Q.Consider the reactions : 2S2O3 2–(aq) + I2(s) → S4O6 2–(aq) + 2I–(aq) S2O3 2–(aq) + 2Br2(l) + 5H2O(l) → 2SO4 2–(aq) + 4Br–(aq) + 10H+(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine ?

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.In winter, polar stratospheric clouds formed over Antarctica provide surface on which compound 'X' responsible for depletion of ozone is formed. What is X? (A) Peroxyacetyl nitrate (B) Acrolein (C) Chlorine nitrate (D) Sulphuryl chloride

   ›Reveal solutionSolution

   The compound formed on polar stratospheric clouds that directly depletes ozone is chlorine nitrate — option (C).

   The key here is understanding the chemistry of ozone depletion over Antarctica. In winter, the polar vortex isolates the air above Antarctica, allowing temperatures to drop low enough for polar stratospheric clouds (PSCs) to form. These clouds are not just passive bystanders — they provide a surface for heterogeneous chemical reactions that convert inactive chlorine reservoirs into active forms that destroy ozone.

   The most important reservoir species is chlorine nitrate, $\text{ClONO}_2$. Normally, chlorine is locked up in this compound and in hydrogen chloride ($\text{HCl}$), which do not react with ozone. But on the surface of PSC particles, a reaction occurs:

   $\text{ClONO}_2 + \text{HCl} \xrightarrow{\text{PSC surface}} \text{Cl}_2 + \text{HNO}_3$

   The molecular chlorine ($\text{Cl}_2$) produced is photolysed by sunlight when spring returns, releasing chlorine radicals that catalytically destroy ozone. So the compound 'X' that is formed on PSCs and is directly responsible for the depletion is chlorine nitrate — it is the key chlorine reservoir that gets activated.

   Let’s walk through the reasoning step by step.

   1. Identify the context. The question is about the Antarctic ozone hole. In winter, the stratosphere over Antarctica becomes extremely cold, and polar stratospheric clouds form. These clouds are composed of ice or nitric acid trihydrate crystals.

   2. Understand the role of PSCs. Normally, chlorine in the stratosphere exists in "reservoir" forms that are harmless to ozone. The two main reservoirs are chlorine nitrate ($\text{ClONO}_2$) and hydrogen chloride ($\text{HCl}$). On their own, these do not react with ozone.

   3. The critical reaction on PSCs. On the surface of PSC particles, a heterogeneous reaction takes place:

   $\text{ClONO}_2 + \text{HCl} \rightarrow \text{Cl}_2 + \text{HNO}_3$

   Here, chlorine nitrate reacts with HCl to produce molecular chlorine ($\text{Cl}_2$) and nitric acid ($\text{HNO}_3$). The nitric acid remains in the cloud particles, while $\text{Cl}_2$ is a gas.

   4. Activation in spring. When sunlight returns to Antarctica in spring, $\text{Cl}_2$ is photolysed:

   $\text{Cl}_2 + \text{UV light} \rightarrow 2\text{Cl}^\bullet$

   These chlorine atoms then catalytically destroy ozone in a cycle:

   $\text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2$

   $\text{ClO}^\bullet + \text{O} \rightarrow \text{Cl}^\bullet + \text{O}_2$ …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Iodine on reaction with concentrated nitric acid gives three products X, Y and Z. The sum of oxidation numbers of the central atoms in X, Y and Z is (A) $5$ (B) $6$ (C) $7$ (D) $4$

   ›Reveal solutionSolution

   $\text{I}_2 + 10\,\text{HNO}_3 \to 2\,\text{HIO}_3 + 10\,\text{NO}_2 + 4\,\text{H}_2\text{O}$; central-atom oxidation numbers $+5\ (\text{I}) + 4\ (\text{N}) + (-2)\ (\text{O}) = 7$.

   Iodine is oxidised by concentrated nitric acid to iodic acid, while the acid is reduced to nitrogen dioxide, and water is formed:

   $\text{I}_2 + 10\,\text{HNO}_3\,(\text{conc.}) \longrightarrow 2\,\text{HIO}_3 + 10\,\text{NO}_2 + 4\,\text{H}_2\text{O}$

   The three products and the oxidation number of their central atoms are:

   • $X = \text{HIO}_3$: central atom $\text{I}$ has oxidation number $+5$. …

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3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Electrolysis of 50% H$_2$SO$_4$ solution at high current density gives a compound ‘A’. Hydrolysis of ‘A’ gives ‘B’. The number of moles of O$_2$ produced when 5 moles of ‘B’ reacts with 2 moles of acidified KMnO$_4$ solution is (A) 4 (B) 5 (C) 6 (D) 7

   ›Reveal solutionSolution

   Electrolysis of concentrated sulfuric acid at high current density produces peroxydisulfuric acid (A). Its hydrolysis yields hydrogen peroxide (B). Hydrogen peroxide then reacts with acidified potassium permanganate, acting as a reducing agent to produce oxygen gas. Given the stoichiometric amounts of reactants, 5 moles of O$_2$ are produced.

   The problem involves a sequence of chemical reactions: an electrolysis reaction, followed by a hydrolysis reaction, and finally a redox reaction. To solve this, we need to identify the compounds formed at each step and then use stoichiometry for the final reaction.

   Concept and Intuition

   1. Electrolysis of Sulfuric Acid: When an aqueous solution of sulfuric acid is electrolyzed, there are competing reactions at the anode (oxidation). Typically, water is oxidized to oxygen gas. However, at high concentrations of sulfuric acid (like 50%) and high current densities, the overpotential for oxygen evolution becomes significant. This kinetic factor favors the oxidation of sulfate ions ($\text{SO}_4^{2-}$) to peroxydisulfate ions ($\text{S}_2\text{O}_8^{2-}$), which then forms peroxydisulfuric acid ($\text{H}_2\text{S}_2\text{O}_8$). This is a key industrial method for producing peroxydisulfuric acid.
   2. Hydrolysis of Peroxydisulfuric Acid: Peroxydisulfuric acid is an unstable compound that readily hydrolyzes in the presence of water. This hydrolysis proceeds in steps, ultimately yielding sulfuric acid and hydrogen peroxide. This reaction is also an important industrial route for hydrogen peroxide production.
   3. Redox Reaction with Permanganate: Hydrogen peroxide ($\text{H}_2\text{O}_2$) is a versatile compound that can act as both an oxidizing and a reducing agent. In the presence of a strong oxidizing agent like acidified potassium permanganate ($\text{KMnO}_4$), hydrogen peroxide acts as a reducing agent, getting oxidized to oxygen gas ($\text{O}_2$). Permanganate, in turn, is reduced to $\text{Mn}^{2+}$ in acidic medium. Balancing this redox reaction is crucial for determining the stoichiometry.

   Step-by-step Solution

   1. Identify Compound 'A' from Electrolysis:

      Electrolysis of 50% $\text{H}_2\text{SO}_4$ solution at high current density leads to the oxidation of sulfate ions at the anode.

      Anode reaction: $2\text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_8 + 2\text{H}^+ + 2\text{e}^-$

      Cathode reaction: $2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$

      Overall reaction: $2\text{H}_2\text{SO}_4 \xrightarrow{\text{electrolysis}} \text{H}_2\text{S}_2\text{O}_8 + \text{H}_2$

      Compound 'A' is peroxydisulfuric acid ($\text{H}_2\text{S}_2\text{O}_8$).

   2. Identify Compound 'B' from Hydrolysis of 'A':

      Peroxydisulfuric acid undergoes hydrolysis in the presence of water. This reaction typically occurs in two steps:

      ›Proof

      Step 1: $\text{H}_2\text{S}_2\text{O}_8 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_5 + \text{H}_2\text{SO}_4$ (Peroxymonosulfuric acid, Caro's acid)

      Step 2: $\text{H}_2\text{SO}_5 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 + \text{H}_2\text{O}_2$

      Adding these two steps gives the overall hydrolysis reaction.

      The overall hydrolysis reaction is:

      [!FORMULA]

      $\text{H}_2\text{S}_2\text{O}_8 + 2\text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4 + \text{H}_2\text{O}_2$

      Compound 'B' is hydrogen peroxide ($\text{H}_2\text{O}_2$).

   3. Balance the Redox Reaction between 'B' ($\text{H}_2\text{O}_2$) and Acidified $\text{KMnO}_4$:

      In this reaction, $\text{H}_2\text{O}_2$ acts as a reducing agent and is oxidized to $\text{O}_2$. $\text{MnO}_4^-$ acts as an oxidizing agent and is reduced to $\text{Mn}^{2+}$ in acidic medium.

      • Oxidation half-reaction: …

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4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Which one of the following statements is not correct regarding interhalogen compounds? (A) IF$_7$ and ClF$_3$ are colourless gases (B) BrF$_5$ and ClF$_5$ are colourless liquids (C) ClF$_3$ on hydrolysis gives HCl and HOF (D) Except F–F bond, X–X' bond of interhalogens is weaker than X–X bond of halogens

   ›Reveal solutionSolution

   The key idea is to recall the physical states, colours, and hydrolysis products of common interhalogen compounds. The incorrect statement is (C), because ClF$_3$ on hydrolysis gives HF, not HCl, and HOF is not a product.

   Interhalogen compounds are formed when two different halogens combine. Their properties — colour, physical state, and chemical reactivity — follow trends based on the size and electronegativity of the constituent halogens. The question tests your ability to spot a factual error in these properties, especially the hydrolysis behaviour of ClF$_3$, which is a classic trap.

   Let’s examine each statement one by one.

   1. Statement (A): IF$_7$ and ClF$_3$ are colourless gases

      IF$_7$ (iodine heptafluoride) is indeed a colourless gas at room temperature. ClF$_3$ (chlorine trifluoride) is also a colourless gas (though it can be condensed to a pale yellow liquid under pressure). Both are gases, and both are colourless. This statement is correct.

   2. Statement (B): BrF$_5$ and ClF$_5$ are colourless liquids

      BrF$_5$ (bromine pentafluoride) is a colourless liquid. ClF$_5$ (chlorine pentafluoride) is also a colourless liquid at room temperature. This matches known data. Statement (B) is correct.

   3. Statement (C): ClF$_3$ on hydrolysis gives HCl and HOF

      This is the suspect one. Hydrolysis of interhalogens typically produces a mixture of the oxyacid of the central halogen and the hydrohalic acid of the more electronegative halogen. For ClF$_3$, the central atom is chlorine (less electronegative than fluorine), and fluorine is more electronegative.

      The actual hydrolysis reaction is:

   $\text{ClF}_3 + 2\text{H}_2\text{O} \rightarrow \text{HClO}_2 + 3\text{HF}$

   (Chlorous acid and hydrogen fluoride are formed, not HCl and HOF).

   HOF (hypofluorous acid) is a rare, unstable compound and does not form here. Also, HCl would require chlorine to be in the –1 oxidation state, but in ClF$_3$, chlorine is in +3 state — hydrolysis doesn’t reduce it to –1. So statement (C) is factually wrong. …

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5. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Observe the following reaction sequence

   [!FORMULA] $\text{Styrene} \xrightarrow{\text{HBr}} X \xrightarrow{\begin{matrix} (i)\\\text{KCN} \\ (ii)\\\text{H}_3\text{O}^+ \end{matrix}} Y \xrightarrow{\begin{matrix} (i)\\\text{Br}_2/\text{red P} \\ (ii)\\\text{H}_2\text{O} \end{matrix}} Z$

   Correct statement regarding Y, Z is (A) Y is stronger acid than benzoic acid (B) Z is stronger acid than Y (C) Y can be reduced with NaBH$_4$ (D) Z on decarboxylation gives benzyl bromide

   ›Reveal solutionSolution

   The reaction sequence converts styrene into a substituted carboxylic acid (Y) and then into an α-bromo acid (Z). The correct statement is that Z is a stronger acid than Y due to the electron-withdrawing effect of the bromine atom.

   The problem is about understanding how functional groups transform step by step, and then comparing acid strengths. The key concept is inductive effect — how electronegative atoms near a carboxylic acid group affect its acidity. Let’s trace the chemistry.

   1. Styrene + HBr follows Markovnikov’s rule. The double bond in styrene is conjugated with the benzene ring, but HBr adds such that the bromine goes to the more substituted carbon (the one attached to the ring). So X is 1-bromo-1-phenylethane (Ph–CHBr–CH₃).

   2. X + KCN does an Sₙ2 substitution. The bromine is on a secondary carbon, but benzylic position makes it reactive. CN⁻ replaces Br, giving a nitrile. Then H₃O⁺ hydrolyzes the nitrile to a carboxylic acid. So Y is 2-phenylpropanoic acid (Ph–CH(CH₃)–COOH).

   3. Y + Br₂/red P is the Hell-Volhard-Zelinsky (HVZ) reaction. Red P generates a small amount of PBr₃, which converts the carboxylic acid to an acyl bromide. Then Br₂ substitutes the α-hydrogen (the one next to the COOH group). Water in the second step hydrolyzes the acyl bromide back to the acid. So Z is 2-bromo-2-phenylpropanoic acid (Ph–CBr(CH₃)–COOH).

   Now evaluate each statement:

   (A) Y is stronger acid than benzoic acid — False. Benzoic acid has the COOH directly attached to the benzene ring, which stabilizes the conjugate base via resonance. In Y, the COOH is separated by a CH(CH₃) group, so no such resonance. Benzoic acid (pKa ≈ 4.2) is stronger than 2-phenylpropanoic acid (pKa ≈ 4.5). …

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Consider the given sequence of reactions.

   [!FORMULA] $\mathrm{C_2H_6} + \frac{3}{2} \mathrm{O_2} \xrightarrow{(\mathrm{CH_3COO})_2 \mathrm{Mn}, \Delta} \mathrm{X} \xrightarrow{\mathrm{Na}} \mathrm{Y}$

   Electrolysis of aqueous solution of Y gives gases P and Q at anode. P and Q are respectively. (A) $\mathrm{C_2H_6}, \mathrm{CO_2}$ (B) $\mathrm{CH_4}, \mathrm{CO_2}$ (C) $\mathrm{C_2H_6}, \mathrm{H_2}$ (D) $\mathrm{CH_4}, \mathrm{CO}$

   ›Reveal solutionSolution

   The reaction sequence converts ethane into sodium acetate, whose electrolysis (Kolbe’s electrolysis) yields ethane and CO₂ at the anode. The correct option is (A).

   Concept & Intuition

   This problem tests your knowledge of two key organic reactions:

   1. Catalytic oxidation of alkanes – Ethane is oxidised to acetic acid using oxygen in the presence of manganese(II) acetate as a catalyst.
   2. Kolbe’s electrolysis – Electrolysis of a concentrated aqueous solution of a sodium salt of a carboxylic acid produces alkanes and CO₂ at the anode.

   The trick is to identify intermediate X (acetic acid) and then Y (sodium acetate). Once you know Y, the anode gases are determined by the Kolbe mechanism.

   ---

   Step-by-step reasoning

   1. Identify X from the first reaction Ethane ($\mathrm{C_2H_6}$) reacts with oxygen in the presence of $(\mathrm{CH_3COO})_2\mathrm{Mn}$ (manganese(II) acetate) and heat. This is a vapour-phase oxidation that converts ethane into acetic acid:

   $\mathrm{C_2H_6} + \frac{3}{2}\mathrm{O_2} \xrightarrow{(\mathrm{CH_3COO})_2\mathrm{Mn}, \Delta} \mathrm{CH_3COOH} \quad (\text{X})$

   The catalyst selectively oxidises the methyl group to a carboxyl group without breaking the C–C bond.

   2. Identify Y from the second reaction Acetic acid (X) reacts with sodium metal. Sodium is a strong reducing agent and reacts with the acidic hydrogen of the carboxyl group:

   $2\,\mathrm{CH_3COOH} + 2\,\mathrm{Na} \rightarrow 2\,\mathrm{CH_3COONa} + \mathrm{H_2}$

   So Y is sodium acetate ($\mathrm{CH_3COONa}$).

   3. Electrolysis of aqueous sodium acetate (Y) In aqueous solution, sodium acetate dissociates:

   $\mathrm{CH_3COONa} \rightarrow \mathrm{CH_3COO^-} + \mathrm{Na^+}$

   At the anode, acetate ions undergo Kolbe electrolysis:

   • Two acetate ions lose electrons to form acetate radicals:

   $2\,\mathrm{CH_3COO^-} \rightarrow 2\,\mathrm{CH_3COO\cdot} + 2e^-$

   • The radicals decarboxylate (lose CO₂) to give methyl radicals:

   $2\,\mathrm{CH_3COO\cdot} \rightarrow 2\,\mathrm{CH_3\cdot} + 2\,\mathrm{CO_2}$

   • Two methyl radicals couple to form ethane: $2\,\mathrm{CH_3\cdot} \rightarrow \mathrm{C_2H_6}$ …

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7. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.Which of the following gives more number of oxides on reacting with HCl? (A) $\mathrm{Na_2CO_3}$ (B) $\mathrm{NaHCO_3}$ (C) $\mathrm{NaNO_2}$ (D) $\mathrm{Na_2SO_3}$

   ›Reveal solutionSolution

   The key is to count the distinct gaseous oxides produced when each salt reacts with HCl. Only $\mathrm{NaNO_2}$ yields two different oxides ($\mathrm{NO}$ and $\mathrm{NO_2}$), so it gives the most.

   The question asks which salt, when treated with hydrochloric acid, produces the greatest number of oxides (meaning gaseous oxide products). We need to think about the chemical reactions: HCl is a strong acid, so it will protonate the anion of each salt, often leading to decomposition that releases oxides of carbon, nitrogen, or sulfur.

   Concept & Intuition

   The number of oxides formed depends on the stability and possible decomposition pathways of the acid formed in situ. For example, carbonates and bicarbonates give only $\mathrm{CO_2}$ (one oxide). Sulfites give $\mathrm{SO_2}$ (one oxide). But nitrites are special: the unstable nitrous acid ($\mathrm{HNO_2}$) can decompose in two ways, producing both $\mathrm{NO}$ and $\mathrm{NO_2}$ — two different oxides. So the winner is the nitrite.

   Let’s check each option step by step.

   1. Option (A): $\mathrm{Na_2CO_3}$

      Reaction: $\mathrm{Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2 \uparrow}$

      Only one gaseous oxide: carbon dioxide ($\mathrm{CO_2}$).

      Count = 1.

   2. Option (B): $\mathrm{NaHCO_3}$

      Reaction: $\mathrm{NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2 \uparrow}$

      Again, only $\mathrm{CO_2}$ is produced.

      Count = 1.

   3. Option (C): $\mathrm{NaNO_2}$

      Reaction: $\mathrm{NaNO_2 + HCl \rightarrow NaCl + HNO_2}$

      Nitrous acid ($\mathrm{HNO_2}$) is unstable and decomposes:

   $2\mathrm{HNO_2} \rightarrow \mathrm{NO} \uparrow + \mathrm{NO_2} \uparrow + \mathrm{H_2O}$

   This yields two different gaseous oxides: nitric oxide ($\mathrm{NO}$) and nitrogen dioxide ($\mathrm{NO_2}$).

   Count = 2. …

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.Which of the following statements is not correct? (A) Carbonates of alkaline earth metals are insoluble in water (B) Beryllium halides are covalent in nature (C) Alkali metal halides have high negative enthalpies of formation (D) The super oxides of alkali metals are colourless

   ›Reveal solutionSolution

   The key idea is to recall the specific anomalous properties of beryllium and the colour of superoxides. The incorrect statement is (D) — alkali metal superoxides are coloured, not colourless.

   The question tests your grasp of periodic trends and exceptions in s-block chemistry. Each statement must be checked against known facts. Let’s go through them one by one.

   1. Statement (A): Carbonates of alkaline earth metals are insoluble in water

      This is correct. The carbonates of Mg, Ca, Sr, Ba, and Ra are all sparingly soluble in water. The lattice energy of these carbonates is high due to the divalent cation, and the hydration energy is insufficient to overcome it. Only beryllium carbonate is slightly different (it decomposes easily), but the statement refers to the group as a whole, and it holds true.

   2. Statement (B): Beryllium halides are covalent in nature

      This is correct. Beryllium is the smallest alkaline earth metal with a high charge density (high polarising power). According to Fajan’s rules, it strongly polarises the halide anion, leading to significant covalent character. BeCl₂, for example, is a covalent polymer in the solid state and exists as a linear molecule in the vapour phase.

   3. Statement (C): Alkali metal halides have high negative enthalpies of formation

      This is correct. The formation of alkali metal halides from their elements is highly exothermic. The large negative enthalpy arises because the alkali metals have low ionisation energies and the halogens have high electron affinities, plus the lattice energy of the resulting ionic crystal is substantial. For example, ΔH_f° for NaCl is about –411 kJ/mol.

   4. Statement (D): The super oxides of alkali metals are colourless …

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9. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.What are X and Y in the following reaction sequence? 2-Methylbutane $\xrightarrow{\mathrm{KMnO_4}}$ X $\xrightarrow{Y}$ $\mathrm{C_5H_{11}Cl}$ (Conc. = concentrated) (A) $\mathrm{(CH_3)_2C(OH)CH_2CH_3}$ (2-methylbutan-2-ol) ; Conc. HCl (B) $(\mathrm{CH_3})_2\mathrm{CHCH_2CH_2OH}$ (3-methylbutan-1-ol) ; Conc. HCl, $\mathrm{ZnCl_2}$ (C) [FIGURE] a branched C$_5$ alcohol with the $-\mathrm{OH}$ drawn below the carbon next to the methyl branch ; Conc. HCl (D) [FIGURE] a branched C$_5$ alcohol with the $-\mathrm{OH}$ drawn at the end of the chain ; NaCl, $\mathrm{H_2SO_4}$

   ›Reveal solutionSolution

   $\mathrm{KMnO_4}$ hydroxylates the lone tertiary C–H of 2-methylbutane to give 2-methylbutan-2-ol, and this tertiary alcohol is converted to the chloride by conc. HCl alone (no $\mathrm{ZnCl_2}$). Option (A).

   The concept first — two selectivity rules

   Rule 1: $\mathrm{KMnO_4}$ oxidises the tertiary C–H of a branched alkane.

   Alkanes are famously inert, but a tertiary C–H is the weakest ($\mathrm{3^\circ < 2^\circ < 1^\circ}$ bond-dissociation energy) because the radical/cation left behind is best stabilised by the three alkyl groups. So a mild oxidant such as $\mathrm{KMnO_4}$ (or $\mathrm{KMnO_4/OH^-}$) selectively inserts an $-\mathrm{OH}$ at the tertiary carbon, converting the alkane into a tertiary alcohol and leaving the primary/secondary C–H bonds untouched.

   Draw 2-methylbutane and find its tertiary carbon:

   $\mathrm{CH_3{-}\underset{\displaystyle |}{\overset{\displaystyle CH_3}{C}}H{-}CH_2{-}CH_3}$

   The C-2 carbon carries three carbon neighbours and only one H — that is the tertiary C–H. Oxidise it:

   $(\mathrm{CH_3})_2\mathrm{CH{-}CH_2CH_3} \xrightarrow{\ \mathrm{KMnO_4}\ } (\mathrm{CH_3})_2\mathrm{C(OH){-}CH_2CH_3}$

   $\boxed{X = \text{2-methylbutan-2-ol, a } 3^\circ \text{ alcohol } (\mathrm{C_5H_{12}O})}$

   Rule 2: the Lucas test — how each class of alcohol meets HCl.

   Conversion $\mathrm{R{-}OH} \rightarrow \mathrm{R{-}Cl}$ by HCl goes through protonation of the $-\mathrm{OH}$ (making $\mathrm{H_2O}$, a good leaving group) and then loss of water to a carbocation ($\mathrm{S_N1}$). The ease therefore tracks carbocation stability:

   Alcohol	Reagent needed	Speed
   Tertiary	conc. HCl alone, room temperature	immediate turbidity
   Secondary	conc. HCl + anhydrous $\mathrm{ZnCl_2}$ (Lucas reagent)	turbidity in $\sim$5 min
   Primary	conc. HCl + $\mathrm{ZnCl_2}$, and heat	no turbidity at RT

   The Lewis acid $\mathrm{ZnCl_2}$ is a crutch: it coordinates the OH and helps a reluctant (2°/1°) alcohol ionise. A tertiary alcohol forms its cation so readily that it needs no crutch at all. …

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10. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
    Q.Permanganate titrations cannot be performed satisfactorily in presence of HCl. The reason is (A) Both HCl and KMnO$_4$ act as oxidising agents (B) KMnO$_4$ is a weaker oxidising agent in presence of HCl (C) KMnO$_4$ oxidises HCl into Cl$_2$ (D) KMnO$_4$ acts as a reducing agent in the presence of HCl

    ›Reveal solutionSolution

    Permanganate (KMnO$_4$) is a strong oxidising agent. In the presence of hydrochloric acid (HCl), it oxidises the chloride ions (Cl$^-$) to chlorine gas (Cl$_2$), leading to inaccurate titration results. The correct option is (C).

    Permanganate titrations are a type of redox titration where potassium permanganate (KMnO$_4$) acts as a powerful oxidising agent. For a titration to be accurate, the titrant (KMnO$_4$ in this case) must react only with the analyte (the substance being measured) and not with any other component of the solution. The problem asks why HCl interferes with these titrations. The core reason is that HCl itself can be oxidised by KMnO$_4$, leading to a side reaction that consumes the titrant unnecessarily.

    Here's a step-by-step explanation:

    1. KMnO$_4$ as a strong oxidising agent: Potassium permanganate (KMnO$_4$) is widely used as an oxidising agent in volumetric analysis. In an acidic medium, the permanganate ion ($MnO_4^-$), where manganese is in the +7 oxidation state, is reduced to the manganese(II) ion ($Mn^{2+}$). This change in oxidation state involves the gain of five electrons:

    $MnO_4^- (aq) + 8H^+ (aq) + 5e^- \rightarrow Mn^{2+} (aq) + 4H_2O (l)$

    This reaction is the basis of its oxidising power. The acidic medium is crucial for this reduction to occur efficiently.
    

    2. The role of acid in permanganate titrations:

    An acidic medium is typically required for permanganate titrations to ensure the complete reduction of $MnO_4^-$ to $Mn^{2+}$. However, the choice of acid is critical. Strong acids like sulfuric acid ($H_2SO_4$) are usually preferred because the sulfate ions ($SO_4^{2-}$) are very stable and are not easily oxidised by $MnO_4^-$.

    3. Interference by hydrochloric acid (HCl): Hydrochloric acid (HCl) provides the necessary $H^+$ ions for the permanganate reduction. However, it also introduces chloride ions ($Cl^-$) into the solution. Chloride ions are susceptible to oxidation by strong oxidising agents like $MnO_4^-$. The chloride ions (oxidation state -1) can be oxidised to elemental chlorine ($Cl_2$), where chlorine has an oxidation state of 0.

    $2Cl^- (aq) \rightarrow Cl_2 (g) + 2e^-$

    4. The side reaction: When KMnO$_4$ is added to a solution containing HCl, it will not only react with the intended analyte but also with the chloride ions from HCl. This constitutes a side reaction:

    $2MnO_4^- (aq) + 10Cl^- (aq) + 16H^+ (aq) \rightarrow 2Mn^{2+} (aq) + 5Cl_2 (g) + 8H_2O (l)$

    This reaction shows that permanganate oxidises chloride ions to chlorine gas. …
    

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11. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.What are X and Y in the following reaction sequence? $\ce{C4H10} \xrightarrow{\ce{KMnO4}} \ce{C4H10O} \xrightarrow[\text{358K}]{\text{20\% H3PO4}} X \xrightarrow[\text{H}^+]{\ce{KMnO4}} Y$ (A) \chemfig{C(-[2]H)(-[6]H)(-[4]CH_3)} ; \ce{O + HCOOH} (B) \chemfig{C(-[2]H)(-[6]H)(-[4]CH_3)} ; \ce{O + CO_2 + H_2O} (C) \chemfig{C(-[2]CH_2OH)(-[6]CH_2OH)} ; \chemfig{C(-[2]OH)(-[6]OH)} (D) \chemfig{C(-[2]CH_2CH_3)(-[6]CH_2CH_3)} ; \ce{CH_3COOH}

    ›Reveal solutionSolution

    Butane is oxidised to $\text{tert}$-butanol, which dehydrates to 2-methylpropene ($X$); hot acidic $\ce{KMnO4}$ cleaves the terminal $\ce{=CH2}$ to give acetone together with $\ce{CO2}$ and $\ce{H2O}$ ($Y$) — option (B).

    Step 1 — $\ce{C4H10 ->[KMnO4] C4H10O}$:

    Butane is oxidised to the alcohol $\ce{C4H10O}$, i.e. 2-methylpropan-2-ol (tert-butyl alcohol), $\ce{(CH3)3C-OH}$.

    Step 2 — dehydration ($\text{20\% } \ce{H3PO4}$, $358\ \text{K}$) gives $X$:

    Acid-catalysed dehydration removes water to form the alkene 2-methylpropene (isobutylene):

    $\ce{(CH3)3C-OH ->[H3PO4][358 K] (CH3)2C=CH2 \; (X) + H2O}$

    Step 3 — oxidative cleavage ($\ce{KMnO4}/\text{H}^+$) gives $Y$: …

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12. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.Identify the oxidizing reactions of KMnO$_4$ in acidic medium I) Liberation of iodine from KI II) Conversion of Fe$^{2+}$ to Fe$^{3+}$ III) Oxidation of nitrite to nitrate IV) Oxidation of iodide to iodate The correct option is (A) II, III, IV only (B) I, II, IV only (C) I, III, IV only (D) I, II, III only

    ›Reveal solutionSolution

    In acidic medium, KMnO₄ acts as a strong oxidizing agent, reducing MnO₄⁻ to Mn²⁺. It oxidizes I⁻ to I₂ (not IO₃⁻), Fe²⁺ to Fe³⁺, and NO₂⁻ to NO₃⁻. Thus, reactions I, II, and III are correct, but IV is not. The correct option is (D).

    Concept & Intuition

    Potassium permanganate (KMnO₄) in acidic medium is one of the most powerful common oxidants. Its half-reaction is:

    $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \quad (E^\circ = +1.51\text{ V})$

    This high reduction potential means it can oxidize many species, but the extent of oxidation depends on the reducing agent and conditions. For iodide (I⁻), the product can be I₂ or IO₃⁻ depending on pH and concentration — but in acidic medium, the typical product is I₂, not IO₃⁻. Let’s check each reaction.

    Step-by-step reasoning

    1. Reaction I: Liberation of iodine from KI KI provides I⁻ ions. In acidic medium, KMnO₄ oxidizes I⁻ to I₂:

    $2\text{MnO}_4^- + 10\text{I}^- + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{I}_2 + 8\text{H}_2\text{O}$

    This is a classic redox reaction — the purple color fades and brown iodine appears. So I is correct.

    2. Reaction II: Conversion of Fe²⁺ to Fe³⁺ Fe²⁺ is easily oxidized to Fe³⁺ by KMnO₄ in acid:

    $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$

    This is a standard titration reaction. So II is correct.

    3. Reaction III: Oxidation of nitrite to nitrate Nitrite (NO₂⁻) is oxidized to nitrate (NO₃⁻) by KMnO₄ in acid:

    $2\text{MnO}_4^- + 5\text{NO}_2^- + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{NO}_3^- + 3\text{H}_2\text{O}$

    The nitrogen goes from +3 to +5 oxidation state. So III is correct.

    4. Reaction IV: Oxidation of iodide to iodate …

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