### Related Formula textNamed Organic Transformations$\text{Named Organic Transformations}$ ### 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** rightarrow$\rightarrow$ **(III)**. - **Reaction (B)**: Conversion of benzene diazonium chloride to aryl halide using copper powder (mathrmCu$\mathrm{Cu}$) in halogen acids like mathrmHCl$\mathrm{HCl}$ is the **Gatterman reaction** rightarrow$\rightarrow$ **(IV)**. - **Reaction (C)**: Substitution of halogen in an alkyl halide with sodium iodide (mathrmNaI$\mathrm{NaI}$) in dry acetone solvent is the classic halogen exchange method called the **Finkelstein reaction** rightarrow$\rightarrow$ **(II)**. - **Reaction (D)**: Replacement of the hydroxyl group in tertiary butyl alcohol with chlorine using conc. mathrmHCl$\mathrm{HCl}$ in the presence of anhydrous mathrmZnCl_2$\mathrm{ZnCl_2}$ catalyst is the **Lucas reaction** rightarrow$\rightarrow$ **(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. mathrmCu_2Cl_2$\mathrm{Cu_2Cl_2}$) and the Gatterman reaction (which uses copper powder, mathrmCu$\mathrm{Cu}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula textActual Yield = textTheoretical Yield times \% text Yield$\text{Actual Yield} = \text{Theoretical Yield} \times \% \text{ Yield}$ ### Core Logic Let us break down each chemical reaction step: - **Step 1**: 2-bromopentane undergoes dehydrohalogenation via an E2 mechanism using alcoholic mathrmKOH$\mathrm{KOH}$. According to Saytzeff's rule, the more substituted alkene is the major product. Thus, **pent-2-ene** is the major product mathrmP$\mathrm{P}$.

- **Step 2**: Pent-2-ene undergoes electrophilic bromination with liquid bromine (mathrmBr_2$\mathrm{Br_2}$) to give **2,3-dibromopentane** (product mathrmQ$\mathrm{Q}$):

### Step 1: Calculate Initial Moles of Reactant Calculate the molar mass of 2-bromopentane (mathrmC_5H_11Br$\mathrm{C_5H_{11}Br}$): textMolar mass = 5(12) + 11(1) + 80 = 60 + 11 + 80 = 151~mathrmg~mol^-1$\text{Molar mass} = 5(12) + 11(1) + 80 = 60 + 11 + 80 = 151~\mathrm{g~mol^{-1}}$ textInitial moles = frac151~mathrmg151~mathrmg~mol^-1 = 1~mathrmmol$\text{Initial moles} = \frac{151~\mathrm{g}}{151~\mathrm{g~mol^{-1}}} = 1~\mathrm{mol}$ ### Step 2: Calculate Moles of Intermediate P and Q Since the yield of mathrmP$\mathrm{P}$ is 80\%$80\%$: textMoles of P formed = 1 times 0.80 = 0.8~mathrmmol$\text{Moles of P formed} = 1 \times 0.80 = 0.8~\mathrm{mol}$ Since the conversion of mathrmP rightarrow mathrmQ$\mathrm{P} \rightarrow \mathrm{Q}$ has a yield of 100\%$100\%$, the mole count remains stoichiometric: textMoles of Q formed = 0.8 times 1.00 = 0.8~mathrmmol$\text{Moles of Q formed} = 0.8 \times 1.00 = 0.8~\mathrm{mol}$ ### Step 3: Calculate Mass of Q Product mathrmQ$\mathrm{Q}$ is 2,3-dibromopentane (mathrmC_5H_10Br_2$\mathrm{C_5H_{10}Br_2}$). Calculate its molar mass: textMolar mass of Q = 5(12) + 10(1) + 2(80) = 60 + 10 + 160 = 230~mathrmg~mol^-1$\text{Molar mass of Q} = 5(12) + 10(1) + 2(80) = 60 + 10 + 160 = 230~\mathrm{g~mol^{-1}}$ textMass of Q = 0.8 times 230 = 184~mathrmg$\text{Mass of Q} = 0.8 \times 230 = 184~\mathrm{g}$ ### Pattern Recognition Saytzeff vs Hofmann: Alcoholic mathrmKOH$\mathrm{KOH}$ 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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes

### Related Formula Dehydrohalogenation via alcoholic KOH follows E2 elimination mechanism: mathrmR-CH_2-CH(Br)-R' xrightarrowtextalc. KOH R-CH=CH-R'$\mathrm{R-CH_2-CH(Br)-R' \xrightarrow{\text{alc. KOH}} R-CH=CH-R'}$ Hydration of alkynes using mathrmHgSO_4/H_2SO_4$\mathrm{HgSO_4/H_2SO_4}$ yields ketones via keto-enol tautomerism. ### Core Logic Let's trace the full sequence line-by-row: 1. **[A]** is an optically active halide with formula mathrmC_4H_9Br rightarrow mathrmCH_3-CH(Br)-CH_2-CH_3$\mathrm{C_4H_9Br} \rightarrow \mathrm{CH_3-CH(Br)-CH_2-CH_3}$ (2-Bromobutane). 2. Reaction of [A] with hot ethanolic KOH produces **[B]** as the major product: mathrmCH_3-CH=CH-CH_3$\mathrm{CH_3-CH=CH-CH_3}$ (But-2-ene). 3. Treatment of [B] with mathrmBr_2$\mathrm{Br_2}$ yields a vicinal dibromide **[C]**: mathrmCH_3-CH(Br)-CH(Br)-CH_3$\mathrm{CH_3-CH(Br)-CH(Br)-CH_3}$ (2,3-Dibromobutane). 4. Reaction of [C] with alcoholic mathrmNaNH_2$\mathrm{NaNH_2}$ converts it via double dehydrohalogenation into gas **[D]**: mathrmCH_3-Cequiv C-CH_3$\mathrm{CH_3-C\equiv C-CH_3}$ (But-2-yne). 5. Hydration of 1 mole of [D] with mathrmH_2O$\mathrm{H_2O}$ in the presence of mathrmHg^2+/H^+$\mathrm{Hg^{2+}/H^+}$ forms an enol intermediate that rapidly tautomerizes to compound **[E]**: mathrmCH_3-CO-CH_2-CH_3$\mathrm{CH_3-CO-CH_2-CH_3}$ (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. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Haloalkanes and Haloarenes Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Aldehydes, Ketones and Carboxylic Acids

### Core Logic To synthesize 4-bromostyrene starting from ethyl benzene, we must carry out ring functionalization prior to developing the side-chain double bond: 1. **Ring Bromination**: Treatment of ethylbenzene with textBr_2$\text{Br}_2$ in the presence of textFe$\text{Fe}$ (or textFeBr_3$\text{FeBr}_3$) acts via electrophilic aromatic substitution. The ethyl group is an ortho/para director, yielding 1-bromo-4-ethylbenzene as the major product owing to steric mitigation. 2. **Side-Chain Halogenation**: Free radical substitution with textCl_2$\text{Cl}_2$ under thermal conditions (Delta$\Delta$) or UV light specifically chlorinates the benzylic position because the benzylic radical is exceptionally stable via resonance. 3. **Elimination**: Heating with alcoholic textKOH$\text{KOH}$ drives an E2$E2$ elimination of the benzylic chloride, cleanly synthesizing the terminal alkene linkage of the styrene system.

### Pattern Recognition If you perform side-chain halogenation/alkene generation first, the ring substitution later would lack para-selectivity control and risk reacting across the alkene path. Hence, ring substitution MUST precede double bond creation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes Class 11 Chemistry: Hydrocarbons

### Core Logic The mechanism of S_N1$S_N1$ substitution proceeds via carbocation intermediate formation. Option (4) gives a triphenylmethyl carbocation (Ph_3C^+$Ph_{3}C^{+}$), which is exceptionally stable due to extensive delocalization of positive charge across three phenyl rings (resonance stabilization via 9 canonical structures).

### Step 1: Stability Comparison Stability sequence: Ph_3C^+ > textbenzylic > textallylic > textalkyl carbocations$Ph_{3}C^{+} > \text{benzylic} > \text{allylic} > \text{alkyl carbocations}$ ### Pattern Recognition Look for maximum phenyl groups attached directly to the carbon bearing the leaving group to maximize resonance contribution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Haloalkanes and Haloarenes