In the given structure, number of sp and sp^2$sp^{2}$ hybridized carbon atoms present respectively are :

### Core Logic Organic compounds react with sodium bicarbonate (mathrmNaHCO_3$\mathrm{NaHCO}_3$) to liberate mathrmCO_2$\mathrm{CO}_2$ gas if they are stronger acids than carbonic acid (mathrmH_2mathrmCO_3$\mathrm{H}_2\mathrm{CO}_3$). 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 mathrmH_2mathrmCO_3$\mathrm{H}_2\mathrm{CO}_3$. - **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 mathrmCO_2$\mathrm{CO}_2$. 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 mathrmCO_2$\mathrm{CO}_2$ from bicarbonate ions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Let us check the statements step-by-step: * Statement A: Pentan-3-one and pentan-2-one have different alkyl groups attached on either side of the divalent polyfunctional carbonyl group (-textCO-$-\text{CO}-$). Hence, they are metamers.

* **Statement B:** Cyanides (-textCN$-\text{CN}$) and Isocyanides (-textNC$-\text{NC}$) contain distinct functional groups, so they are functional isomers.

* **Statement C:** Phenol structures containing a methyl substituent at positions 2 and 3 are structural position isomers. * **Statement D:** The given structures represent members of a homologous series because they differ sequentially by a -textCH_2-$-\text{CH}_2-$ unit. ### Step 1: Verification Evaluating according to standard multi-choice options, statements A and B are perfectly validated. ### Pattern Recognition Shortcut: Metamers require variable alkyl distribution across a polyvalent heteroatom group. Functional isomers require changes like -textCN$-\text{CN}$ vs -textNC$-\text{NC}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Let us check the conjugate bases formed upon losing a proton: * Compounds (A), (B), and (C) generate conjugate bases stabilized by resonance through the aromatic ring or strong electron-withdrawing groups. * Compound (D) represents an ethynyl group in a terminal alkyne structure (EtO_2C-Cequiv CH$EtO_2C-Cequiv CH$). Its conjugate base features a localized negative charge on an sp$sp$-hybridized carbon. Because there is no resonance stabilization present for this anion, it is significantly less stable than the conjugate bases of the other functional groups. ### Step 1: Conclusion Since a less stable conjugate base implies a weaker parent acid, the terminal alkyne compound (D) is the least acidic. ### Pattern Recognition Shortcut: A resonance-stabilized anion is always more stable than a localized one. Look for the alkyne carbon versus oxygen/aromatic-centered acids. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula Using the Principle of Atom Conservation (POAC) for Nitrogen: n_textcompound times textatoms of N per molecule = 2 times n_N_2$$n_{\text{compound}} \times \text{atoms of N per molecule} = 2 \times n_{N_2}$$ ### Core Logic The molecular weight of the given heterocyclic amine organic structure X is calculated as 86text g/mol$86\text{ g/mol}$.

Given mass of compound = 0.42text g$= 0.42\text{ g}$: textMoles of compound X = frac0.4286 $$\text{Moles of compound } X = \frac{0.42}{86} $$ ### Step 1: Calculating STP Volume Using POAC on Nitrogen atoms: n_N_2 = frac0.4286 $$ n_{N_2} = \frac{0.42}{86} $$ textVolume of N_2text at STP = n_N_2 times 22400text mL = frac0.4286 times 22400 approx 110.88text mL $$ \text{Volume of } N_2\text{ at STP} = n_{N_2} \times 22400\text{ mL} = \frac{0.42}{86} \times 22400 \approx 110.88\text{ mL} $$ Rounding to the nearest integer gives 111text mL$111\text{ mL}$. ### Pattern Recognition Shortcut: Always identify the molecular formula from the skeletal grid first. Once M = 86$M = 86$ and total textN = 2$\text{N} = 2$ atoms are established, use the stoichiometric ratio directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Related Formula The percentage of carbon via combustion details is found using: % text C = frac1244 times fractextMass of CO_2textMass of organic compound times 100$$ % \text{ C} = \frac{12}{44} \times \frac{\text{Mass of } CO_2}{\text{Mass of organic compound}} \times 100$$ ### Core Logic Let us substitute the parameters: * Mass of organic compound = 0.5text g$= 0.5\text{ g}$ * Mass of CO_2$CO_2$ collected = 1.46text g$= 1.46\text{ g}$ ### Step 1: Numerical Calculation \% text C = frac1244 times frac1.460.5 times 100$$\% \text{ C} = \frac{12}{44} \times \frac{1.46}{0.5} \times 100$$ % text C = frac12 times 1.4622 times 100 approx 79.63%$$% \text{ C} = \frac{12 \times 1.46}{22} \times 100 \approx 79.63%$$ Rounding to the nearest integer gives 80$80$. ### Pattern Recognition Shortcut: frac1244 approx 0.2727$\frac{12}{44} \approx 0.2727$. Multiply 0.2727 times 1.46$0.2727 \times 1.46$ to find the carbon mass (0.398text g$0.398\text{ g}$). Since 0.398text g$0.398\text{ g}$ out of 0.5text g$0.5\text{ g}$ is practically frac45$\frac{4}{5}$, the value is \right around 80\%$80\%$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques