Isomeric hydrocarbons giving negative Baeyer's test have the molecular formula C_9H_12$C_{9}H_{12}$. The total number of isomers from above with exactly four different non-aliphatic substitution sites is ________.

### 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 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

### Related Formula Orienting effects in electrophilic aromatic substitution: - Bromine (-Br$-Br$) is ortho/para-directing (para-dominated due to steric hindrance). - Sulfonic acid group (-SO_3H$-SO_3H$) is a strong deactivating, meta-directing group.

### Core Logic Analyzing the first step: - Sulfonation of bromobenzene with mathrmSO_3/mathrmH_2mathrmSO_4$\mathrm{SO}_3/\mathrm{H}_2\mathrm{SO}_4$ yields 4-bromobenzenesulfonic acid as the major product (A) due to steric hindrance at the ortho-position. ### Step 1: Determine orientation for the second step In 4-bromobenzenesulfonic acid, we have two substituents: - -Br$-Br$ (ortho/para director) - -SO_3H$-SO_3H$ (meta director) The positions meta to the deactivating -SO_3H$-SO_3H$ group correspond to the positions ortho to the -Br$-Br$ group. Both directing effects align on the same position (carbon-3/carbon-5). Since -Br$-Br$ is activating relative to -SO_3H$-SO_3H$, it controls the orientation. ### Step 2: Identify Product B Halogenation with mathrmBr_2/mathrmFe$\mathrm{Br}_2/\mathrm{Fe}$ introduces a bromine atom ortho to the existing bromine atom (meta to -SO_3H$-SO_3H$): textProduct B = text3,4-dibromobenzenesulfonic acid$$\text{Product B} = \text{3,4-dibromobenzenesulfonic acid}$$ This matches Option (2). ### Pattern Recognition When an activating group (-Br$-Br$) and a deactivating group (-SO_3H$-SO_3H$) compete on a benzene ring, the orienting influence of the activating group wins. Position ortho to the bromine atom is favored over meta positions of the sulfonic acid group. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula textR_1-textCH=textCH-textR_2 xrightarrow[text(ii) Zn, H_2textO]text(i) O_3 textR_1-textCHO + textR_2-textCHO$$\text{R}_1-\text{CH}=\text{CH}-\text{R}_2 \xrightarrow[\text{(ii) Zn, H}_2\text{O}]{\text{(i) O}_3} \text{R}_1-\text{CHO} + \text{R}_2-\text{CHO}$$ ### Core Logic Statement I: Cis-2-butene (CH_3-CH=CH-CH_3$CH_3-CH=CH-CH_3$) undergoes reductive ozonolysis to cleave the double bond and yield two molecules of acetaldehyde (ethanal, CH_3CHO$CH_3CHO$). This statement is true. Statement II: 3,6-dimethyloct-4-ene (CH_3-CH_2-CH(CH_3)-CH=CH-CH(CH_3)-CH_2-CH_3$CH_3-CH_2-CH(CH_3)-CH=CH-CH(CH_3)-CH_2-CH_3$) undergoes reductive ozonolysis to yield 2-methylbutanal (CH_3-CH_2-CH(CH_3)-CHO$CH_3-CH_2-CH(CH_3)-CHO$). This product contains a chiral carbon atom (the C2$C2$ carbon bonded to -H$-H$, -CH_3$-CH_3$, -CHO$-CHO$, and -C_2H_5$-C_2H_5$). Thus, the statement that it has no chiral carbon atom is false. ### Pattern Recognition To check chirality after ozonolysis, draw the cleaved fragment structure first. Any carbon with four different groups is chiral. 2-methylbutanal has four unique groups attached to C2$C2$, making Statement II clearly false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

### Core Logic Statement I is true: In m-xylene, both methyl groups direct incoming electrophiles to position 4 (synergistic ortho/para effect). Nitration gives 4-nitro-m-xylene. Subsequent strong oxidation of both -CH_3$-CH_3$ groups yields 4-nitrobenzene-1,3-dicarboxylic acid.

Statement II is true: Alkyl groups (-CH_3$-CH_3$) act as ortho/para directors via hyperconjugation, whereas nitro groups (-NO_2$-NO_2$) are strongly deactivating meta directors. ### Pattern Recognition When evaluating multi-substituted benzenes, identify whether directors reinforce the same positions. In m-xylene, position 4 is ortho to one methyl and para to the other. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons