Given below are two statements: Statement (I): In partition chromatography, stationary phase is thin film of liquid present in the inert support. Statement (II): In paper chromatography, the material of paper acts as a stationary phase. In the light of the above statements, choose the correct answer from the options given below:

Related Formula

The percentage of carbon via combustion analysis is given by:

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

Rounding to the nearest integer gives 80.

Pattern Recognition

Shortcut: (12)/(44) ≈ 0.2727$\frac{12}{44} \approx 0.2727$. Multiply 0.2727 × 1.46$0.2727 \times 1.46$ to find the mass of carbon (≈ 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

Related Formula

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

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

Related Formula

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

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

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

Related Formula

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:

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