datavorous/entrance-exam-dataset huggingface dataset:
In the part of the circuit shown in the figure, the potential difference between points VA-VB=16 V. The current passing through the 2 Ω resistance will be

Options:
A 2.5 A
B 3.5 A
C 4.0 A
D zero

The correct answer is:
3.5 A

Solution:
VA-VB=16 V
∴4i1+2i1+i2-3+4i1=16 V ......(i)
Using Kirchhoff's second law in the closed loops we have
⇒9-i2-2i1+i2=0 ......(ii)
Solving equations (i) and (ii), we get
i1=1.5 A and i2=2 A
∴ Current through 2 Ω resistor =2+1.5=3.4 A

Topic: Physics, Current Electricity, JEE Main
In the periodic table going down in fluorine group

Options:
A Reactivity will increase
B Electronegativity will increase
C Ionic radius will increase
D Ionization potential will increase

The correct answer is:
Ionic radius will increase

Solution:
Ionic radius will increase as number of shells increases

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
In the periodic table metals usually used as catalysts belong to

Options:
A $s$-block
B $p$-block
C $d$-block
D $f$-block

The correct answer is:
$d$-block

Solution:
In the periodic table metals usually used as catalyst belong to $d$-block as they have variable oxidation state.

Topic: Chemistry, d and f Block Elements, JEE Main
In the periodic table, the metallic character of elements

Options:
A Decreases from left to right across a period and on descending a group
B Decreases from left to right across a period and increases on descending a group
C Increases from left to right across a period and on descending a group
D Increases from right to left across a period and increases on descending a group

The correct answer is:
Increases from right to left across a period and increases on descending a group

Solution:
Hint: Metallic character: An element tends to lose electrons and form positive ions or cations.
Part 1: Metallic character on moving across a period
- Metallic character decreases as we move across a period from left to right or vice versa
- As the size of the atom decreases across a period from left to right, the nucleus attracts the valence electrons more strongly. So, it becomes difficult to lose an electron.
- Hence metallic character decreased on moving left to right over a period or vice versa.
Part 2: Metallic character on moving across a group
- Metallic character increases across a group from top to bottom or vice versa
- As the size of the atom increases from top to bottom across a group, the force of attraction between the nucleus and valence electron becomes weak. So, it becomes easy to lose an electron.
- Hence tendency to lose an electron increases on descending a group i.e metallic character increases or vice versa.
Therefore, the metallic character increases as we move across a period from right to left and increases on descending a group i.e. from top to bottom.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
In the phase diagram shown, the point \( Q \) corresponds to the triple point of water. The regions \( I \), II, and III respectively corresponds to phases:

Options:
A Liquid, solid, vapour
B Solid, liquid, vapour
C Liquid, vapour, solid
D Solid, vapour, liquid

The correct answer is:
Liquid, solid, vapour

Solution:
At typical temperatures and pressures on Earth water is a liquid, but it becomes solid (that is, ice) if its temperature is lowered below zero degrees and gaseous (that is, water vapor) if its temperature is raised above 100 °C, at the same pressure.

In the given P-T diagram,
Region I- Liquid
Region II- Solid
Region III- Vapour

Topic: Physics, Thermal Properties of Matter, JEE Main
In the phenomenon of electric discharge through gases at low pressure, the coloured glow in the tube appears as a result of

Options:
A Collision between different electrons of the atoms of the gas
B Excitation of electrons in the atoms
C Collision between the atoms of the gas
D Collisions between the charged particles emitted from the cathode and the atoms of the gas

The correct answer is:
Excitation of electrons in the atoms

Solution:
Due to excitation of electrons in atoms.

Topic: Physics, Dual Nature of Matter, NEET, NEET 2008 (Mains)
In the plant cell during cytokinesis, cell wall development starts from

Options:
A peripheral and move towards centre of the cell
B centre and move towards centre of the cell
C centre and move towards peripheral of the cell
D peripheral and move towards outside of the cell.

The correct answer is:
centre and move towards peripheral of the cell

Solution:
In plant cell cytokinesis occurs by cell plate formation which is formed by fragments of ER , Golgi body and remains of spindle fibres known as phragmoplast. A small plate develops in the center which grows towards periphery thus touching the existing wall.

Topic: Biology, Cell Cycle and Cell Division, NEET
In the precipitation of the iron group (III) in qualitative analysis, ammonium chloride is added before adding ammonium hydroxide to :

Options:
A increase concentration of \(\mathrm{Cl}^{-}\)ions
B increase concentration of \(\mathrm{NH}_4{ }^{+}\)ions
C prevent interference by phosphate ions
D decrease concentration of \({ }^{-} \mathrm{OH}\) ions

The correct answer is:
decrease concentration of \({ }^{-} \mathrm{OH}\) ions

Solution:
$\begin{aligned}
& \mathrm{NH}_4 \mathrm{OH} \rightleftharpoons \mathrm{NH}_4^{+}+\mathrm{OH}^{-} \\
& \mathrm{NH}_4 \mathrm{Cl} \rightarrow \mathrm{NH}_4^{+}+\mathrm{Cl}
\end{aligned}$
Due to common ion effect of $\mathrm{NH}_4^{+}$,
$\left[\mathrm{OH}^{-}\right]$decreases in such extent that only group-III cation can be precipitated, due to their very low $\mathrm{K}_{\mathrm{sp}}$ in the range of $10^{-38}$.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2024 (04 Apr Shift 1)
In the preparation of baking soda, H2O and CO2 in ratio is used to react with Na2CO3.

Options:
A 3:1
B 1:2
C 2:1
D 1:1

The correct answer is:
1:1

Solution:
Solvay process is used for the production of sodium bicarbonate and sodium carbonate industrially. In this process carbon dioxide, water, ammonia and brine solution in its concentrated form, are used as raw materials. This process is used mainly because it is inexpensive and less number of raw materials are used to produce necessary chemicals. The important chemical reaction that is used in the production of baking soda and sodium carbonate is:

CO2 + H2O + NH3 + NaCl → NaHCO3 + NH4Cl2NaHCO3 → Na2CO3 + CO2 + H2O

Ratio of H2O : CO2=1:1

Topic: Chemistry, p Block Elements (Group 13 & 14), AP EAMCET, AP EAMCET 2021 (19 Aug Shift 2)
In the preparation of glucose from sucrose, on hydrolysis with hydrochloric acid, the alcoholic medium is necessary to

Options:
A get glucose in higher percentage than fructose
B separate the products effectively
C get more yield of products
D make the reaction faster

The correct answer is:
separate the products effectively

Solution:
Glucose is prepared in the laboratory by boiling sucrose (cane sugar) with dilute hydrochloric acid or sulphuric acid for about two hours. This hydrolyzes sucrose to glucose and fructose. In order to separate glucose from fructose, alcohol is added during cooling. Glucose is almost insoluble in alcohol, hence it will crystallize first, whereas fructose is more soluble, it remains in the solution. The solution is filtered and crystals of glucose are obtained. Hence, alcohol is added to separate the product effectively.

Topic: Chemistry, Biomolecules, JEE Main
In the presence of a catalyst, the heat evolved or absorbed during the reaction .........

Options:
A
increases

B
decreases

C
remains unchanged

D
may increase or decrease

The correct answer is:
remains unchanged

Solution:
The heat absorbed or evolved during the reaction remains unchanged in presence of catalyst as there is no change in stability of reactant and product.

Topic: Chemistry, Chemical Kinetics
In the presence of a catalyst, the heat evolved or absorbed during the reacton

Options:
A increases
B decreases
C remains unchanged
D may increase or decrease

The correct answer is:
remains unchanged

Solution:
A catalyst does not alter the enthalpy change, $\Delta H$ of the reaction.

Topic: Chemistry, Surface Chemistry, JIPMER, JIPMER 2013
In the presence of a small amount of phosphorous, aliphatic carboxylic acid reacts with chlorine or bromine to yield a reaction in which, α-hydrogen is been replaced by halogen. This reaction is known as

Options:
A Etard reaction
B Wolff-Kischner reaction
C Rosenmund reaction
D Hell-volhard-zelinsky reaction

The correct answer is:
Hell-volhard-zelinsky reaction

Solution:
This reaction is known as HVZ reaction.

Treatment with bromine and a catalytic amount of phosphorus leads to the selective α-bromination of carboxylic acids.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2015 (10 Apr Online)
In the presence of dilute salt solution, rate of respiration in a respiring tissue

Options:
A Increases
B Decreases
C Remains constant
D None of the above

The correct answer is:
Increases

Solution:
The rate of respiration increases when a plant or tissue is transferred from water to a salt solution. The cause of increased respiration during the absorption of minerals by roots.

Topic: Biology, Respiration in Plants, NEET
In the presence of peroxide, $\mathrm{HCl}$ and $\mathrm{HI}$ do not give anti-Markownikoff's addition of alkenes because:

Options:
A
One of the steps is endothermic in $\mathrm{HCl}$ and $\mathrm{HI}$

B
Both $\mathrm{HCl}$ and $\mathrm{HI}$ are strong acids

C
$\mathrm{HCl}$ is oxidizing and the $\mathrm{HI}$ is reducing

D
All the steps are exothermic is $\mathrm{HCl}$ and $\mathrm{HI}$

The correct answer is:
One of the steps is endothermic in $\mathrm{HCl}$ and $\mathrm{HI}$

Solution:
Anti-Markownikoff addition is possible only in case of $\mathrm{HBr}$ and not in $\mathrm{HCl}$ and $\mathrm{HI}$. In $\mathrm{HBr}$ both the chain initiation and propagation steps are exothermic, while in $\mathrm{HCl}$, first step is exothermic, and second step is endothermic and in HI, no step is exothermic. Hence $\mathrm{HCl}$ and $\mathrm{HI}$ do not undergo anti-Markownikoff's addition.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2014 (12 Apr Online)
In the presence of sunlight, benzene reacts with Cl2 to give product, X. The number of hydrogens in X is

Options:

The correct answer is:
6

Solution:
In the presence of ultraviolet light (but without a catalyst present), hot benzene will also undergo an addition reaction with chlorine. The ring delocalisation is permanently broken, and a chlorine atom adds on to each carbon atom. 1,2,3,4,5,6-hexachlorocyclohexane or benzenehexachloride is formed as a product. It is an example of free radical addition reaction.

Topic: Chemistry, Hydrocarbons, JEE Main, JEE Main 2022 (26 Jul Shift 1)
In the quaternary ammonium salt, the nitrogen atom is attached to other atoms by
bonds.

Options:
A three covalent and one ionic
B three covalent, one coordinate and one ionic
C Three covalent and one coordinate
D four covalent

The correct answer is:
three covalent, one coordinate and one ionic

Solution:
Quaternary ammonium salt are formed by the combination of tertiary amines with alkyl halides.

R3N + R-X → R4N+X-

Here, nitrogen forms three covalent bonds and one coordinate bond with the alkyl group, while the ionic bond with the halide ion.

Topic: Chemistry, Amines, JEE Main
In the radioactive decay of ZXA , which of the following could be considered as incorrect statement?

Options:
A α- decay involves the decrease of both A and Z by 2
B β- decay involves the increase of Z by one, A remaining constant
C K- electron capture results in the decrease of Z by one with no change in A and emission of γ- rays
D γ- ray emission is followed by the emission of α or β- particles

The correct answer is:
α- decay involves the decrease of both A and Z by 2

Solution:
α- decay involves the decrease of mass number i.e. Z by 4unit and atomic number i.e. A by 2 unit.

Topic: Chemistry, Chemical Kinetics, NEET
In the ratio of diameters, lengths and Young's modulus of steel and copper wires shown in the figure are $p, q$ and $s$ respectively, then the corresponding ratio of increase in their length would be:

Options:
A $\frac{7 q}{\left(7 s p^2\right)}$
B $\frac{7 q}{\left(5 s p^2\right)}$
C $\frac{2 q}{(5 s p)}$
D $\frac{7 q}{(5 s p)}$

The correct answer is:
$\frac{7 q}{\left(5 s p^2\right)}$

Solution:
We know that
$$
\text { Young's modulus } \begin{aligned}
\gamma & =\frac{\mathrm{FL}}{\Delta 4 \mathrm{~L}} \\
& =\frac{4 \mathrm{FL}}{\pi \mathrm{D}^2 4 \mathrm{~L}} \\
\Rightarrow \quad \Delta \mathrm{L} & =\frac{4 \mathrm{FL}}{\pi \mathrm{D}^2 \mathrm{~L}}
\end{aligned}
$$
According to questions.
$$
\begin{aligned}
& \frac{\Delta \mathrm{L}_{\mathrm{S}}}{\Delta \mathrm{L}_{\mathrm{C}}}=\frac{\mathrm{F}_{\mathrm{S}}}{\mathrm{F}_{\mathrm{C}}} \cdot \frac{\mathrm{L}_{\mathrm{S}}}{\mathrm{L}_{\mathrm{C}}} \cdot \frac{\mathrm{D}_{\mathrm{C}}^2 \mathrm{~L}_{\mathrm{S}}}{\mathrm{D}_{\mathrm{S}}^2 \mathrm{~L}_{\mathrm{C}}} \\
& \text { Here } \mathrm{F}_{\mathrm{S}}=(5 \mathrm{~m}+2 \mathrm{~m}) \mathrm{g}=7 \mathrm{mg} \\
& \mathrm{F}_{\mathrm{C}}=5 \mathrm{mg},=?, \frac{\mathrm{D}_{\mathrm{S}}}{\mathrm{D}_{\mathrm{C}}}=P, \\
& \frac{\mathrm{L}_{\mathrm{S}}}{\mathrm{L}_{\mathrm{C}}}=\mathrm{S}
\end{aligned}
$$
Putting in eq. (ii)
$$
\begin{aligned}
\frac{\Delta \mathrm{L}_{\mathrm{S}}}{\Delta \mathrm{L}_{\mathrm{C}}} & =\frac{7 \mathrm{mg}}{5 \mathrm{mg}} \times q+\left(\frac{1}{\mathrm{P}}\right)^2\left(\frac{1}{\mathrm{~S}}\right) \\
& =\frac{7 q}{5 p^2 \mathrm{~S}}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
In the ratio of the concentration of electrons that of holes in a semiconductor is $7 / 5$ and the ratio of currents is $7 / 4$ then what is the ratio of their drift velocities?

Options:
A $4 / 7$
B $5 / 8$
C $4 / 5$
D $5 / 4$

The correct answer is:
$5 / 4$

Solution:
Drift velocity, $\mathrm{V}_{\mathrm{d}}=1 / \mathrm{nAe}$
$\left(v_{\mathrm{d}}\right)_{\text {electron }} /\left(v_{\mathrm{d}}\right)_{\text {hole }}=\left(1_e / l_h\right)\left(n_h / n_e\right)=(7 / 4) \times(5 / 7)=5 / 4$

Topic: Physics, Semiconductors, JEE Main
In the reaction
$2 \mathrm{Ag}+2 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{Ag}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2$
$\mathrm{H}_2 \mathrm{SO}_4$ acts as

Options:
A Reducing agent
B Oxidising agent
C Catalytic agent
D Dehydrating agent

The correct answer is:
Oxidising agent

Solution:
$\underset{\text { Reducing agent }}{2 \mathrm{Ag}}+\underset{\substack{\text { Oxidising } \\ \text { agent }}}{2 \mathrm{H}_2 \mathrm{SO}_4} \rightarrow \mathrm{Ag}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}+\mathrm{SO}_2$

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
In the reaction. $2 \mathrm{Al}_{(\mathrm{s})}+6 \mathrm{HCl}_{(\mathrm{s})} \longrightarrow 2 \mathrm{Al}_{(\mathrm{aq})}^{3+}+6 \mathrm{Cl}_{(\mathrm{aq})}^{-}+3 \mathrm{H}_{2(\mathrm{g}),}$

Options:
A
$6 \mathrm{~L} \mathrm{~HCl}_{(\mathrm{aq})}$ is consumed for every $3 \mathrm{~L} \mathrm{~H}_{2(\mathrm{g})}$ produced

B
$33.6 \mathrm{~L} \mathrm{~H}_{2(\mathrm{g})}$ is produced regardless of temperature and pressure for every mole $\mathrm{Al}$ that reacts

C
$67.2 \mathrm{~L~H}_{2(\mathrm{g})}$ at STP is produced for every mole $\mathrm{Al}$ that reacts

D
$11.2 \mathrm{~H}_{2(\mathrm{g})}$ at STP is produced for every mole $\mathrm{HCl}_{(\mathrm{aq})}$ consumed

The correct answer is:
$11.2 \mathrm{~H}_{2(\mathrm{g})}$ at STP is produced for every mole $\mathrm{HCl}_{(\mathrm{aq})}$ consumed

Solution:
$2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{Al}^{3+}(\mathrm{aq})+6 \mathrm{Cl}^{-}(\mathrm{aq})+3 \mathrm{H}_2(\mathrm{g})$
For each mole of $\mathrm{HCl}$ reacted, $0.5$ mole of $\mathrm{H}_2$ gas is formed at STP.
$1$ mole of an ideal gas occupies $22.4$ lit at STP.
Volume of $\mathrm{H}_2$ gas formed at STP per mole of $\mathrm{HCl}$ reacted is $22.4 \times 0.5$ litre.
Hence, (D) is correct.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2007
In the reaction
$3 \mathrm{Br}_{2}+6 \mathrm{CO}_{3}^{2-}+3 \mathrm{H}_{2} \mathrm{O} \rightarrow 5 \mathrm{Br}^{-}+\mathrm{BrO}_{3}^{-}+6 \mathrm{HCO}_{3}^{-}$

Options:
A Bromine is oxidised and carbonate is reduced.
B Bromine is reduced and water is oxidised.
C Bromine is neither reduced nor oxidised.
D Bromine is both reduced and oxidised.

The correct answer is:
Bromine is both reduced and oxidised.

Solution:
$\begin{aligned}
3 \mathrm{Br}_{2}+6 \mathrm{CO}_{3}^{2-}+3 \mathrm{H}_{2} \mathrm{O} & \rightarrow \\
5 \mathrm{Br}^{-} &+\mathrm{BrO}_{3}^{-}+6 \mathrm{HCO}_{3}^{-}
\end{aligned}$
O.N. of $\mathrm{Br}_{2}$ changes from 0 to -1 and +5 , hence it is reduced as well as oxidised.

Topic: Chemistry, Redox Reactions, JEE Main
In the reaction, 4A+2B+3C→A4B2C3 , what will be the number of moles of product formed, starting from 1 mol of A, 0.6 and mol of B and 0.72 and mol of C?

Options:
A 0.25
B 0.3
C 0.24
D 2.32

The correct answer is:
0.24

Solution:
4A1mol+2B0.6mol+3C0.72mol→A4B2C3

14;0.62;0.7230.25;0.3;0.24
In the present case, reactant 'C' will be the limiting reactant because it will give least amount of product on being completely consumed.
As 3 mol C gives 1 mol product,
So 0.72 mol C will give 0.24 mol of product.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
In the reaction, 4A+2B+3CâA4B2C3, what will be the number of moles of product formed, starting from 1 mole of A , 0.6 mole of B and 0.72 mole of C ?

Options:

The correct answer is:
0.24

Solution:
4A+2B+3C→A4B2C3
1 mole 0.6 mole 0.72 mole
In the above reaction, reactant 'C' will be the limiting reactant and it will decide yield product.
∵ from 3 moles of C ; 1 mole product is formed
∴ from 0.72 moles of C=13×0.72
=0.24 moles of product

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
In the reaction, $4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$, When 1 mole of ammonia and 1 mole of $\mathrm{O}_2$ are made to react to completion

Options:
A 1.0 mole of $\mathrm{H}_2 \mathrm{O}$ is produced
B 1.0 mole of $N O$ will be produced
C All the oxygen will be consumed
D All the ammonia will be consumed

The correct answer is:
All the oxygen will be consumed

Solution:
\(\begin{array}{lcccc} & 4 \mathrm{NH}_{3(\mathrm{g})} & + & 5 \mathrm{O}_{2(\mathrm{g})} & \rightarrow & 4 \mathrm{NO}_{(\mathrm{g})} & + & 6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \\ t=0 & 1 & & 1 & & 0 & & 0 \\ t=1 & 1-4x & & 1-5x & & 4x & & 6x \\ \end{array}\)
Oxygen is limiting reagent
So, $\quad X=\frac{1}{5}=0.2$ all oxygen consumed
Left $N H_3=1-4 \times 0.2=0.2$.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
In the reaction, 8Al+3Fe3O4→4Al2O3+9Fe, the number of electrons transferred from reductant to oxidant is

Options:

The correct answer is:
24

Solution:
12H2O+8Al→4Al2O3+24H++24e-
24H++3Fe3O4+24e-→9Fe+12H2O
24 electrons are transferred from reductant to oxidant.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the reaction, $A \rightarrow$ products, If the concentration of the reactant is doubled rate of the reaction remains unchanged. The order of the reaction with respect to $\mathrm{A}$ is

Options:
A $1$
B $2$
C $0.5$
D $0$

The correct answer is:
$0$

Solution:
$\mathrm{r}_1=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}$ and $\mathrm{r}_2=\mathrm{k}[2 \mathrm{~A}]^{\mathrm{n}}$
$\because \quad \mathrm{r}_1=\mathrm{r}_2 \quad \therefore \mathrm{k}[\mathrm{A}]^{\mathrm{n}}=\mathrm{k}(2[\mathrm{~A}])^{\mathrm{n}}$
or, $2^n=1$
or, $2^{\mathrm{n}}=2^{\circ} \Rightarrow \mathrm{n}=0$
It is a zero order reaction.

Topic: Chemistry, Chemical Kinetics, AP EAMCET, AP EAMCET 2022 (08 Jul Shift 1)
In the reaction $\mathrm{B}_2 \mathrm{O}_3+C+\mathrm{Cl}_2 \rightarrow A+C O$. The $A$ is

Options:
A $\mathrm{BCl}_3$
B $\mathrm{BCl}_2$
C $\mathrm{B}_2 \mathrm{Cl}_2$
D $\mathrm{CCl}_{7}$

The correct answer is:
$\mathrm{BCl}_3$

Solution:
$\mathrm{B}_2 \mathrm{O}_3+3 \mathrm{C}+3 \mathrm{Cl}_2 \rightarrow 2 \mathrm{BCl}_3+3 \mathrm{CO}$
$\mathrm{BCl}_3$ is obtained by passing chlorine over the heated mixture of $\mathrm{B}_2 \mathrm{O}_3$ and powdered charcoal.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
In the reaction
$$
\mathrm{BrO}_3^{-}(\mathrm{aq})+5 \mathrm{Br}^{-}(\mathrm{aq})+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{Br}_2(l)+3 \mathrm{H}_2 \mathrm{O}(l)
$$
The rate of appearance of bromine $\left(\mathrm{Br}_2\right)$ is related to rate of disappearance of bromide ions as following

Options:
A $\frac{\mathrm{d}\left(\mathrm{Br}_2\right)}{\mathrm{dt}}=\frac{3}{5} \frac{\mathrm{d}\left(\mathrm{Br}^{-}\right)}{\mathrm{dt}}$
B $\frac{\mathrm{d}\left(\mathrm{Br}_2\right)}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}\left(\mathrm{Br}^{-}\right)}{\mathrm{dt}}$
C $\frac{\mathrm{d}\left(\mathrm{Br}_2\right)}{\mathrm{dt}}=-\frac{5}{3} \frac{\mathrm{d}\left(\mathrm{Br}^{-}\right)}{\mathrm{dt}}$
D $\frac{\mathrm{d}\left(\mathrm{Br}_2\right)}{\mathrm{dt}}=\frac{5}{3} \frac{\mathrm{d}\left(\mathrm{Br}^{-}\right)}{\mathrm{dt}}$

The correct answer is:
$\frac{\mathrm{d}\left(\mathrm{Br}_2\right)}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}\left(\mathrm{Br}^{-}\right)}{\mathrm{dt}}$

Solution:
$$
\begin{aligned}
& -\frac{1}{4} \frac{\mathrm{d}\left(\mathrm{Br}^{-}\right)}{\mathrm{dt}}=-\frac{1}{3} \frac{\mathrm{d}\left(\mathrm{Br}_2\right)}{\mathrm{dt}} \\
& \frac{\mathrm{d}\left(\mathrm{Br}_2\right)}{\mathrm{dt}}=-\frac{3}{5} \frac{\mathrm{d}\left(\mathrm{Br}^{-}\right)}{\mathrm{dt}}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2009 (Mains)
In the reaction,
$\mathrm{C}_2 \mathrm{H}_5 \mathrm{OC}_2 \mathrm{H}_5+\mathrm{CO} \underset{150^{\circ} \mathrm{C}, 500 \text { atm pressure }}{\stackrel{\mathrm{BF}_3}{\longrightarrow}} X$
What is $X$ ?

Options:
A Diethyl carbonate
B Ethyl carbonate
C Diethyl peroxide
D Ethyl propionate

The correct answer is:
Ethyl propionate

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
In the reaction,

CH3·CH2·Cl⟶KCNA⟶LiAlH4P end product P is

Options:
A CH3CH2·NO2
B CH3CH2CH2NO2
C CH3CH2NH2
D CH3·CH2·CH2NH2

The correct answer is:
CH3·CH2·CH2NH2

Solution:
When the reducing agent is lithium aluminium hydride, the aldehyde group is reduced to primary alcohol but the carbon-carbon double bond is unaffected.

In the given reaction the end product is (d). It is obtained
as follows.

CH3CH2·Cl⟶KCNCH3CH2CN(A)→(Reduction)LiAlH4CH3CH2CH2·NH2

Topic: Chemistry, Amines, JEE Main
In the reaction, CH 3 COOH â LiAlH 4 A â I 2 +NaOH B â Î Agâ(dust) C .

The final product (C) is

Options:
A C 2 H 5 I
B C 2 H 5 OH
C C 2 H 2
D CH 3 COCH 3

The correct answer is:
C 2 H 2

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
In the reaction,
CH 3 COOH → LiAlH 4 A → PCl 5 B → Alc. KOH C ,
the product C is :

Options:
A Acetaldehyde
B Acetylene
C Ethylene
D Acetyl chloride

The correct answer is:
Ethylene

Solution:
LiAlH4 is a reducing agent and reduces carboxylic acid to alcohol.

CH 3 COOH → LiAlH 4 CH 3 - CH 2 - OH
(A)
→ PCl 5 CH 3 - CH 2 - Cl → KOH alc CH 2 = CH 2

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2014 (06 Apr)
In the reaction given below

'B' is:

Options:
A
B

C

D

The correct answer is:

Solution:
In the first step protonation of alcohol takes place which leads to the formation of carbocation. Ring expansion takes place in the next step. In the given process five membered ring become six membered ring. Now, the carbocation loses proton in the presence of potassium hydroxide to give alkene.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2023 (13 Apr Shift 1)
In the reaction given below what will be the major product at the end :

Options:
A

B

C

D

The correct answer is:

Solution:
The first step is anti addition of Br2 across the double bond.
In the second step, EtO- (nu- ) gets added by SN1 mechanism. And Br- from the dashed position acts as the leaving group leads to the formation of a stable benzylic carbocation.

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2019 (09 Jan Shift 1)