datavorous/entrance-exam-dataset huggingface dataset:
Cellular totipotency is demonstrated by:

Options:
A only gymnosperm cells
B all plant cells
C all eukaryotic cells
D only bacterial cells

The correct answer is:
all plant cells

Solution:
Totipotency is the process by which a single cell can develop such that all the differentiated cells of the plant. It plays an important role in tissue culture as a single can be cultured to form the complete organism.
Related Theory
Not all plant cells are totipotent, but under appropriate conditions certain cells may become totipotent. They maintain totipotency and developmental plasticity in the differentiated state. They have the ability to dedifferentiate, proliferate, and subsequently regenerate into mature plants under appropriate culture conditions.

Topic: Biology, The Living World, NEET, NEET 2003
Cellulose is a polymer of

Options:
A α - glucose
B α - fructose
C Î² - glucose
D β - fructose

The correct answer is:
β - glucose

Solution:
Cellulose is homopolysaccharide, a polymer of Î² - glucose. The glucose monomers are linked together by Î² - 1, 4 linkages. Cellulose is the main constituent of the plant cell wall.

Topic: Chemistry, Biomolecules, NEET
Cellulose is the major component of cell walls of

Options:
A Pythium
B Xanthomonas
C Pseudomonas
D Saccharomyces

The correct answer is:
Pythium

Solution:
Fungal cell wall contains $80-90 \%$ carbohydrates, the remainder being proteins and lipids. The typical feature of fungal cell wall is presence of chitin but cellulose does occur in cell walls of Oomycetes (eg, Pythium) and Hyphochytridiomycetes.
The cell wall of Saccharomyces is thin, delicate, firm and chitinous in nature. It is made up of two polysaccharides namely glucans and mannan in combination with traces of protein, lipid. The cellulose is absent.
The cell wall of bacteria is formed of peptidoglycan. It is a polymer of $N$-acetyl glucosamine and $N$-acetyl muramic acid.

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2008 (Screening)
Cellulose is the major component of cell walls of

Options:
A Saccharomyces
B Pythium
C Xanthomonas
D Pseudomonas

The correct answer is:
Pythium

Solution:
Pythium is a phycomycetes member having cellulosic cell wall.
Xanthomonas and Pseudomonas are Eubacteria having peptidogycan cell wall.
Saccharomyces - Chitinous cell wall.

Topic: Biology, Biological Classification, NEET, NEET 2008 (Mains)
Centre of hyperbola $9 x^2-16 y^2+18 x+32 y-151=0$ is

Options:
A $(1,-1)$
B $(-1,1)$
C $(-1,-1)$
D $(1,1)$

The correct answer is:
$(-1,1)$

Solution:
Centre is given by
$\left(\frac{h f-b g}{a b-h^2}, \frac{g h-a f}{a b-h^2}\right)=\left(\frac{+16.9}{-9.16}, \frac{-9(16)}{-9(16)}\right)=(-1,1)$

Topic: Mathematics, Hyperbola, JEE Main
Centre of mass (C.M.) of three particles of masses $1 \mathrm{~kg}, 2 \mathrm{~kg}$ and $3 \mathrm{~kg}$ lies at the point $(1,2,3)$ and C.M. of another system of particles of $3 \mathrm{~kg}$ and $2 \mathrm{~kg}$ lies at the point $(-1,3,-2)$. Where should we put a particle of mass $5 \mathrm{~kg}$ so that the C.M. of entire system lies at the C.M. of the first system ?

Options:
A $(3,1,8)$
B $(0,0,0)$
C $(1,3,2)$
D $(-1,2,3)$

The correct answer is:
$(3,1,8)$

Solution:
$\left.\begin{array}{l}6 \mathrm{~kg}[1,2,3] \\ 5 \mathrm{~kg}[-1,3,-2] \\ 5 \mathrm{~kg}[\mathrm{x}, \mathrm{y}, \mathrm{z}]\end{array}\right) \rightarrow(\overline{\mathrm{x}}, \overline{\mathrm{y}}, \overline{\mathrm{z}})=(1,2,3)$
$\bar{x}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}$
$1=\frac{5(-1)+5 \times x}{10}$
$-5+5 x=10 \Rightarrow x=3$
$\bar{y}=2=\frac{5 \times 3+5 y}{10}, 15+5 y=20, y=1$
$\bar{z}=3=\frac{5(-2)+5 . z}{10},-10+5 z=30, z=8$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Centre of mass of the given system of particles will be at

Options:
A $O A$
B $O B$
C $O C$
D $O D$

The correct answer is:
$O B$

Solution:
If all the masses will remain same, then centre of mass will lie at the centre $O$. But as the mass at $B$ is $4 m$, so the centre of mass of the system will shift towards $B$. So centre of mass will be on line $O B$.

Topic: Physics, Center of Mass Momentum and Collision, COMEDK, COMEDK 2021
Centromere is required for:

Options:
A movement of chromosomes towards poles
B cytoplasmic cleavage
C crossing over
D transcription.

The correct answer is:
movement of chromosomes towards poles

Solution:
The main functions include the attachment of sister chromatids, and it is the site for attachment of spindle fibre. Centromeres help in the proper alignment and segregation of the chromosomes during the process of cell division in eukaryotic cells.
Related Theory
The main component of the centromere is the kinetochore and DNA associated proteins. Centromeres are densely packed with the heterogeneous domain capped by the trilaminar kinetochore. The DNA of the centromeres is normally in a heterochromatin state, which is required for attachment of sister chromatids mediated by cohesin complex and also for the separation during anaphase.

Topic: Biology, Cell Cycle and Cell Division, NEET, NEET 2005
Ceric ammonium nitrate and CHCl3/alc. KOH are used for the identification of functional groups present in _ and _ respectively.

Options:
A amine, phenol
B alcohol, amine
C amine, alcohol
D alcohol, phenol

The correct answer is:
alcohol, amine

Solution:
Ceric ammonium nitrate and CHCl3/alc. KOH are used for the identification of functional groups present in Alcohol and amine respectively.

Ceric ammonium nitrate for alcohol and CHCl3/KOH is carbyl amine test for primary amines.

Alcohol give positive test with ceric ammonium nitrate and primary amines gives carbyl amine test with CHCl3, KOH.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Cerium IV has a noble gas configuration. Which of the following is correct statement about it?

Options:
A It will prefer to gain electron and act as an oxidizing agent
B It will prefer to give away an electron and behave as reducing agent
C It will not prefer to undergo redox reactions.
D It acts as both, oxidizing and reducing agent.

The correct answer is:
It will prefer to gain electron and act as an oxidizing agent

Solution:
Formation of CeIV is favoured by its noble gas configuration but it is strong oxidant reverting to the +3 state. The Eo value for Ce4+Ce3+ is ECe4+/Ce3+o=1.74 V is favourable for its oxidising nature.

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Cerium Z=58 is a member of the lanthanoids. Identify the incorrect statements about cerium from the following

Options:
A The +4 oxidation state of cerium is not known in solutions
B The +3 oxidation state of cerium is more stable than the +4 oxidation state
C The common oxidation states of cerium are +3 and +4
D Cerium (IV) acts as an oxidizing agent

The correct answer is:
The +4 oxidation state of cerium is not known in solutions

Solution:
The common oxidation states of cerium are +3 and +4. In which +3 state is more stable.
+ 4 oxidation state of cerium is also known in solution.

Topic: Chemistry, d and f Block Elements, JEE Main
Cerium $(Z=58)$ is an important member of the lanthanoids. Which of the following statements about cerium is incorrect?

Options:
A
The common oxidation states of cerium are $+3$ and $+4$

B
Cerium (IV) acts as an oxidizing agent

C
The $+4$ oxidation state of cerium is not known in solutions

D
The $+3$ oxidation state of cerium is more stable than the $+4$ oxidation state

The correct answer is:
The $+4$ oxidation state of cerium is not known in solutions

Solution:
Oxidation potential of $\mathrm{Ce}(\mathrm{IV})$ in aqueous solution is supposed to be $-$ ve i.e. $-0.784 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2004
Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statements about cerium is incorrect?

Options:
A The common oxidation states of cerium are +3 and +4
B The +3 oxidation state of cerium is more stable than +4 oxidation state
C Cerium (III) acts as an oxidising agent
D The +4 oxidation state of cerium is known in solution

The correct answer is:
Cerium (III) acts as an oxidising agent

Solution:
+4 oxidation state of cerium is also known in solutions.
4f 1 5d 0 6s 0  :  Ce +3
4f 0 5d 0 6s 0  :  Ce +4
The high charge on Ce 4+ leads to high hydration, hydrated Ce 4+ is hydrolysed giving polymeric species and H + . Ce( +4 ) is stable in strongly acidic solutions and used as oxidising agent in volumetric analysis.

Topic: Chemistry, d and f Block Elements, JEE Main
Certain galvanometers have a fixed core made of non magnetic metallic material. The function of this metallic material is

Options:
A to oscillate the coil in magnetic field for longer period of time
B to bring the coil to rest quickly
C to produce large deflecting torque on the coil
D to make the magnetic field radial

The correct answer is:
to bring the coil to rest quickly

Solution:
Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly. Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor according to Faraday’s law of induction. Eddy currents flow in closed loops within conductors, in planes perpendicular to the magnetic field.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Certain organic compound on combustion produces three gaseous oxides A, B and C. A and C turned lime water milky, B turned anhydrous CuSO4 blue and C turned K2Cr2O7 solution green. The elements present in organic compounds are

Options:
A C, N, O
B C, H, S
C C, H only
D C, S only

The correct answer is:
C, H, S

Solution:
The elements present in organic compounds are C, H, S and their oxides are CO2,H2O,SO2 respectively.
CO2 turns lime water milky, water turns anhydrous copper sulphate blue and SO2 being reducing agent can reduce K2Cr2O7 into green coloured Cr3+ compound CrCl3 etc.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Certain substance emits only the wavelengths $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ when it is at a high temperature. When this substance is at a colder temperature, it will absorb only the following wavelengths

Options:
A $\lambda_1$
B $\lambda_2$
C $\lambda_1$ and $\lambda_2$
D $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$

The correct answer is:
$\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$

Solution:
Let e be the emissivity and a be the absorptive power of the body.
From Kirchhoff's law, \(\mathrm{e}=\mathrm{a}\)
It means, if a substance emits a certain wavelength at a high temperature, it absorbs the same wavelengths at considerably lower temperature.
From question, the substance emits only the wavelengths \(\lambda_1, \lambda_2, \lambda_3\) and \(\lambda_4\). This means that the substance will absorb the same wavelengths at lower temperature.
Thus, option (d) is the correct answer.

Topic: Physics, Thermal Properties of Matter, JEE Main
$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{B}_2 \mathrm{H}_6 \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\mathrm{NaOH}}$ Product.
Product in above reaction is

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$
B $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$
C $\mathrm{CH}_3 \mathrm{CHO}$
D None of these

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}$

Solution:
$\mathrm{CH}_2=\mathrm{CH}_2+\mathrm{B}_2 \mathrm{H}_6 \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\mathrm{NaOH}} \mathrm{HC}_2 \mathrm{CH}_2 \mathrm{OH}$

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
$\left(\mathrm{CH}_{2} \mathrm{CO}\right)_{2} \mathrm{O}+\mathrm{RMgX} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} ?$

Options:
A $\mathrm{ROOC}\left(\mathrm{CH}_{2}\right) \mathrm{COOR}$
B $\mathrm{RCOCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$
C RCOOR
D RCOOH

The correct answer is:
$\mathrm{RCOCH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\left(\mathrm{CH}_3\right)_2 \mathrm{CO} \xrightarrow[(\mathrm{HCl})]{\mathrm{NaCN}} A \xrightarrow[\Delta]{\mathrm{H}_{3} \mathrm{O}^{+}} B$

Options:
A $\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{COOH}$
B $\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH})_2$
C $\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_3\right)_2 \mathrm{CHCOOH}$
D $\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{O}$

The correct answer is:
$\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{CN},\left(\mathrm{CH}_3\right)_2 \mathrm{C}(\mathrm{OH}) \mathrm{COOH}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$$
\mathrm{CH}_3 \mathrm{Br}+\mathrm{Nu}^{-} \longrightarrow \mathrm{CH}_3-\mathrm{Nu}+\mathrm{Br}^{-}
$$
The decreasing order of the rate of the above reaction with nucleophiles $\left(\mathrm{Nu}^{-}\right) \mathrm{A}$ to $\mathrm{D}$ is $\left[\mathrm{Nu}^{-}=(\mathrm{A}) \mathrm{PhO}^{-}\right.$, (B) $\mathrm{AcO}^{-}$, (C) $\mathrm{HO}^{-}$, (D) $\left.\mathrm{CH}_3 \mathrm{O}^{-}\right]$

Options:
A
D $>$ C $>$ A $>$ B

B
D $>$ C $>$ B $>A$

C
$A>B>C>D$

D
$B>D>C>A$

The correct answer is:
D $>$ C $>$ A $>$ B

Solution:
No Solution Available

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2006
CH 3 CCl 3 → hydrolysis acid-2
Correct answer is

Options:
A (pKa)HCN > (pKa)acid-2
B (pKa)HCN = (pKa)acid-2
C (pKa)HCN < (pKa)acid-2
D All of these

The correct answer is:
(pKa)HCN > (pKa)acid-2

Solution:
On hydrolysis of CH 3 CCl 3 , acid-2 formed is acetic acid.
(pK a ) HCN  >  (pK a ) acid-2

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br} \underset{\Delta}{\stackrel{\mathrm{aq} \cdot \mathrm{KOH}}{\longrightarrow}} \mathrm{A} \underset{\Delta}{\stackrel{\mathrm{KMnO}_{4} / \mathrm{H}^{+}}{\longrightarrow}}$
$\mathrm{B} \underset{\Delta}{\stackrel{\mathrm{NH}_{3}}{\longrightarrow}} \mathrm{C} \underset{\text { alkali }}{\stackrel{\mathrm{Br}_{2}}{\longrightarrow}} \mathrm{D,}$ 'D' is

Options:
A $\mathrm{CH}_{3} \mathrm{Br}$
B $\mathrm{CH}_{3} \mathrm{CONH}_{2}$
C $\mathrm{CH}_{3} \mathrm{NH}_{2}$
D $\mathrm{CHBr}_{3}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{NH}_{2}$

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
$\mathrm{CH}_{3} \mathrm{CH}_{2}$ Br undergoes Wurtz reaction. We may expect some of the following product
$\mathrm{A}: \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
$\mathrm{B}: \mathrm{CH}_{2}=\mathrm{CH}_{2}$
$\mathrm{C}: \mathrm{CH}_{3}-\mathrm{CH}_{3}$
Select correct product.

Options:
A Only A
B A and B
C $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$
D A and C

The correct answer is:
$\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$

Solution:
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{Na}^{*} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2}^{*}+\mathrm{NaBr}$
Intermediate free radical $\mathrm{CH}_{3} \mathrm{CH}_{2}^{*}$ combines to form $\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{CH}_{3}$ (as a main product) and also $\mathrm{CH}_{2}==\mathrm{CH}_{2}$ and $\mathrm{CH}_{3} \mathrm{CH}_{3}$ by disproportion
$$
\mathrm{CH}_{3} \mathrm{CH}_{2}^{*}+\mathrm{CH}_{3} \mathrm{CH}_{2}^{*} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{CH}_{3}
$$
n-butane
$$
\mathrm{CH}_{3} \mathrm{CH}_{2}^{*}+\mathrm{CH}_{3} \mathrm{CH}_{2}^{*} \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{CH}_{3}-\mathrm{CH}_{3}
$$
(disproportion reaction)

Topic: Chemistry, Haloalkanes and Haloarenes, VITEEE, VITEEE 2011
$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv \mathrm{CH} \xrightarrow[\mathrm{H}_2 \mathrm{O}]{\mathrm{R}}$ Butanone, $R$ is

Options:
A $\mathrm{Hg}^{++}$
B $\mathrm{KMnO}_4$
C $\mathrm{KClO}_3$
D $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$

The correct answer is:
$\mathrm{Hg}^{++}$

Solution:
It is hydration of alkynes.

Topic: Chemistry, Aldehydes and Ketones, JEE Main
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} \underset{\text { red P }}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \mathrm{A} \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{B}$. What is $\mathrm{B}$ ?

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

B
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$

C
$\mathrm{CH}_2=\mathrm{CHCOOH}$

D
$\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{COOH}$

The correct answer is:
$\mathrm{CH}_2=\mathrm{CHCOOH}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{COOH} \underset{\text { red P }}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \mathrm{A} \stackrel{\text { alc. } \mathrm{KOH}}{\longrightarrow} \mathrm{B}$. What is $\mathrm{B}$ ?

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

B
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO}$

C
$\mathrm{CH}_2=\mathrm{CHCOOH}$

D
$\mathrm{ClCH}_2 \mathrm{CH}_2 \mathrm{COOH}$

The correct answer is:
$\mathrm{CH}_2=\mathrm{CHCOOH}$

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2002
CH 3 CH 2 COOH â¶ red P Cl 2 Î â¶ alc.KOH Î
What is B ?

Options:
A CH3CH2COCl
B CH3CH2CHO
C ClCH2CH2COOH
D CH2=CHCOOH

The correct answer is:
CH2=CHCOOH

Solution:
1st step is Hell-Volhard Zelinsky reaction, halogenation of carboxylic acid at Î±â - C
CH 3 CH 2 COOH â red P - HCl Cl 2 CH 3 CHClCOOH â - HCl alc. KOH CH2=CHCOOHAcrylic acid

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{OH} \stackrel{\mathrm{A}}{\longrightarrow}$
$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{Cl} \stackrel{B}{\longrightarrow} \mathrm{CH}_{2}=\mathrm{CH}_{2} \cdot A$ and $B$ in this sequence of reactions are

Options:
A $\mathrm{KOH}(a q)$ and $\mathrm{PCl}_{5}$
B $\mathrm{PCl}_{5}$ and $\mathrm{KOH}(a q)$
C $\mathrm{Cl}_{2}$ and $\mathrm{KOH}($ alc $)$
D $\mathrm{PCl}_{5}$ and $\mathrm{KOH}($ alc $)$

The correct answer is:
$\mathrm{PCl}_{5}$ and $\mathrm{KOH}($ alc $)$

Solution:
The reaction is as follows,
$\therefore \quad A=\mathrm{PCl}_{5}$ and $B=$ alc. $\mathrm{KOH}$.

Topic: Chemistry, Hydrocarbons, JEE Main
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \underset{\text { step-1 }}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \mathrm{CH}_3 \mathrm{CHO}$ $\underset{\text { step-2 }}{\stackrel{3 \mathrm{Cl}_2}{\longrightarrow}} \mathrm{Cl}_3 \mathrm{CCHO}$
In above reactions the role of $\mathrm{Cl}_2$ in step- 1 and step-2 respectively is :

Options:
A oxidation, chlorination
B reduction, chlorination
C oxidation, addition
D reduction, substitution

The correct answer is:
oxidation, chlorination

Solution:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH} \underset{\text { Step 1 }}{\stackrel{\mathrm{Cl}_2}{\longrightarrow}} \mathrm{CH}_3 \mathrm{CHO}$
The above reaction is an example of oxidation. Due to oxidation $-\mathrm{CH}_2 \mathrm{OH}$ group is oxidised to -CHO group.
$\mathrm{CH}_3 \mathrm{CHO} \underset{\text { Step } 2}{\stackrel{3 \mathrm{Cl}_2}{\longrightarrow}} \mathrm{Cl}_3 \cdot \mathrm{C} \cdot \mathrm{CHO}$
In the second step chlorination takes place. In chlorination hydrogen atom changes by chlorine.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
$\mathrm{CH}_{3} \mathrm{CH}_{3}+\mathrm{HNO}_{3} \stackrel{675 \mathrm{~K}}{\longrightarrow}$

Options:
A $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2}$
B $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2}+\mathrm{CH}_{3} \mathrm{NO}_{2}$
C $2 \mathrm{CH}_{3} \mathrm{NO}_{2}$
D $\mathrm{CH}_{2}=\mathrm{CH}_{2}$

The correct answer is:
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2}+\mathrm{CH}_{3} \mathrm{NO}_{2}$

Solution:
Aliphatic nitro compounds are prepared by vapour phase nitration of alkanes at 693 $793 \mathrm{~K}$, under pressure. Alkanes though less reactive do undergo nitration to give a mixture of nitro alkanes resulting through cleavage of carbon-carbon bond alongwith oxidation product like $\mathrm{CO}_{2}, \mathrm{NO}_{2}, \mathrm{H}_{2} \mathrm{O}$ etc.

Topic: Chemistry, Hydrocarbons, JEE Main
$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3 \xrightarrow[\text { reagent }]{\text { Jon's }} X$,
Product $X$ is

Options:
A $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_3$
B $\mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCOCH}_3$
C Both (a) and (b) are correct
D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{COCH}_3$

The correct answer is:
Both (a) and (b) are correct

Solution:
$\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{OH}+\mathrm{HCl} \xrightarrow{\mathrm{Anhyd~} \mathrm{ZnCl}_2+\mathrm{HCl}} \left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \quad 3^{\circ} \text { reacts immediately }$
\(\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{OH}+\mathrm{HCl} \xrightarrow{\mathrm{Anhyd~} \mathrm{ZnCl}_2+\mathrm{HCl}} \left(\mathrm{CH}_3\right)_2 \mathrm{CH}-\mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \quad 2^{\circ}\) reacts after \(5 \mathrm{~min}\).
\(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{OH}+\mathrm{HCl} \xrightarrow{\mathrm{Anhyd~} \mathrm{ZnCl}_2+\mathrm{HCl}} \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2-\mathrm{Cl}+\mathrm{H}_2 \mathrm{O} \quad 1^{\circ}\) reacts only on heating.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
$\mathrm{CH}_{3}-\mathrm{CN} \stackrel{\text { Dil. } \mathrm{HCl}}{\longrightarrow} \mathrm{A} \underset{\mathrm{LiAlH}_{4}}{\longrightarrow} B \stackrel{\mathrm{PCl}_{5}}{\longrightarrow}$ $C \stackrel{\text { Al(c) } \mathrm{KOH}}{\longrightarrow} D$. The product $D$ in the above sequence of reactions is

Options:
A acetaldehyde
B ethyne
C ethene
D acetyl chloride

The correct answer is:
ethene

Solution:
The complete reaction is as follows

Topic: Chemistry, Hydrocarbons, JEE Main
$\mathrm{CH}_3 \mathrm{CO}_2 \mathrm{C}_2 \mathrm{H}_5$ on reaction with sodium ethoxide in ethanol gives $A$, which on heating in the presence of acid gives $B$ compound $B$ is

Options:
A $\mathrm{CH}_3 \mathrm{COCH} \mathrm{C} \mathrm{OOH}$
B $\mathrm{CH}_3 \mathrm{COCH}_3$
C

D $\mathrm{CH}_2=\mathrm{C} \lt \begin{aligned} & \mathrm{OC}_2 \mathrm{H}_3 \\ & \mathrm{OC}_2 \mathrm{H}_3\end{aligned}$

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_3 \mathrm{CONH}_2 \xrightarrow{\mathrm{NaNO}_2 / \mathrm{HG}} \mathrm{X}$

Options:
A $\mathrm{CH}_3 \mathrm{COOH}$
B $\mathrm{CH}_3 \mathrm{CO}^{\oplus} \mathrm{NH}_3^{\ominus} \mathrm{Cl}$
C $\mathrm{CH}_3 \mathrm{NH}_2$
D $\mathrm{CH}_3 \mathrm{CHO}$

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

Solution:
Amide, on treating with $\mathrm{HNO}_2$ give acids.
$\mathrm{CH}_3 \mathrm{CONH}_2 \xrightarrow[\left(\mathrm{HNO}_2\right)]{\mathrm{NaNO}_2 / \mathrm{HCl}} \underset{\text { acetic acid }}{\mathrm{CH}_3 \mathrm{COOH}+\mathrm{N}_2+\mathrm{H}_2 \mathrm{O}}$

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_3$ with excess of $\mathrm{C}_2 \mathrm{H}_3 \mathrm{MgBr}$ and hydrolysis gives

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_3 \mathrm{COOC}_2 \mathrm{H}_5$ with excess of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{MgBr}$ and hydrolysis gives

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main
$\mathrm{CH}_3 \mathrm{COOH}$ is reacted with $\mathrm{CH} \equiv \mathrm{CH}$ in presence of $\mathrm{Hg}^{+2}$, the product is :

Options:
A $\begin{aligned} & \mathrm{CH}_3 \\ & \text { | } \\ & \mathrm{CH}\left(\mathrm{OOC}-\mathrm{CH}_3\right)_2\end{aligned}$
B

C

D None of these

The correct answer is:

Solution:
Correct option is $\mathrm{3}$
$\mathrm{CH} \equiv \mathrm{CH}+\mathrm{CH}_3 \mathrm{COOH} \xrightarrow{\mathrm{Hg}_{g^{2+}}}$
This reaction is an example of addition reaction.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main