $\mathrm{H}_2 / \mathrm{Ni}$
C $\mathrm{Sn} / \mathrm{HCl}$
D $\mathrm{LiAlH}_4$

The correct answer is:
$\mathrm{H}_2 \mathrm{~N}-\mathrm{NH}_2 / \mathrm{KOH}$

Solution:
Wolff-Kishner reduction is done by heating the carbonyl compound with a mixture of hydrazine and $\mathrm{KOH}$ in the presence of ethylene glycol.

Topic: Chemistry, Hydrocarbons, JEE Main
The reagent used to distinguish between acetaldehyde and benzaldehyde is

Options:
A Tollen's reagent
B Fehling's solution
C 2-4-dinitrophenyl hydrazine
D semicarbazide

The correct answer is:
Fehling's solution

Solution:
(i) Reaction with 2,4-dinitrophenylhydrazine (Brady's reagent) : Acetaldehyde and benzaldehyde react with 2,4-dinitrophenylhydrazine (DNP) to form yellow, orange or red ppt of 2,4-dinitrophenylhydrazones (DNP derivatives)
(ii) Reaction with semicarbazide : Both acetaldehyde and benzaldehyde react with semicarbazide to form.
(iii) Reduction ot Tollen's reagent both acetaldehyde and benzaldehyde reduce Tollen's reagent to metallic silver which deposit on the walls of the test tube as bright silver mirror
(iv) Reduction of Fehling's solution: Acetaldehyde reduces Fehling solution to a red ppt of cuprous oxide.
However, benzaldehyde does not reduce Fehling's solution.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reagent used to prepare benzaldehyde by the Stephen reaction is

Options:
A \( \mathrm{CO}+\mathrm{HCl} / \mathrm{AlCl}_{3} \)
B \( \mathrm{SnCl}_{2} / \mathrm{dil} . \mathrm{HCl} \)
C \( \mathrm{CrO}_{2} \mathrm{Cl}_{2} / \mathrm{CS}_{2} \)
D \( \mathrm{Pd} / \mathrm{BaSO}_{4} / \) quinoline

The correct answer is:
\( \mathrm{SnCl}_{2} / \mathrm{dil} . \mathrm{HCl} \)

Solution:
The reagent used to prepare benzaldehyde by the Stephen reduction reaction is SnCl2/dil. HCl.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reagent which does not give acid chloride on treating with a carboxylic acid is

Options:
A \( \mathrm{PCl}_{5} \)
B \( \mathrm{Cl}_{2} / \) red phosphorus
C \( \mathrm{SOCl}_{2} \)
D \( \mathrm{PCl}_{3} \)

The correct answer is:
\( \mathrm{Cl}_{2} / \) red phosphorus

Solution:
PCl3,PCl5 and SOCl2 are used to replace -OH group of an alcohol or an acid by -Cl group
When acid reacts with Cl2 in presence of red phosphorus, ∝-chloro acid is obtained. (Hell-Vohlard-Zelinsky reaction).

CH3CH2COOH+Cl2 → Red P ∆CH3CH(Cl)COOH+HCl
∝-chloropropionic acid

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reagent which forms crystalline osazone derivative when reacted with glucose, is

Options:
A Fehling solution
B Phenylhydrazine
C Benedict solution
D Hydroxylamine

The correct answer is:
Hydroxylamine

Solution:

Topic: Chemistry, Biomolecules, JEE Main
The reagent with which both acetaldehyde and acetone react is

Options:
A Fehling's solution
B $\mathrm{I}_{2} / \mathrm{NaOH}$
C Tollen's reagent
D carbonic acid

The correct answer is:
$\mathrm{I}_{2} / \mathrm{NaOH}$

Solution:
Due to the presence of
group, acetaldehyde
and acetone
react with iodine in sodium hydroxide to give yellow ppt of iodoform.
$\mathrm{CH}_{3} \mathrm{CHO}+3 \mathrm{I}_{2}+3 \mathrm{NaOH} \longrightarrow$
$\mathrm{Cl}_{3} \mathrm{CHO}+3 \mathrm{NaI}+3 \mathrm{H}_{2} \mathrm{O}$
$\mathrm{Cl}_{3} \mathrm{CHO}+\mathrm{NaOH} \longrightarrow \mathrm{CHI}_{3}+\mathrm{HCOONa}$
$\mathrm{CH}_{3} \mathrm{COCH}_{3}+3 \mathrm{I}_{2}+3 \mathrm{NaOH} \longrightarrow$
$\mathrm{Cl}_{3} \mathrm{COCH}_{3}+3 \mathrm{NaI}+3 \mathrm{H}_{2} \mathrm{O}$
$\mathrm{Cl}_{3} \mathrm{COH}_{3}+\mathrm{NaOH} \longrightarrow \mathrm{CHI}_{3}+\mathrm{CH}_{3} \mathrm{COONa}^{-}$
$\mathrm{CHI}_{3}$ is Iodoform.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reagent with which both acetaldehyde and acetophenone react easily are

Options:
A Fehling's solution
B Schiff's reagent
C Tollen's reagent
D 2, 4-dinitrophenylhydrazine

The correct answer is:
2, 4-dinitrophenylhydrazine

Solution:
Aldehydes and ketones both can react with 2, 4-dinitrophenyl hydrazine as

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The reagent(s) for the following conversion is/are?

Options:
A Alcoholic \( \mathrm{KOH} \)
B Alcoholic \( \mathrm{KOH} \) followed by \( \mathrm{NaNH}_{2} \)
C Aqueous \( \mathrm{KOH} \) followed by \( \mathrm{NaNH}_{2} \)
D \( \mathrm{Zn} / \mathrm{CH}_{3} \mathrm{OH} \)

The correct answer is:
Alcoholic \( \mathrm{KOH} \) followed by \( \mathrm{NaNH}_{2} \)

Solution:
Alcoholic KOH can be used to remove one molecule of HBr to generate a double bond .

The removal of second HBr requires more strong base. NaNH2 is strong base as compared to HBr.

Hence for second dehydrohalogenation reaction , NaNH2 is used.

Topic: Chemistry, Hydrocarbons, JEE Main
The reagent(s) required for the following conversion are:

Options:
A i NaBH4 ii Raney Ni/H2 iii H3O+
B i LiAlH4 (ii) H3O+
C i B2H6 iiDIBAL-H iiiH3O+
D i B2H6 iiSnCl2/HCl iii H3O+

The correct answer is:
i B2H6 iiSnCl2/HCl iii H3O+

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2018 (15 Apr)
The reagent(s) required for the following conversion are:

Options:
A
(i) $\mathrm{NaBH}_4$, (ii) Raney $\mathrm{Ni} / \mathrm{H}_2$, (iii) $\mathrm{H}_3 \mathrm{O}^{+}$

B
(i) $\mathrm{LiAlH}_4$, (ii) $\mathrm{H}_3 \mathrm{O}^{+}$

C
(i) $\mathrm{B}_2 \mathrm{H}_6$, (ii) $\mathrm{DIBAL-H}$, (iii) $\mathrm{H}_3 \mathrm{O}^{+}$

D
(i) $\mathrm{B}_2 \mathrm{H}_6$, (ii) $\mathrm{SnCl}_2 / \mathrm{HCl}$, (iii) $\mathrm{H}_3 \mathrm{O}^{+}$

The correct answer is:
(i) $\mathrm{B}_2 \mathrm{H}_6$, (ii) $\mathrm{SnCl}_2 / \mathrm{HCl}$, (iii) $\mathrm{H}_3 \mathrm{O}^{+}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
The reagent(s) required for the following conversion are:

Options:
A
(i) $\mathrm{NaBH}_4$, (ii) Raney $\mathrm{Ni} / \mathrm{H}_2$, (iii) $\mathrm{H}_3 \mathrm{O}^{+}$

B
(i) $\mathrm{LiAlH}_4$, (ii) $\mathrm{H}_3 \mathrm{O}^{+}$

C
(i) $\mathrm{B}_2 \mathrm{H}_6$, (ii) $\mathrm{DIBAL-H}$, (iii) $\mathrm{H}_3 \mathrm{O}^{+}$

D
(i) $\mathrm{B}_2 \mathrm{H}_6$, (ii) $\mathrm{SnCl}_2 / \mathrm{HCl}$, (iii) $\mathrm{H}_3 \mathrm{O}^{+}$

The correct answer is:
(i) $\mathrm{B}_2 \mathrm{H}_6$, (ii) $\mathrm{SnCl}_2 / \mathrm{HCl}$, (iii) $\mathrm{H}_3 \mathrm{O}^{+}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The real forces responsible for the movement of water from one cell to another cell is mainly

Options:
A Osmotic Pressure (OP)
B Turgor Pressure (TP)
C Diffusion Pressure Deficit (DPD)
D imbibition

The correct answer is:
Diffusion Pressure Deficit (DPD)

Solution:
The net force with which water is drawn into a cell is equal to difference of OP and TP, known as diffusion pressure deficit or suction pressure. DPD of pure water is maximum ( $\approx 1236 \mathrm{~atm}$ ) solvent moves from cell of low DPD to high DPD. DPD, or $\mathrm{SP}=\mathrm{OP}-\mathrm{TP}$.

Topic: Biology, Transport in Plants, JIPMER, JIPMER 2004
The real number k for which the equation, 2x3+3x+k=0 has two distinct real roots in [0, 1] belongs to

Options:
A lies between -1 and 0.
B does not exist.
C lies between 1 and 2.
D lies between 2 and 3.

The correct answer is:
does not exist.

Solution:
Given fx=2x3+3x+x
⇒f'x=6x2+3
As f'x>0 for all x.

⇒fx is a strictly increasing function for all x.
Hence fx=0 cannot have more than one root.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2013 (07 Apr)