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Organic Chemistry - Some Basic Principles and Techniques appeared 93 times across 3 years — 10.8% of Chemistry. This question is from Acidity of Organic Compounds.

Year	2026	2025	2024	Total
Questions	19	49	25	93

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The compounds that produce $CO₂$ $\mathrm{CO}_{2}$ with aqueous $NaHCO₃$ $\mathrm{NaHCO}_{3}$ solution are: A.
The prompt lists five structures labeled A through E evaluating structural acidities.
B.
The prompt lists five structures labeled A through E evaluating structural acidities.
C.
The prompt lists five structures labeled A through E evaluating structural acidities.
D.
The prompt lists five structures labeled A through E evaluating structural acidities.
E.
The prompt lists five structures labeled A through E evaluating structural acidities.
Choose the correct answer from the options given below:

A. $A and C only$ $\text{A and C only}$
B. $A, B and E only$ $\text{A, B and E only}$
C. $A, C and D only$ $\text{A, C and D only}$
D. $A and B only$ $\text{A and B only}$

Solution & Explanation

Core Logic

Organic compounds react with sodium bicarbonate ( $NaHCO₃$ $\mathrm{NaHCO}_3$) to liberate $CO₂$ $\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid ( $H₂CO₃$ $\mathrm{H}_2\mathrm{CO}_3$). Evaluating the structures:

A: Benzoic acid, which is significantly more acidic than carbonic acid.
C: Picric acid (2,4,6-trinitrophenol). Due to three strong electron-withdrawing nitro groups, its acidity exceeds typical carboxylic acids and $H₂CO₃$ $\mathrm{H}_2\mathrm{CO}_3$.
D: Benzenesulfonic acid, a highly strong mineral-like organic acid.
B & E: Standard phenols or weakly substituted phenols, which are less acidic than carbonic acid and do not liberate $CO₂$ $\mathrm{CO}_2$.

Therefore, structures A, C, and D give a positive test result.

Pattern Recognition

Sees: Sodium bicarbonate test for organic systems. Shortcut: Only carboxylic acids, sulfonic acids, and highly nitrated phenols like picric acid possess sufficient proton acidity to displace $CO₂$ $\mathrm{CO}_2$ from bicarbonate ions.

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 10

Q jee_main_2025_03_april_morning Quantitative Analysis - Estimation of Carbon

0.5 g of an organic compound on combustion gave 1.46 g of

CO₂

and 0.9 g of

H₂O

. The percentage of carbon in the compound is _____. (Nearest integer) [Given: Molar mass (in

g mol⁻¹

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

Numerical Answer. Answer: 80 to 80

Solution

Related Formula

The percentage of carbon via combustion analysis is given by:

% C = (12)/(44) \times Mass of CO₂Mass of organic compound \times 100

Core Logic

Let us substitute the given parameters:

Mass of organic compound $= 0.5 g$ $= 0.5\text{ g}$
Mass of $CO₂$ $\text{CO}_2$ collected $= 1.46 g$ $= 1.46\text{ g}$

Step 1: Numerical Calculation

% C = (12)/(44) \times (1.46)/(0.5) \times 100

% C = (12 \times 1.46)/(22) \times 100 \approx 79.64%

Rounding to the nearest integer gives 80.

Pattern Recognition

Shortcut: $(12)/(44) \approx 0.2727$ $\frac{12}{44} \approx 0.2727$. Multiply $0.2727 \times 1.46$ $0.2727 \times 1.46$ to find the mass of carbon ( $\approx 0.398 g$ $\approx 0.398\text{ g}$). Since $0.398 g$ $0.398\text{ g}$ out of $0.5 g$ $0.5\text{ g}$ is roughly $(4)/(5)$ $\frac{4}{5}$, the value is right around $80%$ $80\%$.

Evaluation Rubric / Model Answer

80

Chapter Mix

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

Q jee_main_2025_04_april_evening Basicity of Organic Bases

The correct order of basicity for the following molecules is:

Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity of Organic Bases diagram for Q26 - JEE Main 2025 Evening — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

A. $P > Q > R$ $P > Q > R$
B. $R > P > Q$ $R > P > Q$
C. $Q > P > R$ $Q > P > R$
D. $R > Q > P$ $R > Q > P$

Solution

Related Formula

Basicity \propto Availability of lone pair of electrons on Nitrogen

Core Logic

Analyzing the molecules:

In molecule (R), according to Bredt's rule, the bridgehead nitrogen has a localized lone pair which cannot participate in resonance. Thus, it is highly available and most basic.
In molecule (Q), the nitrogen lone pair is involved in cross-conjugation with the carbonyl group, reducing its availability.
In molecule (P), the lone pair on nitrogen is directly conjugated with the carbonyl group (amide resonance), making it the least available.

Therefore, the correct basicity order is: $R > Q > P$ $R > Q > P$

Step 1: Final Identification

Basicity breakdown explanation diagram for Q26 — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity breakdown explanation diagram for Q26 — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Basicity breakdown explanation diagram for Q26 — The diagram displays three nitrogen-containing organic structures labeled P, Q, and R for basicity comparison.

Comparing availability, structure R has localized electrons, Q has cross-conjugation, and P has standard amide resonance. Hence, option (4) is correct.

Pattern Recognition

Look for localized vs delocalized lone pairs on nitrogen. Bridgehead nitrogen lone pairs that violate Bredt's rule for double bond formation remain strictly localized, drastically increasing basicity compared to conjugated amides.

Chapter Mix

Class 11 Chemistry: Some Basic Principles of Organic Chemistry

Q jee_main_2025_04_april_evening Carbocation and Carbanion Stability

In which pairs, the first ion is more stable than the second?

Stability pair A components for Q30 — The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).

Stability pair A components for Q30 — The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).

Stability pair A components for Q30 — The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).

Stability pair A components for Q30 — The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).

Stability pair A components for Q30 — The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).

Stability pair A components for Q30 — The prompt displays visual comparisons of organic ionic species to judge the relative stability within pairs (A), (B), (C), and (D).

A. (B) & (D) only
B. (A) & (B) only
C. (B) & (C) only
D. (A) & (C) only

Solution

Related Formula

Stability \propto Delocalization of charge via resonance, mesomeric, and back-bonding effects

Core Logic

Evaluating each specific structural pair:

Pair (A): The first carbocation is stabilized by strong $+M$ $+M$ back-bonding from the methoxy ( $-OMe$ $-OMe$) oxygen lone pair, making it significantly more stable than the second.
Pair (B): The first carbanion is strongly stabilized by the $-M$ $-M$ and $-I$ $-I$ electronic effects of the nitro ( $-NO₂$ $-NO_2$) group situated at the ortho position, whereas a carbocation in that spot would be destabilized. Thus, the first ion is more stable.
Pair (C): The second cation has extended allylic resonance stabilization, meaning the first is less stable.
Pair (D): Cation with $-OMe$ $-OMe$ backbonding is more stable than tertiary aliphatic carbocation, making the first less stable than the second.

Thus, only in (A) & (B) is the first ion more stable than the second.

Pattern Recognition

Back-bonding from an adjacent oxygen lone pair always triumphs over standard inductive or hyperconjugative alkyl stability templates. For carbanions, ensure electron-withdrawing groups like $-NO₂$ $-NO_2$ match the sign of the charge.

Chapter Mix

Class 11 Chemistry: Some Basic Principles of Organic Chemistry

Q jee_main_2025_04_april_evening Organic Reactions and Mechanisms

Consider the following molecule (X).

Organic reactant hydrocarbon layout for Q43 — The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.

The structure of X is

Organic reactant hydrocarbon layout for Q43 — The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.

Organic reactant hydrocarbon layout for Q43 — The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.

Organic reactant hydrocarbon layout for Q43 — The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.

Organic reactant hydrocarbon layout for Q43 — The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.

Organic reactant hydrocarbon layout for Q43 — The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.

A. Structure (1)
B. Structure (2)
C. Structure (3)
D. Structure (4)

Solution

Core Logic

Analyzing the mechanism:

Protonation ( $H^+$ $H^+$ attack) on the double bond occurs to generate the most stable intermediate carbocation.
The system forms a highly stable tertiary ( $3^°$ $3^\circ$) carbocation at the bridge junction ring site.
Nucleophilic attack by bromide ion ( $Br^-$ $Br^-$) captures this bridge junction center selectively, yielding the major brominated structure designated as Structure (2).

Step 1: Mechanism Breakdown

Carbocation intermediate step illustration for Q43 — The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.

Carbocation intermediate step illustration for Q43 — The prompt visually depicts a polycyclic unsaturated hydrocarbon system reacting to form compound X.

Electrophilic addition follows Markovnikov-like stability rules, directing the halide specifically to the bridgehead junction carbocation position.

Pattern Recognition

Whenever adding hydrohalic acids across cyclic non-conjugated dienes or isolated double bonds, always convert the alkene into the most stable, un-strained tertiary carbocation before attaching the nucleophile.

Chapter Mix

Class 11 Chemistry: Some Basic Principles of Organic Chemistry Class 11 Chemistry: Hydrocarbons

Q33 jee_main_2025_04_april_evening IUPAC Nomenclature

The IUPAC name of the following compound is -

arrayc O H H C \equiv C - C H _ 2 - C H - C H _ 2 - C H = C H _ 2 array

A. 4-Hydroxyhept-1-en-6-yne
B. 4-Hydroxyhept-6-en-1-yne
C. Hept-6-en-1-yn-4-ol
D. Hept-1-en-6-yn-4-ol

Solution

Related Formula

Principal Functional Group Priority: -OH > Double bond (=) \geq Triple bond (\equiv)

Core Logic

Number the seven-carbon parent chain from the side that gives the principal functional group ( $-OH$ $-OH$) and double bond the lowest locants:

Numbering from right-to-left gives locants: alkene at position 1, alcohol at 4, and alkyne at 6.
Numbering from left-to-right gives locants: alkyne at 1, alcohol at 4, alkene at 6.

According to IUPAC rules, when choices are symmetric for the principal group, the lower locant is given to the double bond over the triple bond. Hence, right-to-left numbering is correct:

7H 6C\equiv 5C- 4CH₂- 3CH(OH)- 2CH₂- 1CH=CH₂

This gives: Hept-1-en-6-yn-4-ol.

Pattern Recognition

When terminal unsaturations are tied symmetrically (positions 1 and 6), the alkene ('en') takes prefix allocation priority over the alkyne ('yn'). The secondary alcohol suffix '-ol' forms the principal name ending.

Chapter Mix

Class 11 Chemistry: Some Basic Principles of Organic Chemistry

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