Ph-CH=CH₂ [HBr](PhCOO)₂ Product$\text{Ph-CH=CH}_{2} \xrightarrow[\text{HBr}]{\text{(PhCOO)}_{2}} \text{Product}$
Consider the above reaction
A. The reaction proceeds through a more stable radical intermediate.
B. The role of peroxide is to generate H$\dot{H}$ (Hydrogen radical).
C. During this reaction, benzene is formed as a biproduct.
D. 1-Bromo-2-phenylethane is fanned as the minor product.
E. The same reaction in absence of peroxide proceeds via carbocation intermediate.
Identify the correct statements. Choose the correct answer from the options given below:
A.A & E Only
B.A, B & D Only
C.C, D & E Only
D.A, C & E Only
Solution & Explanation
Core Logic
This is the classic Peroxide Effect (Kharasch effect) showing Anti-Markovnikov addition of HBr$\mathrm{HBr}$ across an alkene.
Ph-CH=CH₂ + HBr Peroxide Ph-CH₂-CH₂-Br$\mathrm{Ph-CH=CH_2 + HBr} \xrightarrow{\text{Peroxide}} \mathrm{Ph-CH_2-CH_2-Br}$ (1-Bromo-2-phenylethane is the MAJOR product).
Step 1: Analyze Radical Mechanism (A, B, C)
$\bullet$ The reaction proceeds via the most stable free radical intermediate (a secondary benzylic radical). Statement A is correct.
$\bullet$ Peroxide (Benzoyl peroxide) undergoes homolytic cleavage to form benzoyloxy radicals, which lose CO₂$CO_2$ to form phenyl radicals (Ph$\dot{Ph}$). Phenyl radical abstracts H from HBr to form a Bromine radical (Br$\dot{Br}$), not a Hydrogen radical (H$\dot{H}$). Statement B is false.
$\bullet$ The abstraction of H from HBr by the phenyl radical (Ph$\dot{Ph}$) yields Benzene (Ph-H$Ph-H$) as a byproduct. Statement C is correct.
Step 2: Product Check (D, E)
$\bullet$ 1-Bromo-2-phenylethane is the major product, not the minor product. Statement D is false.
$\bullet$ In the absence of peroxide, standard electrophilic addition occurs via a stable carbocation intermediate (Markovnikov addition). Statement E is correct.
Final Conclusion
Statements A, C, and E are correct.
Pattern Recognition
Peroxides generate Br$\dot{Br}$, never H$\dot{H}$. The initiating radical pulls H from H-Br since H-Br is weaker than forming C-H bond, leaving Br$\dot{Br}$ to attack the alkene.
Identify major product 'P' formed in the following reaction.
The image shows an aromatic ring undergoing a reaction with an alkyl halide in the presence of anhydrous AlCl3.
A.(1) Product A$\text{(1) Product A}$
B.(2) Product B$\text{(2) Product B}$
C.(3) Product C$\text{(3) Product C}$
D.(4) Product D$\text{(4) Product D}$
Solution
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.
The image shows an aromatic ring undergoing a reaction with an alkyl halide in the presence of anhydrous AlCl3.
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
Qjee_main_2024_31_jan_eveningElectrophilic Addition to Alkenes
Major product of the following reaction is -
The image shows a methylcyclopentene derivative reacting with D-Cl.
A.
B.
C.
D.
Solution
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.
The image shows a methylcyclopentene derivative reacting with D-Cl.
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).
The correct order of reactivity in electrophilic substitution reaction of the following compounds is:
The image displays four aromatic compounds: Benzene (A), Toluene (B), Chlorobenzene (C), and Nitrobenzene (D).
A.B > C > A > D$\text{}B > C > A > D$
B.D > C > B > A$\text{}D > C > B > A$
C.A > B > C > D$\text{}A > B > C > D$
D.B > A > C > D$\text{}B > A > C > D$
Solution
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
Q84jee_main_2024_31_jan_eveningHalogenation of Alkanes (Isomers)
Number of isomeric products formed by mono-chlorination of 2-methylbutane in presence of sunlight is ________
Numerical Answer.Answer: 6 to 6
Solution
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).
Halogenation of Alkanes (Isomers) diagram for Q84 - JEE Main 2024 Evening
Number of alkanes obtained on electrolysis of a mixture of CH₃COONa$CH_3COONa$ and C₂H₅COONa$C_2H_5COONa$ is
Numerical Answer.Answer: 3 to 3
Solution
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:
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