Given below are two statements: Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with CH_3COCl / AlCl_3$CH_{3}COCl / AlCl_{3}$ will give

Statement II: NO_2$NO_{2}$ group is a m-directing, and deactivating group. In the light of the above statements, choose the most appropriate answer from the options given below.

### Core Logic Let's analyze the given reaction sequences. 1. Reaction with 2H_2 / Pt$2H_2 / Pt$: Compound A undergoes hydrogenation with 2 moles of hydrogen, suggesting the presence of two double bonds. 2. Reaction with mathrmKMnO_4 / Delta$\mathrm{KMnO_4} / \Delta$: Compound A undergoes oxidative cleavage to give a substituted dicarboxylic acid (cyclohexane-1,2-dicarboxylic acid) and oxalic acid (HOOC-COOH$HOOC-COOH$). The structure of A must have a diene system that, when cleaved completely at the double bonds, yields these specific acid fragments.

By matching the cleavage points, compound A is identified as 1,2,3,4,4a,5,8,8a-octahydronaphthalene derivative with two isolated double bonds in the ring.

### Pattern Recognition Hot KMnO_4$KMnO_4$ cleaves double bonds entirely, converting =C-H$=C-H$ into -COOH$-COOH$. The formation of oxalic acid indicates a -CH=CH-$-CH=CH-$ fragment caught between two cleavable double bonds in a ring system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons

### Related Formula textMass percentage of Br = fractextMass of BrtextMolar mass of compound times 100$$\text{Mass percentage of Br} = \frac{\text{Mass of Br}}{\text{Molar mass of compound}} \times 100$$ ### Core Logic The given compound is a cycloalkane. Since it consumes one mole of bromine and undergoes addition/substitution giving a C:Br ratio of 3:1, let's analyze the formula. Bromination of an alkane normally requires UV light and is a substitution reaction, yielding mono or dibromo products. But the problem says it "consumes one mole of bromine" to give a product. If it's a cyclopropane or cyclobutane ring, it might undergo ring-opening addition. Let's assume the product has 2 bromine atoms since Br_2$Br_2$ is consumed entirely per mole. If Br = 2$Br = 2$, then from the C:Br = 3:1$C:Br = 3:1$ ratio, Carbon atoms = 3 times 2 = 6$= 3 \times 2 = 6$. So the cycloalkane (X) has 6 carbon atoms. Being a cycloalkane, its formula is C_6H_12$C_6H_{12}$. Wait, the solution provided states C_6H_10 xrightarrowBr_2 C_6H_10Br_2$C_6H_{10} \xrightarrow{Br_2} C_6H_{10}Br_2$. This implies X is cyclohexene (a cycloalkene), not a cycloalkane! Since we must follow the PDF's logic precisely: The PDF assumes X$X$ is C_6H_10$C_6H_{10}$ (cyclohexene) despite the text saying 'cycloalkane'. It undergoes addition of Br_2$Br_2$ to form C_6H_10Br_2$C_6H_{10}Br_2$. For C_6H_10Br_2$C_6H_{10}Br_2$: Number of C = 6, Number of Br = 2. C:Br ratio = 6:2 = 3:1. This perfectly fits the given condition. ### Step 1: Calculate Molar Mass Molar mass of product (Y) C_6H_10Br_2$C_6H_{10}Br_2$: M = (12 times 6) + (1 times 10) + (80 times 2)$M = (12 \times 6) + (1 \times 10) + (80 \times 2)$ M = 72 + 10 + 160 = 242text g mol^-1$M = 72 + 10 + 160 = 242\text{ g mol}^{-1}$ ### Step 2: Calculate Percentage \% text of Br = frac160242 times 100 approx 66.11\%$$\% \text{ of Br} = \frac{160}{242} \times 100 \approx 66.11\%$$ ### Step 3: Rounding Nearest integer is 66. ### Pattern Recognition Be prepared to spot exam typos (like 'cycloalkane' instead of 'cycloalkene') when a fixed atomic ratio constraint immediately points to a specific molecular formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula textVicinal Dibromide xrightarrow2textNaNH_2 textTerminal Alkyne (X) xrightarrow[textalkyl halide]textNaNH_2 textInternal Alkyne (Y)$$\text{Vicinal Dibromide} \xrightarrow{2\text{NaNH}_2} \text{Terminal Alkyne} (X) \xrightarrow[\text{alkyl halide}]{\text{NaNH}_2} \text{Internal Alkyne} (Y)$$ ### Core Logic Step 1: Dehydrohalogenation of dibromo compound using textNaNH_2$\text{NaNH}_2$ yields terminal alkyne sodium acetylide intermediate (X)$(X)$. Step 2: Nucleophilic substitution (S_N2$S_N2$) of acetylide ion with isopropyl bromide yields 2-methylhex-3-yne as final product (Y)$(Y)$.

### Pattern Recognition Sees: Double elimination followed by alkylation of acetylide ion. Shortcut: Count main chain carbon skeleton including the added isopropyl fragment to find IUPAC name: 2-methylhex-3-yne. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula textSymmetric Structure propto textMelting Point (Packing efficiency)$$\text{Symmetric Structure} \propto \text{Melting Point (Packing efficiency)}$$ ### Core Logic **Statement (I) is correct**: Neopentane (2,2-dimethylpropane) contains a quaternary carbon bonded to four methyl groups. There are 12 hydrogen atoms, all of which are primary and chemically equivalent. Halogenation yields exactly one monosubstituted derivative: mathrm(CH_3)_4C + X_2 xrightarrowhnu (CH_3)_3C-CH_2X + HX$$\mathrm{(CH_3)_4C + X_2 \xrightarrow{h\nu} (CH_3)_3C-CH_2X + HX}$$ Statement (II) is correct: Neopentane has a compact, symmetrical, nearly spherical molecular structure. In the solid crystal lattice, these spherical molecules pack much more efficiently compared to the floppy, linear n-pentane. This robust crystalline packing dramatically increases its melting point (256.4~mathrmK$256.4~\mathrm{K}$) compared to that of n-pentane (143.4~mathrmK$143.4~\mathrm{K}$). ### Step 1: Final Verification Since both statements are theoretically and experimentally correct, the correct option is (2).

### Pattern Recognition While branching *decreases* the boiling point (due to decreased surface area and weaker van der Waals forces), branching that creates a highly symmetric structure *increases* the melting point because of close-packing efficiency in the solid phase. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons