datavorous/entrance-exam-dataset huggingface dataset:
Mycorrhizae help in the absorption of

Options:
A

Calcium

B

Metals

C

Nutrients

D

None of these

The correct answer is:

Nutrients

Solution:
The roots of higher plants possess symbiotic mycorrhizae. It is a symbiotic, non-pathogenic association between the roots and soil fungi. Several conifers lack root hair and here the mycorrhizae perform this function. The mycorrhizae are of two types: ectomycorrhizae and endo-mycorrhizae with an intermediate category as ectondomycorrhizae. The most common endomycorrhizae are Vesicular Arbuscular Mycorrhizae (VAM), generally present in herbaceous plants. Mycorrhizae perform the function of root hair, i.e. they absorb essential ions.

Topic: Biology, Organisms and Populations, NEET
Myelin sheath is produced by

Options:
A Schwann Cells and Oligodendrocytes
B Astrocytes and Schwann Cells
C Oligodendrocytes and Osteoclasts
D Osteoclasts and Astrocytes

The correct answer is:
Schwann Cells and Oligodendrocytes

Solution:
Oligodendrocytes are neuroglial cells which produce myelin sheath in the central nervous system while Schwann cell produces myelin sheath in the peripheral nervous system. Myelin is an insulating layer or sheath that forms around nerves, including those in the brain and spinal cord. It is made up of protein and fatty substances. This myelin sheath allows electrical impulses to transmit quickly and efficiently along the nerve cells.

Topic: Biology, Neural Control and Coordination, NEET, NEET 2017
Myelinated nerve fibres are white coloured because of

Options:
A Chromidial substance
B Neurolemma
C Myelin
D None of these

The correct answer is:
Myelin

Solution:
The myelin sheath appears as a tube around the axon. It is filled with the complex mixture of lipids and proteins called myelin, due to which, the myelinated nerve fibres appear white in colour.

Topic: Biology, Structural Organisation in Animals, NEET
Myopia vision is corrected by wearing

Options:
A Convex lenses
B Concave lenses
C Convex mirrors
D Concave mirrors

The correct answer is:
Concave lenses

Solution:
In Myopia or Nearsightedness, the eyeball is anteroposteriorly elongated so that the image of distant objects is formed in front of yellow spot. The defect can be removed by using concave glasses.

Topic: Biology, Neural Control and Coordination, NEET
N0/4 atoms of X(g) are converted into X+(g) by energy E1. N0 /4 atoms of X(g) are converted into Xâ(g) by energy E2. Hence, ionisation potential and electron affinity of X(g) are :

Options:
A 4 E 1 N 0 , 4 E 1 - E 2 N 0
B 4 E 1 N 0 , 4 E 2 N 0
C E 1 - E 2 N 0 , 4 E 2 N 0
D None is correct

The correct answer is:
4 E 1 N 0 , 4 E 2 N 0

Solution:
X g â¶ X + g + e -
If I ionisationh energy then
â´ N 0 4 I = E 1
â´ I = 4 E 1 N 0
If E is electron-affinity then
X g + e - â¶ X - g
N 0 4 E = E 2
â´ E = 4 E 2 N 0

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
N2 and O2 are converted into monoanions N2- and O2- respectively, which of the following statements is wrong?

Options:
A In O2 , bond length increases
B In O2 , the O-O bond order increases
C In N2 , the N-N bond weakens
D N2- becomes paramagnetic

The correct answer is:
In O2 , the O-O bond order increases

Solution:
In the conversion of O2 into O2- bond order decreases.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
N2 and O2 are converted to monocations N2+ and O2+ respectively, which is wrong statement-

Options:
A In N2+ , the N-N bond weakens
B In O2+ , the O-O bond order increases
C In O2+ , the paramagnetism decreases
D N2+ becomes diamagnetic

The correct answer is:
N2+ becomes diamagnetic

Solution:
O2 has bond order =2 while O2+ has bond order 2.5 .
According to MOT, N2⊕ , paramagnetic and O2+ is also paramagnetic

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
N2 and O2 are converted to monopositive cations N2+ and O2+ respectively. Which is incorrect?
(According to Molecular orbital theory)

Options:
A N2+ becomes diamagnetic
B In O2+ , paramagnetism decreases
C In O2+ , the bond order increases
D In N2+ the N-N bond is weakened.

The correct answer is:
N2+ becomes diamagnetic

Solution:
M.O. configuration of N2 is
Ï1s2<Ï*1s2<Ï2s2<Ï*2s2<Ï2px2=Ï2py2<Ï2pz2
B.O. of N2=10-42=3
M.O. configuration of N2+ is
Ï1s2<Ï*1s2<Ï2s2<Ï*2s2<Ï2px2=Ï2py2<Ï2pz1
B.O. of N2+=9-42=2.5
M.O. of configuration of O2 is
Ï1s2<Ï*1s2<Ï2s2<Ï*2s2<Ï2pz2<Ï2px2<Ï2py2<Ï*2px1<Ï*2py1
â´ B.O. of O2=10â62=2
As it is paramagnetic so option (a) is incorrect.
M.O. configuration of O2+ is
Ï1s2<Ï*1s2<Ï2s2<Ï*2s2<Ï2px2<Ï2py2âÏ2pz2<Ï*2py1âÏ*2pz0
B.O. of O2+=10-52=2.5
B.O. â Bond energy â1Bond length

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET
N 2 + H 2 →[X](g) → CO 2 ,  pressure [Y] → Heat [Z]+ H 2 O
In the above sequence of reaction, [Y] and [Z] are respectively

Options:
A urea, ammonium carbonate
B ammonium carbamate, urea
C ammonium carbonate, urea
D urea, hydrazine

The correct answer is:
ammonium carbamate, urea

Solution:
Here Y and Z are ammonium carbamate, urea respectively
N2+3H2→Fe(Mo)NH3(X)→CO2NH2CO2NH4Ammonium&nbspcarbamate→HeatNH2CONH2Urea+H2O

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l)

Options:
A Substance Oxidised: N 2 H 4 Substance Reduced: H 2 O 2 Oxidising Agent: H 2 O 2 Reducing Agent: N 2 H 4
B Substance Oxidised: H 2 O 2 Substance Reduced: N 2 H 4 Oxidising Agent: N 2 H 4 Reducing Agent: H 2 O 2
C Substance Oxidised: H 2 O 2 Substance Reduced: H 2 O Oxidising Agent: H 2 O Reducing Agent: H 2 O 2
D Substance Oxidised: H 2 O 2 Substance Reduced: N 2 Oxidising Agent: H 2 O 2 Reducing Agent: N 2

The correct answer is:
Substance Oxidised: N 2 H 4 Substance Reduced: H 2 O 2 Oxidising Agent: H 2 O 2 Reducing Agent: N 2 H 4

Solution:
Correct Option is : (A)
Substance Oxidised: N 2 H 4 Substance Reduced: H 2 O 2 Oxidising Agent: H 2 O 2 Reducing Agent: N 2 H 4

Topic: Chemistry, Redox Reactions, NEET
N2O4 is dissociated to 33% and 40% at total pressure P1 and P2 atm respectively. Then the ratio P1/P2 is

Options:
A 7/4
B 7/3
C 8/3
D 8/5

The correct answer is:
8/5

Solution:
KP=(nNO2)2nN2O4×PΣn1
For 33% dissociation KP=(2×0.33)20.67×P1.33
For 40% dissociation KP=(2×0.40)20.60×P1.40
So P1P2=1.56≃1.60=85 .

Topic: Chemistry, Chemical Equilibrium, NEET
Na2B4O7.10H2O is correctly represented by

Options:
A Na2[B4O5OH)4.8H2O
B 2NaBO2.Na2B2O3.10H2O
C Na2[B4H2O)4O7.6H2O
D All of the above

The correct answer is:
Na2[B4O5OH)4.8H2O

Solution:
Borax is represented as Na2[B4O5OH)4.8H2O

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET
Na2CO3 can be manufactured by Solvay process but K2CO3 cannot be prepared because:

Options:
A K2CO3 is more soluble
B K2CO3 is less soluble
C KHCO3 is more soluble than NaHCO3
D KHCO3 is less soluble than NaHCO3

The correct answer is:
KHCO3 is more soluble than NaHCO3

Solution:
Na2CO3 can be manufactured but K2CO3 cannot be prepared by Solvay process because KHCO3 is more soluble than NaHCO3 and does not precipitate out.

Topic: Chemistry, s Block Elements, NEET
Na2O2

Options:
A is diamagnetic in nature
B is salt of dibasic acid H2O2
C oxidizes Cr3+green to CrO42-yellow
D all are correct properties of Na2O2

The correct answer is:
all are correct properties of Na2O2

Solution:
It is salt of H2O2 which is in the form of O22- , diamagnetic in nature and oxidises Cr3+ to chromate ion. Na2O2+Cr+3→CrO42-+H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
Na2S2O3.5H2O is used in photography to

Options:
A Reduce AgBr to metallic Ag
B Remove reduced Ag
C Remove undecomposed AgBr as a soluble complex
D Convert metallic Ag to silver salt

The correct answer is:
Remove undecomposed AgBr as a soluble complex

Solution:
Na2S2O3.5H2O is used in photography to remove undecomposed AgBr as a soluble complex as follow
AgBr+2 Na 2 S 2 O 3 → Na 3 [Ag S 2 O 3 ) 2 Soluble complex  +NaBr

Topic: Chemistry, s Block Elements, NEET
n and ℓ for four electrons are given as i  n=4, ℓ=1  ii  n=4, ℓ=0  iii  n=3, ℓ=2  iv  n=3, ℓ=1. Arrange them in order of increasing energy from the lowest to highest:

Options:
A (ii) < (iv) < (i) < (iii)
B (i) < (iii) < (ii) < (iv)
C (iii) < (i) < (iv) < (ii)
D (iv) < (ii) < (iii) < (i)

The correct answer is:
(iv) < (ii) < (iii) < (i)

Solution:
Lower the value of n + â , lower is the energy. When n + â value is the same for two orbitals, lower n value implies lower energy of the electron. Thus, the order will be:
(iv) < (ii) < (iii) < (i).

Topic: Chemistry, Structure of Atom, NEET
NGP (neighbouring group of participation) assistance for S N 2 reaction will not be seen in-

Options:
A
B CH3-S-CH2-CH2-Cl
C
D

The correct answer is:

Solution:
Any type of substitution requires good leaving group. structure (C) do not have leaving group.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET
(NH4)2Cr2O7 on heating liberates a gas. The same gas will be obtained by

Options:
A heating NH4NO3
B heating NH4NO2
C treating H2O2 with NaNO2
D treating Mg3N2 with H2O

The correct answer is:
heating NH4NO2

Solution:
Here nitrogen gas is formed in both the processes as follows
( NH 4 ) 2 Cr 2 O 7 → Δ N 2 + Cr 2 O 3 +4 H 2 O
NH4NO2âÎN2+2H2O

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), NEET
NH4Cl crystallizes in a body-centred cubic lattice with edge length of unit cell equal to 387 pm. If the radius of the Cl- ion is 181 pm, radius for NH4+ ion is

Options:
A 154.1 pm
B 92.6 pm
C 366.3 pm
D None of these

The correct answer is:
154.1 pm

Solution:
In body centered cubic lattice, oppositely charged ions touch each other along the cube
2rc+2ra=3a
rc+ra=32a
=32×387=335.15pm
given ra=181pm
rc=335.15-181.0=154.15pm

Topic: Chemistry, Solid State, NEET
NiCl2[P(C2H5)2(C6H5)]2 exhibits temperature dependent magnetic behaviour (paramagnetic, diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively

Options:
A Square planar and square planar
B Tetrahedral and tetrahedral
C Square planar and tetrahedral
D Tetrahedral and square planar

The correct answer is:
Tetrahedral and square planar

Solution:
For a four coordinated complexes paramagnetic species are generally tetrahedral and diamagnetic species are square planar. As in case of tetrahedral complex weak ligands are used so pairing of inner d-orbitals electrons do not occur so unpaired electrons are present hence there is paramagnetic nature. In case of square planar strong ligands are used so pairing of inner d-orbitals electrons occur so unpaired electrons are not present hence there is diamagnetic nature.
Example: [NiCl4]-2,n=2 , paramagnetic, tetrahedral
[NiCO)4,n=0 , diamagnetic, square planar
Same thing is possible in the given complex according to temperature.

Topic: Chemistry, Coordination Compounds, NEET
n number of waves are produced on a string in 0.5 second. Now the tension in the string is doubled (Assume length and radius constant), the number of waves produced in 0.5 second for the same harmonic will be

Options:
A n
B 2n
C n2
D n5

The correct answer is:
2n

Solution:
n= 1 2L T m
n ′ = 1 2L 2T m
= 2 1 2L T m
n ′ = 2 n

Topic: Physics, Waves and Sound, NEET
$n$-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent?

Options:
A $\mathrm{PCl}_5$
B Reduction
C Oxidation with potassium dichromate
D Ozonolysis

The correct answer is:
Oxidation with potassium dichromate

Solution:
Isopropyl alcohol and n-propyl alcohol can be chemically distinguished by oxidation with potassium dichromate.n-propyl alcohol on oxidation with potassium dichromate gives an aldehyde. Isopropyl alcohol on oxidation with potassium dichromate gives a ketone. During oxidation of potassium dichromate, solution of potassium dichromate turns bluegreen in the reactions. For isopropyl alcohol this colour change is fast and for $n$-propyl alcohol the colour change is slow.
Isopropyl alcohol and n-propyl alcohol are examples of position isomers.In isopropyl alcohol, the hydroxyl group is present in the second carbon atom whereas in n-propyl alcohol the hydroxyl group is present in the first carbon atom.

Topic: Chemistry, Alcohols Phenols and Ethers, NEET, NEET 2002
$\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\text { heat }}{\longrightarrow} \mathrm{X}+\mathrm{NaBO}_2$ in the above reaction the product " $\mathrm{X}$ " is :

Options:
A $\mathrm{H}_3 \mathrm{BO}_3$
B $\mathrm{B}_2 \mathrm{O}_3$
C $\mathrm{Na}_2 \mathrm{~B}_2 \mathrm{O}_5$
D $\mathrm{NaB}_3 \mathrm{O}_5$

The correct answer is:
$\mathrm{B}_2 \mathrm{O}_3$

Solution:
$\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \cdot 10 \mathrm{H}_2 \mathrm{O} \underset{-10 \mathrm{H}_2 \mathrm{O}}{\stackrel{\Delta}{\longrightarrow}} \mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7$
$\begin{array}{r}
\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaBO}_2+\mathrm{B}_2 \mathrm{O}_3 \\
(\mathrm{X})
\end{array}$
$\mathrm{X}=$ Boric anhydride

Topic: Chemistry, p Block Elements (Group 13 & 14), NEET, NEET 2022 (Phase 2)
Na and Mg crystallize in BCC and FCC crystals respectively, then the number of atoms of Na and Mg presenting the unit cell of their respective crystal is

Options:
A 4 and 2
B 9 and 14
C 14 and 9
D 2 and 4

The correct answer is:
2 and 4

Solution:
Rank of BCC = 2
Rank of FCC = 4
The number of atoms of Na and Mg in the unit cell of their respective crystal is 2 and 4.

Topic: Chemistry, Solid State, NEET
NaCl is doped with 2×10-3 mol % SrCl2 , the concentration of cation vacancies is

Options:
A 3.01×1018mol-1
B 12.04×1018mol-1
C 6.02×1018mol-1
D 12.04×1020mol-1

The correct answer is:
6.02×1018mol-1

Solution:
For each Sr2+ added , one cationic vacancy is created, because two Na+ ions are removed and only one Sr2+ is added.
So Number of cation vacancy =2Ã10-3Ã1 mol % of NaCl
=2Ã10-3100 mol-1 of NaCl
Total cationic vacancies =2Ã10-5Ã6.02Ã1023mol-1 of NaCl
=12.04Ã1018mol-1 of NaCl

Topic: Chemistry, Solid State, NEET
NaCl, MgCl2 and CaSO4 are known as

Options:
A 1 – 1, 2 – 1 and 2 – 2 type electrolytes respectively
B strong, weak and strong electrolytes respectively
C electrolytes with different value of A
D electrolytes with same molar conductivity

The correct answer is:
1 – 1, 2 – 1 and 2 – 2 type electrolytes respectively

Solution:
Depending on the charges on the cation and anion produced on dissociation of the electrolyte in the solution, the type of electrolytes are known as 1 – 1, 2 – 1 and 2 – 2 electrolytes respectively.
NaCl→Na++Cl-(1-1)
MgCl2→Mg2++2Cl-(2-1)
CaSO4→Ca2++SO42-(2-2)

Topic: Chemistry, Ionic Equilibrium, NEET
NADP is converted into NADPH2 in

Options:
A Photosystem-I
B Non-cyclic photophosphorylation
C Calvin cycle
D Photosystem-II

The correct answer is:
Non-cyclic photophosphorylation

Solution:
Non-cyclic photophosphorylation involved both PS-I and PS-II. The flow of electrons is unidirectional. Here electrons are not cycled back and are used in the reduction of NADP to NADPH2.

Topic: Biology, Photosynthesis in Higher Plants, NEET
NADP+ is reduced to NADPH in

Options:
A Cyclic photophosphorylation
B In PS II
C Calvin cycle
D Non-cyclic photophosphorylation

The correct answer is:
Non-cyclic photophosphorylation

Solution:
Non-cyclic photophosphorylation is not only connected with ATP synthesis but also the production of NADPH. It involves two photosystems - PSI and PSII. PSII (680) receive electrons from photolysis of water and transfer it to PSI (700) to form NADPH from NADP+. The hydrogen required for reduction comes from the breaking of water molecules.

Topic: Biology, Photosynthesis in Higher Plants, NEET
NADP+ is reduced to NADPH in

Options:
A HMP
B Calvin Cycle
C Glycolysis
D EMP

The correct answer is:
HMP

Solution:
The complete degradation of a molecule of glucose produces 12 molecules of NADPH2 in the Hexose monophosphate pathway.

Topic: Biology, Respiration in Plants, NEET
NADP reductase enzyme is present

Options:
A In the mitochondrial matrix
B In the lumen of thylakoid
C On the stroma side of the thylakoid membrane.
D On the inner side of the thylakoid membrane.

The correct answer is:
On the stroma side of the thylakoid membrane.

Solution:
The NADP reductase enzyme is located on the stromal side of the membrane. Along with electrons that come from the acceptor of electrons of PS I, protons are necessary for the reduction of NADP+ to NADPH + H+. These protons are also removed from the stroma.

Topic: Biology, Photosynthesis in Higher Plants, NEET
Naked cytoplasm, multinucleated and saprophytic is the characteristics of

Options:
A Monera
B Protista
C Fungi
D Slime moulds

The correct answer is:
Slime moulds

Solution:
Naked cytoplasm, multinucleated and saprophytic are the characteristics of Slime moulds.

Topic: Biology, Biological Classification, NEET
Naked (wall-less), multinucleated and saprophytic organisms are grouped as ______________.

Options:
A monerans
B protists
C euglenoids
D slime moulds

The correct answer is:
slime moulds

Solution:
Kingdom Protista includes all single-celled eukaryotes. Protozoans, Euglenoids, Dinoflagellates are included in this group. Organisms in the kingdom are primarily aquatic and some show presence of flagella and cilia. photosynthetic. It also includes the Slime moulds which are the saphrophytic protists. Slime moulds are found engulfing organic matter from the dead decaying plant body. These groups are generally multi-nucleated, without any proper cell divider and stripped cytoplasm. They are known to form an aggregate called plasmodium under favarouble conditions that can grow upto several feet. The plasmodium breaks and differentiates into fruiting bodies forming spores under unfavarouble conditions.
Euglenoids are unicellular protists with flagella but no cell wall. They can perform photosynthesis as their cell structure contains chlorophyll.
The kingdom Monera includes all the prokaryotic celled organisms. Prokaryotes are organisms whose cells lack a well-defined nucleus. All the bacterial groups are included in Monera as well since bacteria have a prokaryotic cell structure. Cyanobacteria are a group of free-living photosynthetic bacteria that perform oxygenic photosynthetic process which are included in the kingdom Monera.

Topic: Biology, Biological Classification, NEET
Nalorphene (C19H21NO3), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of Nalorphene generally given is 1.5 mg. Calculate the mass of 1.5×10–3m aqueous solution required for the above dose.

Options:
A 1.3g of water
B 6.3g of water
C 3.2g of water
D 9.8g of water

The correct answer is:
3.2g of water

Solution:
Correct Option is : (C)
3.2g of water

Topic: Chemistry, Solutions, NEET
Name a species that will be isoelectronic with each of the following atoms or ions: (i) F-and (ii) Ar.

Options:
A (i) Cl–and (ii) K+
B (i) Ne and (ii) K–
C (i) Ne and (ii) Cl–
D (i) Cl and (ii) K

The correct answer is:
(i) Ne and (ii) Cl–

Solution:
Correct Option is : (C)
(i) Ne and (ii) Cl–

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Name a species that will be isoelectronic with each of the following atoms or ions: (i) Mg2+and (ii) Rb+

Options:
A (i) Cl–and (ii) Na
B (i) Ne and (ii) Br+
C (i) Ne and (ii) Sr²+
D (i) Cl and (ii) I–

The correct answer is:
(i) Ne and (ii) Sr²+

Solution:
Correct Option is : (C)
(i) Ne and (ii) Sr²+

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
Name a stain commonly used to colour chromosomes.

Options:
A Acetocarmine
B Basic fuchsin
C Chromatin
D Both a and b

The correct answer is:
Both a and b

Solution:
Basic fuchsin, acetocarmine etc are stains used to colour chromosomes.

Topic: Biology, Cell Cycle and Cell Division, NEET
Name a substance which can be used as an antiseptic as well as disinfectant.

Options:
A Hydrogen peroxide
B Alcohol
C Iodine
D Bleach

The correct answer is:
Hydrogen peroxide

Solution:
Correct Option is : (A)
Hydrogen peroxide

Topic: Chemistry, Chemistry in Everyday Life, NEET
Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Options:
A Titration
B Chromatography
C Distillation
D Sublimation

The correct answer is:
Sublimation

Solution:
Correct Option is : (D)
Sublimation

Topic: Chemistry, General Organic Chemistry, NEET
Name an artificial sweetener which is derivative of sucrose.

Options:
A Amylose
B Saccharin
C Amylopectin
D Sucralose

The correct answer is:
Sucralose

Solution:
Correct Option is : (D)
Sucralose

Topic: Chemistry, Chemistry in Everyday Life, NEET
Name an example of a molecule that exhibits abnormal behavior due to the presence of a conjugated system.

Options:
A Water
B Methane
C 1,3-butadiene
D Ethanol

The correct answer is:
1,3-butadiene

Solution:
The paragraph provides 1,3-butadiene as an example of a molecule that exhibits abnormal behavior due to the presence of a conjugated system.

Topic: Chemistry, General Organic Chemistry, NEET
Name (CH3)3CCH2COOH according to IUPAC system of nomenclature.

Options:
A 1,3-Dimethylbutanoic acid
B 1,2-Dimethylbutanoic acid
C 2,3-Dimethylbutanoic acid
D 3,3-Dimethylbutanoic acid

The correct answer is:
3,3-Dimethylbutanoic acid

Solution:
Correct Option is : (D)
3,3-Dimethylbutanoic acid

Topic: Chemistry, Aldehydes and Ketones, NEET
Name CH3CH2COCH(C2H5)CH2CH2Cl according to IUPAC system of nomenclature.

Options:
A 6-Chloro-2-ethylhexan-3-one
B 5-Chloro-4-ethylhexan-3-one
C 6-Chloro-4-ethylhexan-3-one
D Chloro-ethylhexan-one

The correct answer is:
6-Chloro-4-ethylhexan-3-one

Solution:
Correct Option is : (C)
6-Chloro-4-ethylhexan-3-one

Topic: Chemistry, Aldehydes and Ketones, NEET
Name CH3CH(CH3)CH2C(CH3)2COCH3according to IUPAC system of nomenclature.

Options:
A 3,4,5-Trimethylhexan-2-one
B 3,3,5-Trimethylhexan-2-one
C 1,3,5-Trimethylhexan-2-one
D 3,3,6-Trimethylhexan-2-one

The correct answer is:
3,3,5-Trimethylhexan-2-one

Solution:
Correct Option is : (B)
3,3,5-Trimethylhexan-2-one

Topic: Chemistry, Aldehydes and Ketones, NEET
Name CH3CH(CH3)CH2CH2CHO according to IUPAC system of nomenclature.

Options:
A 4-Methylpentanal
B 3-Methylpentanal
C 2-Methylpentanal
D 1-Methylpentanal

The correct answer is:
4-Methylpentanal

Solution:
Correct Option is : (A)
4-Methylpentanal

Topic: Chemistry, Aldehydes and Ketones, NEET
Name CH3CH=CHCHO according to the IUPAC system of nomenclature.

Options:
A But-1-enal
B But-2-enal
C Butene
D Butanoic acid

The correct answer is:
But-2-enal

Solution:
Correct Option is : (B)
But-2-enal

Topic: Chemistry, Aldehydes and Ketones, NEET
Name CH3COCH2COCH3according to IUPAC system of nomenclature.

Options:
A Pentane-1,2-dione
B Pentane-2,2-dione
C Pentane-2,3-dione
D Pentane-2,4-dione

The correct answer is:
Pentane-2,4-dione

Solution:
Correct Option is : (D)
Pentane-2,4-dione

Topic: Chemistry, Aldehydes and Ketones, NEET
Name of some compounds are given. Which one is not in accordance with IUPAC system?

Options:
A

B

C

D

The correct answer is:

Solution:
As per the IUPAC nomenclature, substituents are written in alphabetical order. Thus, correct IUPAC name for (A) is 4-ethyl-3methyl heptane.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2005
Name of the compound given below is

Options:
A 4-ethyl-3-methyloctane
B 3-methyl-4-ethyloctane
C 2, 3-diethylheptane
D 5-ethyl-6-methyloctane

The correct answer is:
4-ethyl-3-methyloctane

Solution:

Topic: Chemistry, Hydrocarbons, NEET
Name of the compound given below is:

Options:
A 4-ethyl-3-methyloctane
B 3-methyl-4-ethyloctane
C 2,3-diethylheptane
D 5-ethyl-6-methyloctane

The correct answer is:
4-ethyl-3-methyloctane

Solution:

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2003
Name of the compound given below is

Options:
A 5-ethyl-6-methyloctane
B 4-ethyl-3-methyloctane
C 3-methyl-4-ethyloctane
D 2, 3-diethylheptane

The correct answer is:
4-ethyl-3-methyloctane

Solution:

Topic: Chemistry, General Organic Chemistry, NEET
Name OHCC6H4CHO-p according to IUPAC system of nomenclature.

Options:
A Benzene –1,1-dicarbaldehyde
B Benzene –1,3-dicarbaldehyde
C Benzene –1,4-dicarbaldehyde
D Benzene –2,4-dicarbaldehyde

The correct answer is:
Benzene –1,4-dicarbaldehyde

Solution:
Correct Option is : (C)
Benzene –1,4-dicarbaldehyde

Topic: Chemistry, Aldehydes and Ketones, NEET