Given below are two statements: Statement I: Benzene is nitrated to give nitrobenzene, which on further treatment with CH₃COCl / AlCl₃$CH_{3}COCl / AlCl_{3}$ will give

Statement II: NO₂$NO_{2}$ 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 map out this complete aromatic synthesis pathway:

• Benzene + Oleum: Sulfonation steps take place to form Benzene Sulfonic Acid (C₆H₅SO₃H$C_6H_5SO_3H$, Compound X).
• Fusion with molten NaOH$NaOH$ followed by H^+$H^+$ activation: The sulfonic group is displaced, passing through a sodium phenoxide intermediate to yield Phenol (C₆H₅OH$C_6H_5OH$, Compound Y).
• Phenol + Zinc dust distillation: Phenol undergoes clean deoxygenation reduction when heated with Zinc metal, stripping the hydroxyl group away to reform Benzene (Compound Z).

Pattern Recognition

Zinc dust distillation is a highly reliable reduction tool designed explicitly to strip phenolic hydroxyl groups away, leaving a clean unsubstituted aromatic ring behind.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Alcohols, Phenols and Ethers

Core Logic

Let's analyze the steps of the reaction sequence:

• Step 1 (Br₂ / hν$Br_2 / h\nu$): Light-induced free radical substitution targeted at the most stable tertiary position, producing 1-bromo-1-methylcyclohexane.
• Step 2 (Alcoholic KOH, Δ$KOH, \Delta$): Dehydrohalogenation occurs via an E2 mechanism. Following Saytzeff's rule, elimination favors the formation of the more highly substituted, stable alkene: 1-methylcyclohexene.
• Step 3 (HBr / R-O-O-R, hν$HBr / R-O-O-R, h\nu$): Radical hydrobromination across the unsymmetrical alkene. The presence of peroxide shifts addition toward the Anti-Markovnikov path, placing the bromine atom cleanly at the less-substituted secondary carbon to yield 1-bromo-2-methylcyclohexane.

Pattern Recognition

Combining Saytzeff elimination with a peroxide-promoted HBr$HBr$ addition allows you to reposition functional groups from highly substituted tertiary carbons to adjacent secondary positions.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes

Related Formula

Core Logic

Complete oxidative cleavage of all double bonds via reductive ozonolysis breaks the molecule into smaller fragments:

Let's trace the fragmentation logic mapping visually:

Step 1: Tracking Fragment Structures

The chemical reaction outputs two major molecular product types:

• CH3-CHO$\text{CH}3\text{-CHO}$ (Acetaldehyde): Optically inactive as it lacks a chiral carbon.
• OHC-CH(CH₃)-CHO$\text{OHC-CH}(\text{CH}_3)\text{-CHO}$ (2-methylpropanedial): Let's inspect the substituted central carbon. It is bonded to: a hydrogen atom (-H$-\text{H}$), a methyl group (-CH₃$-\text{CH}_3$), and two identical formyl groups (-CHO$-\text{CHO}$).

Because two of the groups are identical (-CHO$-\text{CHO}$), this molecule does not have a chiral center and is entirely optically inactive.

Step 2: Total Summation

Since every single product formed is achiral, the number of optically active products is zero.

Pattern Recognition

Symmetry check shortcut: When a symmetrical dialkene is cleaved, it yields symmetric fragments. The central carbon is attached to identical flanking aldehyde units post-cleavage, destroying any prior asymmetry and leaving 0 optically active compounds.

Chapter Mix

Class 11 Chemistry: Hydrocarbons Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques

Related Formula

Core Logic

Let's track the molecular changes at every intermediate junction:

• Step 1: Acetophenone (Ph-CO-CH₃$\text{Ph-CO-CH}_3$) reacts with PCl₅$\text{PCl}_5$ to generate a gem-dichloride intermediate [A]: Ph-CCl₂-CH₃$\text{Ph-CCl}_2\text{-CH}_3$.
• Step 2: Reaction with 3$3$ equivalents of the incredibly strong base NaNH₂$\text{NaNH}_2$ triggers dual elimination to form a terminal sodium acetylide salt [B]: Ph-C ^-Na^+$\text{Ph-C}\equiv\text{C}^-\text{Na}^+$.
• Step 3: Acidification yields phenylacetylene [C]: Ph-C$\text{Ph-C}\equiv\text{CH}$.
• Step 4: Hydroboration-oxidation of phenylacetylene leads to anti-Markovnikov water addition forming an enol structure, which immediately tautomerizes to [D] phenylacetaldehyde: Ph-CH₂-CHO$\text{Ph-CH}_2\text{-CHO}$.

Step 1: Counting Hybridized Carbons

The step transformations match the sequential tracking map:

Let's locate all sp²$sp^2$ hybridised carbon environments in product D (Ph-CH₂-CHO$\text{Ph-CH}_2\text{-CHO}$):

• The aromatic benzene ring contains 6 sp²$sp^2$ carbon atoms.
• The aldehyde carbonyl carbon (-CHO$-\text{CHO}$) is double-bonded to oxygen, adding 1 sp²$sp^2$ carbon atom.
• The aliphatic link carbon (-CH₂-$-\text{CH}_2-$) is entirely sp³$sp^3$ hybridized.

Total sp²$sp^2$ carbon count = 6 + 1 = 7$= 6 + 1 = 7$.

Pattern Recognition

Alkyne oxidation mapping: Hydroboration-oxidation transforms a terminal alkyne into an aldehyde carbonyl group, while oxymercuration-demercuration yields a ketone carbonyl. Both introduce precisely one extra carbonyl sp²$sp^2$ site on top of the original aromatic framework.

Chapter Mix

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

Related Formula

Core Logic

Addition of H-Br(aq)$H-Br(aq)$ follows an electrophilic addition pathway where a carbocation intermediate is formed in the rate-determining step. Among the given structures, compound Q forms a resonance-stabilized allylic/benzylic carbocation, rendering it highly stable compared to the others.

Pattern Recognition

Look for conjugated or allylic systems that stabilize the positive charge dynamically via resonance.

Chapter Mix

Class 11 Chemistry: Hydrocarbons