### 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

### Core Logic A negative Baeyer's test confirms that the hydrocarbon structural isomers contain no aliphatic alkene or alkyne unsaturations, establishing that they are purely aromatic benzene derivatives with side alkyl chains.

To find structures possessing four distinct ring positions available for electrophilic substitution, we examine the symmetries of specific tri-substituted configurations. There are exactly 2$2$ such structural isomers satisfying these spatial conditions. ### Pattern Recognition Negative test = aromatic ring constraint. Calculate positional substitution patterns meticulously to ensure ring symmetry matches the required counts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons