### 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
More Haloalkanes and Haloarenes Previous-Year Questions
Q54jee_main_2026_21_jan_morningReactions of Haloalkanes
A hydrocarbon ‘P’ (C_4H_8$C_{4}H_{8}$) on reaction with HCl gives an optically active compound ‘Q’ (C_4H_9Cl$C_{4}H_{9}Cl$) which on reaction with one mole of ammonia gives compound ‘R’ (C_4H_11N$C_{4}H_{11}N$). ‘R’ on diazotization followed by hydrolysis gives ‘S’. Identify P, Q, R and S.
### Core Logic
Since P (C_4H_8$C_4H_8$) reacts with HCl to give an optically active compound Q (C_4H_9Cl$C_4H_9Cl$), P must be But-2-ene. Addition of HCl yields 2-chlorobutane, which possesses a chiral center.
mathrmCH_3-mathrmCH=mathrmCH-mathrmCH_3 xrightarrowmathrmHCl mathrmCH_3-mathrmCH_2-mathrmC^*mathrmH(mathrmCl)-mathrmCH_3 quad text(P is But-2-ene, Q is 2-chlorobutane)$$\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3 \xrightarrow{\mathrm{HCl}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C}^*\mathrm{H}(\mathrm{Cl})-\mathrm{CH}_3 \quad \text{(P is But-2-ene, Q is 2-chlorobutane)}$$Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Compound Q reacts with mathrmNH_3$\mathrm{NH}_3$ to undergo nucleophilic substitution forming a primary amine R:
mathrmCH_3-mathrmCH_2-mathrmCH(mathrmCl)-mathrmCH_3 xrightarrowmathrmNH_3 mathrmCH_3-mathrmCH_2-mathrmCH(mathrmNH_2)-mathrmCH_3 quad text(R is Butan-2-amine)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Cl})-\mathrm{CH}_3 \xrightarrow{\mathrm{NH}_3} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \quad \text{(R is Butan-2-amine)}$$Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
Diazotization of primary aliphatic amines followed by hydrolysis yields an alcohol via a carbocation intermediate (S):
mathrmCH_3-mathrmCH_2-mathrmCH(mathrmNH_2)-mathrmCH_3 xrightarrowmathrmNaNO_2 / mathrmHCl / mathrmH_2mathrmO mathrmCH_3-mathrmCH_2-mathrmCH(mathrmOH)-mathrmCH_3 quad text(S is Butan-2-ol)$$\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{NH}_2)-\mathrm{CH}_3 \xrightarrow{\mathrm{NaNO}_2 / \mathrm{HCl} / \mathrm{H}_2\mathrm{O}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}_3 \quad \text{(S is Butan-2-ol)}$$Reactions of Haloalkanes diagram for Q54 - JEE Main 2026 Morning
### Pattern Recognition
An optically active alkyl halide formed from a C_4H_8$C_4H_8$ alkene with HX is a classic pointer to 2-halobutane originating from But-2-ene or But-1-ene. Diazotization of primary aliphatic amines (R-NH_2$R-NH_2$) with NaNO_2/HCl$NaNO_2/HCl$ yields alcohols (R-OH$R-OH$) with possible rearrangements, though here a secondary carbocation is already stable.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Haloalkanes and Haloarenes
Class 12 Chemistry: Amines
Q48jee_main_2025_02_april_eveningElimination and Addition Reaction Sequences
Consider the following sequence of reactions:
mathrmCH_3-CH_2-CH_2-CH(Br)-CH_3 xrightarrowtextalcoholic KOH mathrmP text (Major Product) xrightarrowmathrmBr_2 mathrmQ$$\mathrm{CH_3-CH_2-CH_2-CH(Br)-CH_3} \xrightarrow{\text{alcoholic KOH}} \mathrm{P} \text{ (Major Product)} \xrightarrow{\mathrm{Br_2}} \mathrm{Q}$$
Consider the above sequence of reactions. 151~mathrmg$151~\mathrm{g}$ of 2-bromopentane is made to react. Yield of major product mathrmP$\mathrm{P}$ is 80\%$80\%$ whereas mathrmQ$\mathrm{Q}$ is 100\%$100\%$.
Mass of product mathrmQ$\mathrm{Q}$ obtained is _______ g.
Given molar mass in mathrmg~mol^-1$\mathrm{g~mol^{-1}}$ H: 1, C: 12, O: 16, Br: 80
Numerical Answer.Answer: 184 to 184
Solution
### 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$\mathrm{P}$.
Chemical reaction showing elimination of 2-bromopentane to form pent-2-ene
- **Step 2**: Pent-2-ene undergoes electrophilic bromination with liquid bromine (mathrmBr_2$\mathrm{Br_2}$) to give **2,3-dibromopentane** (product mathrmQ$\mathrm{Q}$):
Chemical reaction showing elimination of 2-bromopentane to form pent-2-ene
### 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
Qjee_main_2025_02_april_morningReactions of Alkyl Halides and Alkyne Hydration
An optically active alkyl halide mathrmC_4H_9Br$\mathrm{C_4H_9Br}$ [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic mathrmNaNH_2$\mathrm{NaNH_2}$. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333mathrmK$333\mathrm{K}$ to form compound [E]. The IUPAC name of compound [E] is :
A.(1)\ textBut-2-yne$(1)\ \text{But-2-yne}$
B.(2)\ textButan-2-ol$(2)\ \text{Butan-2-ol}$
C.(3)\ textButan-2-one$(3)\ \text{Butan-2-one}$
D.(4)\ textButan-1-al$(4)\ \text{Butan-1-al}$
Solution
### 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
Reaction roadmap step verification for Q27
### 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
Qjee_main_2025_08_april_eveningPreparation and Reactions of Styrene derivatives
Choose the correct set of reagents for the following conversion:
textEthyl benzene longrightarrow text4-bromostyrene$$\text{Ethyl benzene} \longrightarrow \text{4-bromostyrene}$$
{{Q_IMG1}}
The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.
### 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. The figure illustrates the multi-step conversion starting from ethylbenzene to yield a brominated styrene derivative.
### 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
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.svg "Aromatic substitution, Side chain bromination, Dehydrohalogenation, Styrene synthesis")
