Arrange the following alkenes in decreasing order of stability. I.

II.

III.

IV.

Choose the correct 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 Nitrobenzene + Friedel Crafts Acylation rightarrow$\rightarrow$ No Reaction. ### Core Logic Statement I claims that nitrobenzene undergoes Friedel-Crafts acylation with CH_3COCl/AlCl_3$CH_3COCl/AlCl_3$ to yield a product. This is strictly false. The -NO_2$-NO_2$ group is highly electron-withdrawing, meaning it severely deactivates the benzene ring toward electrophilic aromatic substitution. Nitrobenzene does not undergo Friedel-Crafts alkylation or acylation. Statement II correctly states that the NO_2$NO_2$ group is a meta-directing and deactivating group. Because it withdraws electron density, the ring is less reactive than benzene itself, and any substitution that does occur (under harsh conditions) happens at the meta position. ### Step 1: Final Conclusion Statement I is incorrect but Statement II is correct. ### Pattern Recognition Highly deactivated rings (e.g., nitrobenzene) act as a solvent in Friedel-Crafts reactions precisely because they are completely unreactive to the alkylating/acylating conditions. ### 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

### Core Logic Identify the geometrical isomers formed by different hydrogenation conditions. Pd/C$Pd/C$ (or Lindlar's catalyst) provides syn-addition forming the cis-alkene (X). Na/liq. NH_3$Na/liq. NH_3$ (Birch reduction) provides anti-addition forming the trans-alkene (Y). ### Step 1: Structural Analysis X = cis-But-2-ene. Y = trans-But-2-ene.

### Step 2: Checking Statements A. X and Y are geometrical isomers, a subset of stereoisomers. (Correct statement) B. Dipole moment of X (cis) is non-zero because the dipole moments of the two CH_3$CH_3$ groups reinforce each other, whereas Y (trans) has a zero dipole moment due to symmetry cancellation. (Incorrect statement) C. Cis-isomers generally have higher boiling points than trans-isomers because they are more polar, leading to stronger intermolecular dipole-dipole forces. (Correct statement) D. Both cis and trans-But-2-ene yield the exact same product upon reductive ozonolysis (O_3 / Zn + H_2O$O_3 / Zn + H_2O$): 2 moles of ethanal (acetaldehyde). (Incorrect statement) ### Step 3: Final Conclusion Statements B and D are incorrect. ### Pattern Recognition Lindlar = cis (polar, higher BP), Birch = trans (non-polar symmetric, zero dipole). Ozonolysis breaks the double bond structurally and ignores original stereochemistry. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids