Choose the correct set of reagents for the following conversion: Ethyl benzene 4-bromostyrene$$\text{Ethyl benzene} \longrightarrow \text{4-bromostyrene}$$ {{Q_IMG1}}

Core Logic

Analyzing each option: (1) Alkyl substitution reaction with Br₂, hν$Br_2, h\nu$ (free radical substitution) targets the most stable free radical. The allylic or benzylic position is preferred. In the given structure

(2) Reaction involves diazotization followed by Sandmeyer reaction with Cu₂Br₂/HBr$Cu_2Br_2/HBr$

(3) Free radical halogenation on toluene side-chain

(4) Electrophilic aromatic substitution of toluene with Br₂/Fe$Br_2/Fe$ in the dark correctly

Step 1: Final Conclusion

Reaction (1) is incorrectly represented as it gives a 1°$1^{\circ}$ radical product rather than the more stable 3°$3^{\circ}$ substituted major product.

Pattern Recognition

Free radical halogenation favors 3° > 2° > 1°$3^{\circ} > 2^{\circ} > 1^{\circ}$ substitution due to intermediate stability. Allylic and benzylic are even more favored. Always check if the halogen landed on the most substituted available carbon.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Related Formula

Core Logic

Let us systematically match each reaction in List-I to its standardized organic reaction name in List-II:

• Reaction (A): Coupling of two aryl halides with sodium metal in dry ether to form biaryl is the classic Fittig reaction arrow$\rightarrow$ (III).
• Reaction (B): Conversion of benzene diazonium chloride to aryl halide using copper powder (Cu$\mathrm{Cu}$) in halogen acids like HCl$\mathrm{HCl}$ is the Gatterman reaction arrow$\rightarrow$ (IV).
• Reaction (C): Substitution of halogen in an alkyl halide with sodium iodide (NaI$\mathrm{NaI}$) in dry acetone solvent is the classic halogen exchange method called the Finkelstein reaction arrow$\rightarrow$ (II).
• Reaction (D): Replacement of the hydroxyl group in tertiary butyl alcohol with chlorine using conc. HCl$\mathrm{HCl}$ in the presence of anhydrous ZnCl₂$\mathrm{ZnCl_2}$ catalyst is the Lucas reaction arrow$\rightarrow$ (I).

Step 1: Selection

Combining the selections, the correct sequence is: (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

This maps directly to option (2).

Pattern Recognition

Named reaction matching questions are very straightforward. Keep a clear distinction between the Sandmeyer reaction (which uses cuprous halide, e.g. Cu₂Cl₂$\mathrm{Cu_2Cl_2}$) and the Gatterman reaction (which uses copper powder, Cu$\mathrm{Cu}$).

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Related Formula

Core Logic

Let us break down each chemical reaction step:

• Step 1: 2-bromopentane undergoes dehydrohalogenation via an E2 mechanism using alcoholic KOH$\mathrm{KOH}$. According to Saytzeff's rule, the more substituted alkene is the major product. Thus, pent-2-ene is the major product P$\mathrm{P}$.



• Step 2: Pent-2-ene undergoes electrophilic bromination with liquid bromine (Br₂$\mathrm{Br_2}$) to give 2,3-dibromopentane (product Q$\mathrm{Q}$):

Step 1: Calculate Initial Moles of Reactant

Calculate the molar mass of 2-bromopentane (C₅H₁₁Br$\mathrm{C_5H_{11}Br}$):

Step 2: Calculate Moles of Intermediate P and Q

Since the yield of P$\mathrm{P}$ is 80%$80\%$:

Since the conversion of P arrow Q$\mathrm{P} \rightarrow \mathrm{Q}$ has a yield of 100%$100\%$, the mole count remains stoichiometric:

Step 3: Calculate Mass of Q

Product Q$\mathrm{Q}$ is 2,3-dibromopentane (C₅H₁₀Br₂$\mathrm{C_5H_{10}Br_2}$). Calculate its molar mass:

Pattern Recognition

Saytzeff vs Hofmann: Alcoholic KOH$\mathrm{KOH}$ is a small, non-bulky base, which selectively targets the internal secondary proton to yield the thermodynamic trans-alkene (pent-2-ene) as the major product rather than the terminal 1-alkene.

Chapter Mix

Class 12 Chemistry: Haloalkanes and Haloarenes

Related Formula

Dehydrohalogenation via alcoholic KOH follows E2 elimination mechanism:

Hydration of alkynes using HgSO₄/H₂SO₄$\mathrm{HgSO_4/H_2SO_4}$ yields ketones via keto-enol tautomerism.

Core Logic

Let's trace the full sequence line-by-row:

• [A] is an optically active halide with formula C₄H₉Br arrow CH₃-CH(Br)-CH₂-CH₃$\mathrm{C_4H_9Br} \rightarrow \mathrm{CH_3-CH(Br)-CH_2-CH_3}$ (2-Bromobutane).
• Reaction of [A] with hot ethanolic KOH produces [B] as the major product: CH₃-CH=CH-CH₃$\mathrm{CH_3-CH=CH-CH_3}$ (But-2-ene).
• Treatment of [B] with Br₂$\mathrm{Br_2}$ yields a vicinal dibromide [C]: CH₃-CH(Br)-CH(Br)-CH₃$\mathrm{CH_3-CH(Br)-CH(Br)-CH_3}$ (2,3-Dibromobutane).
• Reaction of [C] with alcoholic NaNH₂$\mathrm{NaNH_2}$ converts it via double dehydrohalogenation into gas [D]: CH₃-C≡ C-CH₃$\mathrm{CH_3-C\equiv C-CH_3}$ (But-2-yne).
• Hydration of 1 mole of [D] with H₂O$\mathrm{H_2O}$ in the presence of Hg²⁺/H^+$\mathrm{Hg^{2+}/H^+}$ forms an enol intermediate that rapidly tautomerizes to compound [E]: CH₃-CO-CH₂-CH₃$\mathrm{CH_3-CO-CH_2-CH_3}$ (Butan-2-one).

Step 1: Visualization

Pattern Recognition

Whenever you see a 4-carbon chain undergoing terminal/internal dehydrohalogenation followed by hydration of the resulting alkyne, look closely at the configuration: symmetric or unsymmetric alkyne hydration both systematically lead to Butan-2-one because a stable ketone cannot form on position 1 via standard Kucherov hydration of an internal chain.

Chapter Mix

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