datavorous/entrance-exam-dataset huggingface dataset:
The correct IUPAC name of the given compound is

Options:
A 3-Chloro-1-fluoro-1-iodo-4-methoxybut-1-en-3-yne
B 4-Methoxy-2-chloro-1-fluoro-1-iodobutenyne
C 3-Chloro-4-fluoro-4-iodo-1-methoxybutenyne
D 2-Chloro-1-fluoro-1-iodo-4-methoxybut-1-en-3-yne

The correct answer is:
2-Chloro-1-fluoro-1-iodo-4-methoxybut-1-en-3-yne

Solution:
The correct numbering to the longest chain is given by \(\mathrm{I}-\underset{\substack{| \\ \mathrm{F}}}{\stackrel{1}{\mathrm{C}}}=\underset{\substack{| \\ \mathrm{Cl}}}{\stackrel{2}{\mathrm{C}}}-\stackrel{3}{\mathrm{C}} \equiv \stackrel{4}{\mathrm{C}}-\mathrm{OCH}_3\)
parent chain is of four carbon thus root word is butane
\(-X(F, C l, B r, I),-O R\) do not have any priority order and considered as alkyl substituents while numbering.
the correct IUPAC name is 2-Chloro-1-fluoro-1-iodo-4-methoxybut-1-en-3-yne

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct IUPAC nomenclature for the following compound is:

Options:
A 2-Methyl-5-oxohexanoic acid
B 2-Formyl-5-methylhexan-6-oic acid
C 5-Methyl-2-oxohexan-6-oic acid
D 5-Formyl-2-methylhexanoic acid

The correct answer is:
2-Methyl-5-oxohexanoic acid

Solution:
IUPAC name of the following compound is
2-Methyl-5-oxo-hexanoic acid

Identify the longest continuous carbon chain that contains the carboxyl group (COOH). In this case, it is a six-carbon chain.

Number the carbon chain starting from the end closest to the carboxyl group. The carboxyl carbon is assigned the number 1.

Identify and name any substituents attached to the carbon chain. In this case, there is a methyl group attached to the second carbon atom of the chain.

Combine the substituent name with the parent chain name, with a hyphen in between. The resulting name is "2-methylhexanoic acid".

Finally, add the prefix "oxo" to indicate the presence of a ketone group (C=O) on the fifth carbon of the chain.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The correct Lewis acid order for boron halides is

Options:
A BBr 3  >  BCl 3  >  Bl 3  >  BF 3
B BI 3  >  BF 3  >  BBr 3  >  BCl 3
C BF 3  >  BCl 3  >  BBr 3  >  Bl 3
D BI 3  >  BBr 3  >  BCl 3  >  BF 3

The correct answer is:
BI 3  >  BBr 3  >  BCl 3  >  BF 3

Solution:
(i) On the basis of back bonding, smaller be the halide atom, more effective be the back bonding and show less tendency to accept pair of electrons
(ii) Size of halides increases down the group hence ability to accept pair of electrons increases down the group.
Hence, correct order of Lewis acid for boron-halides is
Bl 3  >  BBr 3  >  BCl 3  >  BF 3

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The correct match between column-I and column-II is

Column-I (drug)	Column-II (test)

(A) Chloroxylenol	(P) Carbylamine test

(B) Norethindrone	(Q) Sodium hydrogen carbonate test

(C) Sulphapyridine	(R) Ferric chloride test

(D) Penicillin	(S) Baeyer’s test

Options:
A A→R, B→P, C→S, D→Q
B A→Q, B→S, C→P, D→R
C A→Q, B→P, C→S, D→R
D A→R, B→S, C→P, D→Q

The correct answer is:
A→R, B→S, C→P, D→Q

Solution:
Chloroxylenol is a phenol derivative and hence, can be identified with FeCl3 test.
The primary amine of sulphapyridine can be identified with carbylamine test.
The aliphatic alkyne of Norethindrone can be identified by the Baeyer's reagent.
Penicillin gives test with sodium hydrogen carbonate.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The correct match between item (I) and item (II) is :

Options:
A (A) $\rightarrow$ (Q); (B) $\rightarrow$ (R); (C) $\rightarrow$ (S)
B (A) $\rightarrow(\mathrm{Q}) ;(\mathrm{B}) \rightarrow(\mathrm{P}) ;(\mathrm{C}) \rightarrow(\mathrm{R})$
C (A) $\rightarrow$ (R); (B) $\rightarrow$ (P); (C) $\rightarrow$ (S)
D (A) $\rightarrow$ (R); (B) $\rightarrow$ (P); (C) $\rightarrow$ (R)

The correct answer is:
(A) $\rightarrow(\mathrm{Q}) ;(\mathrm{B}) \rightarrow(\mathrm{P}) ;(\mathrm{C}) \rightarrow(\mathrm{R})$

Solution:
(A) Norethindrone $-$ Antifertility (Q)

(B) Ofloxacin - Antibiotics (P)

(C) Equanil-used for hypertension (R)

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The correct match between Item I and Item II is:

Options:
A $(\mathrm{A}) \rightarrow(\mathrm{Q}, \mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{S}) ;(\mathrm{C}) \rightarrow(\mathrm{P})$
B $(\mathrm{A}) \rightarrow(\mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{Q}) ;(\mathrm{C}) \rightarrow(\mathrm{P})$
C $(\mathrm{A}) \rightarrow(\mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{S}) ;(\mathrm{C}) \rightarrow(\mathrm{Q})$
D $(\mathrm{A}) \rightarrow(\mathrm{Q}) ;(\mathrm{B}) \rightarrow(\mathrm{S}) ;(\mathrm{C}) \rightarrow(\mathrm{R})$

The correct answer is:
$(\mathrm{A}) \rightarrow(\mathrm{Q}, \mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{S}) ;(\mathrm{C}) \rightarrow(\mathrm{P})$

Solution:

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The correct match between Item I and Item II is:
\(\begin{array}{|l||l|l|l|}\hline & \text{ Item-I } & & \text{ Item-II } \\\hline (A) & \text{Allosteric effect} & (P) & \text{Molecule binding to the active site of enzyme} \\\hline (B) & \text{Competitive inhibitor} & (Q) & \text{Molecule crucial for communication in the body} \\\hline (C) & \text{Receptor} & (R) & \text{Molecule binding to a site other than the active site of enzyme} \\\hline (D) & \text{Poison} & (S) & \text{Molecule binding to the enzyme covalently} \\\hline\end{array}\)

Options:
A \( \mathrm{A} \rightarrow \mathrm{P} \quad \mathrm{B} \rightarrow \mathrm{R} \quad \mathrm{C} \rightarrow \mathrm{S} \quad \mathrm{D} \rightarrow \mathrm{Q} \)
B \( \mathrm{A} \rightarrow \mathrm{R} \quad \mathrm{B} \rightarrow \mathrm{P} \quad \mathrm{C} \rightarrow \mathrm{Q} \quad \mathrm{D} \rightarrow \mathrm{S} \)
C \( \mathrm{A} \rightarrow \mathrm{R} \quad \mathrm{B} \rightarrow \mathrm{P} \quad \mathrm{C} \rightarrow \mathrm{S} \quad \mathrm{D} \rightarrow \mathrm{Q} \)
D \( \mathrm{A} \rightarrow \mathrm{P} \quad \mathrm{B} \rightarrow \mathrm{R} \quad \mathrm{C} \rightarrow \mathrm{Q} \quad \mathrm{D} \rightarrow \mathrm{S} \)

The correct answer is:
\( \mathrm{A} \rightarrow \mathrm{R} \quad \mathrm{B} \rightarrow \mathrm{P} \quad \mathrm{C} \rightarrow \mathrm{Q} \quad \mathrm{D} \rightarrow \mathrm{S} \)

Solution:
When enzymes bind to a site other than the active site, the site is called allosteric site. Poison binds at the active site through covalent bonds. Receptors are important for communication. In competitive inhibition, an inhibitor that resembles the normal substrate binds to the enzyme, usually at the active site, and prevents the substrate from binding.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The correct match between item 'I' and item 'II' is

Item I (Compound)	Item II (Reagent)

(1) Lysine	(P) 1-naphthol

(2) Furfural	(Q) ninhydrin

(3) Isopropyl alcohol	(R) KMnO4

(4) Vinyl benzene (Styrene)	(S) Ceric ammonium Nitrate

Options:
A 1-R; 2-P; 3-Q; 4-S
B 1-Q; 2-P; 3-R; 4-S
C 1-Q; 2-R; 3-S; 4-P
D 1- Q; 2-P; 3-S; 4-R

The correct answer is:
1- Q; 2-P; 3-S; 4-R

Solution:
Lysine (amino acid) can be tested by ninhydrin as ninhydrin test is used for testing amino acids.
Furfural can be tested by 1-naphthol.
Isopropyl alcohol can be tested by Ceric ammonium nitrate which is a yellow reagent and on reaction with alcohol changes to red.
Vinyl benzene (Styrene) can be tested by KMnO4 as styrene decolourises alk. KMnO4 due to presence of unsaturation (double bond).

Topic: Chemistry, Practical Chemistry, JEE Main
The correct match between item I and item II is

Item I (Compound)	Item II (Reagent)

a.	Lysine	p.	1-naphthol

b.	Furfural	q.	Ninhydrin

c.	Benzyl alcohol	r.	KMnO4

d.	Styrene	s.	Ceric ammonium nitrate

Options:
A a→q, b→r, c→s, d→p
B a→q, b→p, c→s, d→r
C a→r, b→p, c→q, d→s
D a→q, b→p, c→r, d→s

The correct answer is:
a→q, b→p, c→s, d→r

Solution:
Lysine→Ninhydrin (this is given out by all the amino acids and Lysine is an amino acid.)
Furfural→1-naphthol.

Rapid furfural test is used to distinguish between glucose and fructose. In this test, the sugar solution is added to 1-naphthol and concentrated HCl.
Benzyl alcohol → ceric ammonium nitrate.
Ceric ammonium nitrate test is used for testing of alcohols.

Styrene undergoes oxidation when treated with KMnO4.

Topic: Chemistry, Biomolecules, JEE Main
The correct match between item I and item II is

Item I (Compound)	Item II (Reagent)

a.	Lysine	p.	1-naphthol

b.	Furfural	q.	Ninhydrin

c.	Benzyl alcohol	r.	KMnO4

d.	Styrene	s.	Ceric ammonium nitrate

Options:
A a→q, b→r, c→s, d→p
B a→q, b→p, c→s, d→r
C a→r, b→p, c→q, d→s
D a→q, b→p, c→r, d→s

The correct answer is:
a→q, b→p, c→s, d→r

Solution:
Lysine→Ninhydrin (this is given out by all the amino acids and Lysine is an amino acid.)
Furfural→1-naphthol.

Rapid furfural test is used to distinguish between glucose and fructose. In this test, the sugar solution is added to 1-naphthol and concentrated HCl.
Benzyl alcohol → ceric ammonium nitrate.
Ceric ammonium nitrate test is used for testing of alcohols.

Styrene undergoes oxidation when treated with KMnO4.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2019 (10 Jan Shift 2)
The correct match between Item I and Item II is

Item I	Item II

(1) Norethindrone	(P) Anti-biotic

(2) Ofloxacin	(Q) Anti-fertility

(3) Equanil	(R) Hypertension

(4) Aspirin	(S) Analgesics

Options:
A 1-R; 2-P; 3-S; 4-Q
B 1-Q; 2-R; 3-S; 4-P
C 1-R; 2-P; 3-R; 4-S
D 1-Q; 2-P; 3-R, 4-S

The correct answer is:
1-Q; 2-P; 3-R, 4-S

Solution:
(1) Norethindrone Antifertility drug
(2) Ofloxacin Anti â Biotic
(3) Equanil Hypertension (tranquailizer)
(4) Aspirin Analgesics

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The correct match between Item I and Item II is:

Item I	Item II

A Benzaldehyde	P Mobile phase

B Alumina	Q Adsorbent

C Acetonitrile	R Adsorbate

Options:
A A→P;B→R;C→Q
B A→R;B→Q;C→P
C A→Q;B→P;C→R
D A→Q;B→R;C→P

The correct answer is:
A→R;B→Q;C→P

Solution:
In chromatography;

Benzaldehyde is a substance that is absorbed. It acts as an adsorbate.

Benzaldehyde:- Adsorbate

Alumina is a highly porous substance that adsorbs another substance. It acts as an adsorbent.

Alumina:- adsorbent

Acetonitrile is used as a mobile phase in chromatography.

Acetonitrile:- mobile phase

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The correct match between Item - I and Item - II is:

Item - I	Item - II
(a)	High density polythene	(I)	Peroxide catalyst
(b)	Polyacrylonitrile	(II)	Condensation at high temperature & pressure
(c)	Novolac	(III)	Ziegler-Natta catalyst
(d)	Nylon 6	(IV)	Acid or base catalyst

Options:
A a→IV, b→(II), (c)→(I), (d)→(III)
B (a)→III, (b)→(I), (c)→(IV), (d)→(II)
C (a)→III, (b)→ (I), (c)→(II), (d)→(IV)
D (a)→II, b→IV, c→I, d→III

The correct answer is:
(a)→III, (b)→(I), (c)→(IV), (d)→(II)

Solution:
Ziegler-Natta is used in the production of high-density polymer.
The addition polymerisation of Acrylonitrite in presence of a peroxide catalyst leads to the formation of Polyacrylonitrile
Nylon - 6 is a condensation polymer made from caprolactum
Novolac is the linear product of phenol-formaldehyde which is produced by condensation reactions in the presence of either acid or base.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The correct match between item - I and item - II is :

Item - I	Item - II

(a)	Natural rubber	(I)	1, 3-butadiene + styrene

(b)	Neoprene	(II)	1, 3-butadiene

(c)	Buna-N	(III)	chloroprene

(d)	Buna-S	(IV)	Isoprene

Options:
A (a)-(III), (b)-(IV), (c)-(I), (d)-(II)
B (a)-(III), (b)-(IV), (c)-(II), (d)-(I)
C (a)-(IV), (b)-(III), (c)-(II), (d)-(I)
D (a)-(IV), (b)-(III), (c)-(I), (d)-(II)

The correct answer is:
(a)-(IV), (b)-(III), (c)-(II), (d)-(I)

Solution:

Topic: Chemistry, Polymers, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The correct match between Item I and Item II is:

Options:
A $(\mathrm{A}) \rightarrow(\mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{P}) ;(\mathrm{C}) \rightarrow(\mathrm{Q}) ;(\mathrm{D}) \rightarrow(\mathrm{S})$
B $(\mathrm{A}) \rightarrow(\mathrm{P}) ;(\mathrm{B}) \rightarrow(\mathrm{R}) ;(\mathrm{C}) \rightarrow(\mathrm{Q}) ;(\mathrm{D}) \rightarrow(\mathrm{S})$
C $(\mathrm{A}) \rightarrow(\mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{P}) ;(\mathrm{C}) \rightarrow(\mathrm{S}) ;(\mathrm{D}) \rightarrow(\mathrm{Q})$
D $(\mathrm{A}) \rightarrow(\mathrm{P}) ;(\mathrm{B}) \rightarrow(\mathrm{R}) ;(\mathrm{C}) \rightarrow(\mathrm{S}) ;(\mathrm{D}) \rightarrow(\mathrm{Q})$

The correct answer is:
$(\mathrm{A}) \rightarrow(\mathrm{R}) ;(\mathrm{B}) \rightarrow(\mathrm{P}) ;(\mathrm{C}) \rightarrow(\mathrm{Q}) ;(\mathrm{D}) \rightarrow(\mathrm{S})$

Solution:

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The correct match between Item-I (starting material) and Item-II (reagent) for the preparation of benzaldehyde is :

Item-I	Item-II

(I)	Benzene	(P)	HCl and SnCl2, H3O+

(II)	Benzonitrile	(Q)	H2, Pd-BaSO4,S and quinoline

(III)	Benzoyl Chloride	(R)	Co, HCl and AlCl3

Options:
A (I)-(Q), (II)-(R) and (III)-(P)
B (I)-(P), (II)-(Q) and (III)-(R)
C (I)-(R), (II)-(P) and (III)-(Q)
D (I)-(R), (II)-(Q) and (III)-(P)

The correct answer is:
(I)-(R), (II)-(P) and (III)-(Q)

Solution:
Gatterman-Koach reaction

Reduction	Stephan Reduction

Rosenmund Reduction

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The correct match between items I and II is :

Options:
A $(\mathrm{A}) \rightarrow(\mathrm{S}) ;(\mathrm{B}) \rightarrow(\mathrm{R}) ;(\mathrm{C}) \rightarrow(\mathrm{P})$
B (A) $\rightarrow(\mathrm{Q}) ;(\mathrm{B}) \rightarrow(\mathrm{R}) ;(\mathrm{C}) \rightarrow(\mathrm{S})$
C (A) $\rightarrow$ (R); (B) $\rightarrow$ (P); (C) $\rightarrow$ (S)
D (A) $\rightarrow(\mathrm{Q}) ;(\mathrm{B}) \rightarrow(\mathrm{R}) ;(\mathrm{C}) \rightarrow(\mathrm{P})$

The correct answer is:
(A) $\rightarrow(\mathrm{Q}) ;(\mathrm{B}) \rightarrow(\mathrm{R}) ;(\mathrm{C}) \rightarrow(\mathrm{S})$

Solution:
$\mathrm{H}_{2} \mathrm{O}$ : Sugar - Recrystallisation

$\mathrm{H}_{2} \mathrm{O}:$ Aniline - Separation by steam distillation

$\mathrm{H}_{2} \mathrm{O}:$ Toluene - Differential extraction

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (11 Jan Shift 1)
The correct match between items of List-I and List-II is

List - I	List - II

(A)	Coloured impurity	(P)	Steam distillation

(B)	Mixture of o â nitrophenol and p ânitrophenol	(Q)	Fractional distillation

(C)	Crude Naphtha	(R)	Charcoal treatment

(D)	Mixture of glycerol and sugars	(S)	Distillation under reduced pressure

Options:
A (A)-(R),(B)-(S),(C)-(P),(D)-(Q)
B (A)-(P),(B)-(S),(C)-(R),(D)-(Q)
C (A)-(R),(B)-(P),(C)-(Q),(D)-(S)
D (A)-(R),(B)-(P),(C)-(S),(D)-(Q)

The correct answer is:
(A)-(R),(B)-(P),(C)-(Q),(D)-(S)

Solution:
(A)-(R),(B)-(P),(C)-(Q),(D)-(S)
1. Coloured impurity → Charcoal treatment
2. mixture of o – nitrophenol and p –nitrophenol → Steam distillation(Both form intermolecular H-Bond)
3. Crude Naphtha → Fractional distillation
4. mixture of glycerol and sugars → Distillation under reduced pressure

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2018 (15 Apr)
The correct match between items of List-I and List-II is:

List –I	List - II

(A) Phenelzine	(P) Pyrimidine

(B) Chloroxylenol	(Q) Furan

(C) Uracil	(R) Hydrazine

(D) Ranitidine	(S) Phenol

Options:
A A-S, B-R, C-Q, D-P
B A-R, B-S, C-P, D-Q
C A-R, B-S, C-Q, D-P
D A-S, B-R, C-P, D-Q

The correct answer is:
A-R, B-S, C-P, D-Q

Solution:
→ Phenelzine contains hydrazine(2-phenylethylhydrazine)
→ Chloroxylenol contains phenol(4-Chloro-3,5-dimethylphenol)
→ Uracil is the pyrimidine base(2,4-Dihydroxypyrimidine)
→ Ranitidine contains furan ring

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2018 (16 Apr Online)
The correct match between items of List-I and List-II is :

Options:
A
(A)-(r),(B)-(s), (C)-(p),(D)-(q)

B
(A)-(p),(B)-(s), (C)-(r),(D)-(q)

C
(A)-(r), (B)-(p), (C)-(q), (D)-(s)

D
(A)-(r),(B)-(p), (C)-(s),(D)-(q)

The correct answer is:
(A)-(r), (B)-(p), (C)-(q), (D)-(s)

Solution:
(a) Charcoal treatment removes coloured impurity through adsorption.
(b) Steam distillation separates the mixture of $o$-nitrophenol and $p$-nitrophenol. The $o$-nitrophenol is steam volatile (due to intramolecular hydrogen bonding), and the para isomer is not volatile.
(c) Fractional distillation separates crude naphtha. Naphtha is a flammable liquid hydrocarbon mixture.
(d) Distillation under reduced pressure separates mixture of glycerol and sugars. Vacuum distillation lowers the boiling point and prevents decomposition.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The correct match between the entries in column I and column II are :

I	II

Radiation	Wavelength

a	Microwave	i	100 m

b	Gamma rays	ii	10-15m

c	A.M. radio	iii	10-10m

d	X–rays	iv	10-3m

Options:
A (a)–(ii), (b) – (i), (c) – (iv), (d) – (iii)
B (a)–(i), (b) – (iii), (c) – (iv), (d) – (ii)
C (a)–(iii), (b) – (ii), (c) – (i), (d) – (iv)
D (a)–(iv), (b) – (ii), (c) – (i), (d) – (iii)

The correct answer is:
(a)–(iv), (b) – (ii), (c) – (i), (d) – (iii)

Solution:
According to the EM Wave Spectrum, wavelength range for,

(a) Gamma-rays is fm.

(b) X-rays is A0

(c) Microwave is in mm

(d) A. M. radio wave is in m

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The correct matching of List I from List-II is :

Options:
A

B

C

D

The correct answer is:

Solution:
We know that
(a) $(1-x)^{-n}=1+n x+\frac{n(n+1)}{2 !} x^2+\ldots$ if $|x| < 1$
(b) $(1+x)^{-n}=1-n x+\frac{n(n+1)}{2 !} x^2-\ldots \quad$ if $|x| < 1$
(c) $1+\frac{1}{x}+\frac{1}{x^2}+\ldots \quad(\because x>1)$
$=\frac{1}{1-\frac{1}{x}}=\frac{x}{x-1}$
(d) $\quad 1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\ldots$
$=\frac{x^4}{\left(x^2+1\right)^2}$

Topic: Mathematics, Binomial Theorem, JEE Main
The correct name of PtNH34Cl2 PtCl4 is

Options:
A Tetraammine dichloro platinum (iv) tetrachloro platinate (ii)
B Dichloro tetra ammine platinium (iv) tetrachloro platinate (ii)
C Tetrachloro platinum (ii) tetraammine platinate (iv)
D Tetrachloro platinum (ii) dichloro tetraammine platinate (iv)

The correct answer is:
Tetraammine dichloro platinum (iv) tetrachloro platinate (ii)

Solution:
In the naming of complex ion ligands are named first alphabatically followed by name of metal ion along with its oxidation number.

the ligands are Amine and Chloride. Since there are 4 amine groups weâll name it as tetramine and 2 chloride groups will be named as dichloro. Also, we have platinum metal as the central metal atom in a cation, and itâs oxidation state is +4.

Now for the other (anionic part):

Here the ligand is chloride. Since there are 4 chlorides so weâll call it tetrachloride. We have the central metal atom as platinum again, but since it is in the anionic coordination sphere.

Topic: Chemistry, Coordination Compounds, JEE Main
The correct name of 'sucrose' is

Options:
A $\alpha-D$ - glucopyrarosyl $-\beta-D$ - fructofurmoside
B $\beta-D$ - glucopyrarosyl $-\beta-D-$ fructofurmoside
C $\alpha-D$ - glucopyramsyl $-\alpha-D$ - fructofurmoside
D $\beta-D$ - glucopyramsyl $-\alpha-L-$ fructofuranoside

The correct answer is:
$\alpha-D$ - glucopyrarosyl $-\beta-D$ - fructofurmoside

Solution:
The correct name of sucrose is $\alpha-D$-glucopyramsyl $-\beta-D$ - fructofuraoside

Topic: Chemistry, Biomolecules, JEE Main
The correct name of the structure
is

Options:
A (2E), (4E)-2, 4-hexadiene
B (2Z), (4Z)-2, 4-hexadiene
C (2E), (4Z)-3, 5-hexadiene
D (2Z), (4E)-2, 4-hexadiene

The correct answer is:
(2Z), (4E)-2, 4-hexadiene

Solution:
The correct name of this compound is 2-Z 4-E hexa 2, 4-di-ene.

Topic: Chemistry, General Organic Chemistry, JEE Main
The correct nomenclature for the following compound is :

Options:
A 2-formyl-4-hydroxyhept-7-enoic acid
B 2-formyl-4-hydroxyhept-6-enoic acid
C 2-carboxy-4-hydroxyhept-7-enal
D 2-carboxy-4-hydroxyhept-6-enal

The correct answer is:
2-formyl-4-hydroxyhept-6-enoic acid

Solution:
2-formly-4-hydroxyhept-6-enoic acid

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (05 Apr Shift 2)
The correct option among the following is.

Options:
A Brownian motion in colloidal solution is faster if the viscosity of the solution is very high.
B Colloidal particles in lyophobic sols can be precipitated by electrophoresis.
C Colloidal medicines are more effective because they have small surface area.
D Addition of alum to water makes it unfit for drinking.

The correct answer is:
Colloidal particles in lyophobic sols can be precipitated by electrophoresis.

Solution:
Brownian motion of colloidal particle depends on size of particles.
Colloidal medicines are better because of their large surface area.
Electrophoresis is used for ppt. of lyophobic colloid. In electrophoresis precipitation occurs at the electrode which is oppositely charged.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The correct option among the following is

Options:
A Collodial particles in lyophobic sols can be precipitated by electrophoresis
B Brownian motion in collodial solution is faster if the viscosity of the solution is very high.
C Collodial medicines are more effective because they have small surface area
D Addition of alum to water makes it unfit for drinking.

The correct answer is:
Collodial particles in lyophobic sols can be precipitated by electrophoresis

Solution:
1. Lyophilic colloids may migrate in either direction (or) may not migrate at all where as lyophobic particles migrate in only one direction in electric field.
2. Brownian movement is faster if viscosity of medium is low.
3. Colloidal medicines are more effective due to more surface area.
4. Addition of alum removes impurities from water due to coagulation.

Topic: Chemistry, Surface Chemistry, JEE Main
The correct option for the above reaction is

Options:
A

B

C

D

The correct answer is:

Solution:
Hint :
In acidic medium, only monohalogenation takes place.
\(\begin{aligned}
& \mathrm{Y}: \mathrm{CHBr}_3 \\
& \mathrm{Z}: \mathrm{CH}_3 \mathrm{COONa}
\end{aligned}\)
Second step is bromoform reaction that takes place here on \(\mathrm{CH}_3 \mathrm{COCH}_2-\mathrm{Br}\)

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
The correct option with respect to the Pauling electronegativity values of the elements is:

Options:
A $\mathrm{Te}>\mathrm{Se}$
B $\mathrm{Ga} < \mathrm{Ge}$
C $\mathrm{Si} < \mathrm{Al}$
D $\mathrm{P}>\mathrm{S}$

The correct answer is:
$\mathrm{Ga} < \mathrm{Ge}$

Solution:
Correct order of electronegativity values of the elements is

$\mathrm{Si}>\mathrm{Al} ; \mathrm{S}>\mathrm{P} ; \mathrm{Se}>\mathrm{Te} ; \mathrm{Ge}>\mathrm{Ga}$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2019 (11 Jan Shift 2)
The correct option(s) to distinguish nitrate salts of Mn2+ and Cu2+ taken separately is (are)

(1) Mn2+ shows the characteristic green colour in the flame test

(2) only Cu2+ shows the formation of precipitate by passing H2S in acidic medium

(3) only Mn2+ shows the formation of precipitate by passing H2S in faintly basic medium

(4) Cu 2+ |Cu has higher reduction potential than Mn 2+ |Mn (measured under similar conditions)

Options:
A 1, 2
B 1, 3
C 2, 4
D 1, 2, 4

The correct answer is:
2, 4

Solution:
A Manganese show pale purple colour in flame test.
B Cu2+ââH2SHClâCuSâBlack ppt.
C Both Cu2+ and Mn2+ form precipitate with H2S in basic medium.
D E Cu 2+ |Cu o =+0.34V, E Mn 2+ |Mn o =−1.18V

Topic: Chemistry, Practical Chemistry, JEE Main
The correct order for acid strength of compounds CH≡CH, CH3-C≡CH and CH2=CH2 is as follows:

Options:
A CH3-C≡CH>CH≡CH>CH2=CH2
B HC≡CH>CH3-C≡CH>CH2=CH2
C CH≡CH>CH2=CH2>CH3-C≡CH
D CH2-C≡CH>CH2=CH2>HC≡CH

The correct answer is:
HC≡CH>CH3-C≡CH>CH2=CH2

Solution:
Because acidic character ∝ stability of conjugate base (Anion)
(a) CH=C−anion(−ve on Csp)
(b) CH3−C≡C−(−ve on Csp and +I of −CH3)
(c) CH2=C−H(−ve on Csp2)
Acidic character order CH≡C−Csp>CH3−C≡C−Cspwith+I of CH3>CH2=C−Csp2

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2019 (12 Jan Shift 1)