Identify product A and product B :

Core Logic

The given reaction is an intramolecular Friedel-Crafts alkylation.

• The alkyl chloride reacts with anhydrous AlCl₃$AlCl_3$ to form a carbocation intermediate.
• The carbocation generated will act as an electrophile and attack the adjacent phenyl ring.
• The intermediate carbocation will undergo electrophilic aromatic substitution to form a new six-membered ring, as a 6-membered ring is highly stable and preferred over other ring sizes.

Pattern Recognition

Intramolecular Friedel-Crafts usually prefers forming 5 or 6 membered rings due to lesser angle strain. Here, the tether length is perfect for closing into a 6-membered tetralin-like system.

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Core Logic

Addition of D-Cl$D-Cl$ across the double bond takes place via an electrophilic addition mechanism following Markovnikov's rule.

• The electrophile D^+$D^+$ attacks the double bond to generate the most stable carbocation. The tertiary carbocation formed at the methyl-substituted carbon is more stable than the secondary carbocation.
• The nucleophile Cl^-$Cl^-$ then attacks the planar tertiary carbocation from either face (top or bottom), resulting in a racemic mixture if a new chiral center is fully free, but here the stereochemistry depends on the relative anti/syn addition logic or thermodynamic stability. A mixture of diastereomers can be formed, but standard electrophilic additions often yield predominantly the trans-product due to steric reasons or via a bridged intermediate depending on conditions, though pure HCl/DCl$HCl/DCl$ addition is non-stereospecific.

However, based on the official answer key (Option 3), we identify the correct stereochemical representation provided by the examining body.

Note on Discrepancy

According to the official NTA key, Option 3 is correct. According to our experts, both options 3 and 4 can be formed as a mixture since the carbocation is planar and Cl^-$Cl^-$ can attack from both sides. We proceed with the officially accepted answer (3).

Chapter Mix

Class 11 Chemistry: Hydrocarbons

Core Logic

Electrophilic substitution reactivity depends on the electron density of the aromatic ring, which is influenced by the inductive (I$I$) and mesomeric (M$M$) effects of the substituents.

• Compound A (Benzene): Standard reference.
• Compound B (Toluene): The -CH₃$-CH_3$ group shows +I$+I$ and hyperconjugation effects, activating the ring. Most reactive.
• Compound C (Chlorobenzene): The -Cl$-Cl$ group shows +M$+M$ and -I$-I$ effects, but the -I$-I$ effect dominates, mildly deactivating the ring.
• Compound D (Nitrobenzene): The -NO₂$-NO_2$ group shows strong -M$-M$ and -I$-I$ effects, heavily deactivating the ring. Least reactive.

Order of reactivity: Toluene (B) > Benzene (A) > Chlorobenzene (C) > Nitrobenzene (D).

Step 1: Final Order

Reactivity: B > A > C > D$B > A > C > D$. This matches option (4).

Chapter Mix

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

Core Logic

The structure of 2-methylbutane is CH₃-CH(CH₃)-CH₂-CH₃$CH_3-CH(CH_3)-CH_2-CH_3$. It has four different types of hydrogen atoms, which can be substituted to form structural isomers:

• 1-chloro-2-methylbutane: Chlorination at terminal CH₃$CH_3$ near branch. Yields a chiral center at C2 (2 enantiomers).
• 2-chloro-2-methylbutane: Chlorination at the tertiary carbon (1 achiral product).
• 2-chloro-3-methylbutane: Chlorination at the CH₂$CH_2$ group. Yields a chiral center at C2 (2 enantiomers).
• 1-chloro-3-methylbutane: Chlorination at the far terminal CH₃$CH_3$. No chiral center (1 achiral product).

Step 1: Counting Stereoisomers

Total isomeric products = 2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$.

Pattern Recognition

When asked for "isomeric products" in halogenation without specifying "structural isomers", you must count stereoisomers (enantiomers) as well.

Chapter Mix

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

Core Logic

Kolbe's electrolytic method generates free radicals at the anode, which then combine to form alkanes.

The given mixture yields two types of carboxylate radicals which decarboxylate to form alkyl radicals: CH₃COONa arrow CH₃$CH_3COONa \rightarrow \dot{C}H_3$ C₂H₅COONa arrow C₂H₅$C_2H_5COONa \rightarrow \dot{C}_2H_5$

These radicals can couple in three different ways:

• Cross coupling: CH₃ + C₂H₅ arrow CH₃-CH₂-CH₃$\dot{C}H_3 + \dot{C}_2H_5 \rightarrow CH_3-CH_2-CH_3$ (Propane)
• Self-coupling 1: CH₃ + CH₃ arrow CH₃-CH₃$\dot{C}H_3 + \dot{C}H_3 \rightarrow CH_3-CH_3$ (Ethane)
• Self-coupling 2: C₂H₅ + C₂H₅ arrow CH₃-CH₂-CH₂-CH₃$\dot{C}_2H_5 + \dot{C}_2H_5 \rightarrow CH_3-CH_2-CH_2-CH_3$ (Butane)

Thus, a total of 3 different alkanes are formed.

Chapter Mix

Class 11 Chemistry: Hydrocarbons