### 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

### Related Formula textBasicity propto textAvailability of lone pair of electrons on Nitrogen$\text{Basicity} \propto \text{Availability of lone pair of electrons on Nitrogen}$ ### 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

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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry

### Related Formula textStability propto textDelocalization of charge via resonance, mesomeric, and back-bonding effects$\text{Stability} \propto \text{Delocalization of charge via resonance, mesomeric, and back-bonding effects}$ ### 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 (-NO_2$-NO_2$) 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 -NO_2$-NO_2$ match the sign of the charge. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry

### Related Formula textPrincipal Functional Group Priority: -OH > textDouble bond (=) ge textTriple bond (equivtext)$\text{Principal Functional Group Priority: } -OH > \text{Double bond (=)} \ge \text{Triple bond (}\equiv\text{)}$ ### 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: overset7textHoverset6textCequiv overset5textC-overset4textCtextH_2-overset3textCtextH(OH)-overset2textCtextH_2-overset1textCtextH=textCH_2$\overset{7}{\text{H}}\overset{6}{\text{C}}\equiv \overset{5}{\text{C}}-\overset{4}{\text{C}}\text{H}_2-\overset{3}{\text{C}}\text{H(OH)}-\overset{2}{\text{C}}\text{H}_2-\overset{1}{\text{C}}\text{H}=\text{CH}_2$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Principles of Organic Chemistry