Q.Write the reactions of Williamson synthesis of 2-ethoxy-3-methylpentane starting from ethanol and 3-methylpentan-2-ol.

Showing the 12 most recent of 27 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Which of the following statements are correct? I. Liquid sodium metal is used as coolant in fast breeder nuclear reactor II. LiCl is deliquescent III. LiF and CsI have low solubility in water (A) I, III only (B) II, III only (C) I, II only (D) I, II, III

   ›Reveal solutionSolution

   All three statements are correct: liquid sodium is used as a coolant in fast breeder reactors, LiCl is deliquescent, and both LiF and CsI have low solubility in water. The correct option is (D).

   The question asks us to identify the correct statements regarding properties of alkali metals and their compounds. We will evaluate each statement based on fundamental chemical principles.

   Concept and Intuition

   1. Coolants in Nuclear Reactors: A good coolant needs to efficiently transfer heat, remain liquid over a wide temperature range, and ideally have a low neutron absorption cross-section in certain reactor types.
   2. Deliquescence: This property describes a substance that absorbs moisture from the atmosphere until it dissolves in the absorbed water to form a solution. It is related to hygroscopy (absorbing moisture) and the ability to form stable hydrates.
   3. Solubility in Water: The solubility of an ionic compound in water is determined by the balance between its lattice energy (energy required to break the ionic lattice) and its hydration energy (energy released when ions are surrounded by water molecules). For a compound to dissolve, the hydration energy must be sufficient to overcome the lattice energy.

   Step-by-Step Evaluation

   1. Statement I: Liquid sodium metal is used as coolant in fast breeder nuclear reactor

   • Reasoning: Fast breeder reactors operate at very high temperatures and require a coolant that can efficiently remove heat without moderating (slowing down) the fast neutrons. Liquid sodium is an excellent choice for this purpose due to several key properties:
     • High Thermal Conductivity: It can transfer heat very efficiently.
     • Wide Liquid Range: Sodium has a relatively low melting point ($97.8^\circ C$) and a very high boiling point ($883^\circ C$), allowing it to remain liquid over a broad and useful temperature range for reactor operation.
     • Low Neutron Absorption Cross-section: It does not significantly absorb neutrons, which is crucial for maintaining the neutron economy in a fast breeder reactor.
     • Low Viscosity: This allows for easy pumping and circulation.
   • Conclusion: Statement I is correct.

   2. Statement II: LiCl is deliquescent

   • Reasoning: Deliquescence is the property of a substance to absorb moisture from the atmosphere and dissolve in it. This property is strongly linked to the hygroscopic nature of a compound and its ability to form stable hydrates.
     • Lithium chloride (LiCl) is known to be highly hygroscopic and readily forms hydrates, such as $LiCl \cdot H_2O$, $LiCl \cdot 2H_2O$, and $LiCl \cdot 3H_2O$.
     • The small size and high charge density of the $Li^+$ ion lead to a very high hydration energy. This strong interaction with water molecules drives the absorption of moisture from the air.
     • Among the alkali metal chlorides, LiCl is the most deliquescent.
   • Conclusion: Statement II is correct.

   3. Statement III: LiF and CsI have low solubility in water

   • Reasoning: The solubility of an ionic compound in water depends on the relative magnitudes of its lattice energy and hydration energy.
     • For LiF:
       • $Li^+$ is the smallest alkali metal ion, and $F^-$ is the smallest halide ion.
       • The very small sizes of both ions lead to a very short inter-ionic distance and, consequently, an exceptionally high lattice energy for LiF.
       • While $Li^+$ and $F^-$ both have high hydration energies due to their small sizes, the extremely high lattice energy of LiF dominates. The energy released upon hydration is not sufficient to overcome the energy required to break the strong ionic bonds in the crystal lattice.
       • Therefore, LiF has very low solubility in water.
     • For CsI:
       • $Cs^+$ is the largest alkali metal ion, and $I^-$ is the largest halide ion.
       • The large sizes of both ions lead to a relatively long inter-ionic distance and thus a relatively low lattice energy compared to compounds with smaller ions.
       • However, the hydration energies of very large ions like $Cs^+$ and $I^-$ are also very low because their charge densities are low, leading to weak interactions with water molecules.
       • In the case of CsI, the low hydration energy is not sufficient to overcome even the relatively low lattice energy. The energy released upon hydration is less than the energy required to break the lattice.
       • Therefore, CsI also has low solubility in water.
   • Conclusion: Statement III is correct.

   Since all three statements (I, II, and III) are correct, the option that includes all of them is the answer.

   ✓Final answer

   All the given statements are correct, so the correct option is (D).

2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Which of the following are Position isomers? [FIGURE] (A) I, III (B) II, IV (C) II, III (D) I, IV

   ›Reveal solutionSolution

   Among these $\mathrm{C_6H_{12}}$ alkenes, only I (4-methylpent-1-ene) and III (4-methylpent-2-ene) have the same carbon skeleton with the double bond in different places — option (A).

   The concept first

   All four compounds are $\mathrm{C_6H_{12}}$, so they are all isomers of one another. The question is which kind. Two structural isomers are:

   • Chain (skeletal) isomers if the carbon framework itself differs (branching pattern, length of the main chain);
   • Position isomers if the framework is identical and only the location of the functional group (here the C=C) changes.

   So the working method is mechanical: hydrogenate each structure in your head (i.e. ignore the double bond and look only at the carbon skeleton). Any two that collapse to the same alkane are position isomers; ones that collapse to different alkanes are chain isomers.

   Step-by-step

   1. Reduce each to its skeleton.
      • I: $\mathrm{(CH_3)_2CH\!-\!CH_2\!-\!CH{=}CH_2} \to$ 2-methylpentane skeleton (5-carbon chain, methyl branch on the carbon next to one end).
      • II: $\mathrm{CH_2{=}C(CH_3)\!-\!CH(CH_3)_2} \to$ 2,3-dimethylbutane skeleton (4-carbon chain, two methyl branches).
      • III: $\mathrm{(CH_3)_2CH\!-\!CH{=}CH\!-\!CH_3} \to$ 2-methylpentane skeleton — same as I.
      • IV: unbranched hexene $\to$ n-hexane skeleton (no branch).
   2. Group by skeleton.
      • 2-methylpentane skeleton: {I, III}
      • 2,3-dimethylbutane skeleton: {II}
      • n-hexane skeleton: {IV}
   3. Check the double-bond positions within the matching pair. Numbering the main chain of I and III so the double bond gets the lower locant:
      • I = 4-methylpent-1-ene
      • III = 4-methylpent-2-ene Same skeleton, C=C shifted from position 1 to position 2 → position isomers ✓.
   4. Reject the rest. II vs anything, and IV vs anything, involve a change in the carbon framework, which makes them chain isomers, not position isomers.

   ✓Final answer

   The position-isomeric pair is I and III, so the correct option is (A).

   ANSWER: A

3. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.What are X, Y, Z respectively in the following set of reactions? (A) [benzene ring]-Br ; [ethanol] ; [benzene ring]-O-[ethyl group]-Br (B) [benzene ring]-OH ; [ethyl bromide] ; [benzene ring]-O-[ethyl group]-Br (C) [benzene ring]-OH ; [ethyl bromide] ; [benzene ring]-(Br)$_2$-O-[ethyl group] (D) [benzene ring]-Br ; [ethyl bromide] ; [benzene ring]-O-[ethyl group]-Br

   ›Reveal solutionSolution

   The reaction sequence is the Williamson ether synthesis: phenol (X) reacts with ethyl bromide (Y) in the presence of a base to give phenetole, which then undergoes bromination at the para position to yield p-bromophenetole (Z). The correct option is (B).

   The question presents a set of reactions where three unknown compounds X, Y, and Z are to be identified. The key is to recognize the classic Williamson ether synthesis followed by electrophilic aromatic substitution. Let's walk through the logic.

   1. Identify the first reaction: X + NaOH → sodium phenoxide

      The product after treatment with NaOH is a sodium salt of an aromatic alcohol. This is a characteristic reaction of phenol ($C_6H_5OH$). Phenol reacts with NaOH to form sodium phenoxide ($C_6H_5ONa$). So X must be phenol, not bromobenzene (which does not react with NaOH under these conditions). This immediately eliminates options (A) and (D), which list X as bromobenzene.

   2. Identify the second reaction: sodium phenoxide + Y → an ether

      Sodium phenoxide is a strong nucleophile. It reacts with an alkyl halide in an $S_N2$ reaction to form an ether — this is the Williamson ether synthesis. The product shown is an aromatic ether with an ethyl group attached to the oxygen. Therefore Y must be ethyl bromide ($C_2H_5Br$). The product is ethyl phenyl ether, commonly called phenetole ($C_6H_5OC_2H_5$).

   3. Identify the third reaction: phenetole + $Br_2$ → Z

      Phenetole has an activated benzene ring (the -OCH$_2$CH$_3$ group is strongly activating and ortho/para-directing). Bromination occurs predominantly at the para position due to steric hindrance at the ortho positions. The product Z is p-bromophenetole, which has the structure $Br-C_6H_4-OC_2H_5$ (with bromine para to the ethoxy group). This matches the description in option (B): [benzene ring]-O-[ethyl group]-Br.

   Watch out

   A common mistake is to think that bromobenzene could be X because it contains bromine. But bromobenzene does not react with NaOH to form a sodium salt — it requires much harsher conditions (high pressure, high temperature) for nucleophilic substitution. The mild NaOH step here is a dead giveaway for phenol.

   Tip

   The Williamson ether synthesis is the most reliable method for preparing unsymmetrical ethers. The alkoxide (from the alcohol/phenol) must be the nucleophile, and the alkyl halide should be primary to avoid elimination. Here, ethyl bromide is primary, so $S_N2$ proceeds cleanly.

   ✓Final answer

   The correct option is (B): X = phenol, Y = ethyl bromide, Z = p-bromophenetole.

4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Consider the following reactions Borax $\xrightarrow{\text{H}_2\text{O}}$ A + B Borax $\xrightarrow{\text{H}_2\text{O}/\text{HCl}}$ B + C Which of the following are the properties of A, B, C? I. Both A, B are strong bases II. B is a white crystalline solid with soapy touch III. C is water soluble (A) I, II, III (B) II, III only (C) I, III only (D) I, II only

   ›Reveal solutionSolution

   Borax hydrolyses in water to give boric acid and a strong base; with HCl it gives boric acid and NaCl. The correct properties are II and III only — option (B).

   The key is to understand what borax (sodium tetraborate decahydrate, $\text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O}$) does in water and in acidic water. Borax is not a simple salt — it reacts with water to produce boric acid and a strong base. When you add HCl, the base gets neutralised, leaving a different set of products.

   Let’s write the reactions clearly.

   1. Borax in pure water Borax dissolves and hydrolyses:

   $\text{Na}_2\text{B}_4\text{O}_7 + 7\text{H}_2\text{O} \rightarrow 2\text{NaOH} + 4\text{H}_3\text{BO}_3$

   Here, A is $\text{NaOH}$ (sodium hydroxide) and B is $\text{H}_3\text{BO}_3$ (boric acid).

   So A is a strong base, B is a weak acid — not a base at all.

   2. Borax in water with HCl The same hydrolysis occurs, but the HCl neutralises the NaOH formed:

   $\text{Na}_2\text{B}_4\text{O}_7 + 2\text{HCl} + 5\text{H}_2\text{O} \rightarrow 2\text{NaCl} + 4\text{H}_3\text{BO}_3$

   Here, B is again $\text{H}_3\text{BO}_3$, and C is $\text{NaCl}$ (sodium chloride).

   Now evaluate each statement:

   • I. Both A, B are strong bases

     A is NaOH (strong base), but B is boric acid — a very weak acid, not a base. So I is false.

   • II. B is a white crystalline solid with soapy touch

     Boric acid ($\text{H}_3\text{BO}_3$) is indeed a white crystalline solid. It feels slippery or soapy to the touch (a property of many boron compounds). This is correct.

   • III. C is water soluble

     C is NaCl, common salt — highly soluble in water. This is correct.

   Only statements II and III are true.

   Watch out

   A common mistake is to think boric acid is a base because it feels soapy. In fact, it is a weak acid (it accepts $\text{OH}^-$ from water, not donates $\text{H}^+$ in the usual sense). Soapy touch does not imply basicity.

   ✓Final answer

   The correct option is (B) — only II and III are true.

5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Which of the following is most reactive towards $\mathrm{S_N2}$ reaction? (A) $\mathrm{(CH_3)_3C-Br}$ (tert-butyl bromide) (B) $\mathrm{CH_3CH_2CH_2CH_2-Br}$ (1-bromobutane) (C) $\mathrm{(CH_3)_2CH-CH_2-Br}$ (1-bromo-2-methylpropane) (D) $\mathrm{CH_3CH_2-CH(Br)-CH_3}$ (2-bromobutane) ![The four options are skeletal (line) formulae of bromoalkanes:

   (1) A central carbon bearin](/figures/tg-eapcet/eng-2026-05-10-FN-chemistry-q155.svg)

   ›Reveal solutionSolution

   $\mathrm{S_N2}$ needs an unobstructed backside approach, so steric bulk kills it. The unbranched primary halide 1-bromobutane is the least hindered and therefore the most reactive — option (B).

   The concept first

   In the $\mathrm{S_N2}$ mechanism, bond-making and bond-breaking happen simultaneously in one step. The nucleophile attacks the electrophilic carbon from the side directly opposite the leaving group (backside attack), passing through a transition state in which that carbon is momentarily bonded to five groups:

   $\mathrm{Nu^{-}}+\ \mathrm{R\!-\!Br}\ \longrightarrow\ \left[\mathrm{Nu}\cdots\mathrm{C}\cdots\mathrm{Br}\right]^{\ddagger}\ \longrightarrow\ \mathrm{Nu\!-\!R}+\mathrm{Br^{-}}$

   The carbon is $sp^2$-like in that transition state with the three remaining groups splayed out in a plane. Every extra alkyl group attached to that carbon (or even to the carbon next to it) makes the transition state more crowded, raises its energy, and slows the reaction:

   $\text{Rate: }\ \mathrm{CH_3X}\ >\ 1^\circ\ >\ 2^\circ\ \gg\ 3^\circ$

   This is the exact opposite of the $\mathrm{S_N1}$ order (which follows carbocation stability, $3^\circ>2^\circ>1^\circ$). Knowing which mechanism the question names is half the battle.

   Step-by-step through the options

   (A) $\mathrm{(CH_3)_3C\!-\!Br}$, tert-butyl bromide — tertiary. Three methyl groups completely block the backside of the C–Br carbon. Nucleophiles simply cannot get in; this halide reacts almost exclusively by $\mathrm{S_N1}$ (it forms a very stable $3^\circ$ carbocation). Slowest for $\mathrm{S_N2}$. ✗

   (B) $\mathrm{CH_3CH_2CH_2CH_2\!-\!Br}$, 1-bromobutane — primary and unbranched. The carbon bearing $\mathrm{Br}$ carries only two hydrogens and one straight chain, so the backside is wide open. Fastest of the four. ✓

   (C) $\mathrm{(CH_3)_2CH\!-\!CH_2\!-\!Br}$, 1-bromo-2-methylpropane (isobutyl bromide) — primary but $\beta$-branched. Although the carbon holding the $\mathrm{Br}$ is primary, the bulky isopropyl group on the very next carbon leans over the reaction site and hinders the incoming nucleophile. Isobutyl bromide reacts several times more slowly than n-butyl bromide in $\mathrm{S_N2}$. ✗

   (D) $\mathrm{CH_3CH_2\!-\!CH(Br)\!-\!CH_3}$, 2-bromobutane — secondary. Two alkyl groups on the reacting carbon; distinctly slower than any primary halide. ✗

   The ordering

   $\underbrace{\text{1-bromobutane}}_{1^\circ,\ \text{unbranched}}\ >\ \underbrace{\text{1-bromo-2-methylpropane}}_{1^\circ,\ \beta\text{-branched}}\ >\ \underbrace{\text{2-bromobutane}}_{2^\circ}\ \gg\ \underbrace{\text{tert-butyl bromide}}_{3^\circ}$

   A one-line way to remember it: $\mathrm{S_N2}$ hates crowds.

   ✓Final answer

   The unhindered primary halide 1-bromobutane undergoes $\mathrm{S_N2}$ fastest, so the correct option is (B).

   ANSWER: B

6. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.In the formation of photochemical smog, two oxides of nitrogen are significantly formed. The oxidation states of nitrogen in these oxides are respectively (A) $+1,\ +4$ (B) $+2,\ +4$ (C) $+2,\ +5$ (D) $+1,\ +2$

   ›Reveal solutionSolution

   Photochemical smog forms when nitrogen oxides (NO and NO₂) from vehicle exhaust react with sunlight. The oxidation states of nitrogen in these two key oxides are +2 and +4, making option (B) correct.

   The question is about the chemistry of photochemical smog — a type of air pollution that forms when sunlight triggers reactions between pollutants, especially from car engines. The two oxides of nitrogen that play the starring role here are nitric oxide (NO) and nitrogen dioxide (NO₂). They are produced in high-temperature combustion (e.g., in a car engine) when nitrogen and oxygen from the air combine.

   To find the oxidation states, we apply the standard rule: oxygen almost always has an oxidation state of –2 (except in peroxides, which aren't relevant here). For a neutral molecule, the sum of oxidation states must be zero.

   1. For NO: Let the oxidation state of nitrogen be $x$.

      $x + (-2) = 0 \implies x = +2$.

      So nitrogen is in the +2 oxidation state in nitric oxide.

   2. For NO₂: Let the oxidation state of nitrogen be $y$.

      $y + 2(-2) = 0 \implies y - 4 = 0 \implies y = +4$.

      So nitrogen is in the +4 oxidation state in nitrogen dioxide.

   Watch out

   A common mistake is to confuse NO₂ (nitrogen dioxide, +4) with N₂O (nitrous oxide, +1) or N₂O₅ (dinitrogen pentoxide, +5). But in photochemical smog, the two oxides formed directly from combustion are NO and NO₂ — not these others.

   These two oxides are the primary precursors: NO is released first, then quickly oxidised to NO₂ in the air. NO₂ absorbs sunlight and breaks down, starting a chain of reactions that produce ozone and other components of photochemical smog.

   ✓Final answer

   The correct option is (B) $+2,\ +4$.

7. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.What are X, Y, Z respectively in the following reaction sequence? (Conc. = concentrated) Chlorobenzene $\xrightarrow[\text{conc. } \mathrm{H_2SO_4}]{\mathrm{HNO_3}} X \xrightarrow{Y} Z$ (A) $X$ = 1-chloro-4-nitrobenzene ($\mathrm{O_2N-C_6H_4-Cl}$, para) ; $Y$ =

   (i) $\mathrm{NaOH}\,|\,443\ \mathrm{K}$

   (ii) $\mathrm{H^+}$ ; $Z$ = 4-nitrophenol ($\mathrm{O_2N-C_6H_4-OH}$, para) (B) $X$ = 1-chloro-3-nitrobenzene (Cl and $\mathrm{NO_2}$ meta) ; $Y$ =

   (i) $\mathrm{NaOH}\,|\,368\ \mathrm{K}$

   (ii) $\mathrm{H^+}$ ; $Z$ = 3-nitrophenol (OH and $\mathrm{NO_2}$ meta) (C) $X$ = 1-chloro-4-nitrobenzene (para) ; $Y$ =

   (i) $\mathrm{NaOH}\,|\,300\ \mathrm{K}$

   (ii) $\mathrm{H^+}$ ; $Z$ = 4-chlorophenol ($\mathrm{HO-C_6H_4-Cl}$, para) (D) $X$ = 1-chloro-4-nitrobenzene (para) ; $Y$ = $\mathrm{H_2O}\,|\,300\ \mathrm{K}$ ; $Z$ = 4-nitrophenol (para)

   ›Reveal solutionSolution

   Nitration of chlorobenzene gives mainly 1-chloro-4-nitrobenzene ($X$); the para nitro group makes the $\mathrm{C\!-\!Cl}$ bond susceptible to nucleophilic attack, so NaOH at 443 K followed by $\mathrm{H^+}$ converts it to 4-nitrophenol ($Z$). That is option (A).

   The concept first

   1. Electrophilic step. Halogens are deactivating but ortho–para directing: the $-I$ effect withdraws electron density (slowing the reaction), while the $+R$ (lone-pair) effect places negative charge at the $o$ and $p$ positions, so the electrophile goes there. Steric crowding makes para the major product.
   2. Nucleophilic step. Aryl halides resist $\mathrm{S_N}$ reactions (partial double-bond character of $\mathrm{C\!-\!Cl}$, resonance stabilisation). Chlorobenzene itself needs $623\ \mathrm{K}/300\ \mathrm{atm}$ NaOH (Dow process). A strongly electron-withdrawing $-\mathrm{NO_2}$ at the ortho or para position changes everything: it stabilises the negatively-charged carbanion (Meisenheimer) intermediate by resonance, so the reaction now runs under much milder conditions ($\mathrm{NaOH}$, $443\ \mathrm{K}$). A meta nitro group gives no such stabilisation.

   Step-by-step

   1. $\mathrm{C_6H_5Cl} \xrightarrow[\text{conc. } \mathrm{H_2SO_4}]{\mathrm{HNO_3}}$ chiefly $p$-nitrochlorobenzene (with a little ortho).

   $X = \mathrm{1\text{-}chloro\text{-}4\text{-}nitrobenzene}$

   This already eliminates the meta option.

   2. Treat $X$ with aqueous $\mathrm{NaOH}$ at $443\ \mathrm{K}$: $\mathrm{OH^-}$ adds at the carbon bearing $\mathrm{Cl}$; the resulting carbanion is delocalised onto the para $-\mathrm{NO_2}$ oxygen. Loss of $\mathrm{Cl^-}$ restores aromaticity, giving sodium $p$-nitrophenoxide.

   3. Acidify: $\mathrm{H^+}$ protonates the phenoxide.

   $Z = \mathrm{4\text{-}nitrophenol},\ \mathrm{O_2N\!-\!C_6H_4\!-\!OH}$

   4. Why the other options fail.
      • 300 K NaOH giving 4-chlorophenol: the $\mathrm{NO_2}$, not the $\mathrm{Cl}$, would have to leave — nitro is a very poor leaving group here, and 300 K is far too mild for aryl-halide substitution.
      • $\mathrm{H_2O}$ at 300 K: water is far too weak a nucleophile; nothing happens.
      • meta isomer: contradicts the $o,p$-directing power of $\mathrm{Cl}$.

   ✓Final answer

   The sequence is chlorobenzene → $p$-nitrochlorobenzene → (NaOH, 443 K; then $\mathrm{H^+}$) → $p$-nitrophenol, so the correct option is (A).

   ANSWER: A

8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The major products P and Q from the following reactions are $\text{P} \xleftarrow[\text{(ii) H}_2\text{O}]{\text{(i) LiAlH}_4} \ \text{C}_6\text{H}_5\text{CONH}_2 \ \xrightarrow{\text{Br}_2\ /\ \text{NaOH}} \text{Q}$ (A) $P = C_6H_5NH_2$ ; $Q = C_6H_5CH_2NH_2$ (B) $P = C_6H_5CH_2NH_2$ ; $Q = C_6H_5NH_2$ (C) $P = C_6H_5-\underset{\overset{|}{OH}}{CH}-NH_2$ ; $Q = C_6H_5COONa$ (D) $P = C_6H_5CN$ ; $Q = C_6H_5Br$

   ›Reveal solutionSolution

   $\mathrm{LiAlH_4}$ reduces benzamide to benzylamine ($P$), while $\mathrm{Br_2/NaOH}$ degrades it to aniline ($Q$), one carbon shorter. That is option (B).

   The concept first

   Benzamide, $\mathrm{C_6H_5CONH_2}$, has two very different fates depending on whether you attack the carbon or the nitrogen.

   • $\mathrm{LiAlH_4}$ delivers $\mathrm{H^-}$ to the electrophilic carbonyl carbon. The C–N bond is never broken; the oxygen leaves and the carbon ends up as a $\mathrm{CH_2}$ bridging the ring and the nitrogen. Amides are the only carbonyl compounds reduced all the way to an amine rather than an alcohol — that is worth remembering.
   • $\mathrm{Br_2/NaOH}$ instead brominates the nitrogen. The resulting $\mathrm{N}$-bromoamide is deprotonated, loses $\mathrm{Br^-}$ to give a nitrene-like species, and the phenyl group migrates from carbon to nitrogen. The old carbonyl carbon ends up as an isocyanate carbon and is finally hydrolysed away as $\mathrm{Na_2CO_3}$. Hence a one-carbon shortening.

   Step-by-step

   1. Reduction ($P$):

   $\mathrm{C_6H_5CONH_2 \xrightarrow[\ (ii)\ H_2O\ ]{(i)\ LiAlH_4} C_6H_5CH_2NH_2}$

   Benzylamine, a primary amine with 7 carbons (same as benzamide).

   2. Hofmann degradation ($Q$):

   $\mathrm{C_6H_5CONH_2 + Br_2 + 4NaOH \rightarrow C_6H_5NH_2 + Na_2CO_3 + 2NaBr + 2H_2O}$

   Aniline, a primary amine with 6 carbons (one less).

   3. Mechanistic checkpoints for $Q$: $\mathrm{C_6H_5CONHBr} \rightarrow \mathrm{C_6H_5CO\bar{N}Br} \rightarrow$ phenyl migration $\rightarrow \mathrm{C_6H_5N{=}C{=}O}$ (phenyl isocyanate) $\rightarrow$ hydrolysis $\rightarrow$ aniline.

   4. Match to the options: $P=\mathrm{C_6H_5CH_2NH_2}$, $Q=\mathrm{C_6H_5NH_2}$ — exactly option (B). Option (A) swaps them; (C) invents an unstable hemiaminal; (D) has no reagent that could make a nitrile or an aryl bromide here.

   ✓Final answer

   Benzamide gives benzylamine on $\mathrm{LiAlH_4}$ reduction and aniline on Hofmann degradation, so the correct option is (B).

   ANSWER: B

9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Identify the product 'Y' in the given sequence of reactions. Chlorobenzene $\xrightarrow[\text{Conc. H}_2\text{SO}_4]{\text{HNO}_3}$ X (Major) $\xrightarrow[\text{(ii) H}^+]{\text{(i) NaOH, 443 K}}$ Y (A) $p$-Nitrophenol (benzene ring with $-\mathrm{OH}$ and $-\mathrm{NO_2}$ para to each other) (B) $o$-Nitrophenol (benzene ring with $-\mathrm{OH}$ and $-\mathrm{NO_2}$ ortho to each other) (C) $p$-Hydroxybenzenesulphonic acid (benzene ring with $-\mathrm{OH}$ and $-\mathrm{SO_3H}$ para to each other) (D) 2,4-Dinitrophenol (benzene ring with $-\mathrm{OH}$ and $-\mathrm{NO_2}$ groups at positions 2 and 4)

   ›Reveal solutionSolution

   Chlorobenzene nitrates mainly at the para position to give $p$-nitrochlorobenzene ($X$); the para $-\mathrm{NO_2}$ activates the C–Cl carbon, so $\mathrm{NaOH}$ at 443 K followed by $\mathrm{H^+}$ gives $p$-nitrophenol — option (A).

   The concept first

   Chlorine on a ring plays a double game. Through its $-I$ effect it deactivates the ring, but through lone-pair $+R$ donation it directs an incoming electrophile to the ortho and para positions. So nitration of chlorobenzene is slower than that of benzene, yet its major product is the para isomer (ortho is crowded).

   The second step is the interesting one. Aryl halides are normally inert to nucleophiles — the C–Cl bond has partial double-bond character and the ring is electron-rich. Nucleophilic aromatic substitution becomes easy only when a strong electron-withdrawing group sits ortho or para to the halogen, because then the negative charge of the intermediate carbanion (the Meisenheimer complex) can be delocalised onto that group.

   Step-by-step

   Step 1 — nitration ($X$).

   $\mathrm{C_6H_5Cl}\ \xrightarrow[\text{conc. }\mathrm{H_2SO_4}]{\mathrm{HNO_3}}\ p\text{-}\mathrm{O_2N\text{-}C_6H_4\text{-}Cl}\ (\text{major})\ +\ o\text{-isomer (minor)}$

   The electrophile is $\mathrm{NO_2^{+}}$, formed as $\mathrm{HNO_3+2H_2SO_4\rightarrow NO_2^{+}+H_3O^{+}+2HSO_4^{-}}$. So $X=p$-nitrochlorobenzene. Note $\mathrm{H_2SO_4}$ acts only as the catalyst — it does not sulphonate the ring, which removes option (C).

   Step 2 — why 443 K suffices. After $\mathrm{OH^-}$ attacks the C–Cl carbon, the resulting carbanion can push its charge onto the nitro oxygens (only possible from the ortho or para position). That extra stabilisation drops the activation energy enormously, so the harsh Dow conditions (623 K, 300 atm) are no longer needed: 443 K is enough.

   Step 3 — substitution and acidification ($Y$).

   $p\text{-}\mathrm{O_2N\text{-}C_6H_4\text{-}Cl}\ \xrightarrow[443\ \mathrm{K}]{\mathrm{NaOH}}\ p\text{-}\mathrm{O_2N\text{-}C_6H_4\text{-}O^{-}Na^{+}}\ \xrightarrow{\mathrm{H^{+}}}\ p\text{-}\mathrm{O_2N\text{-}C_6H_4\text{-}OH}$

   Step 4 — eliminate the rest.

   • (B) $o$-Nitrophenol would come from the minor ortho nitration product, not the major one.
   • (C) A sulphonic acid group cannot form under nitrating conditions.
   • (D) 2,4-Dinitrophenol needs two nitro groups; only one nitration is shown.

   ✓Final answer

   The sequence runs chlorobenzene $\to$ $p$-nitrochlorobenzene $\to$ $p$-nitrophenol, so the correct option is (A).

   ANSWER: A

10. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.What is ‘Z’ in the given set of reactions?

    [!FORMULA] $\begin{aligned} \text{C}_6\text{H}_5\text{OCH}_3 & \xrightarrow{\text{HI}} X + Y \\ X & \xrightarrow{\text{Zn},\, \Delta} \text{(benzene ring)} \\ Y & \xrightarrow[\text{Anhy. AlCl}_3]{\text{C}_6\text{H}_6} Z \\ \end{aligned}$

    (A) \chemfig{*6(-=-(-CH_3)-=-)} (B) \chemfig{*6(-=-(-CH_2Cl)-=-)} (C) \chemfig{*6(-=-(-C_2H_5)-=-)} (D) \chemfig{*6(-=-(-Cl)-=-)}

    ›Reveal solutionSolution

    The reaction sequence converts anisole into iodobenzene and methanol, then iodobenzene reacts with benzene in a Friedel–Crafts alkylation to give diphenylmethane, which is toluene — so Z is methylbenzene, option (A).

    Concept & Intuition

    This problem tests your understanding of ether cleavage by HI and subsequent aromatic substitution. Anisole (C₆H₅OCH₃) is an aryl methyl ether. When treated with HI, the C–O bond breaks selectively because the aryl–O bond is stronger (partial double-bond character from resonance with the ring). The products are iodobenzene (X) and methanol (Y). Then, X (iodobenzene) is reduced by Zn dust to benzene. Meanwhile, Y (methanol) reacts with benzene in the presence of anhydrous AlCl₃ — a Friedel–Crafts alkylation — to give toluene (methylbenzene). The key insight: methanol generates a methyl carbocation under these conditions, which attacks benzene.

    Step-by-step reasoning

    1. First reaction — Cleavage of anisole with HI Anisole (C₆H₅OCH₃) reacts with excess HI. The mechanism: protonation of the ether oxygen, then nucleophilic attack by I⁻ at the methyl carbon (since the methyl–O bond is weaker than the aryl–O bond). This yields iodobenzene (C₆H₅I) as X and methanol (CH₃OH) as Y.

    $\text{C}_6\text{H}_5\text{OCH}_3 + \text{HI} \rightarrow \text{C}_6\text{H}_5\text{I} + \text{CH}_3\text{OH}$

    2. Second reaction — Reduction of iodobenzene X (iodobenzene) is treated with Zn dust and heated (Δ). Zinc reduces the C–I bond, replacing iodine with hydrogen. The product is benzene (C₆H₆).

    $\text{C}_6\text{H}_5\text{I} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnI}_2$

    3. Third reaction — Friedel–Crafts alkylation of benzene Y is methanol (CH₃OH). In the presence of anhydrous AlCl₃ (a Lewis acid), methanol forms a complex that generates a methyl carbocation (CH₃⁺). This electrophile attacks benzene, yielding toluene (methylbenzene, C₆H₅CH₃).

    $\text{CH}_3\text{OH} + \text{C}_6\text{H}_6 \xrightarrow{\text{Anhy. AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3 + \text{H}_2\text{O}$

    The product Z is therefore toluene.

    4. Matching with options Option (A) shows a benzene ring with a –CH₃ substituent — that is toluene. Option (B) has –CH₂Cl, (C) has –C₂H₅, (D) has –Cl. Only (A) matches.

    Watch out

    A common mistake is to think that HI cleaves the aryl–O bond, giving phenol and methyl iodide. But HI is a strong acid and I⁻ is a good nucleophile; it attacks the less hindered methyl carbon, not the aromatic ring. So the products are iodobenzene and methanol, not phenol.

    Tip

    Remember: For aryl alkyl ethers, HI always cleaves the alkyl–O bond, not the aryl–O bond. This is a classic “hard-soft” acid-base principle — the soft I⁻ prefers the softer alkyl carbon.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

11. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.What are X and Y respectively in the following reactions? (A) $\mathrm{CH_3Cl, \ CH_3COCl}$ (B) $\mathrm{C_2H_5Cl, \ CH_3COCl}$ (C) $\mathrm{CH_3COCl, \ CH_3Cl}$ (D) $\mathrm{C_2H_5COCl, \ CH_3Cl}$

    ›Reveal solutionSolution

    The first step is a Friedel–Crafts alkylation ($X=\text{CH}_3\text{Cl}$) and the second a Friedel–Crafts acylation ($Y=\text{CH}_3\text{COCl}$). Correct option: (A).

    Both steps are carried out on the aromatic ring with anhydrous $\text{AlCl}_3$, so each reagent is a Friedel–Crafts electrophile.

    • $X$ (alkylation). An alkyl chloride substitutes an alkyl group onto the ring:

    $\text{C}_6\text{H}_6+\text{CH}_3\text{Cl}\xrightarrow{\text{AlCl}_3}\text{C}_6\text{H}_5\text{CH}_3+\text{HCl}.$

    So $X=\text{CH}_3\text{Cl}$.

    • $Y$ (acylation). An acyl chloride introduces a $\text{–COCH}_3$ group:

    $\text{ArH}+\text{CH}_3\text{COCl}\xrightarrow{\text{AlCl}_3}\text{Ar–COCH}_3+\text{HCl}.$

    So $Y=\text{CH}_3\text{COCl}$.

    Alkylation activates the ring and must precede acylation (an acyl group deactivates the ring), which fixes the order $X=\text{CH}_3\text{Cl}$, $Y=\text{CH}_3\text{COCl}$.

    ✓Final answer

    (A) $X=\text{CH}_3\text{Cl}$, $Y=\text{CH}_3\text{COCl}$.

12. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.In which of the following reactions, hydrogen is evolved? I. Reaction of sodium borohydride with iodine II. Oxidation of diborane III. Reaction of boron trifluoride with sodium hydride IV. Hydrolysis of diborane (A) I, II only (B) I, II, IV only (C) III, IV only (D) I, IV only

    ›Reveal solutionSolution

    Hydrogen gas is evolved when diborane is hydrolysed and when sodium borohydride reacts with iodine — the key is to check whether the reaction produces free $H_2$ or only hydrogen-containing byproducts. The correct set is I and IV only.

    The question tests your understanding of the chemistry of boron hydrides and related hydride-transfer reactions. Hydrogen evolution means free $H_2$ gas is released, not just that hydrogen atoms appear in a product. Many boron compounds are electron-deficient and react with protic sources or oxidising agents to liberate hydrogen, but not every reaction that involves a hydride does so.

    Let’s examine each reaction carefully.

    1. Reaction of sodium borohydride with iodine Sodium borohydride ($\text{NaBH}_4$) is a source of hydride ions. Iodine ($I_2$) is a mild oxidising agent. They react to give diborane and hydrogen gas:

    $2\text{NaBH}_4 + I_2 \rightarrow \text{B}_2\text{H}_6 + 2\text{NaI} + H_2$

    Hydrogen is clearly evolved here. So I is correct.

    2. Oxidation of diborane Diborane ($\text{B}_2\text{H}_6$) burns in oxygen to form boric oxide and water:

    $\text{B}_2\text{H}_6 + 3\text{O}_2 \rightarrow \text{B}_2\text{O}_3 + 3\text{H}_2\text{O}$

    No free hydrogen is produced — the hydrogen ends up in water. So II is incorrect.

    3. Reaction of boron trifluoride with sodium hydride Sodium hydride ($\text{NaH}$) is a strong hydride donor. With $\text{BF}_3$, it forms diborane and sodium fluoride:

    $2\text{BF}_3 + 6\text{NaH} \rightarrow \text{B}_2\text{H}_6 + 6\text{NaF}$

    All the hydrogen from NaH goes into diborane; no $H_2$ gas is released. So III is incorrect.

    4. Hydrolysis of diborane Diborane reacts vigorously with water to give boric acid and hydrogen gas:

    $\text{B}_2\text{H}_6 + 6\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{BO}_3 + 6\text{H}_2$

    This is a classic reaction — the electron-deficient B–H bonds are attacked by water, and each B–H unit ultimately yields one $H_2$ molecule. So IV is correct.

    Watch out

    A common mistake is to think that any reaction producing a boron hydride (like diborane) must evolve hydrogen. But in reaction III, the hydride from NaH is completely consumed to form $\text{B}_2\text{H}_6$ — no free $H_2$ appears. Always check the balanced equation for $H_2$ as a product.

    Tip

    For boron hydride reactions, remember the pattern: if a hydride source meets a protic source (like water, alcohol, or an acid), $H_2$ is usually evolved. If two hydride sources react (like $\text{NaH}$ and $\text{BF}_3$), they combine to form a higher borane without releasing $H_2$.

    Only reactions I and IV produce hydrogen gas.

    ✓Final answer

    The correct option is (D) I, IV only.