datavorous/entrance-exam-dataset huggingface dataset:
Percentage loss in mass, when NaHCO3(s) is heated in open vessel
2  NaHCO 3 (s)→ Na 2 CO 3 s + CO 2 g + H 2 O g

Options:
A 21.12%
B 36.9%
C 30%
D 32.23%

The correct answer is:
36.9%

Solution:
Let the number of moles of NaHCO 3 is x.
Wt. loss =  x 2  × 44 +  x 2  × 18
% Weight loss= x 2 ×62 x ×84 ×100
=36.9%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
Percentage of free space in a body centred cubic unit cell is

Options:
A $28 \%$
B $30 \%$
C $32 \%$
D $34 \%$

The correct answer is:
$32 \%$

Solution:
$32 \%$
In BCC, packing fraction is $68 \%$.

Topic: Chemistry, Solid State, NEET, NEET 2008 (Mains)
Percentage of free space in cubic close packed structure and in body centred packed structure are respectively

Options:
A
$30 \%$ and $26 \%$

B
$26 \%$ and $32 \%$

C
$32 \%$ and $48 \%$

D
$48 \%$ and $26 \%$

The correct answer is:
$26 \%$ and $32 \%$

Solution:
Packing fraction of cubic close packing and body centred packing are $0.74$ and $0.68$ respectively.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2010
Pericycle of root produces

Options:
A Mechanical support
B Lateral roots
C Vascular bundles
D Adventitious buds

The correct answer is:
Lateral roots

Solution:
Pericycle in root is active in the formation of root branches or lateral roots. The main function of the pericycle cells is to provide support, structure and protection to the plant.

Topic: Biology, Anatomy of Flowering Plants, NEET
Perigynous flowers are found in:

Options:
A Guava
B Cucumber
C China rose
D Rose

The correct answer is:
Rose

Solution:
Perigynous flower means thalamus is either disc/cup/flasked shaped. Thalamus of rose is cup shaped, ovary lies in the centre of thalamus, all the whorls arise from periphery and remain at the same level. Flowers of guava and cucumber are epigynous (i.e. gynoecium is completely inserted within thalamus).

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2015 (Phase 1)
Peripatus is a connecting link between :

Options:
A Coelenterata and Porifera
B Ctenophora and Platyhelminthes
C Mollusca and Echinodermata
D Annelida and Arthropoda

The correct answer is:
Annelida and Arthropoda

Solution:
No Solution Available

Topic: Biology, Animal Kingdom, NEET, NEET 2009 (Mains)
Peripatus is known as a connecting link because it has the characteristics of both

Options:
A Aves and Fishes
B Reptiles and Birds
C Fishes and Amphibians
D Arthropods and Annelids

The correct answer is:
Arthropods and Annelids

Solution:
It acts as a connecting link between Annelids and Arthropods. Like annelids, it has continuous muscle layers in the body wall, unjointed legs like parapodia, nephridia for excretion and simple gut. Main arthropod characters are claves on the legs, haemocoel. tracheae for respiration, dorsal hearts with ostia etc.

Topic: Biology, Evolution, NEET
Perisperm differs from endosperm in

Options:
A being a haploid tissue
B having no reserve food
C being a diploid tissue
D its formation by fusion of secondary nucleus with several sperms

The correct answer is:
being a diploid tissue

Solution:
In some seeds like black pepper and beet, remnants of nucellus are also persistent this persistent nucellus is called perisperm. The cells of endosperm tissue are filled with reserve food materials and are used for the nutrition of the developing embryo. The main difference is perisperm is present in seeds and endosperm is present in developing embryo as its reserved food which is completely consumed by is during development.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2013 (All India)
Peritoneum of coelom is an example of

Options:
A Stratified squamous epithelium
B Non-ciliated columnar epithelium
C Cuboidal epithelium
D Simple squamous epithelium

The correct answer is:
Simple squamous epithelium

Solution:
Epithelial tissue can be classified as simple or compound epithelium depending on the number of layers resting on the basement membrane. The simple epithelium consists of only one layer resting on the basement membrane. This is further divided into various types depending on the shape of cells. These are squamous epithelium when the cells are flat, cuboidal when the cells are square in shape, columnar when the cells are tall and pillar like and ciliated when cells show the presence of cilia. The main function of simple epithelium is the transport of substances. Peritoneum shows flat cells and hence is an example of simple squamous epithelium.

Topic: Biology, Structural Organisation in Animals, NEET
Permanent hardness can be removed by adding

Options:
A \( \mathrm{Cl}_{2} \)
B \( \mathrm{Na}_{2} \mathrm{CO}_{3} \)
C \( \mathrm{Ca}(\mathrm{OCl}) \mathrm{Cl} \)
D \( \mathrm{K}_{2} \mathrm{CO}_{3} \)

The correct answer is:
\( \mathrm{Na}_{2} \mathrm{CO}_{3} \)

Solution:
Permanent hardness is removed by precipitating carbonates of Ca2+ and Mg2+.
CaCl2+ Na2CO3→CaCO3↓+2NaCl

Topic: Chemistry, Hydrogen, JEE Main
Permanent hardness in water cannot be cured by:

Options:
A Treatment with washing soda
B Ion exchange method
C Calgon's method
D Boiling

The correct answer is:
Boiling

Solution:
Permanent hardness due to SO42-, Cl- of Ca2+ and Mg2+ cannot be removed by boiling.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2015 (10 Apr Online)
Perovskite is a mineral composed of Ca, Ti and oxygen, cations of titanium lie at the centre, oxides ions at the face centres and calcium ions lie at corners. In this compound the oxidation number of Titanium is +x . Find the value of x?

Options:

The correct answer is:
4

Solution:
Number of Ca2+ ions =8×18=1
Number of Ti =1
Number of O-2 ions =6×12=3
So the formula of perovskite =CaTiO3
In CaTiO3 let oxidation number of Ti = x
2+x-6=0
x=+4

Topic: Chemistry, Solid State, JEE Main
Peroxide bond is present in

Options:
A $\mathrm{MgO}$
B $\mathrm{CaO}$
C $\mathrm{Li}_2 \mathrm{O}$
D $\mathrm{BaO}_2$

The correct answer is:
$\mathrm{BaO}_2$

Solution:
Oxide bond is present in $\mathrm{MgO}, \mathrm{CaO} \& \mathrm{Li}_2 \mathrm{O}$, whereas peroxide bond is
present in $\mathrm{BaO}_2$.
Hence, Option "4" is the correct answer.

Topic: Chemistry, s Block Elements, JEE Main
Peroxide ion is present in:

Options:
A MgO
B CaO
C Li2O
D BaO2

The correct answer is:
BaO2

Solution:
BaO2→Ba+2+O2–2
Peroxide ion.

Topic: Chemistry, s Block Elements, JEE Main
Peroxyacetyl nitrate (PAN), an eye irritant, is produced by:

Options:
A Classical smog
B Acid rain
C Organic waste
D Photochemical smog

The correct answer is:
Photochemical smog

Solution:
P.A.N is produced by photochemical smog.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Persons $A$ and $B$ are standing on the opposite sides of a $3.5 \mathrm{~m}$ wide water stream which they wish to cross. Each one of them has a rigid wooden plank whose mass can be neglected. However, each plank is only slightly longer than $3 \mathrm{~m}$. So they decide to arrange them together as shown in the figure schematically. With B (mass $17 \mathrm{~kg}$ ) standing, the maximum mass of $\mathrm{A}$, who can walk over the plank is close to,

Options:
A $17 \mathrm{~kg}$
B $65 \mathrm{~kg}$.
C $80 \mathrm{~kg}$.
D $105 \mathrm{~kg}$.

The correct answer is:
$80 \mathrm{~kg}$.

Solution:
$N=10 \mathrm{~m}$
$N \times 0.5=170 \mathrm{x}$
$10 \mathrm{~m} \times \frac{5}{10}=170 \mathrm{x} \Rightarrow \mathrm{m}=\frac{170 \mathrm{x}}{5}$
Maximum value is near to $2.5$ infact shift by less than $2.5$
Maximum vlaue of $m=\frac{170}{5} \times 2.5=85$ Opction (C) is correct

Topic: Physics, Rotational Motion, JEE Main
$\mathrm{PH}_3$ forms bubbles when passed slowly in water but $\mathrm{NH}_3$ dissolves. Explain why?

Options:

None

Solution:
$\mathrm{NH}_3$ forms $\mathrm{H}$-bond with water so it is soluble but $\mathrm{PH}_3$ does not form $\mathrm{H}$-bond with water so it remains as gas and forms bubbles in water.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18)
pH of 0.1 M aqueous solution of weak monoprotic acid HA is 2. Calculate osmotic pressure of this solution at 27   o C. Take solution constant R = 0.082 L atm/K-mol

Options:
A 2.46 atm
B 3.5 atm
C 3.05 atm
D 2.7 atm

The correct answer is:
2.7 atm

Solution:
H+=0.1Ãâ=10-2
â´ â=0.1
Ïâ=âiCRTâ=â1.1âÃâ0.1âÃâ0.082âÃâ300â=â2.7âatm

Topic: Chemistry, Solutions, JEE Main
pH of a 0.1 M monobasic acid (HA) is measured to be 2. Its osmotic pressure at temperature T K is

Options:
A 0.1 RT
B 0.11 RT
C 1.1 RT
D 0.01 RT

The correct answer is:
0.11 RT

Solution:
[H+]=Cα;[H+]=10-pH
So Cα=10-2
0.1 α=10-2
α=0.1
α=i-1n-1;0.1=i-12-1
i=1.1
So π=iCRT=1.1×0.1×RT=0.11RT

Topic: Chemistry, Solutions, JEE Main
pH of a saturated solution of BaOH2 is 12 . The value of its Ksp is

Options:
A 3.3×10-7 M3
B 5.0×10-7 M3
C 4.0×10-6 M3
D 5.0×10-6 M

The correct answer is:
5.0×10-7 M3

Solution:
Given that,

pH of BaOH2=12

∵ pH+pOH=14

∴ pOH=14-pH=14-12=2

Now, pOH=-logOH-

⇒ 2=-logOH

OH-=10-2 M

BaOH2sS mol L-1⇌Ba2+S+2OH-2S

OH-=2S=10-2 M

∴ S=0.5×10-2=5×10-3 M

Ksp=Ba2+OH-2

⇒(5×10-3)(10-2)2=5×10-7M3

Therefore, the value of Ksp is 5.0×10-7 M3.

Topic: Chemistry, Ionic Equilibrium, JEE Main
pH of human blood is 7.4. Then $H^{+}$concentration will be

Options:
A $4 \times 10^{-8}$
B $2 \times 10^{-8}$
C $4 \times 10^{-4}$
D $2 \times 10^{-4}$

The correct answer is:
$4 \times 10^{-8}$

Solution:
$p H=-\log \left[H^{+}\right] ; 7.4=-\log \left[H^{+}\right] ;\left[H^{+}\right]=4 \times 10^{-8} \mathrm{M}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
$\mathrm{pH}$ of soft drink is 3.6. Calculate the concentration of hydrogen ions in it.

Options:
A $2.51 \times 10^{-4} \mathrm{M}$
B $2.3 \times 10^{-3} \mathrm{M}$
C $2.0 \times 10^{-3} \mathrm{M}$
D $2.81 \times 10^{-4} \mathrm{M}$

The correct answer is:
$2.51 \times 10^{-4} \mathrm{M}$

Solution:
$\begin{aligned} & \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\ & {\left[\mathrm{H}^{+}\right]=10^{-\mathrm{pH}}} \\ & =10^{-3.6} \\ & =10^{-4+0.4} \quad\left(10^{0.4}=2.51\right) \\ & =10^{0.4} \times 10^{-4} \\ & {\left[\mathrm{H}^{+}\right]=2.51 \times 10^{-4} \mathrm{M}}\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
pH of the anodic solution of the following cell is
Pt,â H 2 (1 atm) H + xM H + 1 M H 2 (1 atm),âPt
if E cell = 0.2364 V.

Options:

The correct answer is:
4

Solution:
At Cathode:
2H+1M+2e-âH2g
At Anode:
H2gâ2H+xM+2e-
Now:
E Cell = E Cell o â 0.0591 2 log [H + ] anode 2 [H + ] cathode 2
ECell=ECellÎ¿-0.05912logâ¡[H+]anode[H+]cathode2
ECell=ECellÎ¿-0.05912logâ¡x12
0.2364=0-0.05912logâ¡x12
0.2364×2=−0.0591 log x 1 2
0.2364Ã20.0591=-logâ¡x12
-logâ¡x12=0.2364Ã20.0591=8
â2âlog x 1 â=â8
logâ¡x1=-4
x=10-4
pH=-logâ¡(x)
pH=4

Topic: Chemistry, Electrochemistry, JEE Main
Phage genome site on bacterial chromosome resulted in the structure

Options:
A nucleic acid
B heterocyst
C prophage
D None of these

The correct answer is:
prophage

Solution:
Prophage is the DNA of bacteriophage that is repressed for lytic functions and is maintained in the host bacterium in a stable state. The phage genome is integrated into DNA of its bacterial host and may be replicated along with the host DNA, as in the case or bacteriophage lamda or may be maintained as extra chromosomal DNA, as in the case for bacteriophage, lambda $P_1$.

Topic: Biology, Biomolecules (B), JIPMER, JIPMER 2014
Phase difference $(\phi)$ and path difference $(\delta)$ are related by :

Options:
A $\frac{2 \pi}{\lambda} \delta$
B $\frac{\pi}{2 \lambda} \delta$
C $\frac{\lambda}{2 \pi} \delta$
D $\frac{2 \lambda}{\pi} \delta$

The correct answer is:
$\frac{2 \pi}{\lambda} \delta$

Solution:
Relation between phase difference and path difference is
$\phi=\frac{2 \pi}{\lambda} \delta$
where $\lambda$ is the wavelength of the wave.
For constructive interference:
Path difference $\delta=n \lambda \rightarrow \phi=2 n \pi$ where $n=0,1,2, \ldots$
For destructive interference :
Path difference $\delta=\left(n+\frac{1}{2}\right) \lambda \rightarrow \phi=(2 n+1) \pi$

Topic: Physics, Waves and Sound, JEE Main
PhCH2CH3âii H3O+i CrO2Cl2/CCl4PhCH2CHO

The above reaction is an example of which name reaction?

Options:
A Rossenmund reduction
B Birch reduction
C Mendius reduction
D Etard reduction

The correct answer is:
Etard reduction

Solution:
The terminal alkyl group is partially oxidised to -CHO group by Etard reaction by CrO2Cl2/CCl4+H3O+
So, the given is Etard reduction

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Phenol can be distinguished from ethanol by the following reagents, except:

Options:
A Sodium
B Neutral FeCl3
C Phthalic anhydride/conc. H2SO4 and NaOH
D Br 2 /H 2 O

The correct answer is:
Sodium

Solution:
With neutral FeCl 3 solution, phenol gives violet colour solution but CH 3 CH 2 OH does not give this test.
6 C 6 H 5 OH + FeCl 3 → Fe O C 6 H 5 6 3 - + 3 H + + 3 HCl Violet colour complex
Phthalic anhydride/Conc. H 2 SO 4 and NaOH
Phenol cannot be distinguished from ethanol by sodium because both evolve hydrogen with sodium.
2C6H5OH+2Na→2C6H5ONa+H2↑
2C2H5OH+2Na→2C2H5ONa+H2↑

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenol is heated with a solution of mixture of $\mathrm{KBr}$ and $\mathrm{KBrO}_3$. The major product obtained in the above reaction is

Options:
A
3-Bromophenol

B
4-Bromophenol

C
2, 4, 6- Tribromophenol

D
2-Bromophenol

The correct answer is:
2, 4, 6- Tribromophenol

Solution:
In acidic medium, $\mathrm{KBr}+\mathrm{KBrO}_3$ in turn produces $\mathrm{Br}_2$. Phenol reacts with $\mathrm{Br}_2$ (aq) to give 2, 4, 6 trinitrophenol

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2011
Phenol is less acidic than

Options:
A ethanol
B o- nitrophenol
C o-methylphenol
D o-methoxyphenol

The correct answer is:
o- nitrophenol

Solution:
In o-nitrophenol, nitro group is present at ortho position. Presence of electron withdrawing group at ortho position increases the acidic strength. On the other hand, in omethylphenol and in o-methoxyphenol, electron releasing group $\left(-\mathrm{CH}_3,-\mathrm{OCH}_3\right)$ are present.
Presence of these groups at ortho or para positions of phenol decreases the acidic strength of phenols. So, phenol is less acidic than o - nitrophenol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenol is mainly manufactured from a compound $\mathrm{X}$ by subjecting it to oxidation in air followed by treating with dilute acid. Identify the compound $\mathrm{X}$

Options:
A

B

C

D

The correct answer is:

Solution:
Phenol is prepared by oxidation of cumene followed by treatment with an acid.

Topic: Chemistry, Alcohols Phenols and Ethers, TS EAMCET, TS EAMCET 2023 (12 May Shift 1)
Phenol is obtained by heating aqueous solution of

Options:
A Aniline
B Benzene diazonium chloride
C Benzoic acid
D None of these

The correct answer is:
Benzene diazonium chloride

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenol $\xrightarrow{\mathrm{NaNO}_2 / \mathrm{H}_2 \mathrm{SO}_4} B \xrightarrow{\mathrm{H}_2 \mathrm{O}} C \xrightarrow{\mathrm{NaOH}} D$
Name of the above reaction is

Options:
A Liebermann's reaction
B Phthalein fusion test
C Reimer-Tiemann reaction
D Schottenf-Baumann reaction

The correct answer is:
Liebermann's reaction

Solution:
Liebermann's reaction.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenol on heating with $\mathrm{CHCl}_3$ and $\mathrm{NaOH}$ gives salicylaldehyde. The reaction is called:

Options:
A
Reimer - Tiemann reaction

B
Claisen reaction

C
Cannizzaro's reaction

D
Hell - Volhard - Zelinsky reaction

The correct answer is:
Reimer - Tiemann reaction

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (25 Apr Online)
Phenol on reaction with dilute nitric acid, gives two products. Which method will be most efficient for large scale separation?

Options:
A Chromatographic separation
B Steam distillation
C Fractional Crystallisation
D Sublimation

The correct answer is:
Steam distillation

Solution:
The compound A is steam volatile. It is due to the presence of intramolecular hydrogen bonding in the compound A. Hence, the above mixture can be separated by steam distillation.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2022 (25 Jun Shift 1)
Phenol on reaction with H2SO4 at high temperature gives mainly

Options:
A o-phenol sulfonic acid
B para-phenol sulfonic acid
C meta-phenol sulfonic acid
D none of these

The correct answer is:
para-phenol sulfonic acid

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, NEET
Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with CH3CO2O in the presence of catalytic amount of H2SO4 produces:

Options:
A
B
C
D

The correct answer is:

Solution:
Two different reactions are taking place here Kolbe–Schmitt reaction and esterification reaction (treatment of the product of Kolbe–Schmitt reaction with (CH3CO)2O (ester).

The two reactions can be written as follows:

1.Kolbe–Schmitt reaction: reaction of phenol with carbon dioxide in the presence of sodium hydroxide, followed by acidification converts phenol to 1−1− hydroxybenzoic acid (Salicylic acid).

2.Esterification reaction: The salicylic acid is then reacted with acetic anhydride (CH3CO)2O) to form aspirin.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2018 (08 Apr)
Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with CH3CO2O in the presence of catalytic amount of H2SO4 produces:

Options:
A
B
C
D

The correct answer is:

Solution:
Two different reactions are taking place here Kolbe–Schmitt reaction and esterification reaction (treatment of the product of Kolbe–Schmitt reaction with (CH3CO)2O (ester).

The two reactions can be written as follows:

1.Kolbe–Schmitt reaction: reaction of phenol with carbon dioxide in the presence of sodium hydroxide, followed by acidification converts phenol to 1−1− hydroxybenzoic acid (Salicylic acid).

2.Esterification reaction: The salicylic acid is then reacted with acetic anhydride (CH3CO)2O) to form aspirin.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X as the major product. X on treatment with CH3CO2O in the presence of catalytic amount of H2SO4 produces

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenol reacts with $\mathrm{CCl}_4$ in presence of aqueous alkali and forms a product which on hydrolysis gives

Options:
A Salicylaldehyde
B Salicylic acid
C Benzaldehyde
D Benzoic acid

The correct answer is:
Salicylic acid

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenol reacts with $\mathrm{CHCl}_3$ and $\mathrm{NaOH}$ (at $340 \mathrm{~K}$ ) to give

Options:
A o-chlorophenol
B Salicylaldehyde
C Benzaldehyde
D Chlorobenzene

The correct answer is:
Salicylaldehyde

Solution:

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
Phenol reacts with dilute $\mathrm{HNO}_3$ at normal temperature to form

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with Br2 to form product B. A and B are respectively

Options:
A
B
C
D

The correct answer is:

Solution:
This step involves acid-base reaction at first followed by nucleophilic addition elimination.
This step is called bromination.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2018 (08 Apr)
Phenol treated with chloroform in presence of sodium hydroxide, which further hydrolysed in presence of an acid results

Options:
A Salicyclic acid
B Benzene-1,2-diol
C Benzene-1, 3-diol
D 2-Hydroxybenzaldehyde

The correct answer is:
2-Hydroxybenzaldehyde

Solution:
Chloroform CHCl3 in the presence of an alkali NaOH undergoes formylation of phenol at the ortho-position. This is also called as the Reimer-Tiemann reaction.

Hence, the answer is option D.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives

Options:
A
2,4,6-trinitrobenzene

B
o-nitrophenol

C
p-nitrophenol

D
nitrobenzene

The correct answer is:
o-nitrophenol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2008
Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives

Options:
A 6-trinitrotoluene
B Nitrophenol
C 2, 4, 6-trinitrophenol
D None of the above

The correct answer is:
2, 4, 6-trinitrophenol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenolic group can be identified by a positive:

Options:
A Phthalein dye test
B Lucas test
C Tollen’s test
D Carbylamine test

The correct answer is:
Phthalein dye test

Solution:
Carbylamine Test-Identification of primary amines

Lucas Test - Differentiation between 1°,2° and 3° alcohols

Tollen's Test - Identification of Aldehydes

Phthalein Dye Test - Identification of phenols

Phenol on heating with phthalic anhydride in the presence of concentrated sulphuric acid forms a colourless condensation compound called phenolphthalein. On further reaction with dilute sodium hydroxide solution gives a pink colour fluorescent compound called fluorescein. Characteristic colours are produced by different phenolic compounds which can be viewed under white background.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Phenolphthalein is obtained by heating phthalic anhydride with conc. H 2 SO 4 and

Options:
A Benzyl alcohol
B Benzene
C Phenol
D Benzoic acid

The correct answer is:
Phenol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Phenyl magnesium bromide reacts with methanol to give

Options:
A
a mixture of anisole and $\mathrm{Mg}(\mathrm{OH}) \mathrm{Br}$

B
a mixture of benzene and $\mathrm{Mg}(\mathrm{OMe}) \mathrm{Br}$

C
a mixture of toluene and $\mathrm{Mg}(\mathrm{OH}) \mathrm{Br}$

D
a mixture of phenol and $\mathrm{Mg}(\mathrm{Me}) \mathrm{Br}$

The correct answer is:
a mixture of benzene and $\mathrm{Mg}(\mathrm{OMe}) \mathrm{Br}$

Solution:
No Solution Available

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2006
Phenylacetylene on treatment with HgSO 4 /H 2 SO 4 ,  H 2 O produces

Options:
A Acetophenone
B Phenylacetic acid
C 1-Phenylethanol
D 2-phenyletharol

The correct answer is:
Acetophenone

Solution:
On reaction with HgSO 4 /H 2 SO 4 ,  H 2 O phenyl acetylene gives acetophenone as a product:

Topic: Chemistry, Hydrocarbons, JEE Main
Phloem sap is mainly composed of

Options:
A Water & sucrose
B Water and minerals
C Sucrose and minerals
D Minerals and hormones

The correct answer is:
Water & sucrose

Solution:
Phloem sap is a complex mixture of organic and inorganic substances. Sucrose is the main sugar in phloem sap and its concentration varies between species. Other sugars such as hexoses (e.g., glucose and fructose), raffinose-oligosaccharides, and polyols (mannitol and sorbitol) are also found in phloem sap.

Topic: Biology, Transport in Plants, NEET
Phoenix dactylifera is an example of

Options:
A 2000 years old viable seed
B Flowers once in a lifetime
C Showing 10,000 years of dormancy
D Releasing largest pollen

The correct answer is:
2000 years old viable seed

Solution:
The longevity of the seeds to remain alive after they are dispersed varies greatly. In few species, the seeds lose viability within a few months. Seeds of a large number of species live for several years. Some seeds can remain alive for hundreds of years. There are several records of very old yet viable seeds. A recent record of 2000 years old viable seed is of the date palm, Phoenix dactylifera, discovered during the archeological excavation at King Herodâs palace near the Dead Sea.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET