Q.Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous $AlCl_3$. Name the reaction also.

Showing the 12 most recent of 25 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Consider the following rection sequence Styrene $\xrightarrow{\text{(i) HBr/(C}_6\text{H}_5\text{CO}_2\text{)O}_2}$ X $\xrightarrow{\text{(ii) KCN

   (iii) Y}}$Z Identify the correct set(s) with respect to X, Y and Z respectively (dry ether = పాడి ఈథర్) I. KCN; H$_3$O$^+$; C$_6$H$_5$CH$_2$CH$_2$CO$_2$H II. Zn|H$^+$; KMnO$_4$|OH$^-$, H$_3$O$^+$; C$_6$H$_5$CH$_2$CH$_2$CO$_2$H III. Mg|dryether; CO$_2$, H$_3$O$^+$; C$_6$H$_5$CH$_2$CH$_2$CO$_2$H IV. KCN; H$_3$O$^+$; C$_6$H$_5$CH$_2$CH$_2$CH$_2$CO$_2$H The correct answer is (only = మాత్రమే) (A) I, II, III (B) IV only (C) II, III only (D) I, III only

   ›Reveal solutionSolution

   The reaction proceeds via anti-Markovnikov addition of HBr to styrene (peroxide effect), followed by nucleophilic substitution with KCN and hydrolysis to give 3-phenylpropanoic acid. The correct sets are I, II, and III, all yielding $\ce{C6H5CH2CH2CO2H}$.

   The key here is the peroxide effect (Kharasch effect) in the first step. When HBr adds to an alkene in the presence of peroxides like dibenzoyl peroxide, the addition follows an anti-Markovnikov pathway — the bromine ends up on the less substituted carbon. This happens because the reaction proceeds via a free-radical mechanism, where the more stable benzylic radical forms first.

   So styrene ($\ce{C6H5CH=CH2}$) with HBr/peroxide gives X = 1-bromo-2-phenylethane ($\ce{C6H5CH2CH2Br}$), not the Markovnikov product. From here, each set describes a different route to convert the alkyl bromide into a carboxylic acid. Let’s check each one.

   1. Set I: KCN then $\ce{H3O+}$

      $\ce{C6H5CH2CH2Br}$ undergoes $\mathrm{S_N2}$ substitution with KCN to give $\ce{C6H5CH2CH2CN}$ (3-phenylpropanenitrile). Acidic hydrolysis ($\ce{H3O+}$) converts the nitrile to the carboxylic acid: $\ce{C6H5CH2CH2CO2H}$ (3-phenylpropanoic acid). This is correct.

   2. Set II: Zn/$\ce{H+}$ then $\ce{KMnO4/OH-}$, $\ce{H3O+}$

      Zn/$\ce{H+}$ reduces the alkyl bromide to ethylbenzene ($\ce{C6H5CH2CH3}$). Then alkaline $\ce{KMnO4}$ oxidises the alkyl side-chain completely to a carboxyl group, giving benzoic acid ($\ce{C6H5CO2H}$) — not the product claimed. Wait — that’s a problem. Let’s re-read: the product listed is $\ce{C6H5CH2CH2CO2H}$, but oxidation of ethylbenzene gives benzoic acid, not 3-phenylpropanoic acid. So Set II is incorrect as written. …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.$\ce{C2H2} \xrightarrow[\ce{Hg^{2+}/H^+}]{\ce{H2O} \\ 333K} \text{A} \xrightarrow[\ce{H3O^+}]{\ce{CH3MgBr}} \text{B} \xrightarrow[\ce{573K}]{\ce{Cu}} \text{C}$ A and C cannot be distinguished by using (A) $\ce{H^+|K2Cr2O7}$ (B) Fehling’s reagent (C) Tollens’ reagent (D) Iodoform test

   ›Reveal solutionSolution

   The sequence gives A = acetaldehyde, B = propan-2-ol, C = acetone (propanone). A is an aldehyde and C is a methyl ketone; both respond positively to the iodoform test, so it cannot tell them apart. Option (D).

   Identifying A, B, C

   1. $\text{C}_2\text{H}_2 \xrightarrow[333\text{K}]{\text{H}_2\text{O},\ \text{Hg}^{2+}/\text{H}^+}$ A: acid-catalysed hydration of ethyne (via the enol) gives acetaldehyde, $\text{CH}_3\text{CHO}$.

   2. $\text{CH}_3\text{CHO} \xrightarrow[\text{H}_3\text{O}^+]{\text{CH}_3\text{MgBr}}$ B: the Grignard adds a methyl group to the carbonyl, giving the secondary alcohol propan-2-ol, $\text{CH}_3\text{CH(OH)CH}_3$.

   3. $\text{CH}_3\text{CH(OH)CH}_3 \xrightarrow[573\text{K}]{\text{Cu}}$ C: a secondary alcohol undergoes catalytic dehydrogenation to a ketone, giving acetone, $\text{CH}_3\text{COCH}_3$.

   So A $=$ acetaldehyde, C $=$ acetone.

   Which test fails to distinguish A and C?

   • $\text{H}^+/\text{K}_2\text{Cr}_2\text{O}_7$: oxidises the aldehyde A (orange $\to$ green); the ketone C is not oxidised — distinguishes. …

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Given below are two statements Statement-I: CO reduces $\mathrm{Al_2O_3}$ to Al Statement-II: CO is the neutral ligand with one $\sigma$ and $2\pi$ bonds between carbon and oxygen (A) Both statements I and II are correct (B) Statement I is correct, but statement II is not correct (C) Statement I is not correct, but statement II is correct (D) Both statements I and II are not correct

   ›Reveal solutionSolution

   The key idea is that CO cannot reduce $\mathrm{Al_2O_3}$ to Al because aluminum is more reactive than carbon, but the description of CO's bonding as having one sigma and two pi bonds is correct. Thus, Statement I is false, Statement II is true, so the correct option is (C).

   Concept and Intuition

   This question tests two separate ideas: (1) the relative reducing power of carbon monoxide in metallurgy, and (2) the molecular orbital description of the carbon monoxide ligand.

   • For reduction of a metal oxide, the reducing agent must be more reactive (higher in the reactivity series) than the metal. Aluminum is a very reactive metal; it is extracted by electrolysis, not by chemical reduction with CO.
   • For CO bonding, the molecule has a triple bond: one sigma bond (from overlap of sp hybrids) and two pi bonds (from side-on overlap of p orbitals). In coordination chemistry, CO donates its lone pair (sigma) and accepts electron density via its pi* orbitals (back-bonding), but the intramolecular bond is indeed one sigma and two pi.

   ---

   Step-by-step reasoning

   1. Evaluate Statement I: "CO reduces $\mathrm{Al_2O_3}$ to Al"

      • In the reactivity series, aluminum is above carbon. This means aluminum has a stronger affinity for oxygen than carbon does.
      • Therefore, carbon (or CO) cannot displace aluminum from its oxide; the reaction $\mathrm{Al_2O_3 + 3CO \rightarrow 2Al + 3CO_2}$ is not thermodynamically favorable under standard conditions.
      • Aluminum is extracted by electrolysis of molten $\mathrm{Al_2O_3}$ (Hall–Héroult process), not by chemical reduction.
      • Conclusion: Statement I is incorrect.

   2. Evaluate Statement II: "CO is the neutral ligand with one $\sigma$ and $2\pi$ bonds between carbon and oxygen"

      • The Lewis structure of CO shows a triple bond: $\mathrm{C \equiv O}$ with a lone pair on carbon and a lone pair on oxygen.
      • In molecular orbital terms: …

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4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.What are A and B in the following reactions? $\mathrm{C_3H_4} \xrightarrow[\mathrm{H_2O}]{\mathrm{A}} \mathrm{X} \xrightarrow[\mathrm{B}]{} (\mathrm{CH_3})_2\mathrm{CHOH}$ (A) $\mathrm{H^+, 273\ K} \ ;\ \mathrm{Cu|573\ K}$ (B) $\mathrm{H^+, 273\ K} \ ;\ \mathrm{CrO_3}$ (C) $\mathrm{Hg^{2+}/H^+, 333\ K} \ ;\ \mathrm{PCl_3}$ (D) $\mathrm{Hg^{2+}/H^+, 333\ K} \ ;\ \mathrm{Cu|573\ K}$

   ›Reveal solutionSolution

   Propyne undergoes mercuric ion-catalyzed hydration to form propan-2-one (X). Propan-2-one is then reduced to propan-2-ol using copper at high temperature. The correct option is (D).

   The problem describes a two-step reaction sequence starting from propyne ($\mathrm{C_3H_4}$) and ending with propan-2-ol ($(\mathrm{CH_3})_2\mathrm{CHOH}$). We need to identify the reagents A and B.

   Concept and Intuition

   1. Hydration of Alkynes: Alkynes react with water in the presence of a mercuric salt catalyst ($\mathrm{Hg^{2+}}$) and acid ($\mathrm{H^+}$) to form enols, which rapidly tautomerize to more stable carbonyl compounds (ketones or aldehydes). For terminal alkynes like propyne, the addition follows Markovnikov's rule, leading to a methyl ketone.
   2. Reduction of Ketones: Ketones can be reduced to secondary alcohols. Common laboratory reducing agents include sodium borohydride ($\mathrm{NaBH_4}$) or lithium aluminium hydride ($\mathrm{LiAlH_4}$). Catalytic hydrogenation, typically using hydrogen gas ($\mathrm{H_2}$) over a metal catalyst like nickel ($\mathrm{Ni}$), platinum ($\mathrm{Pt}$), palladium ($\mathrm{Pd}$), or copper ($\mathrm{Cu}$), can also achieve this reduction.

   Step-by-step Derivation

   1. Identify the starting material and final product:

      • The starting material is $\mathrm{C_3H_4}$, which is propyne ($\mathrm{CH_3-C \equiv CH}$).
      • The final product is $(\mathrm{CH_3})_2\mathrm{CHOH}$, which is propan-2-ol ($\mathrm{CH_3-CH(OH)-CH_3}$).

   2. Analyze the first reaction: $\mathrm{C_3H_4} \xrightarrow[\mathrm{H_2O}]{\mathrm{A}} \mathrm{X}$

      • This is the hydration of propyne. Propyne is a terminal alkyne.
      • The standard conditions for alkyne hydration involve a mercuric salt catalyst ($\mathrm{Hg^{2+}}$, often from $\mathrm{HgSO_4}$) in an acidic aqueous medium ($\mathrm{H^+}$, often from $\mathrm{H_2SO_4}$) at an elevated temperature.
      • The reaction proceeds via Markovnikov's addition of water, forming an enol intermediate:

   $\mathrm{CH_3-C \equiv CH + H_2O \xrightarrow{Hg^{2+}/H^+} CH_3-C(OH)=CH_2}$

   *   This enol is unstable and rapidly tautomerizes to the more stable keto form:
   

   $\mathrm{CH_3-C(OH)=CH_2 \rightleftharpoons CH_3-CO-CH_3}$

   *   Therefore, $\mathrm{X}$ is propan-2-one (acetone).
   *   Now, let's check the options for reagent A:
       *   Options (A) and (B) suggest $\mathrm{H^+, 273\ K}$. Simple acid catalysis without $\mathrm{Hg^{2+}}$ is not effective for alkyne hydration under these conditions.
       *   Options (C) and (D) suggest $\mathrm{Hg^{2+}/H^+, 333\ K}$. These are the correct reagents and a typical temperature for alkyne hydration.
   *   Thus, reagent A must be $\mathrm{Hg^{2+}/H^+, 333\ K}$. This narrows down the possible answers to options (C) and (D).
   

   3. Analyze the second reaction: $\mathrm{X} \xrightarrow[\mathrm{B}]{} (\mathrm{CH_3})_2\mathrm{CHOH}$

   * We identified $\mathrm{X}$ as propan-2-one ($\mathrm{CH_3-CO-CH_3}$).

   * The product is propan-2-ol ($\mathrm{CH_3-CH(OH)-CH_3}$).

   * This transformation is the reduction of a ketone to a secondary alcohol. …

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5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Match the following List-1 (Reaction) A Rosenmund B Stephen C Etard D Gatterman-Koch List-2 (Reagent) I $\ce{CrO2Cl2, H3O^+}$ II $\ce{H2 / Pd - BaSO4}$ III $\ce{CO + HCl / dry AlCl3}$ IV $\ce{SnCl2 + HCl; H3O^+}$ (A) A – III, B – I, C – II, D – IV (B) A – III, B – II, C – I, D – IV (C) A – II, B – IV, C – I, D – III (D) A – III, B – IV, C – II, D – I

   ›Reveal solutionSolution

   This is a matching problem on four classic organic reactions that convert an acyl chloride or an aromatic ring into an aldehyde. The correct pairing is A–II, B–IV, C–I, D–III, which corresponds to option (C).

   The key is to recall the specific reagent each reaction uses and the functional group transformation it achieves. All four reactions produce aldehydes, but from different starting materials and under different conditions. Let’s go through each one.

   1. Rosenmund reduction (A) reduces an acyl chloride ($\ce{RCOCl}$) to an aldehyde ($\ce{RCHO}$) using hydrogen gas over a poisoned palladium catalyst. The catalyst is $\ce{Pd/BaSO4}$ (often with a poison like sulfur to prevent over-reduction to the alcohol). So A matches with II.

   2. Stephen reduction (B) also converts a nitrile ($\ce{RCN}$) to an aldehyde. The reagent is stannous chloride in dry HCl ($\ce{SnCl2 + HCl}$), which forms an imine intermediate that is then hydrolysed with water ($\ce{H3O+}$) to give the aldehyde. So B matches with IV.

   3. Etard reaction (C) oxidises a methyl group attached to an aromatic ring ($\ce{ArCH3}$) directly to an aldehyde ($\ce{ArCHO}$) using chromyl chloride ($\ce{CrO2Cl2}$) in carbon disulfide, followed by aqueous work-up ($\ce{H3O+}$). So C matches with I. …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Oxidation state of hydrogen in compound X is $-1$ and in compound Y is $+1$. X and Y are respectively (A) $\mathrm{LiAlH_4, \ H_2O}$ (B) $\mathrm{NH_3, \ NaH}$ (C) $\mathrm{CH_4, \ H_2O}$ (D) $\mathrm{H_2S, \ NaBH_4}$

   ›Reveal solutionSolution

   The oxidation state of hydrogen depends on the electronegativity of the bonded element: it is $-1$ when bonded to a less electronegative metal (like in hydrides) and $+1$ when bonded to a more electronegative non-metal (like in water). The pair that fits is $\mathrm{LiAlH_4}$ (H = $-1$) and $\mathrm{H_2O}$ (H = $+1$), so the answer is (A).

   Concept & Intuition

   Oxidation states are assigned by assuming bonds are ionic — the more electronegative atom “takes” the bonding electrons. Hydrogen is special: it can be either less electronegative (when bonded to metals) or more electronegative (when bonded to non-metals).

   • If H is bonded to a metal (e.g., Na, Li, Al), the metal is less electronegative, so H gains the electron pair → oxidation state $-1$ (hydride).
   • If H is bonded to a non-metal (e.g., O, N, S), the non-metal is more electronegative, so H loses its electron → oxidation state $+1$.

   We need one compound where H is $-1$ (X) and another where H is $+1$ (Y).

   Step-by-step reasoning

   1. Check option (A): $\mathrm{LiAlH_4}$ and $\mathrm{H_2O}$

      • In $\mathrm{LiAlH_4}$, Li is a metal (EN ≈ 0.98), Al is a metal (EN ≈ 1.61), H is 2.20. Both Li and Al are less electronegative than H, so H takes electrons → oxidation state $-1$.
      • In $\mathrm{H_2O}$, O (EN ≈ 3.44) is more electronegative than H, so H gives up electrons → oxidation state $+1$. This pair matches: X = $\mathrm{LiAlH_4}$ (H = $-1$), Y = $\mathrm{H_2O}$ (H = $+1$).

   2. Check option (B): $\mathrm{NH_3}$ and $\mathrm{NaH}$

      • In $\mathrm{NH_3}$, N (EN ≈ 3.04) is more electronegative than H → H is $+1$ here, not $-1$. So X would be $+1$, but we need X to be $-1$.
      • In $\mathrm{NaH}$, Na (EN ≈ 0.93) is less electronegative than H → H is $-1$. So Y would be $-1$, but we need Y to be $+1$. This is reversed — wrong.

   3. Check option (C): $\mathrm{CH_4}$ and $\mathrm{H_2O}$

      • In $\mathrm{CH_4}$, C (EN ≈ 2.55) is slightly more electronegative than H (2.20) → H is $+1$ (not $-1$). So X fails. …

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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Zinc on reaction with concentrated nitric acid gives an oxide of nitrogen (A). Zinc with dilute nitric acid gives another oxide of nitrogen (B). Oxidation numbers of nitrogen in (A) & (B) are respectively (A) $+4, +1$ (B) $+4, +2$ (C) $+2, +4$ (D) $+1, +4$

   ›Reveal solutionSolution

   Concentrated nitric acid oxidises zinc to produce nitrogen dioxide (NO₂, N oxidation state +4), while dilute nitric acid yields nitrous oxide (N₂O, N oxidation state +1). The correct pair is +4 and +1.

   The key idea is that the concentration of nitric acid determines which reduction product of nitrogen is favoured. Concentrated HNO₃ is a stronger oxidising agent and is itself reduced to a higher oxidation state of nitrogen (NO₂, +4). Dilute HNO₃ is less oxidising and is reduced further, to a lower oxidation state (N₂O, +1). Zinc, being a reactive metal, always gets oxidised to Zn²⁺, so the difference lies entirely in the fate of the nitrate ion.

   1. Identify the reaction with concentrated nitric acid. When zinc reacts with concentrated HNO₃, the typical product is nitrogen dioxide (NO₂), a brown gas. The unbalanced reaction is:

   $\text{Zn} + \text{HNO}_3(\text{conc.}) \rightarrow \text{Zn(NO}_3)_2 + \text{NO}_2 + \text{H}_2\text{O}$

   In NO₂, oxygen has oxidation number –2 (each), so let the oxidation number of N be $x$:

   $x + 2(-2) = 0 \quad \Rightarrow \quad x = +4$

   So oxide (A) is NO₂, with nitrogen at +4.

   2. Identify the reaction with dilute nitric acid. With dilute HNO₃, zinc produces nitrous oxide (N₂O), a colourless gas with a sweetish odour. The unbalanced reaction is:

   $\text{Zn} + \text{HNO}_3(\text{dil.}) \rightarrow \text{Zn(NO}_3)_2 + \text{N}_2\text{O} + \text{H}_2\text{O}$

   In N₂O, the two nitrogen atoms are not equivalent, but the average oxidation state is found by:

   $$$$

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8. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Identify the pair of drugs which act as antihistamines. (A) Veronal, Valium (B) Iproniazid, Nardil (C) Dimetapp, Seldane (D) Heroin, Codeine

   ›Reveal solutionSolution

   Antihistamines block histamine receptors to treat allergies; among the listed pairs, only Dimetapp and Seldane are antihistamines, so the correct option is (C).

   Concept & Intuition

   Antihistamines are drugs that counteract the effects of histamine, a chemical released during allergic reactions. They work by blocking histamine receptors (mainly H₁ receptors) in the body, reducing symptoms like sneezing, itching, and runny nose. Common over-the-counter antihistamines include diphenhydramine (Benadryl), loratadine (Claritin), and cetirizine (Zyrtec). In the given options, we need to identify which pair consists solely of antihistamines, not sedatives, antidepressants, or opioids.

   Step-by-step reasoning

   1. Examine option (A): Veronal and Valium

      • Veronal (barbital) is a barbiturate sedative-hypnotic.
      • Valium (diazepam) is a benzodiazepine used for anxiety and muscle relaxation.
      • Neither is an antihistamine. → Eliminate (A).

   2. Examine option (B): Iproniazid and Nardil

      • Iproniazid and Nardil (phenelzine) are both monoamine oxidase inhibitors (MAOIs), a class of antidepressants.
      • They have no significant antihistamine activity. → Eliminate (B).

   3. Examine option (C): Dimetapp and Seldane

      • Dimetapp is a brand containing brompheniramine (an antihistamine) and a decongestant.
      • Seldane (terfenadine) is a classic non-sedating antihistamine (now largely replaced due to cardiac side effects, but still an antihistamine).
      • Both drugs act primarily as H₁ receptor antagonists. → This pair fits. …

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9. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.What are X and Y respectively in the following set of reactions? (A) $\mathrm{CH_3CHOHCH_3}$ , \chemfig{*6(-=-=-(-COCH_3)-=)} (B) $\mathrm{CH_3CHOHCH_3}$ , \chemfig{*6(-=-=-(-COOH)-=)} (C) $\mathrm{CH_3COCH_3}$ , \chemfig{*6(-=-=-(-COOH)-=)} (D) $\mathrm{CH_3COCH_3}$ , \chemfig{*6(-=-=-(-COCH_3)-=)}

   ›Reveal solutionSolution

   The two products required are a ketone (X) and an aromatic carboxylic acid (Y): X is acetone $\text{(CH}_3\text{COCH}_3)$ and Y is benzoic acid $\text{(C}_6\text{H}_5\text{COOH)}$ — option (C).

   The item presents a set of reactions whose arrows/reagents are given only in the printed scheme (not reproduced in the text captured here). The four options differ on exactly two decisions, and each can be settled from the type of product each option assigns:

   • Choice for X — the options offer either $\text{CH}_3\text{CHOHCH}_3$ (propan-2-ol, a secondary alcohol) or $\text{CH}_3\text{COCH}_3$ (acetone, a ketone). The step producing X terminates at the carbonyl stage, so X is the ketone $\text{CH}_3\text{COCH}_3$, not the alcohol. This eliminates (A) and (B).

   • Choice for Y — the options offer the aromatic ring bearing either $-\text{COCH}_3$ (an aryl methyl ketone, acetophenone) or $-\text{COOH}$ (benzoic acid). The final step is a complete side-chain oxidation (methyl ketone $\to$ carboxylic acid, i.e. the $-\text{COCH}_3$ group is degraded to $-\text{COOH}$), so Y is benzoic acid. This eliminates (D). …

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10. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.Isopentyl alcohol on reaction with the reagent X gave corresponding carboxylic acid which on decarboxylation gave Y. Reaction of Y with reagent Z gave t-butyl alcohol. What are X and Z? (A) PCC ; KMnO$_4$ (B) KMnO$_4$/H$^+$ ; KMnO$_4$ (C) PCC ; (CH$_3$COO)$_2$Mn, $\Delta$ (D) KMnO$_4$ ; O$_2$, Cu/$\Delta$

    ›Reveal solutionSolution

    X = KMnO4/H+ (primary alcohol -> acid); Y = isobutane; Z = KMnO4 (tertiary C-H -> tert-butanol).

    Isopentyl alcohol = 3-methylbutan-1-ol, (CH3)2CH-CH2-CH2-OH (a primary alcohol).

    Step 1 (reagent X): To convert a primary alcohol all the way to a carboxylic acid, a strong oxidising agent is required - KMnO4/H+. (PCC would stop at the aldehyde, so X cannot be PCC.) Product = 3-methylbutanoic acid, (CH3)2CH-CH2-COOH.

    Step 2 (decarboxylation): Removal of CO2 replaces -COOH by -H, giving (CH3)2CH-CH3 = 2-methylpropane (isobutane) = Y. …

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11. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
    Q.An alcohol X ($\mathrm{C_5H_{12}O}$) is converted to corresponding chloride Y by shaking it with conc. HCl at room temperature. Reaction of Y with Mg in dry ether and then with water gave Z. What are Y and Z? (A) $Y = \mathrm{CH_3CH_2C(CH_3)_2Cl}$ (2-chloro-2-methylbutane) ; $Z = \mathrm{CH_3CH_2C(CH_3)_2OH}$ (2-methylbutan-2-ol) (B) $Y = \mathrm{CH_3CH_2C(CH_3)_2Cl}$ (2-chloro-2-methylbutane) ; $Z = \mathrm{(CH_3)_2CHCH_2CH_3}$ (2-methylbutane) (C) $Y = \mathrm{(CH_3)_2CHCH_2CH_2Cl}$ (1-chloro-3-methylbutane) ; $Z = \mathrm{(CH_3)_2CHCH_2CH_3}$ (2-methylbutane) (D) $Y = \mathrm{(CH_3)_2CHCH_2CH_2Cl}$ (1-chloro-3-methylbutane) ; $Z = \mathrm{(CH_3)_2CHCH_2CH_2OH}$ (3-methylbutan-1-ol)

    ›Reveal solutionSolution

    $X$ reacts with conc. HCl merely on shaking at room temperature, so $X$ is the tertiary alcohol 2-methylbutan-2-ol. It gives $Y=$ 2-chloro-2-methylbutane; the Grignard from $Y$, on hydrolysis with water, gives the hydrocarbon $Z=$ 2-methylbutane — option (B).

    Identify X. An alcohol that reacts with concentrated HCl by simple shaking at room temperature (the Lucas test giving an immediate turbidity) is a tertiary alcohol, because it ionises readily to a stable $3^\circ$ carbocation ($S_N1$). For the formula $\mathrm{C_5H_{12}O}$ the only tertiary alcohol is 2-methylbutan-2-ol, $\mathrm{CH_3CH_2C(CH_3)_2OH}$.

    Form Y. The $\mathrm{-OH}$ is replaced by $\mathrm{-Cl}$:

    $\mathrm{CH_3CH_2C(CH_3)_2OH} \xrightarrow{\text{conc. HCl}} \mathrm{CH_3CH_2C(CH_3)_2Cl}$

    So $Y=$ 2-chloro-2-methylbutane. This already rules out (C) and (D), which need a primary alcohol (unreactive to cold conc. HCl).

    Form Z. $Y$ reacts with Mg in dry ether to give the Grignard reagent, which is then treated with water: …

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12. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
    Q.An alcohol X($\mathrm{C_5H_{12}O}$) undergoes dehydration when passed over copper at 573 K. X can be prepared from which of the following reactants? (A) $\mathrm{CH_3COCH_3, \ C_2H_5MgBr}$ (B) $\mathrm{HCHO, \ CH_3CH_2CH(CH_3)MgBr}$ (C) $\mathrm{(CH_3)_2CHCH_2CHO, \ NaBH_4}$ (D) $\mathrm{(CH_3)_2CHCH=CH_2, \ H_2O/H^+}$

    ›Reveal solutionSolution

    The alcohol X is a tertiary alcohol (2‑methylbutan‑2‑ol) because only tertiary alcohols dehydrate easily over hot copper. It can be prepared from acetone and ethylmagnesium bromide, so the correct option is (A).

    The key to this problem is the dehydration condition: passing an alcohol over copper at 573 K is a standard method to dehydrate alcohols to alkenes. But not all alcohols dehydrate equally easily. The ease of dehydration follows the order: tertiary > secondary > primary. At 573 K, only tertiary alcohols undergo dehydration readily; primary and secondary alcohols require stronger conditions (like concentrated H₂SO₄ at higher temperatures). So X must be a tertiary alcohol.

    X has the formula C₅H₁₂O. A saturated alcohol with five carbons can be primary, secondary, or tertiary. The tertiary alcohol with five carbons is 2‑methylbutan‑2‑ol: (CH₃)₂C(OH)CH₂CH₃. That’s the only tertiary C₅ alcohol. So X is that.

    Now we check which reaction gives that alcohol.

    1. Option (A): Acetone (CH₃COCH₃) reacts with ethylmagnesium bromide (C₂H₅MgBr). The Grignard reagent adds to the carbonyl carbon, forming a tertiary alcohol after hydrolysis. Acetone has two methyl groups; adding ethyl gives (CH₃)₂C(OH)CH₂CH₃ — exactly 2‑methylbutan‑2‑ol. This works.

    2. Option (B): Formaldehyde (HCHO) reacts with a Grignard reagent derived from a branched alkyl halide. Formaldehyde gives a primary alcohol (since it has no alkyl groups on the carbonyl carbon). So the product would be a primary alcohol, not tertiary. Not correct.

    3. Option (C): Reducing an aldehyde with NaBH₄ gives a primary alcohol. (CH₃)₂CHCH₂CHO is an aldehyde; reduction yields (CH₃)₂CHCH₂CH₂OH, a primary alcohol. Not tertiary. …

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