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Organic Chemistry - Some Basic Principles and Techniques appeared 101 times across 3 years — 11.7% of Chemistry. This question is from Stability of Carbanions.

Year	2026	2025	2024	Total
Questions	22	49	30	101

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CORRECT order of stability for the following is $CH₂=CH⁻$ $CH_{2}=CH^{-}$, $CH₃-CH₂⁻$ $CH_{3}-CH_{2}^{-}$, $CH\equiv C⁻$ $CH\equiv C^{-}$

A. $CH₃-CH₂⁻>CH₂=CH⁻>CH\equiv C⁻$
B. $CH₂=CH⁻>CH\equiv C⁻>CH₃-CH₂⁻$
C. $CH\equiv C⁻>CH₂=CH⁻>CH₃-CH₂⁻$
D. $CH\equiv C⁻>CH₃-CH₂⁻>CH₂=CH⁻$

Solution & Explanation

Core Logic

The stability of a carbanion is directly proportional to the electronegativity of the carbon atom bearing the negative charge. The electronegativity of carbon increases with the increase in s-character of its hybridization state.

Step 1: Hybridization Analysis

$CH\equiv C⁻$ $\bullet \; CH\equiv C^{-}$ (sp hybridized, $50%$ $50\%$ s-character) $arrow$ $\rightarrow$ Highest electronegativity, most stable. $CH₂=CH⁻$ $\bullet \; CH_{2}=CH^{-}$ (sp $²$ $^{2}$ hybridized, $33.3%$ $33.3\%$ s-character) $arrow$ $\rightarrow$ Intermediate electronegativity. $CH₃-CH₂⁻$ $\bullet \; CH_{3}-CH_{2}^{-}$ (sp $³$ $^{3}$ hybridized, $25%$ $25\%$ s-character) $arrow$ $\rightarrow$ Lowest electronegativity, least stable.

Final Conclusion

Order of stability: $CH\equiv C⁻ > CH₂=CH⁻ > CH₃-CH₂⁻$

Pattern Recognition

Stability of carbanion $\propto % s-character$ $\propto \% \; s\text{-character}$. More s-character pulls the electron pair closer to the nucleus, stabilizing the negative charge.

Chapter Mix

Class 11 Chemistry: Organic Chemistry Some Basic Principles and Techniques

Reference Study Guides

Organic Chemistry Some Basic Principles And Techniques

More Organic Chemistry Some Basic Principles and Techniques Previous-Year Questions — Page 5

Q60 jee_main_2026_28_january_evening Hyperconjugation

The cyclic cations having the same number of hyperconjugation are : A.

Hyperconjugation

B.

Hyperconjugation

C.

Hyperconjugation

D.

Hyperconjugation

Choose the correct answer from the options given below :

A. $(1) A and C Only$ $(1)\text{ A and C Only}$
B. $(2) B and C Only$ $(2)\text{ B and C Only}$
C. $(3) A and B Only$ $(3)\text{ A and B Only}$
D. $(4) A, C and D only$ $(4)\text{ A, C and D only}$

Solution

Core Logic

Count the number of alpha hydrogens ( $α-H$ $\alpha-H$) adjacent to the carbocation in each structure. (A)

Hyperconjugation

α-H = 6

(B)

Hyperconjugation

α-H = 7

(C)

Hyperconjugation

α-H = 6

(D)

Hyperconjugation

α-H = 5

Step 1: Final Conclusion

Both cations (A) and (C) have 6 $α$ $\alpha$-hydrogens, meaning they share the same number of hyperconjugative structures.

Pattern Recognition

Hyperconjugation count corresponds strictly to the number of $C-H$ $C-H$ bonds on the carbon atoms immediately adjacent ( $sp³$ $sp^3$) to the positive center.

Chapter Mix

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

Q70 jee_main_2026_28_january_evening Quantitative Analysis Of Elements

A student has been given

0.314 g

of an organic compound and asked to estimate Sulphur. During the experiment, the student has obtained

0.4813 g

of barium sulphate. The percentage of sulphur present in the compound is (Given Molar mass in

g mol⁻¹

S:32

,

BaSO₄:233

)

A. $(1) 42.10%$ $(1)\text{ 42.10\%}$
B. $(2) 63.15%$ $(2)\text{ 63.15\%}$
C. $(3) 21.05%$ $(3)\text{ 21.05\%}$
D. $(4) 48.24%$ $(4)\text{ 48.24\%}$

Solution

Related Formula

% of S = (32)/(233) \times Mass of BaSO₄Mass of organic compound \times 100

Core Logic

Mass of organic compound = $0.314 g$ $0.314\text{ g}$ Mass of $BaSO₄$ $BaSO_4$ formed = $0.4813 g$ $0.4813\text{ g}$ Substituting the values:

% S = (32)/(233) \times (0.4813)/(0.314) \times 100

% S = 0.1373 \times 1.5328 \times 100 \approx 21.052%

Step 1: Final Conclusion

The percentage of sulphur is approximately $21.05%$ $21.05\%$.

Pattern Recognition

Straight application of Carius method for Sulphur estimation formula.

Chapter Mix

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

Q jee_main_2025_02_april_evening Hybridization in Organic Compounds

In 3, 3-dimethylhex-1-ene-4-yne, there are

sp³

,

sp²

and

sp

hybridised carbon atoms respectively:

A. $4, 2, 2$ $4, 2, 2$
B. $3, 3, 2$ $3, 3, 2$
C. $2, 4, 2$ $2, 4, 2$
D. $2, 2, 4$ $2, 2, 4$

Solution

Related Formula

Hybridization of Carbon = cases sp³ & 4~σ-bonds sp² & 3~σ-bonds, 1~π-bond sp & 2~σ-bonds, 2~π-bonds cases

Core Logic

Let's first draw the structural formula of 3,3-dimethylhex-1-ene-4-yne:

Hybridization in Organic Compounds

6CH₃ - 5C \equiv 4C - 3C(CH₃)₂ - 2CH = 1CH₂

Step 1: Identify Hybridization of Each Carbon

We count the sigma ( $σ$ $\sigma$) bonds or pi ( $π$ $\pi$) bonds on each carbon:

$C₁$ $\mathrm{C}_1$: Involved in a double bond ( $CH₂ =$ $\mathrm{CH_2 =}$) $sp²$ $\implies \mathrm{sp^2}$
$C₂$ $\mathrm{C}_2$: Involved in a double bond ( $=CH-$ $=\mathrm{CH-}$) $sp²$ $\implies \mathrm{sp^2}$
$C₃$ $\mathrm{C}_3$: Single bonds only (bonded to $C₂$ $\mathrm{C}_2$, $C₄$ $\mathrm{C}_4$, and two methyl carbons) $sp³$ $\implies \mathrm{sp^3}$
Two methyl carbons attached to $C₃$ $\mathrm{C}_3$: Single bonds only $2 \times sp³$ $\implies 2 \times \mathrm{sp^3}$
$C₄$ $\mathrm{C}_4$: Involved in a triple bond ( $-C \equiv$ $-\mathrm{C} \equiv$) $sp$ $\implies \mathrm{sp}$
$C₅$ $\mathrm{C}_5$: Involved in a triple bond ( $\equiv C-$ $\equiv \mathrm{C}-$) $sp$ $\implies \mathrm{sp}$
$C₆$ $\mathrm{C}_6$: Single bonds only ( $-CH₃$ $-\mathrm{CH_3}$) $sp³$ $\implies \mathrm{sp^3}$

Step 2: Calculate the Count

Summing the hybridization counts:

$sp³$ $\mathrm{sp^3}$ carbons: $C₃$ $\mathrm{C}_3$, $C₆$ $\mathrm{C}_6$, and $2 \times CH₃$ $2 \times \mathrm{CH_3}$ on $C₃$ $\mathrm{C}_3$ $= 4$ $= 4$ carbon atoms
$sp²$ $\mathrm{sp^2}$ carbons: $C₁$ $\mathrm{C}_1$ and $C₂$ $\mathrm{C}_2$ $= 2$ $= 2$ carbon atoms
$sp$ $\mathrm{sp}$ carbons: $C₄$ $\mathrm{C}_4$ and $C₅$ $\mathrm{C}_5$ $= 2$ $= 2$ carbon atoms

Thus, the number of $sp³$ $\mathrm{sp^3}$, $sp²$ $\mathrm{sp^2}$ and $sp$ $\mathrm{sp}$ hybridized carbons is $4, 2, 2$ $4, 2, 2$ respectively.

Pattern Recognition

Quick Tip: Always draw side substituents (like methyl groups) explicitly. A common mistake is to skip counting the methyl substituent carbons as $sp³$ $\mathrm{sp^3}$.

Chapter Mix

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

Q jee_main_2025_02_april_evening Methods of Purification of Organic Compounds

Match List-I with List-II:

array|l|l| arrayc List-I (Purification technique) array & arrayc List-II (Mixture of organic compounds) array (A) Distillation (simple) & (I) Diesel + Petrol (B) Fractional distillation & (II) Aniline + Water (C) Distillation under reduced pressure & (III) Chloroform + Aniline (D) Steam distillation & (IV) Glycerol + Spent-lye array

Choose the correct answer from the options given below:

A. $(A)-(II), (B)-(III), (C)-(IV), (D)-(I)$
B. $(A)-(II), (B)-(IV), (C)-(I), (D)-(III)$
C. $(A)-(III), (B)-(IV), (C)-(II), (D)-(I)$
D. $(A)-(III), (B)-(I), (C)-(IV), (D)-(II)$

Solution

Related Formula

Method Selection = f( Δ Tb, decomposition threshold, volatility with steam )

Core Logic

Let's align each purification technique to its designated mixtures based on NCERT guidelines:

(A) Simple distillation: Used for liquids having a significant difference in their boiling points ( $>30~K$ $>30~\mathrm{K}$ or $30^$ $30^\circ\mathrm{C}$). Chloroform (b.p. $334~K$ $334~\mathrm{K}$) and aniline (b.p. $457~K$ $457~\mathrm{K}$) are separated easily using simple distillation $arrow$ $\rightarrow$ (III).
(B) Fractional distillation: Used if boiling point differences of the components are very close (less than $25~K$ $25~\mathrm{K}$). Separation of petrochemical fractions such as diesel and petrol uses this technique $arrow$ $\rightarrow$ (I).
(C) Distillation under reduced pressure: Used for liquids that tend to decompose at or below their normal boiling points. Glycerol is separated from spent-lye in soap manufacturing industry using this vacuum method to prevent glycerol decomposition $arrow$ $\rightarrow$ (IV).
(D) Steam distillation: Applied to substances which are steam-volatile and completely immiscible in water. Aniline and water are separated using this technique $arrow$ $\rightarrow$ (II).

Step 1: Conclusion

Thus, the correct match is: (A)-(III), (B)-(I), (C)-(IV), (D)-(II) This corresponds perfectly to option (4).

Pattern Recognition

Glycerol from spent-lye is a highly tested practical chemistry concept. Remember that vacuum distillation lowers the boiling point, permitting evaporation without decomposition.

Chapter Mix

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

Q jee_main_2025_02_april_evening Quantitative Estimation (Dumas Method)

In Dumas' method for estimation of nitrogen, 0.5 gram of an organic compound gave

60~mL

of nitrogen collected at

300K

temperature and

715~mmHg

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

300K = 15~mmHg

) is

A. 1.257
B. 20.87
C. 18.67
D. 12.57

Solution

Related Formula

pN₂ = ptotal - paq

nN₂ = pN₂ VR T

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

Core Logic

Dumas' method estimates nitrogen by collecting dry nitrogen gas ( $N₂$ $N_2$). We must subtract the aqueous tension (vapor pressure of water) to find the pressure exerted solely by the dry nitrogen gas.

Step 1: Calculate Pressure of Dry Nitrogen

pN₂ = 715~mmHg - 15~mmHg = 700~mmHg

Converting pressure to atmospheres:

pN₂ = (700)/(760)~atm

Step 2: Calculate Moles of Nitrogen Gas

Using the ideal gas law with $R = 0.0821~ L~atm~mol⁻¹~K⁻¹$ , $T = 300~K$ $T = 300~\mathrm{K}$, and $V = 60~mL = 60 \times 10⁻³~L$ :

nN₂ = ((700)/(760)) \times 60 \times 10⁻³0.0821 \times 300

nN₂ = (0.92105 \times 0.060)/(24.63) \approx 2.244 \times 10⁻³~mol

Step 3: Calculate Mass and Percentage of Nitrogen

The molar mass of $N₂$ $\mathrm{N_2}$ is $28~ g~mol⁻¹$ $28~\mathrm{g~mol^{-1}}$:

Mass of N₂ = nN₂ \times 28 = 2.244 \times 10⁻³ \times 28 \approx 0.06283~g

Now find the percentage in $0.5~g$ $0.5~\mathrm{g}$ of organic compound:

% N = 0.06283~g0.5~g \times 100 = 12.566% \approx 12.57%

Pattern Recognition

Watch out! Always subtract the aqueous tension from the wet gas pressure first to find the dry gas pressure. Forgetting this step is the most common source of error in Dumas calculations.

Chapter Mix

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

More Organic Chemistry Some Basic Principles and Techniques Questions — jee_main_2026_28_january_morning

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