Number of geometrical isomers possible for the given structure is/are

Year	2026	2025	2024	Total
Questions	22	49	30	101

Q jee_main_2025_03_april_evening Stoichiometry of Nitration

X~g

of nitrobenzene on nitration gave

4.2~g

of m-dinitrobenzene. The value of

X

is ________

g

. (nearest integer) [Given: molar mass (in

g~mol⁻¹

) C: 12, H: 1, O: 16, N: 14]

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

Balanced reaction for nitration of nitrobenzene:

C₆H₅NO₂ + HNO₃ arrow C₆H₄(NO₂)₂ + H₂O

Moles = MassMolar Mass

Core Logic

From the balanced stoichiometry:

1 mole of nitrobenzene yields 1 mole of m-dinitrobenzene.

Step 1: Determine molar masses

Molar mass of Nitrobenzene ( $C₆H₅NO₂$ $\mathrm{C_6H_5NO_2}$):

M₁ = 6(12) + 5(1) + 14 + 2(16) = 72 + 5 + 14 + 32 = 123~g/mol

Molar mass of m-Dinitrobenzene ( $C₆H₄(NO₂)₂$ $\mathrm{C_6H_4(NO_2)_2}$):

M₂ = 6(12) + 4(1) + 2(14) + 4(16) = 72 + 4 + 28 + 64 = 168~g/mol

Stoichiometry of Nitration Step 2: Calculate moles and find X Moles of m-dinitrobenzene produced:

n = 4.2~g168~g/mol = 0.025~mol

Since stoichiometry is$ Since stoichiometry is $1:1, the moles of nitrobenzene required is also$, the moles of nitrobenzene required is also $0.025\mathrm{~mol}:

Mass of nitrobenzene X = 0.025~mol \times 123~g/mol = 3.075~g

Rounding to the nearest integer gives $ Rounding to the nearest integer gives $3$. Pattern Recognition Electrophilic aromatic substitution stoichiometry is straightforward: each aromatic precursor ring converts to exactly one product ring. Finding moles from the heavier substituted product and converting back using the reactant's molecular weight quickly yields the answer. Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Amines

Q jee_main_2025_03_april_evening Isomerism in Benzene Derivatives

The total number of structural isomers possible for the substituted benzene derivatives with the molecular formula

C₉H₁₂

is ________.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

Degrees of Unsaturation (Double Bond Equivalents, DBE):

DBE = C + 1 - (H)/(2) + (N)/(2)

For formula $C₉H₁₂$ $\mathrm{C_9H_{12}}$:

DBE = 9 + 1 - (12)/(2) = 4

These 4 degrees of unsaturation match a benzene ring exactly (one ring + three double bonds).

Core Logic

Since the question specifies 'substituted benzene derivatives', we must keep the benzene core ( $C₆H₅-$ $\mathrm{C_6H_5-}$ or similar) intact. This leaves 3 carbon atoms to be distributed as alkyl substituents.

Step 1: Categorize by substitution patterns

Mono-substituted benzene (one propyl group containing 3 carbons):
n-Propylbenzene: $C₆H₅-CH₂-CH₂-CH₃$ $\mathrm{C_6H_5-CH_2-CH_2-CH_3}$ (Isomer 1)
Isopropylbenzene (Cumene): $C₆H₅-CH(CH₃)₂$ $\mathrm{C_6H_5-CH(CH_3)_2}$ (Isomer 2)
Di-substituted benzene (one ethyl group and one methyl group):
1-Ethyl-2-methylbenzene (ortho-ethylmethylbenzene) (Isomer 3)
1-Ethyl-3-methylbenzene (meta-ethylmethylbenzene) (Isomer 4)
1-Ethyl-4-methylbenzene (para-ethylmethylbenzene) (Isomer 5)

Step 2: Tri-substituted benzenes

Tri-substituted benzene (three methyl groups):
1,2,3-Trimethylbenzene (Hemimellitene) (Isomer 6)
1,2,4-Trimethylbenzene (Pseudocumene) (Isomer 7)
1,3,5-Trimethylbenzene (Mesitylene) (Isomer 8)

Step 3: Total Count

Summing all options:

Total structural isomers = 2 + 3 + 3 = 8

Pattern Recognition

For alkyl benzenes with $N$ $N$ extra carbons, systematically group them as single chain substituents down to multiple methyl substituents. This hierarchical sorting prevents duplicates or missing patterns.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 11 Chemistry: Hydrocarbons

Q36 jee_main_2025_03_april_evening Dumas' Method for Nitrogen Estimation

In Dumas' method for estimation of nitrogen

0.4~g

of an organic compound gave

60~mL

of nitrogen collected at

300~K

temperature and

715~mm~Hg

pressure. The percentage composition of nitrogen in the compound is : (Given: Aqueous tension at

300~K = 15~mm~Hg

)

A. 15.71%
B. 20.95%
C. 17.46%
D. 7.85%

Solution

Related Formula

Pressure of dry nitrogen gas:

PN₂ = Ptotal - Aqueous tension

Using Ideal Gas Law:

nN₂ = PN₂ VR T

%N = Mass of nitrogenMass of organic compound \times 100

Core Logic

Given parameters:

Mass of compound $m = 0.4~g$ $m = 0.4\mathrm{~g}$
Volume of nitrogen $V = 60~mL = 0.060~L$ $V = 60\mathrm{~mL} = 0.060\mathrm{~L}$
Total pressure $Ptotal = 715~mm~Hg$ $P_{\text{total}} = 715\mathrm{~mm~Hg}$
Temperature $T = 300~K$ $T = 300\mathrm{~K}$
Aqueous tension $= 15~mm~Hg$ $= 15\mathrm{~mm~Hg}$

Step 1: Calculate dry nitrogen pressure

PN₂ = 715~mm~Hg - 15~mm~Hg = 700~mm~Hg

PN₂ = (700)/(760)~atm \approx 0.921~atm

Step 2: Calculate moles of nitrogen gas Using$ Step 2: Calculate moles of nitrogen gas Using $R = 0.0821\mathrm{~L\cdot atm\cdot K^{-1}\cdot mol^{-1}}:

nN₂ = (((700)/(760)) \times 0.060)/(0.0821 \times 300) = (0.05526)/(24.63) \approx 2.2436 \times 10⁻³~mol

Mass of$ Mass of $\mathrm{N}_2gas:

Mass = 2.2436 \times 10⁻³ \times 28~g \approx 0.06282~g

Step 3: Calculate percentage of Nitrogen

\%\mathrm{N} = \frac{0.06282\mathrm{~g}}{0.4\mathrm{~g}} \times 100 \approx 15.71\%$$

This matches Option (1).

Pattern Recognition

In Dumas' method calculations, always subtract the aqueous tension to obtain the pressure of dry nitrogen gas. Do not use the raw moist gas pressure, as doing so will overestimate the nitrogen content.

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q40 jee_main_2025_03_april_evening Hyperconjugation and Cation Stability

Given below are two statements: Statement I: Hyperconjugation is not a permanent effect. Statement II: In general, greater the number of alkyl groups attached to a positively charged C-atom, greater is the hyperconjugation interaction and stabilization of the cation. In the light of the above statements, choose the correct answer from the options given below :

A. Statement I is true but Statement II is false
B. Both Statement I and Statement II are false
C. Statement I is false but Statement II is true
D. Both Statement I and Statement II are true

Solution

Related Formula

The number of hyperconjugation structures is directly related to the count of $α$ $\alpha$-hydrogen atoms:

Number of hyperconjugative structures = Number of α-hydrogens

Core Logic

Statement I Analysis:

Hyperconjugation (no-bond resonance) involves the delocalization of $σ$ $\sigma$ electrons of $C-H$ $\mathrm{C-H}$ bonds of an alkyl group directly attached to an atom of unsaturated system or a positively charged carbon atom. This is a permanent ground-state electronic effect, not dependent on external reagents. Thus, Statement I is False.

Step 1: Analyze Statement II

Statement II states that more alkyl groups attached to a carbocation center increase hyperconjugative stabilization. Each alkyl group brings additional $σC-H$ $\sigma_{\mathrm{C-H}}$ bonds adjacent to the empty p-orbital, increasing the total count of $α$ $\alpha$-hydrogens and enhancing charge delocalization. Thus, Statement II is True.

Step 2: Conclusion

Therefore, Statement I is False but Statement II is True, matching Option (3).

Pattern Recognition

Permanent organic effects include: Inductive, Mesomeric (Resonance), and Hyperconjugation effects. Temporary electronic effects include: Electromeric and Inductomeric effects (which require an attacking reagent to manifest).

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Q33 jee_main_2025_07_april_morning IUPAC Nomenclature

Which of the following is the correct IUPAC name of given organic compound (X)?

Organic haloalkene structure X for Q33 - JEE Main 2025 — The image shows structural representation of compound X with a double bond and a bromine substituent.

A. $2-Bromo-2-methylbut-2-ene$ $\text{2-Bromo-2-methylbut-2-ene}$
B. $3-Bromo-3-methylprop-2-ene$ $\text{3-Bromo-3-methylprop-2-ene}$
C. $1-Bromo-2-methylbut-2-ene$ $\text{1-Bromo-2-methylbut-2-ene}$
D. $4-Bromo-3-methylbut-2-ene$ $\text{4-Bromo-3-methylbut-2-ene}$

Solution

Core Logic

To determine the IUPAC name of the compound shown in

:

Identify the principal functional group, which is the double bond (alkene).
Find the longest carbon chain containing the double bond:

C1(H₂Br) - C2(CH₃) = C3(H) - C4(H₃)

The longest chain has $4$ $4$ carbons, which means the parent alkane is butane, and with a double bond it's "but-2-ene".

Number the chain from the end that gives lower locants to the double bond. Starting from left or right both give the double bond at position $2$ $2$. However, starting from left gives substituent locants as $1$ $1$ (for bromo) and $2$ $2$ (for methyl), whereas starting from right gives substituent locants as $3$ $3$ and $4$ $4$.
Hence, correct numbering is:
$C1$ $\text{C1}$: bonded to Bromine ( $-Br$ $-\text{Br}$)
$C2$ $\text{C2}$: bonded to Methyl ( $-CH₃$ $-\text{CH}_3$)
$C3$ $\text{C3}$: alkene carbon
$C4$ $\text{C4}$: terminal methyl group

IUPAC numbered chain diagram for Q33 — The image shows structural representation of compound X with a double bond and a bromine substituent.

Combining these rules, the name is: 1-Bromo-2-methylbut-2-ene.

Pattern Recognition

Double bond takes precedence over halogen substituent in numbering direction. If double bond is symmetrical (at position $2$ $2$ in a $4$ $4$-carbon chain), use the substituent positions to break the tie, choosing lowest possible locants ( $1$ $1$ and $2$ $2$ vs $3$ $3$ and $4$ $4$).

Chapter Mix

Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques Class 12 Chemistry: Haloalkanes and Haloarenes

Number of geometrical isomers possible for the given structure is/are The diagram shows a highly substituted polyene containing deuterium isotopes.

Solution & Explanation

Core Logic

Step 1: Calculate Number of Isomers

Pattern Recognition

Chapter Mix

More Organic Chemistry Some Basic Principles and Techniques Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Determine molar masses

Step 2: Calculate moles and find X

Step 2: Calculate moles and find X

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Categorize by substitution patterns

Step 2: Tri-substituted benzenes

Step 3: Total Count

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate dry nitrogen pressure

Step 2: Calculate moles of nitrogen gas

Step 2: Calculate moles of nitrogen gas

Step 3: Calculate percentage of Nitrogen

Step 3: Calculate percentage of Nitrogen

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Analyze Statement II

Step 2: Conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Pattern Recognition

Chapter Mix

More Organic Chemistry Some Basic Principles and Techniques Questions — jee_main_2024_30_january_evening

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Number of geometrical isomers possible for the given structure is/are
The diagram shows a highly substituted polyene containing deuterium isotopes.