Q.Predict the major product of acid catalysed dehydration of

Showing the 12 most recent of 36 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.Observe the following reaction sequence and sets (I to IV) given for X, Y, Z and choose the correct set(s) (A) I only (B) I, III only (C) I, II only (D) III, IV only

   ›Reveal solutionSolution

   The key is to identify the correct products X, Y, and Z in the reaction sequence by applying the principles of the Hunsdiecker reaction, decarboxylation, and Friedel-Crafts acylation. The correct set is I, III only, corresponding to option (B).

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   Concept and Intuition

   This problem tests your ability to predict the products of a multi-step organic synthesis. The sequence involves three classic transformations:

   1. Hunsdiecker reaction (silver salt of a carboxylic acid + Br₂) → alkyl halide (X).
   2. Decarboxylation (sodium salt of a carboxylic acid + sodalime) → hydrocarbon (Y).
   3. Friedel-Crafts acylation (acyl chloride + benzene + AlCl₃) → ketone (Z).

   The trick is to recognize that the starting material is a dicarboxylic acid (succinic acid, HOOC–CH₂–CH₂–COOH). Each step acts on a different functional group, and the sets (I–IV) give possible structures for X, Y, and Z. We must match the correct products.

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   Step-by-Step Reasoning

   1. Identify the starting material and the first reaction (X).

   The starting compound is succinic acid:

   $\text{HOOC–CH}_2\text{–CH}_2\text{–COOH}$

   It is first converted to its silver salt (by reaction with Ag₂O or AgNO₃), then treated with Br₂. This is the Hunsdiecker reaction, which replaces the carboxyl group (–COOAg) with a bromine atom, but only one carboxyl group reacts per molecule because the reaction is typically performed on a mono-silver salt. However, here the problem implies the reaction is done on the di-silver salt, so both carboxyl groups undergo the reaction.

   The product X is:

   $\text{Br–CH}_2\text{–CH}_2\text{–Br} \quad (\text{1,2-dibromoethane})$

   This matches Set I and Set II (both show Br–CH₂–CH₂–Br). So X is the same in I and II.

   2. Determine Y from the second reaction.

   The starting material is treated with NaOH (to form the sodium salt) and then heated with sodalime (CaO + NaOH). This is decarboxylation — each –COONa group is replaced by –H.

   For succinic acid, the sodium salt is NaOOC–CH₂–CH₂–COONa. Decarboxylation removes both carboxyl groups, yielding:

   $\text{CH}_3\text{–CH}_3 \quad (\text{ethane})$

   So Y is ethane (CH₃–CH₃).

   Looking at the sets:

   • Set I: Y = CH₃–CH₃ (correct)
   • Set II: Y = CH₃–CH₂–CH₃ (propane) — wrong
   • Set III: Y = CH₃–CH₃ (correct)
   • Set IV: Y = CH₃–CH₂–CH₃ (propane) — wrong Thus, only Sets I and III have the correct Y.

   3. Determine Z from the third reaction.

   The product Z is formed by reacting X (1,2-dibromoethane) with benzene in the presence of AlCl₃. This is a Friedel-Crafts alkylation, not acylation (the problem says "acylation" but the reagent is an alkyl halide, so it's alkylation). However, note that 1,2-dibromoethane can undergo double alkylation to form a bridged product, but under typical conditions, it reacts twice to give 1,2-diphenylethane:

   $\text{Br–CH}_2\text{–CH}_2\text{–Br} + 2\text{C}_6\text{H}_6 \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{–CH}_2\text{–CH}_2\text{–C}_6\text{H}_5 + 2\text{HBr}$

   This is dibenzyl (1,2-diphenylethane).

   Now check the sets:

   • Set I: Z = C₆H₅–CH₂–CH₂–C₆H₅ (correct)
   • Set II: Z = C₆H₅–CH₂–CH₂–C₆H₅ (correct, but Y is wrong)
   • Set III: Z = C₆H₅–CH₂–CH₂–C₆H₅ (correct)
   • Set IV: Z = C₆H₅–CH₂–CH₂–C₆H₅ (correct, but Y is wrong)

   So only Sets I and III have both Y and Z correct.

   4. Conclusion.

   The correct sets are I and III only.

   Watch out

   A common mistake is to think that the Hunsdiecker reaction only works on mono-carboxylic acids, but here the di-silver salt of succinic acid undergoes the reaction at both ends, giving a dihalide. Also, decarboxylation of a dicarboxylic acid with sodalime removes both –COOH groups, yielding the parent hydrocarbon with the same number of carbons — not a chain-shortened product.

   Tip

   Remember: In the Hunsdiecker reaction, the product has one fewer carbon than the original acid. For a dicarboxylic acid, each carboxyl group loses one carbon, so the product has (n-2) carbons total. Succinic acid (C4) gives C2 product (1,2-dibromoethane).

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Observe the following reaction sequence and the statements given about the product Y $\mathrm{C_3H_6} \xrightarrow{\text{(i) HBr / (C}_6\mathrm{H}_5\mathrm{CO}_2), \text{O}_2} \mathrm{X} \xrightarrow{\text{(i) V}_2\mathrm{O}_5, 773\,\mathrm{K}, 10-20\,\mathrm{atm}} \xrightarrow{\text{(ii) Na / dry ether}} \xrightarrow{\text{(ii) CH}_3\mathrm{Cl} / \text{anhy. AlCl}_3} \mathrm{Y}$ I. Reaction of ‘Y’ with Br$_2$ / UV light forms aryl bromide II. Reaction of ‘Y’ with CrO$_3$ / (CH$_3$CO)$_2$O followed by hydrolysis gives a compound which is acidic in nature (A) Both statements I and II are correct (B) Both statements I and II are not correct (C) Statement I is correct, but statement II is not correct (D) Statement I is not correct, but statement II is correct

   ›Reveal solutionSolution

   The sequence builds up to toluene ($Y$). Br$_2$/UV gives a side-chain benzyl bromide (not an aryl bromide), and CrO$_3$/(CH$_3$CO)$_2$O followed by hydrolysis gives neutral benzaldehyde (not acidic) — so both statements are wrong.

   Following the sequence, propene is converted through anti-Markovnikov HBr addition, Wurtz coupling, aromatisation over V$_2$O$_5$, and Friedel–Crafts methylation with CH$_3$Cl/AlCl$_3$ to give $Y = \text{toluene}\ (\text{C}_6\text{H}_5\text{CH}_3)$.

   Statement I: $\text{C}_6\text{H}_5\text{CH}_3 + \text{Br}_2$ under UV light is a free-radical substitution that occurs at the benzylic (side-chain) position, giving benzyl bromide $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ — a benzylic halide, not an aryl bromide (ring bromination would require Br$_2$/FeBr$_3$). Statement I is incorrect.

   Statement II: $\text{C}_6\text{H}_5\text{CH}_3$ with CrO$_3$/(CH$_3$CO)$_2$O forms benzylidene diacetate, which on hydrolysis gives benzaldehyde $\text{C}_6\text{H}_5\text{CHO}$ — a neutral compound, not acidic. Statement II is incorrect.

   ✓Final answer

   Both statements I and II are not correct — option (B)

3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.The major end product Z in the given sequence of reactions is Phenol ($\mathrm{C_6H_5OH}$) $\xrightarrow[\Delta]{\mathrm{Zn}}$ (A) $\xrightarrow[\text{anhy. AlCl}_3]{\mathrm{(CH_3)_2CHCl}}$ (B) $\xrightarrow[\text{(ii) } \mathrm{H^+,\,H_2O}]{\text{(i) } \mathrm{O_2}}$ (C) $\xrightarrow{\mathrm{Br_2\,|\,CS_2}}$ (Z) (A) Bromobenzene ($\mathrm{C_6H_5Br}$) (B) 3-Bromophenol (benzene ring with OH and Br in meta positions) (C) 4-Bromophenol (benzene ring with OH and Br para to each other) (D) 2,4,6-Tribromophenol (phenol with Br at positions 2, 4 and 6)

   ›Reveal solutionSolution

   Zn/$\Delta$ takes phenol to benzene, Friedel–Crafts gives cumene, the $\mathrm{O_2}$/$\mathrm{H^+}$ cumene process returns phenol, and $\mathrm{Br_2}$ in the non-polar solvent $\mathrm{CS_2}$ gives mainly 4-bromophenol. Option (C).

   Step 1 — Phenol $\xrightarrow{\mathrm{Zn},\ \Delta}$ (A)

   Heating phenol with zinc dust reduces it: the zinc takes the oxygen away as zinc oxide and the ring is left bare.

   $\mathrm{C_6H_5OH} + \mathrm{Zn} \xrightarrow{\ \Delta\ } \mathrm{C_6H_6} + \mathrm{ZnO}$

   $\therefore\ \textbf{(A)} = \text{benzene}$

   This is the standard laboratory conversion of a phenol to the parent arene.

   Step 2 — Benzene $\xrightarrow{(\mathrm{CH_3})_2\mathrm{CHCl},\ \text{anhy. } \mathrm{AlCl_3}}$ (B)

   This is a Friedel–Crafts alkylation. The Lewis acid $\mathrm{AlCl_3}$ abstracts $\mathrm{Cl^-}$ from isopropyl chloride to give the isopropyl carbocation, which attacks the electron-rich benzene ring:

   $(\mathrm{CH_3})_2\mathrm{CHCl} + \mathrm{AlCl_3} \rightarrow (\mathrm{CH_3})_2\overset{+}{\mathrm{C}}\mathrm{H} + \mathrm{AlCl_4^-}$

   $\mathrm{C_6H_6} + (\mathrm{CH_3})_2\overset{+}{\mathrm{C}}\mathrm{H} \rightarrow \mathrm{C_6H_5{-}CH(CH_3)_2}$

   $\therefore\ \textbf{(B)} = \text{isopropylbenzene} = \textbf{cumene}$

   Step 3 — Cumene $\xrightarrow{\text{(i) } \mathrm{O_2},\ \text{(ii) } \mathrm{H^+}/\mathrm{H_2O}}$ (C): the cumene process

   This is the industrial manufacture of phenol, and the naming of the reagents is a giveaway.

   1. Air oxidation. The benzylic C–H of cumene is tertiary and easily abstracted, so $\mathrm{O_2}$ converts cumene into cumene hydroperoxide, $\mathrm{C_6H_5C(CH_3)_2{-}O{-}O{-}H}$.
   2. Acid treatment. Dilute acid triggers a rearrangement in which the phenyl group migrates to oxygen; hydrolysis then cleaves the molecule into two useful products:

   $\mathrm{C_6H_5C(CH_3)_2OOH} \xrightarrow{\ \mathrm{H^+},\ \mathrm{H_2O}\ } \mathrm{C_6H_5OH} + \mathrm{CH_3COCH_3}$

   $\therefore\ \textbf{(C)} = \textbf{phenol} \quad (\text{acetone is the valuable by-product})$

   So the sequence has taken us in a full circle back to phenol — a deliberate touch by the examiner to check whether you really know the cumene process.

   Step 4 — Phenol $\xrightarrow{\mathrm{Br_2}/\mathrm{CS_2}}$ (Z): the solvent decides the answer

   The $-\mathrm{OH}$ group is a powerful activating, ortho/para-directing group ($+\mathrm{R}$ effect), so phenol brominates far more readily than benzene. How many bromines go on depends entirely on the medium:

   Conditions	Product
   $\mathrm{Br_2}$ in water (a polar solvent)	phenol ionises appreciably to the even more activated phenoxide, so bromination runs away $\rightarrow$ 2,4,6-tribromophenol (a white precipitate)
   $\mathrm{Br_2}$ in a non-polar solvent — $\mathrm{CS_2}$ or $\mathrm{CHCl_3}$ — at low temperature	ionisation is suppressed, the ring is less activated, so only monobromination occurs, giving mainly the para isomer, 4-bromophenol (with a little ortho)

   The reaction here is run in $\mathrm{CS_2}$, so:

   $\therefore\ \textbf{(Z)} = \textbf{4-bromophenol}$

   Option (D), 2,4,6-tribromophenol, is the trap for anyone who ignores the solvent; option (B), the meta isomer, is impossible because $-\mathrm{OH}$ is an o/p director; option (A) discards the $-\mathrm{OH}$ altogether.

   ✓Final answer

   The sequence regenerates phenol via the cumene process, and bromination in $\mathrm{CS_2}$ gives the mono-substituted para product, 4-bromophenol.

   ANSWER: C

4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Match the following List-1 (Reaction) List-2 (End product) A. Reimer-Tiemann reaction I. A benzene ring bearing one $-\mathrm{CHO}$ group (benzaldehyde) B. Etard reaction II. A benzene ring bearing $-\mathrm{CHO}$ and $-\mathrm{OH}$ on adjacent carbons (salicylaldehyde) C. Kolbe reaction III. A benzene ring bearing one $-\mathrm{Cl}$ group (chlorobenzene) D. Sandmeyer reaction IV. A benzene ring bearing $-\mathrm{OH}$ and $-\mathrm{COOH}$ on adjacent carbons (salicylic acid) V. A benzene ring bearing one $-\mathrm{COOH}$ group (benzoic acid) The correct answer is (A) A – V, B – III, C – II, D – I (B) A – III, B – IV, C – I, D – II (C) A – II, B – III, C – V, D – IV (D) A – II, B – I, C – IV, D – III

   ›Reveal solutionSolution

   Each named reaction transforms a benzene derivative in a characteristic way: Reimer–Tiemann formylates phenol ortho to the hydroxyl, Etard oxidizes a methyl group to aldehyde, Kolbe carboxylates phenoxide ortho to the oxygen, and Sandmeyer replaces a diazonium group with halide. The correct match is (D).

   These four reactions are classic transformations in aromatic chemistry, each installing a functional group onto a benzene ring under specific conditions. The key is to remember what substrate each requires and where the new group ends up.

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   The reactions and their products

   A. Reimer–Tiemann reaction

   This reaction treats phenol (a benzene ring with $-\mathrm{OH}$) with chloroform ($\mathrm{CHCl_3}$) and strong base (usually aqueous $\mathrm{NaOH}$ or $\mathrm{KOH}$). The base generates dichlorocarbene ($:\mathrm{CCl_2}$), which attacks the activated aromatic ring ortho to the hydroxyl group. After hydrolysis, you obtain an aldehyde group at the ortho position.

   The product is salicylaldehyde: a benzene ring bearing $-\mathrm{OH}$ and $-\mathrm{CHO}$ on adjacent carbons.

   Match: A → II

   ---

   B. Etard reaction

   This oxidizes a methyl group attached to benzene (toluene, $\mathrm{C_6H_5CH_3}$) to an aldehyde using chromyl chloride ($\mathrm{CrO_2Cl_2}$) in an inert solvent like carbon disulfide or carbon tetrachloride. The reaction stops at the aldehyde stage, unlike harsher oxidations that push all the way to carboxylic acid.

   The product is benzaldehyde: a benzene ring bearing one $-\mathrm{CHO}$ group.

   Match: B → I

   ---

   C. Kolbe (Kolbe–Schmitt) reaction

   This treats sodium phenoxide ($\mathrm{C_6H_5O^-Na^+}$, the sodium salt of phenol) with carbon dioxide under pressure and moderate heat (around 125 °C), then acidifies. The $\mathrm{CO_2}$ inserts ortho to the phenoxide oxygen, introducing a carboxyl group.

   The product is salicylic acid: a benzene ring bearing $-\mathrm{OH}$ and $-\mathrm{COOH}$ on adjacent carbons.

   Match: C → IV

   ---

   D. Sandmeyer reaction

   This converts an aryldiazonium salt ($\mathrm{C_6H_5N_2^+X^-}$, prepared from aniline and nitrous acid) into an aryl halide by treatment with a cuprous halide ($\mathrm{CuCl}$, $\mathrm{CuBr}$, or $\mathrm{CuCN}$). The diazonium group is replaced by the halogen.

   When you use $\mathrm{CuCl}$, the product is chlorobenzene: a benzene ring bearing one $-\mathrm{Cl}$ group.

   Match: D → III

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   Tip

   A mnemonic: Reimer gives Ring + OH + CHO (salicylaldehyde), Kolbe gives Karboxyl next to OH (salicylic acid), Etard stops at aldEhyde, Sandmeyer Substitutes the diazonium.

   Putting it all together:

   Reaction	Product
   A. Reimer–Tiemann	II. Salicylaldehyde ($-\mathrm{OH}$ and $-\mathrm{CHO}$ ortho)
   B. Etard	I. Benzaldehyde ($-\mathrm{CHO}$)
   C. Kolbe	IV. Salicylic acid ($-\mathrm{OH}$ and $-\mathrm{COOH}$ ortho)
   D. Sandmeyer	III. Chlorobenzene ($-\mathrm{Cl}$)

   This is A – II, B – I, C – IV, D – III.

   ✓Final answer

   The correct option is (D).

5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.A carbonyl compound (A) on reaction with Grignard reagent (B) followed by hydrolysis gives an alcohol. This on reaction with

   (i) PBr$_3$ and

   (ii) Zn | dil. H$^+$ gives an alkane. Photochemical chlorination of alkane gives only one monochloro derivative. A and B respectively are (A) $\begin{array}{c} \chemfig{C(=O)C} \\ \text{,} \\ \chemfig{C-CMgBr} \end{array}$ (B) $\begin{array}{c} \chemfig{C(=O)H} \\ \text{,} \\ \chemfig{C(-C)(-C)MgBr} \end{array}$ (C) $\begin{array}{c} \chemfig{C(=O)CH} \\ \text{,} \\ \chemfig{C-CMgBr} \end{array}$ (D) $\begin{array}{c} \text{HCHO} \\ \text{,} \\ \chemfig{C(-C)(-C)MgBr} \end{array}$

   ›Reveal solutionSolution

   The key is that the alkane formed after the reaction sequence gives only one monochloro derivative on photochemical chlorination, meaning it must be neopentane (2,2-dimethylpropane). Working backwards, the carbonyl compound (A) is formaldehyde (HCHO) and the Grignard reagent (B) is tert-butylmagnesium bromide, which corresponds to option (D).

   The problem is a classic retrosynthesis puzzle in organic chemistry. You are given a sequence of reactions and a final constraint — the alkane produced yields only one monochloro derivative. That constraint is the anchor: it tells you the alkane has all its hydrogen atoms equivalent, or at least that only one type of hydrogen can be replaced by chlorine. The only alkane that gives a single monochloro product is neopentane (2,2-dimethylpropane), because all 12 hydrogens are identical (three equivalent methyl groups attached to a quaternary carbon). Any other alkane would give at least two different monochloro isomers.

   Now we trace the sequence backwards from the alkane.

   1. The alkane is neopentane.

      Neopentane has the structure $(\text{CH}_3)_3\text{C}-\text{CH}_3$. It is formed in the last step: reaction of an alkyl halide with Zn and dilute acid. That step is a reduction — Zn/dil. H⁺ converts an alkyl halide (here, a bromide) into the corresponding alkane. So the alkyl bromide just before that step must be neopentyl bromide, $(\text{CH}_3)_3\text{C}-\text{CH}_2\text{Br}$.

   2. How was neopentyl bromide made?

      The step before that is: alcohol + PBr$_3$ → alkyl bromide. PBr$_3$ replaces the –OH group with –Br, with inversion of configuration if the carbon is chiral, but here the alcohol is neopentyl alcohol, $(\text{CH}_3)_3\text{C}-\text{CH}_2\text{OH}$. So the alcohol in the sequence is neopentyl alcohol.

   3. How was neopentyl alcohol formed?

      It came from a Grignard reaction: a carbonyl compound (A) reacts with a Grignard reagent (B), followed by hydrolysis, to give an alcohol. The alcohol here is a primary alcohol (neopentyl alcohol has the –OH on a primary carbon).

      A Grignard reaction that yields a primary alcohol must involve formaldehyde (HCHO) as the carbonyl compound. Why? Because when a Grignard reagent R–MgX reacts with formaldehyde, the product after hydrolysis is R–CH$_2$OH — a primary alcohol with one more carbon than the Grignard reagent. If the carbonyl were any other aldehyde or a ketone, you would get a secondary or tertiary alcohol.

      So (A) must be HCHO (formaldehyde).

   4. What Grignard reagent gives neopentyl alcohol from HCHO?

      Neopentyl alcohol is $(\text{CH}_3)_3\text{C}-\text{CH}_2\text{OH}$. The alkyl group attached to the –CH$_2$OH is $(\text{CH}_3)_3\text{C}-$ (tert-butyl). Therefore the Grignard reagent must be tert-butylmagnesium bromide, $(\text{CH}_3)_3\text{C}-\text{MgBr}$. That is (B).

   5. Check the options.

      Option (D) lists HCHO and $(\text{CH}_3)_3\text{C}-\text{MgBr}$ — exactly what we deduced. The other options involve different carbonyls (like acetone or acetaldehyde) or different Grignard reagents, which would give secondary or tertiary alcohols, and those would not lead to neopentane after the sequence.

   Watch out

   A common mistake is to think that any alkane with only one type of hydrogen works. But even ethane gives only one monochloro derivative — however, ethane has only 2 carbons, and the sequence here builds a larger molecule. The alcohol formed is neopentyl alcohol, which has 5 carbons, so the alkane must have 5 carbons. Neopentane is the only C5 alkane with all equivalent hydrogens.

   Tip

   The key shortcut: "only one monochloro derivative" almost always points to neopentane or a symmetrically substituted alkane. In exam problems, neopentane is the most common answer because it's the smallest branched alkane with all hydrogens equivalent.

   ✓Final answer

   The correct option is (D).

6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The correct statements about the compounds of boron are I. In borax bead test, the colour of cobalt metaborate is blue II. Diborane is prepared by the oxidation of sodium borohydride with iodine III. In diborane oxidation state of hydrogen is $+1$ IV. Boric acid is a tribasic acid (A) I & II (B) III & IV (C) I & III (D) II & IV

   ›Reveal solutionSolution

   The key is to recall the specific chemistry of boron compounds: the borax bead test gives a blue colour with cobalt, diborane is made from NaBH₄ and I₂, hydrogen in diborane is hydridic (−1), and boric acid is monobasic, not tribasic. Only statements I and II are correct, so the answer is option (A).

   Let’s examine each statement carefully, because boron chemistry is full of subtle but important distinctions.

   Concept & Intuition

   Boron is electron-deficient, which governs its bonding and reactions. In borax, the bead test relies on transition metal borates forming characteristic colours. Diborane (B₂H₆) has a unique “banana bond” structure where hydrogen bridges are hydridic (H⁻), not protonic. Boric acid (H₃BO₃) is actually a weak monobasic acid because it accepts a hydroxide ion rather than donating a proton. Knowing these patterns helps you avoid common traps.

   ---

   1. Statement I: “In borax bead test, the colour of cobalt metaborate is blue”

      • Borax (Na₂B₄O₇·10H₂O) on heating forms a glassy bead of sodium metaborate (NaBO₂) and B₂O₃. When a cobalt salt is added, cobalt metaborate (Co(BO₂)₂) forms.
      • Cobalt metaborate is indeed blue (a classic test for cobalt).
      • ✓ True.

   2. Statement II: “Diborane is prepared by the oxidation of sodium borohydride with iodine”

      • The reaction is:

   $2\text{NaBH}_4 + \text{I}_2 \rightarrow \text{B}_2\text{H}_6 + 2\text{NaI} + \text{H}_2$

   • Iodine oxidises BH₄⁻, coupling two boron centres to form diborane. This is a standard laboratory preparation.
   • ✓ True.

   3. Statement III: “In diborane oxidation state of hydrogen is +1”

      • In B₂H₆, the terminal hydrogens are slightly hydridic (H⁻), and the bridging hydrogens are even more so.
      • Electronegativity: B (2.04) < H (2.20), so hydrogen is slightly more electronegative. In practice, the oxidation state of H in B₂H₆ is −1 (hydride), not +1.
      • ✗ False.

   4. Statement IV: “Boric acid is a tribasic acid”

      • Boric acid, H₃BO₃, does not donate protons directly. Instead, it acts as a Lewis acid:

   $\text{B(OH)}_3 + \text{H}_2\text{O} \rightleftharpoons \text{B(OH)}_4^- + \text{H}^+$

   • It is monobasic (only one H⁺ is released per molecule). The three hydroxyl groups are not all acidic.
   • ✗ False.

   Watch out

   A common mistake is to count the three OH groups in boric acid and assume it’s tribasic. But boron’s electron deficiency makes it accept an OH⁻, not donate H⁺. Only one proton is liberated.

   ---

   Conclusion: Only statements I and II are correct.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

7. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Which of the following reactions is an example of Clemmensen reduction? (A) $\mathrm{R{-}C(=O){-}Cl} + \mathrm{H_2} \xrightarrow{\mathrm{Pd}/\mathrm{BaSO_4}} \mathrm{R{-}C(=O){-}H} + \mathrm{HCl}$ (B) $\mathrm{R{-}C(=O){-}H} \xrightarrow{\begin{array}{l} (i)\\\mathrm{NH_2{-}NH_2} \\ (ii)\\\mathrm{KOH} \left| \begin{array}{c} \mathrm{CH_2{-}CH_2} \\ \mathrm{OH\\\\\\OH} \end{array} \right. \end{array}} \mathrm{R{-}CH_3}$ (C) $\mathrm{R{-}C(=O){-}OC_2H_5} \xrightarrow{\begin{array}{l} 1.\\\mathrm{DIBAL{-}H} \\ 2.\\\mathrm{H_2O} \end{array}} \mathrm{R{-}C(=O){-}H} + \mathrm{C_2H_5OH}$ (D) $\mathrm{R{-}C(=O){-}CH_3} \xrightarrow{\mathrm{Zn{-}Hg}/\mathrm{HCl}} \mathrm{R{-}CH_2{-}CH_3}$

   ›Reveal solutionSolution

   Clemmensen reduction uses Zn‑Hg amalgam and concentrated HCl to reduce a carbonyl group (C=O) directly to a methylene group (CH₂). The reaction that matches this reagent set is option (D).

   Concept & Intuition

   The Clemmensen reduction is a classic method for converting a ketone or aldehyde into an alkane. The key idea is that the strongly acidic, reducing environment (Zn‑Hg in HCl) strips the oxygen atom from the carbonyl and replaces it with two hydrogens. This is especially useful when the molecule also contains acid‑sensitive groups that would be destroyed by the basic conditions of the alternative Wolff‑Kishner reduction.

   Here, we simply need to identify which option shows a carbonyl compound being treated with Zn‑Hg / HCl to give a hydrocarbon.

   Step‑by‑Step Reasoning

   1. Recall the defining reagents of Clemmensen reduction.

      The reaction uses zinc amalgam (Zn‑Hg) and concentrated hydrochloric acid (HCl). The substrate is typically a ketone or aldehyde, and the product is the corresponding alkane (the C=O becomes CH₂).

   2. Examine each option for the correct reagent set.

      • (A) Uses Pd/BaSO₄ (Lindlar’s catalyst) with H₂ — that’s a hydrogenation of an acid chloride to an aldehyde (Rosenmund reduction), not Clemmensen.
      • (B) Shows hydrazine (NH₂‑NH₂) followed by KOH in ethylene glycol — that’s the Wolff‑Kishner reduction (basic conditions), not Clemmensen.
      • (C) Uses DIBAL‑H then water — that’s a partial reduction of an ester to an aldehyde, not Clemmensen.
      • (D) Shows a ketone (R‑CO‑CH₃) treated with Zn‑Hg / HCl and yields R‑CH₂‑CH₃ — exactly the Clemmensen reduction.

   3. Confirm the transformation in (D).

      The carbonyl group in the starting ketone is reduced to a methylene group:

   $\mathrm{R{-}C(=O){-}CH_3} \xrightarrow{\mathrm{Zn{-}Hg}/\mathrm{HCl}} \mathrm{R{-}CH_2{-}CH_3}$

   This matches the textbook definition of Clemmensen reduction.

   Watch out

   A common mistake is to confuse Clemmensen (acidic, Zn‑Hg/HCl) with Wolff‑Kishner (basic, NH₂NH₂/KOH). Option (B) is the Wolff‑Kishner reduction, not Clemmensen.

   Tip

   If you ever see “Zn‑Hg / conc. HCl” in a reaction scheme, you can immediately think: “This is Clemmensen reduction — the C=O is being turned into CH₂.”

   ✓Final answer

   The correct option is (D).

   ANSWER: D

8. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.An alcohol X ($\mathrm{C_4H_{10}O}$) reacts with Conc.HCl|ZnCl$_2$ to give corresponding chloride. X on dehydration forms Y, which reacts with Baeyer's reagent to give Z. What is Z? (A) \chemfig{*6(-=-=-(-[::+60]C(=[::+60]O)-[::-60]CO_2)-=)} (B) $\mathrm{CH_3COOH}$ (C) \chemfig{CH_3-CH_2-CH_2-CHO} (D) \chemfig{CH_3-CH(OH)-CH_2-CH_2OH}

   ›Reveal solutionSolution

   Alcohol X, identified as butan-1-ol, undergoes dehydration to form but-1-ene (Y). This alkene then reacts with Baeyer's reagent to yield butane-1,2-diol (Z). The final product is $\boxed{\mathrm{CH_3-CH(OH)-CH_2-CH_2OH}}$.

   The problem describes a sequence of reactions starting from an alcohol X with the molecular formula $\mathrm{C_4H_{10}O}$. We need to identify the final product Z. Let's break down the reactions and the properties of the compounds involved.

   Concept and Intuition

   1. Identifying Alcohol X: The formula $\mathrm{C_4H_{10}O}$ indicates a saturated monohydric alcohol. There are four possible structural isomers. The first reaction, with Conc. HCl/ZnCl$_2$ (Lucas reagent), is a diagnostic test for the class of alcohol (primary, secondary, or tertiary). While all alcohols can react, their rates differ significantly. Tertiary alcohols react fastest, followed by secondary, and then primary alcohols. The problem states X "reacts" to give the chloride, implying the reaction occurs, which helps narrow down possibilities when combined with subsequent steps.

   2. Dehydration to Alkene Y: Alcohols undergo dehydration in the presence of strong acids or heat to form alkenes. This reaction typically follows Zaitsev's rule, where the most substituted (and thus most stable) alkene is the major product.

   3. Reaction with Baeyer's Reagent to Diol Z: Baeyer's reagent (cold, dilute, alkaline $\mathrm{KMnO_4}$) is used to test for unsaturation (carbon-carbon double or triple bonds). It causes syn-dihydroxylation of alkenes, meaning two hydroxyl groups are added to the same face of the double bond, forming a vicinal diol (a 1,2-diol). The purple color of $\mathrm{KMnO_4}$ disappears, and a brown precipitate of $\mathrm{MnO_2}$ forms.

   By systematically applying these reactions to the possible isomers of X and checking against the given options for Z, we can determine the correct pathway.

   Step-by-Step Solution

   1. Identify possible isomers of alcohol X ($\mathrm{C_4H_{10}O}$):

      The molecular formula $\mathrm{C_4H_{10}O}$ corresponds to the following saturated alcohols:

      • Butan-1-ol (1° alcohol): $\mathrm{CH_3CH_2CH_2CH_2OH}$
      • Butan-2-ol (2° alcohol): $\mathrm{CH_3CH_2CH(OH)CH_3}$
      • 2-Methylpropan-1-ol (1° alcohol): $\mathrm{(CH_3)_2CHCH_2OH}$
      • 2-Methylpropan-2-ol (3° alcohol): $\mathrm{(CH_3)_3COH}$

   2. Analyze the reaction of X with Conc. HCl/ZnCl$_2$ (Lucas reagent):

      This reaction converts an alcohol to an alkyl chloride.

      $\mathrm{R-OH + HCl \xrightarrow{ZnCl_2} R-Cl + H_2O}$

      The reactivity order is 3° > 2° > 1°. While primary alcohols react very slowly at room temperature, they do react, especially if heated or given sufficient time. The problem states X "reacts... to give corresponding chloride," which means all these alcohols are potential candidates. We will use the subsequent reactions to pinpoint X.

   3. Analyze the dehydration of X to form Y:

      Alcohols undergo elimination of water to form alkenes.

      $\mathrm{R-CH_2-CH_2-OH \xrightarrow{H_2SO_4/\Delta} R-CH=CH_2 + H_2O}$

      • If X is Butan-1-ol, Y would be but-1-ene ($\mathrm{CH_3CH_2CH=CH_2}$).
      • If X is Butan-2-ol, Y would be a mixture of but-1-ene (minor) and but-2-ene (major, due to Zaitsev's rule).
      • If X is 2-Methylpropan-1-ol, Y would be 2-methylpropene ($\mathrm{(CH_3)_2C=CH_2}$).
      • If X is 2-Methylpropan-2-ol, Y would be 2-methylpropene ($\mathrm{(CH_3)_2C=CH_2}$).

   4. Analyze the reaction of Y with Baeyer's reagent to form Z:

      Baeyer's reagent (cold, dilute, alkaline $\mathrm{KMnO_4}$) performs syn-dihydroxylation on alkenes, forming vicinal diols.

      $\mathrm{R-CH=CH_2 \xrightarrow{KMnO_4/OH^-/H_2O} R-CH(OH)-CH_2OH}$

   5. Evaluate each possible alcohol X against the given options for Z:

      Let's consider the options for Z:

      (A) A complex aromatic compound - not possible from a 4-carbon aliphatic alcohol.

      (B) $\mathrm{CH_3COOH}$ (acetic acid) - a 2-carbon carboxylic acid, not a 4-carbon diol.

      (C) $\mathrm{CH_3-CH_2-CH_2-CHO}$ (butanal) - a 4-carbon aldehyde, not a diol.

      (D) $\mathrm{CH_3-CH(OH)-CH_2-CH_2OH}$ (butane-1,2-diol) - a 4-carbon vicinal diol. This is a plausible product from the dihydroxylation of a 4-carbon alkene.

      Now, let's work backward or forward to see which X leads to Z as butane-1,2-diol:

      • If Z is butane-1,2-diol ($\mathrm{CH_3CH_2CH(OH)CH_2OH}$):

        This diol is formed from the syn-dihydroxylation of but-1-ene ($\mathrm{CH_3CH_2CH=CH_2}$).

        Therefore, Y must be but-1-ene.

      • Which alcohol X dehydrates to form but-1-ene (Y)?

        • Butan-1-ol ($\mathrm{CH_3CH_2CH_2CH_2OH}$): Dehydrates to form but-1-ene. This is a direct and unambiguous path.
        • Butan-2-ol ($\mathrm{CH_3CH_2CH(OH)CH_3}$): Dehydrates to form a mixture of but-1-ene (minor product) and but-2-ene (major product). If Y were predominantly but-2-ene, Z would be butane-2,3-diol ($\mathrm{CH_3CH(OH)CH(OH)CH_3}$), which is not option (D). While but-1-ene is a minor product, the question implies a specific product Z.
        • 2-Methylpropan-1-ol ($\mathrm{(CH_3)_2CHCH_2OH}$): Dehydrates to form 2-methylpropene. This would lead to 2-methylpropane-1,2-diol, not butane-1,2-diol.
        • 2-Methylpropan-2-ol ($\mathrm{(CH_3)_3COH}$): Dehydrates to form 2-methylpropene. This would also lead to 2-methylpropane-1,2-diol, not butane-1,2-diol.

      Considering all steps, butan-1-ol is the only alcohol X that consistently leads to butane-1,2-diol (option D) as product Z. Although butan-1-ol is a primary alcohol and reacts slowly with Lucas reagent, the problem statement only says it "reacts," which is true. The subsequent reactions confirm this choice.

      The reaction sequence is:

      $\mathrm{CH_3CH_2CH_2CH_2OH}$ (X, Butan-1-ol)

      $\xrightarrow{\text{Dehydration}}$ $\mathrm{CH_3CH_2CH=CH_2}$ (Y, But-1-ene)

      $\xrightarrow{\text{Baeyer's reagent}}$ $\mathrm{CH_3CH_2CH(OH)CH_2OH}$ (Z, Butane-1,2-diol)

   The final product Z is butane-1,2-diol.

   ✓Final answer

   The final product Z is $\boxed{\mathrm{CH_3-CH(OH)-CH_2-CH_2OH}}$.

9. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.What are X, Y, Z in the following reaction sequence?

   [!FORMULA] $\ce{CH3-CH=CH2} \xrightarrow{\ce{H2O}/\ce{H+}} \ce{C3H8O} \xrightarrow{\text{X}} \text{Y} \xrightarrow[\text{anhy.}~\ce{AlCl3}]{\ce{C6H6}} \text{Z (major)}$

   (A) \ce{Cl2},~\ce{Cl},~\raisebox{0.5em}{\chemfig{*6(-=-=-(-[::-60]C(-[::60]CH_3)(-[::-60]CH_3))-=)}} (B) \text{conc. HCl},~\ce{Cl},~\raisebox{0.5em}{\chemfig{*6(-=-=-(-[::-60]C(-[::60]CH_2CH_3))-=)}} (C) \ce{HCl}/\ce{ZnCl2},~\ce{Cl},~\raisebox{0.5em}{\chemfig{*6(-=-=-(-[::-60]C(-[::60]CH_2CH_2CH_3))-=)}} (D) \ce{HCl}/\ce{ZnCl2},~\ce{Cl},~\raisebox{0.5em}{\chemfig{*6(-=-=-(-[::-60]C(-[::60]CH_2CH_3))-=)}}

   ›Reveal solutionSolution

   The reaction sequence involves acid-catalysed hydration of propene to isopropyl alcohol, conversion to isopropyl chloride using Lucas reagent ($\ce{HCl/ZnCl2}$), and then Friedel–Crafts alkylation of benzene to give cumene (isopropylbenzene). The correct option is (D).

   The key to this problem is recognising each transformation in the sequence and matching the reagents and products to the given options. Let’s walk through it step by step.

   1. First step: hydration of propene

      Propene ($\ce{CH3-CH=CH2}$) reacts with water in the presence of an acid catalyst ($\ce{H2O/H+}$). This is acid-catalysed hydration, which follows Markovnikov’s rule: the hydrogen adds to the less substituted carbon of the double bond, and the hydroxyl group adds to the more substituted carbon.

      The product is propan-2-ol (isopropyl alcohol), $\ce{CH3-CH(OH)-CH3}$, with molecular formula $\ce{C3H8O}$.

      So the intermediate $\ce{C3H8O}$ is isopropyl alcohol.

   2. Second step: conversion of alcohol to alkyl halide

      The alcohol is now treated with reagent X to give product Y, which is an alkyl chloride (since the next step involves benzene and $\ce{AlCl3}$, a Friedel–Crafts alkylation).

      To convert an alcohol to an alkyl chloride, common reagents include $\ce{SOCl2}$, $\ce{PCl5}$, $\ce{PCl3}$, or $\ce{HCl}$ with a catalyst. Among the options, we see $\ce{Cl2}$ (which would not directly convert an alcohol to an alkyl chloride — it would give a hypochlorite or chlorinate elsewhere), conc. $\ce{HCl}$ (which works but slowly for secondary alcohols), and $\ce{HCl/ZnCl2}$ (Lucas reagent).

      Lucas reagent ($\ce{HCl/ZnCl2}$) is specifically used to convert secondary alcohols to alkyl chlorides rapidly at room temperature. For isopropyl alcohol, it gives isopropyl chloride, $\ce{CH3-CHCl-CH3}$.

      Thus X is $\ce{HCl/ZnCl2}$ and Y is isopropyl chloride ($\ce{Cl}$ in the options refers to the isopropyl group attached to chlorine, i.e., $\ce{CH3-CHCl-CH3}$).

   3. Third step: Friedel–Crafts alkylation

      Y (isopropyl chloride) reacts with benzene ($\ce{C6H6}$) in the presence of anhydrous $\ce{AlCl3}$, a Lewis acid catalyst. This is a classic Friedel–Crafts alkylation.

      The mechanism: $\ce{AlCl3}$ abstracts the chlorine from isopropyl chloride, generating an isopropyl carbocation ($\ce{CH3-CH+-CH3}$). This carbocation attacks the benzene ring, and after deprotonation, the product is isopropylbenzene, commonly known as cumene.

      The structure of cumene is: a benzene ring with an isopropyl group ($\ce{-CH(CH3)2}$) attached. In the options, this is represented as a benzene ring with a branched three-carbon chain: $\ce{C(-[::60]CH3)(-[::-60]CH3)}$ — that is, a carbon attached to two methyl groups.

   4. Matching with the options

      • Option (A): $\ce{Cl2}$ is wrong for the first step; product is a different alkylbenzene.
      • Option (B): conc. $\ce{HCl}$ is possible but less specific; the product shown has an ethyl group ($\ce{-CH2CH3}$), not isopropyl.
      • Option (C): $\ce{HCl/ZnCl2}$ is correct for X, but the product shown has an n-propyl group ($\ce{-CH2CH2CH3}$), not isopropyl.
      • Option (D): $\ce{HCl/ZnCl2}$ for X, and the product is isopropylbenzene (cumene) — exactly what we expect.

   Watch out

   A common mistake is to think that $\ce{HCl}$ alone is sufficient for the alcohol-to-chloride step. While it works, the reaction is slow for secondary alcohols; Lucas reagent ($\ce{HCl/ZnCl2}$) is the standard reagent used in such sequences. Also, be careful with the alkyl group in the final product — Friedel–Crafts alkylation with isopropyl chloride gives a branched isopropyl group, not a straight-chain propyl group.

   Tip

   In Friedel–Crafts alkylation, the carbocation rearranges if a more stable carbocation is possible. Here, the isopropyl carbocation is already secondary and stable, so no rearrangement occurs. If you started with n-propyl chloride, you would get rearrangement to isopropylbenzene anyway — but here we start with isopropyl chloride, so the product is directly cumene.

   ✓Final answer

   The correct option is (D): X is $\ce{HCl/ZnCl2}$, Y is isopropyl chloride, and Z is cumene (isopropylbenzene).

10. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.The functional groups present in the product 'X' of the reaction given below are Phenyl benzoate — a benzene ring bearing $\mathrm{-O-C(=O)-Ph}$ — $\xrightarrow{\mathrm{AlCl_3}}$ X (A) $-\mathrm{OH}$, $-\mathrm{C(=O)-H}$ (B) $-\mathrm{OH}$, $-\mathrm{C(=O)-}$ (C) $-\mathrm{O-ph}$, $-\mathrm{C(=O)-}$ (D) $-\mathrm{C(=O)-OH}$, $-\mathrm{OH}$

    ›Reveal solutionSolution

    The reaction is a Fries rearrangement of phenyl benzoate under $\mathrm{AlCl_3}$, which migrates the acyl group to the ortho or para position of the phenol ring. The product 'X' contains $-\mathrm{OH}$ and $-\mathrm{C(=O)-}$ (ketone) groups, making option (B) correct.

    The key here is recognising the reaction type. Phenyl benzoate is an ester of phenol and benzoic acid. When treated with a Lewis acid like $\mathrm{AlCl_3}$, it undergoes the Fries rearrangement — not a simple hydrolysis or acylation. The $\mathrm{AlCl_3}$ coordinates to the ester oxygen, weakening the $\mathrm{C-O}$ bond, and the acyl group ($\mathrm{-C(=O)-Ph}$) migrates to the ortho or para position of the phenol ring. After work-up, you get a hydroxy ketone.

    Let’s walk through it step by step.

    1. Identify the starting material. Phenyl benzoate has the structure $\mathrm{Ph-O-C(=O)-Ph}$. The left benzene ring is attached to oxygen (phenolic part), the right benzene ring is part of the acyl group ($\mathrm{-C(=O)-Ph}$). So the molecule is an aryl ester.

    2. Recall the Fries rearrangement. Under $\mathrm{AlCl_3}$, aryl esters rearrange to give ortho- and para-hydroxy ketones. The $\mathrm{AlCl_3}$ coordinates to the carbonyl oxygen, making the ester more electrophilic. The acyl group then breaks off and attacks the electron-rich ortho/para positions of the phenol ring. This is an intramolecular rearrangement — the acyl group doesn’t leave entirely; it migrates.

    3. What happens to the ester linkage? The $\mathrm{-O-}$ bond to the original acyl group breaks. The oxygen that was part of the ester becomes a phenolic $\mathrm{-OH}$ after work-up (protonation). The acyl group ($\mathrm{-C(=O)-Ph}$) attaches directly to the benzene ring that originally held the oxygen. So the product is $\mathrm{HO-C_6H_4-C(=O)-Ph}$ — a hydroxybenzophenone.

    4. Identify the functional groups in the product. The product has:

       • A phenolic hydroxyl group ($\mathrm{-OH}$) attached to the benzene ring.
       • A ketone carbonyl group ($\mathrm{-C(=O)-}$) linking the same benzene ring to the phenyl group. This is not an aldehyde ($\mathrm{-C(=O)-H}$) and not a carboxylic acid ($\mathrm{-C(=O)-OH}$). It’s simply a ketone.

    5. Match with the options.

       • (A) $-\mathrm{OH}$, $-\mathrm{C(=O)-H}$ — aldehyde, wrong.
       • (B) $-\mathrm{OH}$, $-\mathrm{C(=O)-}$ — correct: hydroxyl and ketone.
       • (C) $-\mathrm{O-ph}$, $-\mathrm{C(=O)-}$ — still has an ether link, not formed.
       • (D) $-\mathrm{C(=O)-OH}$, $-\mathrm{OH}$ — carboxylic acid, not formed.

    Watch out

    A common mistake is to think the ester simply hydrolyses under $\mathrm{AlCl_3}$ to give phenol and benzoic acid. That would require water, which is not present. The Fries rearrangement is specific to aryl esters and Lewis acids — the product is a hydroxy ketone, not a mixture of acid and phenol.

    Tip

    The Fries rearrangement is the reverse of the Baeyer–Villiger oxidation in a conceptual sense — here an ester becomes a hydroxy ketone, whereas Baeyer–Villiger turns a ketone into an ester. Remembering this pair can help in exams.

    ✓Final answer

    The correct option is (B), as the product contains a phenolic $-\mathrm{OH}$ group and a ketone $-\mathrm{C(=O)-}$ group.

11. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
    Q.Hydrolysis of an alkyl halide X $(\mathrm{C_3H_7Br})$ follows second order kinetics. Reaction of X with $\mathrm{C_6H_5Cl}$ in the presence of Na/dry ether gave Y. Oxidation of Y in the presence of $\mathrm{KMnO_4/\overline{OH}}$ gave Z. What are Y and Z respectively? (A) $\mathrm{C_6H_5CH_2CH_2CH_3}$ (n-propylbenzene) and $\mathrm{C_6H_5COOK}$ (potassium benzoate) (B) $\mathrm{C_6H_5CH(CH_3)_2}$ (isopropylbenzene) and $\mathrm{C_6H_5COOK}$ (potassium benzoate) (C) $\mathrm{C_6H_5CH_2CH_2CH_3}$ (n-propylbenzene) and $\mathrm{C_6H_5CH_2CH_2COOK}$ (D) $\mathrm{C_6H_5CH(CH_3)_2}$ (isopropylbenzene) and $\mathrm{C_6H_5COCH_3}$ (acetophenone)

    ›Reveal solutionSolution

    Second-order hydrolysis $\Rightarrow \mathrm{S_N2} \Rightarrow$ X is the primary halide, n-propyl bromide. Wurtz–Fittig with chlorobenzene gives Y = n-propylbenzene, and $\mathrm{KMnO_4/OH^-}$ chops the whole side chain to give Z = potassium benzoate. Option (A).

    The concept first — three ideas, one chain

    1. Kinetics tells you the structure. $\mathrm{S_N1}$ is unimolecular (rate $=k[\mathrm{RX}]$, favoured by 3° > 2° > 1°), while $\mathrm{S_N2}$ is bimolecular (rate $=k[\mathrm{RX}][\mathrm{Nu}]$, favoured by 1° > 2° > 3°, because the backside attack needs an uncrowded carbon). So "second-order kinetics" is the paper telling you X is primary.
    2. Wurtz–Fittig. Aryl halide + alkyl halide + $\mathrm{2Na}$ in dry ether $\to$ alkylbenzene:

       $\mathrm{C_6H_5Cl + 2Na + R{-}Br \xrightarrow{\text{dry ether}} C_6H_5{-}R + NaCl + NaBr}$

    3. Side-chain oxidation. Hot $\mathrm{KMnO_4}$ attacks the benzylic carbon. Whatever the length of the alkyl chain, as long as there is at least one hydrogen on the benzylic carbon, the entire chain is degraded to a single –COOH group attached to the ring. (Only a chain with no benzylic H, like tert-butylbenzene, resists.)

    Step-by-step

    1. Identify X. $\mathrm{C_3H_7Br}$ has two isomers. Isopropyl bromide (2°) would hydrolyse largely by $\mathrm{S_N1}$ (first-order). Since the observed hydrolysis is second order, the mechanism is $\mathrm{S_N2}$, so

    $\mathbf{X = CH_3CH_2CH_2Br}\quad(\text{n-propyl bromide, }1^\circ)$

    2. Form Y (Wurtz–Fittig).

    $\mathrm{C_6H_5Cl + CH_3CH_2CH_2Br + 2Na \xrightarrow{dry\ ether} C_6H_5CH_2CH_2CH_3 + NaCl + NaBr}$

    $\mathbf{Y = n\text{-}propylbenzene}$

    Note the alkyl group keeps its skeleton — no rearrangement — so we get the n-propyl, not the isopropyl, product.

    3. Oxidise to Z. Alkaline $\mathrm{KMnO_4}$ (then the alkaline work-up) oxidises the propyl side chain completely:

    $\mathrm{C_6H_5CH_2CH_2CH_3 \xrightarrow[\Delta]{KMnO_4/\overline{O}H} C_6H_5COO^-K^+ \;+\;} \text{(2 C lost as carbonate/}\mathrm{CO_2)}$

    $\mathbf{Z = C_6H_5COOK}\ (\text{potassium benzoate})$

    Because the medium is alkaline, the product isolated is the potassium salt, not the free acid (acidification would be needed to get benzoic acid).

    4. Match the options. Y = n-propylbenzene, Z = potassium benzoate → (A). Option (B) wrongly assumes X is isopropyl bromide (which would need first-order kinetics); (C) wrongly keeps two carbons of the side chain (side-chain oxidation never stops half-way); (D) is wrong on both counts.

    ✓Final answer

    X = n-propyl bromide ($\mathrm{S_N2}$), Y = n-propylbenzene (Wurtz–Fittig), Z = potassium benzoate (complete side-chain oxidation).

    ANSWER: A

12. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.Phenatole can be prepared from which of the following reactants? (A) $\begin{array}{cc} \chemfig{*6(-=-(-ONa)-=-)} & ,\\\mathrm{CH_3CH_2Br} \end{array}$ (B) $\begin{array}{cc} \chemfig{*6(-=-(-Br)-=-)} & ,\\\mathrm{CH_3CH_2ONa} \end{array}$ (C) $\begin{array}{cc} \chemfig{*6(-=-(-Br)-=-)} & ,\\\mathrm{CH_3CH_2CH_2ONa} \end{array}$ (D) $\begin{array}{cc} \chemfig{*6(-=-(-ONa)-=-)} & ,\\\mathrm{CH_3CH_2CH_2Br} \end{array}$

    ›Reveal solutionSolution

    Phenatole (phenetole) is ethyl phenyl ether (C₆H₅OCH₂CH₃). It is best prepared by the Williamson ether synthesis: reacting sodium phenoxide with ethyl bromide. The correct reactants are sodium phenoxide and ethyl bromide, which corresponds to option (A).

    Concept & Intuition

    Phenatole (often spelled phenetole) is the common name for ethyl phenyl ether, C₆H₅–O–CH₂CH₃. The classic laboratory method for making unsymmetrical ethers like this is the Williamson ether synthesis: an alkoxide (or phenoxide) ion attacks an alkyl halide in an Sₙ2 reaction.

    The key is to choose the combination that gives the desired ether in high yield. The best route uses the more nucleophilic oxygen (the phenoxide ion) and the less hindered alkyl halide (primary halide). If you reverse the roles — using a phenoxide with a secondary or tertiary halide — elimination dominates. If you use an alkoxide with an aryl halide, the reaction fails because aryl halides are unreactive toward Sₙ2.

    Step-by-step reasoning

    1. Identify the target molecule

       Phenatole = ethyl phenyl ether = C₆H₅–O–CH₂CH₃. The oxygen is bonded to a phenyl group and an ethyl group.

    2. Recall the Williamson ether synthesis

       The general reaction:

    $\mathrm{RONa + R'\!X \rightarrow ROR' + NaX}$

    The alkoxide (or phenoxide) acts as the nucleophile; the alkyl halide must be primary (or methyl) to favor Sₙ2 over elimination.

    3. Examine each option

       • (A) Sodium phenoxide (C₆H₅ONa) + ethyl bromide (CH₃CH₂Br) Phenoxide is a good nucleophile; ethyl bromide is a primary halide. Sₙ2 proceeds cleanly:

    $\mathrm{C_6H_5ONa + CH_3CH_2Br \rightarrow C_6H_5OCH_2CH_3 + NaBr}$

     This gives phenatole directly. ✓
    

    • (B) Bromobenzene (C₆H₅Br) + sodium ethoxide (CH₃CH₂ONa)

      Aryl halides do not undergo Sₙ2 reactions (the carbon–halogen bond has partial double‑bond character; backside attack is impossible). This would require harsh conditions (e.g., Ullmann coupling) and is not a simple Williamson synthesis. ✗

    • (C) Bromobenzene + sodium propoxide (CH₃CH₂CH₂ONa)

      Same problem as (B): aryl halide is unreactive in Sₙ2. Even if it reacted, the product would be propyl phenyl ether, not ethyl phenyl ether. ✗

    • (D) Sodium phenoxide + propyl bromide (CH₃CH₂CH₂Br)

      This would give propyl phenyl ether (C₆H₅OCH₂CH₂CH₃), not phenatole. ✗

    4. Confirm the correct choice Only option (A) pairs the correct nucleophile (phenoxide) with the correct primary alkyl halide (ethyl bromide) to yield ethyl phenyl ether.

    Watch out

    A common mistake is to think that an aryl halide can be used in a Williamson synthesis. Aryl halides are inert to Sₙ2; you must always put the aryl group on the oxygen (as the phenoxide) and the alkyl group on the halide.

    Tip

    The Williamson synthesis is most reliable when the alkyl halide is primary. If you need an ether like methyl phenyl ether (anisole), use methyl iodide + sodium phenoxide — never bromobenzene + sodium methoxide.

    ✓Final answer

    The correct option is (A).

    ANSWER: A