Predict the major product of the following reaction sequence:

Core Logic

Let's analyze the given reaction sequences.

• Reaction with 2H₂ / Pt$2H_2 / Pt$: Compound A undergoes hydrogenation with 2 moles of hydrogen, suggesting the presence of two double bonds.
• Reaction with KMnO₄ / Δ$\mathrm{KMnO_4} / \Delta$: 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 KMnO₄$KMnO_4$ 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.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Related Formula

Nitrobenzene + Friedel Crafts Acylation arrow$\rightarrow$ No Reaction.

Core Logic

Statement I claims that nitrobenzene undergoes Friedel-Crafts acylation with CH₃COCl/AlCl₃$CH_3COCl/AlCl_3$ to yield a product. This is strictly false. The -NO₂$-NO_2$ 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 NO₂$NO_2$ 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.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Related Formula

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 Br₂$Br_2$ 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 × 2 = 6$= 3 \times 2 = 6$. So the cycloalkane (X) has 6 carbon atoms. Being a cycloalkane, its formula is C₆H₁₂$C_6H_{12}$. Wait, the solution provided states C₆H₁₀ Br₂ C₆H₁₀Br₂$C_6H_{10} \xrightarrow{Br_2} C_6H_{10}Br_2$. 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 C₆H₁₀$C_6H_{10}$ (cyclohexene) despite the text saying 'cycloalkane'. It undergoes addition of Br₂$Br_2$ to form C₆H₁₀Br₂$C_6H_{10}Br_2$. For C₆H₁₀Br₂$C_6H_{10}Br_2$: 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) C₆H₁₀Br₂$C_6H_{10}Br_2$: M = (12 × 6) + (1 × 10) + (80 × 2)$M = (12 \times 6) + (1 \times 10) + (80 \times 2)$ M = 72 + 10 + 160 = 242 g mol⁻¹$M = 72 + 10 + 160 = 242\text{ g mol}^{-1}$

Step 2: Calculate Percentage

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.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 11 Chemistry: Some Basic Concepts of Chemistry

Related Formula

Core Logic

Step 1: Dehydrohalogenation of dibromo compound using NaNH₂$\text{NaNH}_2$ yields terminal alkyne sodium acetylide intermediate (X)$(X)$.

Step 2: Nucleophilic substitution (SN2$S_N2$) 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.

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. NH₃$Na/liq. NH_3$ (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 CH₃$CH_3$ 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 (O₃ / Zn + H₂O$O_3 / Zn + H_2O$): 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.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Aldehydes Ketones and Carboxylic Acids