Given below are two statements I and II. Statement I: Dumas method is used for estimation of "Nitrogen" in an organic compound. Statement II: Dumas method involves the formation of ammonium sulphate by heating the organic compound with conc mathrmH_2mathrmSO_4$\mathrm{H}_2\mathrm{SO}_4$ In the light of the above statements, choose the correct answer from the options given below

### Related Formula Direct phase transition without passing through an intermediate liquid phase defines sublimation: textSolid rightleftharpoons textVapour$$\text{Solid} \rightleftharpoons \text{Vapour}$$ ### Core Logic The schematic diagram represents a solid turning directly into vapor on heating, which then reverts back to a solid phase upon cooling. This distinct behavior isolates sublimable solids from non-sublimable impurities. ### Step 1: Identification This transformation perfectly defines the laboratory purification process known as **Sublimation**. ### Pattern Recognition Look for the skipping of the liquid state entirely: Solid rightarrow$\rightarrow$ Vapour rightarrow$\rightarrow$ Solid. Common examples include camphor, naphthalene, benzoic acid, and iodine. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula Isomers share an identical molecular formula but differ in structural arrangement or functional groups: textSame M_f neq textSame structural layout$$\text{Same } M_f \neq \text{Same structural layout}$$ ### Core Logic Evaluating the structural parameters: - **Statement I**: Oxacyclobutane (a cyclic ether) and prop-2-en-1-ol (an unsaturated alcohol) both possess the molecular formula C_3H_6O$C_3H_6O$. They are functional/ring-chain isomers, so Statement I is true. - **Statement II**: Propan-1-amine (1^circ$1^\circ$ amine) and N-methylethanamine (2^circ$2^\circ$ amine) both share the molecular formula C_3H_9N$C_3H_9N$. Because primary, secondary, and tertiary amines contain different functional groups, they act as functional group isomers. Thus, Statement II is true. ### Step 1: Conclusion Match Since both structural statements are valid, both Statement I and Statement II are true.

### Pattern Recognition Always remember that 1^circ$1^\circ$, 2^circ$2^\circ$, and 3^circ$3^\circ$ amines are classified as *different functional groups* in IUPAC nomenclature. Consequently, structural shifts between them with a constant carbon count represent functional group isomerism. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula Nucleophiles are electron-rich species containing lone pairs of electrons or pi$\pi$-bonds that can donate an electron pair to an electrophilic center. ### Core Logic Let us examine each species: * mathrmNH_3$\mathrm{NH}_3$: Contains a lone pair on nitrogen rightarrow$\rightarrow$ Nucleophile * mathrmPhSH$\mathrm{PhSH}$: Contains lone pairs on sulfur rightarrow$\rightarrow$ Nucleophile * (mathrmH_3mathrmC)_2mathrmS$(\mathrm{H}_3\mathrm{C})_2\mathrm{S}$: Contains lone pairs on sulfur rightarrow$\rightarrow$ Nucleophile * mathrmH_2mathrmC=mathrmCH_2$\mathrm{H}_2\mathrm{C}=\mathrm{CH}_2$: Contains a nucleophilic pi$\pi$-bond rightarrow$\rightarrow$ Nucleophile * stackrelominusmathrmOmathrmH$\stackrel{\ominus}{\mathrm{O}}\mathrm{H}$: Negatively charged with lone pairs rightarrow$\rightarrow$ Nucleophile * mathrmH_3mathrmO^oplus$\mathrm{H}_3\mathrm{O}^{\oplus}$: Electron deficient, positively charged oxygen cannot donate electrons rightarrow$\rightarrow$ Electrophile * (mathrmCH_3)_2mathrmCO$(\mathrm{CH}_3)_2\mathrm{CO}$: Carbonyl carbon is electrophilic * >=mathrmNCH_3$>=\mathrm{NCH}_3$: Imine carbon is electrophilic Thus, the total number of nucleophiles is 5. ### Pattern Recognition Neutral molecules with lone pairs (N, S) or alkenes/alkynes with available pi$\pi$-electrons operate as good nucleophiles, along with full anions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### 1. RELATED FORMULA IUPAC Rules select the longest principal carbon chain and number it to give substituents the lowest possible locants. ### 2. EXECUTION **CORE LOGIC** Let's systematic decode each name : * (A) Longest chain contains 7 carbons (heptane) with an ethyl group at position 3 and a methyl group at position 5 rightarrow$\rightarrow$ 3-Ethyl-5-methylheptane (II) . * (B) C expanded yields a 7 carbon main chain with two methyl groups at carbon-4 rightarrow$\rightarrow$ 4,4-Dimethylheptane (III) . * (C) 5-carbon diene numbered from the left double bond side rightarrow$\rightarrow$ 2-Methyl-1,3-pentadiene (IV) . * (D) 5-carbon alkene starting from the double bond end rightarrow$\rightarrow$ 4-Methylpent-1-ene (I) . Therefore, matching sequence: (A)-(II), (B)-(III), (C)-(IV), (D)-(I). ### 3. PATTERN RECOGNITION Expanding compressed groupings such as mathrm(C_3H_7)_2$\mathrm{(C_3H_7)_2}$ prevents errors regarding parent chain carbon counts. ### 4. EVALUATION RUBRIC / MODEL ANSWER null ### 5. CHAPTER MIX Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques