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

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

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:

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



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

Related Formula

Core Logic

Given:

• Mass of organic compound taken = 500 mg = 500 × 10⁻³ g$= 500 \text{ mg} = 500 \times 10^{-3} \text{ g}$
• Mass of CO₂$\mathrm{CO}_2$ produced = 220 mg = 220 × 10⁻³ g$= 220 \text{ mg} = 220 \times 10^{-3} \text{ g}$

Using the formula:

Thus, the percentage of carbon is 12$12$.

Pattern Recognition

Carbon dioxide has exactly 12/44 ≈ 27.27%$12/44 \approx 27.27\%$ carbon by mass. Multiply the mass fraction of CO₂$\mathrm{CO}_2$ (220/500 = 0.44$220/500 = 0.44$) by 12/44$12/44$ to directly get 0.12$0.12$ or 12%$12\%$.

Evaluation Rubric / Model Answer

A perfect step-by-step conversion of organic compound mass and combustion carbon dioxide mass to obtain a precise 12$12$ percent carbon composition.

Chapter Mix

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

Core Logic

Let us apply official IUPAC priority indexing rules:

• Principal Functional Group: The hydroxyl group (-OH$-\text{OH}$) possesses higher naming priority over double bonds and simple alkyl side chains. Thus, the carbon bearing the -OH$-\text{OH}$ group is assigned position C-1.
• Numbering Direction: We must number through the ring towards the double bond to assign it the lowest possible locant. Hence, the alkene carbons are given coordinates C-2 and C-3.
• Locating Side Chains: Proceeding with this direction puts the ethyl group at position C-4.

  

Assembling the structural parts alphabetically:

• Substituent: `4-Ethyl`
• Parent root: `cyclopent-2-en`
• Suffix: `1-ol`

Combined IUPAC format: 4-Ethylcyclopent-2-en-1-ol.

Pattern Recognition

Principal suffix priority hierarchy: -OH > Double bond > Alkyl side-chain$-\text{OH} > \text{Double bond} > \text{Alkyl side-chain}$. Always fix the highest priority suffix at index 1 and head instantly towards the alkene bond to safely restrict locant numbers.

Chapter Mix

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

Core Logic

Let us analyze each term carefully:

• A. Carbocation: Features a positively charged trivalent carbon atom. It represents an sp²$sp^2$ hybridized carbon with an empty unhybridized p-orbital.

  

• B. Carbon Free Radical: Contains a trivalent carbon carrying a single unpaired lone electron. It typically exhibits sp²$sp^2$ or sp³$sp^3$ hybridization depending on structural environments.

  

• C. Nucleophile: An electron-rich chemical species containing a lone pair or negative charge capable of donating/supplying a pair of electrons.
• D. Electrophile: An electron-deficient chemical species possessing empty low-lying orbitals capable of accepting/receiving a pair of electrons.

Step 1: Alignment Matrix

Matching each item yields:

• A arrow III$\text{A} \rightarrow \text{III}$
• B arrow IV$\text{B} \rightarrow \text{IV}$
• C arrow I$\text{C} \rightarrow \text{I}$
• D arrow II$\text{D} \rightarrow \text{II}$

This sequence aligns flawlessly with Option (4).

Pattern Recognition

Nucleophiles donate ('nucleo-loving' = seeks positive sites with its electrons), Electrophiles accept ('electro-loving' = seeks electron density). Carbocations explicitly harbor a vacant p-orbital because of their positive charge configuration, making identification extremely swift.

Chapter Mix

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