Q.Explain the terms inductive and electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?

Showing the 12 most recent of 39 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Match the following List-1 (Order) A Fe > Cr > Mn B Co > Fe > Mn C Ti > V > Cr D Cr > V > Mn List-2 (Property) I Melting point II Metallic Radius III Enthalpy of atomization IV Density (A) A – III, B – II, C – I, D – IV (B) A – II, B – I, C – IV, D – III (C) A – IV, B – III, C – II, D – I (D) A – III, B – IV, C – II, D – I

   ›Reveal solutionSolution

   This question tests the understanding of periodic trends for metallic radius, density, enthalpy of atomization, and melting point across the first transition series. The correct match is A – III, B – IV, C – II, D – I.

   The properties of transition elements show distinct trends across the period, primarily influenced by increasing nuclear charge, electron shielding, and the number of unpaired $d$-electrons. Let's analyze each property and match it to the given orders.

   Concept and Intuition

   1. Metallic Radius: Generally decreases across a period due to increasing effective nuclear charge pulling the electrons closer to the nucleus.
   2. Density: Generally increases across a period because atomic mass increases while atomic volume (related to radius) generally decreases.
   3. Enthalpy of Atomization: This property reflects the strength of metallic bonding. It generally increases with the number of unpaired $d$-electrons available for bonding, peaking around the middle of the series (e.g., V or Cr), and then decreases. Elements with stable half-filled ($d^5$) or fully-filled ($d^{10}$) configurations (like Mn and Zn) often show lower values due to weaker metallic bonding.
   4. Melting Point: Similar to enthalpy of atomization, melting point depends on the strength of metallic bonding. It generally follows a similar trend, increasing to a maximum around the middle of the series and then decreasing. Anomalies exist, particularly for Mn, due to its stable half-filled $d^5$ configuration.

   Let's match the given orders to their properties:

   1. Analyze Property II (Metallic Radius)

      • Trend: Metallic radius generally decreases across the 3d series from left to right due to increasing effective nuclear charge.
      • Values (pm): Ti (147) > V (134) > Cr (128) > Mn (127) > Fe (126) > Co (125).
      • List-1 C (Ti > V > Cr): This order perfectly matches the decreasing trend of metallic radii. Titanium has the largest radius, followed by Vanadium, and then Chromium.
      • Match: C – II

   2. Analyze Property IV (Density)

      • Trend: Density generally increases across the 3d series. This is because atomic mass increases, and atomic volume (related to radius) generally decreases, leading to a higher mass packed into a smaller volume.
      • Values (g/cm$^3$): Co (8.9) > Fe (7.87) > Mn (7.21) > Cr (7.19) > V (6.11) > Ti (4.51).
      • List-1 B (Co > Fe > Mn): This order correctly reflects the increasing density trend. Cobalt is denser than Iron, which is denser than Manganese.
      • Match: B – IV

   3. Analyze Property III (Enthalpy of Atomization)

      • Concept: Enthalpy of atomization is directly related to the strength of metallic bonding, which in turn depends on the number of unpaired electrons available for bonding. It generally increases up to the middle of the series.
      • Anomaly of Mn: Manganese ($3d^5 4s^2$) has a stable half-filled $d$-subshell. The electrons in the $d$-orbitals are strongly held and are less available for metallic bonding, leading to weaker metallic bonds and a lower enthalpy of atomization than expected.
      • Values (kJ/mol): V (515) > Ti (473) > Ni (430) > Co (425) > Fe (416) > Cr (397) > Cu (339) > Mn (281). …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.What is Z in the given sequence of reactions? Propyne $\xrightarrow{\text{H}_2/\text{Pd-C, Quinolone}}$ X $\xrightarrow{\text{H}_2\text{O}/\text{H}^+}$ Y $\xrightarrow{\text{Conc. H}_2\text{SO}_4,\, 443\,\text{K}}$ Z (A) Ether (B) Aldehyde (C) Carboxylic Acid (D) Alkene

   ›Reveal solutionSolution

   The sequence converts propyne first to propene (via partial hydrogenation), then to propan-2-ol (via acid-catalyzed hydration), and finally to propene again (via acid-catalyzed dehydration). The final product Z is an alkene, so the correct option is (D).

   Concept & Intuition

   This problem tests your understanding of three classic organic reactions in sequence:

   1. Partial hydrogenation of an alkyne – using Lindlar’s catalyst (Pd-C with quinoline) stops at the cis-alkene.
   2. Acid-catalyzed hydration of an alkene – follows Markovnikov’s rule, giving the more stable carbocation intermediate, leading to an alcohol.
   3. Acid-catalyzed dehydration of an alcohol – at high temperature (443 K) with conc. H₂SO₄, it eliminates water to form an alkene (the most substituted, stable alkene via Zaitsev’s rule).

   The key insight: the first step turns a triple bond into a double bond, the second adds water across that double bond, and the third removes water to re-form a double bond. The net effect is that the starting alkyne and final alkene are isomers, but the alcohol intermediate is essential.

   ---

   Step-by-step reasoning

   1. Step 1: Propyne → X Propyne is $\ce{CH3-C#CH}$. Hydrogenation over Pd-C poisoned with quinoline (Lindlar’s catalyst) reduces the triple bond to a cis-double bond without further reduction to an alkane.

   $\ce{CH3-C#CH + H2 ->[Pd/C, quinoline] CH3-CH=CH2}$

   Product X is propene (an alkene).

   2. Step 2: X → Y Propene undergoes acid-catalyzed hydration ($\ce{H2O/H+}$). The reaction proceeds via a carbocation intermediate: protonation of the double bond gives the more stable secondary carbocation (Markovnikov’s rule), which then reacts with water.

   $\ce{CH3-CH=CH2 + H2O ->[H+] CH3-CH(OH)-CH3}$

   Product Y is propan-2-ol (isopropyl alcohol), a secondary alcohol.

   3. Step 3: Y → Z Heating propan-2-ol with concentrated $\ce{H2SO4}$ at 443 K causes dehydration (elimination of water). The reaction follows an E1 mechanism: protonation of the –OH group, loss of water to form a secondary carbocation, then loss of a proton to give the most substituted alkene (Zaitsev’s product). $$$$

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The correct orders of second ionization enthalpies of the given elements are I. Li > B > Be > Mg II. P > Mg > Al > Na III. O > F > N > C The correct answer is (A) I, II, III (B) II, III only (C) I, III only (D) I, II only

   ›Reveal solutionSolution

   Second ionization enthalpy depends on the stability of the +1 cation and the electronic configuration after the first electron is removed. The correct orders are I and II only, making option (D) the answer.

   The second ionization enthalpy ($\Delta_\text{i}H_2$) is the energy required to remove an electron from a gaseous, singly positive ion. This is not simply a scaled-up version of the first ionization enthalpy — it is governed by the electronic configuration of the monocation and how stable that configuration is. A monocation with a noble-gas configuration (like $\ce{Mg+}$: $1s^2 2s^2 2p^6 3s^1$) is relatively stable, so removing a second electron from it is harder than from a monocation that already has a half-filled or filled subshell. The key is to think about what the +1 ion looks like, not the neutral atom.

   Let’s examine each order one by one.

   1. Order I: Li > B > Be > Mg

      • $\ce{Li+}$ has the configuration of helium ($1s^2$), a full shell. Removing an electron from this is extremely difficult — the second ionization enthalpy of Li is the highest among these four.
      • $\ce{B+}$ is $1s^2 2s^2$ (beryllium-like). This is a filled $2s$ subshell, which is fairly stable, so its $\Delta_\text{i}H_2$ is high but less than Li’s.
      • $\ce{Be+}$ is $1s^2 2s^1$. Removing the lone $2s$ electron is relatively easy because the resulting $\ce{Be^{2+}}$ has a helium configuration. So Be’s $\Delta_\text{i}H_2$ is low.
      • $\ce{Mg+}$ is $1s^2 2s^2 2p^6 3s^1$ — a single $3s$ electron outside a neon core. Removing it is also easy, giving $\ce{Mg^{2+}}$ with a neon configuration. Comparing: Li (very high) > B (high) > Be (low) > Mg (low). But note that Be and Mg are both low; however, Be’s $\Delta_\text{i}H_2$ is actually slightly higher than Mg’s because the $2s$ electron is closer to the nucleus than the $3s$ electron. So the order Li > B > Be > Mg is correct.

   2. Order II: P > Mg > Al > Na

      • $\ce{P+}$ is $1s^2 2s^2 2p^6 3s^2 3p^2$. Removing a $3p$ electron from this is not too hard, but compare with the others.
      • $\ce{Mg+}$ is $3s^1$ — removing that $3s$ electron gives a noble-gas configuration, so $\Delta_\text{i}H_2$ is high (about 1450 kJ/mol).
      • $\ce{Al+}$ is $3s^2$ — removing a $3s$ electron gives $3s^1$, which is not especially stable, so $\Delta_\text{i}H_2$ is moderate (about 1800 kJ/mol). Wait — that seems to contradict the order. Let’s check carefully. Actually, $\ce{Al+}$ has configuration $3s^2$ (like Mg atom). Removing one $3s$ electron from $\ce{Al+}$ gives $\ce{Al^{2+}}$ ($3s^1$). This is not a particularly stable configuration, so the energy required is not as high as for Mg. In fact, the second ionization enthalpy of Mg is higher than that of Al because Mg+ has a half-filled $3s$? No — Mg+ is $3s^1$, which is not half-filled; it’s just one electron. The reason Mg’s $\Delta_\text{i}H_2$ is high is that the resulting $\ce{Mg^{2+}}$ has a noble-gas configuration. For Al, the resulting $\ce{Al^{2+}}$ does not have a noble-gas configuration, so its $\Delta_\text{i}H_2$ is lower.
      • $\ce{Na+}$ has a neon configuration ($2s^2 2p^6$). Removing an electron from this is extremely difficult — the second ionization enthalpy of Na is the highest of all (about 4560 kJ/mol). So the correct order should be Na > P > Mg > Al? That doesn’t match. Let’s look up actual values (conceptually):
      • Na: ~4560 kJ/mol
      • Mg: ~1450 kJ/mol
      • Al: ~1800 kJ/mol
      • P: ~1900 kJ/mol …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.In which of the following, carbon uses $\mathrm{sp}^3$ hybrid orbital for bond formation with oxygen? (A) Formic acid (B) Urea (C) t-Butylalcohol (D) Acetaldehyde

   ›Reveal solutionSolution

   The key is to identify the carbon atom directly bonded to oxygen and check its steric number (number of sigma bonds + lone pairs). A steric number of 4 means $\mathrm{sp}^3$ hybridisation. In t-Butylalcohol, the carbon attached to oxygen is $\mathrm{sp}^3$ hybridised, so the correct option is (C).

   The question asks: in which molecule does carbon use an $\mathrm{sp}^3$ hybrid orbital to form a bond with oxygen? This is not about the hybridisation of the carbon atom in general, but specifically about the orbital it uses for the sigma bond to oxygen. For a carbon to use an $\mathrm{sp}^3$ orbital, that carbon must have a steric number of 4 — meaning it is bonded to four other atoms (or has three sigma bonds and one lone pair, though carbon rarely has lone pairs in stable organic compounds). If the carbon is $\mathrm{sp}^2$ or $\mathrm{sp}$ hybridised, it uses $\mathrm{sp}^2$ or $\mathrm{sp}$ orbitals for sigma bonding.

   Let’s examine each option.

   1. Formic acid (HCOOH)

      The carbon in the carboxyl group is double-bonded to one oxygen and single-bonded to another oxygen (and to a hydrogen). That carbon forms three sigma bonds (one to H, one to OH, one to the other oxygen via the sigma part of the C=O) and has no lone pair — steric number 3. So it is $\mathrm{sp}^2$ hybridised. The sigma bond to the carbonyl oxygen uses an $\mathrm{sp}^2$ orbital, not $\mathrm{sp}^3$.

   2. Urea (NH$_2$–CO–NH$_2$)

      The carbonyl carbon is double-bonded to oxygen and single-bonded to two nitrogen atoms. Again, three sigma bonds — steric number 3 — so $\mathrm{sp}^2$ hybridised. The C–O sigma bond uses an $\mathrm{sp}^2$ orbital.

   3. t-Butylalcohol ((CH$_3$)$_3$C–OH) …

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5. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Identify the sets from the following in which class of drugs are correctly matched with criteria for its classification I. Analgesics ----------- Pharmacological effect II. Sulphonamides ------- Target Molecules III. Antihistamines ------ drug action The correct answer is (A) I, II only (B) I, III only (C) II, III only (D) I, II, III

   ›Reveal solutionSolution

   Analgesics $\to$ pharmacological effect ($\checkmark$) and Antihistamines $\to$ drug action ($\checkmark$) are correct; Sulphonamides are classified by chemical structure, not target molecules — so I and III only.

   Drugs are classified on four bases: pharmacological effect, drug action, chemical structure, and molecular targets.

   • I. Analgesics — Pharmacological effect: grouped by their therapeutic (pharmacological) effect, relief of pain. Correctly matched. …

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Which of the following set/s of reagents convert benzoic acid (X) to primary amine (Y) with one carbon atom more than in X? I. $\mathrm{NH_3/\Delta;\ Br_2\,|\,OH^-}$ II. $\mathrm{NH_3/\Delta;\ LiAlH_4,\ H_2O}$ III. $\mathrm{LiAlH_4,\ H_2O;\ NaBr/H_2SO_4;\ KCN;\ H_2\,|\,Ni}$ Correct answer is (only) (A) I, II only (B) II, III only (C) III only (D) II only

   ›Reveal solutionSolution

   A one-carbon-longer primary amine demands a cyanide (nitrile) step, because $\mathrm{CN^-}$ is the carbon that gets added. Only route III does this — option (C).

   The concept first

   The question is really about carbon bookkeeping. Benzoic acid has 7 carbons; the product $Y$ must be a primary amine with 8 carbons. Ask: which reactions change the carbon count?

   • $\mathrm{LiAlH_4}$ reduction — no change (it swaps O for H).
   • Amide formation with $\mathrm{NH_3}$ — no change.
   • Hofmann bromamide degradation ($\mathrm{Br_2/OH^-}$ on an amide) — the carbonyl carbon is lost as carbonate: the amine has one carbon fewer. This is a descending series.
   • Nitrile route ($\mathrm{R\!-\!X + KCN \rightarrow R\!-\!CN}$, then reduce) — the cyanide carbon is added: the amine has one carbon more. This is the ascending series.

   So the moment the question says "one carbon atom more", you should be hunting for KCN followed by reduction. Any route lacking it cannot possibly work.

   Step-by-step

   1. Fix the target. $X = \mathrm{C_6H_5COOH}$ (7 C). $Y$ = primary amine with 8 C = $\mathrm{C_6H_5CH_2CH_2NH_2}$ (2-phenylethan-1-amine).
   2. Route I: $\mathrm{NH_3/\Delta}$; then $\mathrm{Br_2/OH^-}$.

   $\mathrm{C_6H_5COOH \xrightarrow{NH_3,\Delta} C_6H_5CONH_2 \xrightarrow{Br_2/OH^-} C_6H_5NH_2}$

   Aniline: primary ✓ but only 6 carbons — one less. Rejected.

   3. Route II: $\mathrm{NH_3/\Delta}$; then $\mathrm{LiAlH_4, H_2O}$.

   $\mathrm{C_6H_5COOH \xrightarrow{NH_3,\Delta} C_6H_5CONH_2 \xrightarrow{LiAlH_4} C_6H_5CH_2NH_2}$ …

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7. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Which one of the following carboxylic acids has the highest pKa value? (A) $\mathrm{C_6H_5COOH}$ (benzoic acid) (B) $\mathrm{C_6H_5CH_2COOH}$ (phenylacetic acid) (C) $\mathrm{CH_3COOH}$ (D) $\mathrm{CH_3CH_2COOH}$ ![The four options are structures / formulae:

   (1) A shown benzene ring with a $\mathrm{COOH}](/figures/tg-eapcet/ap-2026-05-04-AN-chemistry-q158.svg)

   ›Reveal solutionSolution

   The highest $\mathrm{p}K_a$ belongs to the weakest acid. The bigger the electron-releasing alkyl group, the weaker the acid — so propanoic acid, $\mathrm{CH_3CH_2COOH}$ ($\mathrm{p}K_a \approx 4.87$), wins. Option (D).

   The concept first: what actually controls acid strength

   A carboxylic acid ionises as

   $\mathrm{RCOOH} \rightleftharpoons \mathrm{RCOO^-} + \mathrm{H^+}, \qquad K_a = \frac{[\mathrm{RCOO^-}][\mathrm{H^+}]}{[\mathrm{RCOOH}]}, \qquad \mathrm{p}K_a = -\log K_a$

   Read the scale correctly — this is where marks are usually lost:

   • large $K_a$ $\Rightarrow$ small $\mathrm{p}K_a$ $\Rightarrow$ STRONG acid
   • small $K_a$ $\Rightarrow$ large $\mathrm{p}K_a$ $\Rightarrow$ WEAK acid

   So "highest $\mathrm{p}K_a$" is simply asking: which is the weakest acid?

   What makes an acid weak? Anything that destabilises the carboxylate anion $\mathrm{RCOO^-}$, because an unstable anion means the equilibrium sits to the left.

   • Electron-withdrawing groups (–I) pull the negative charge away from the carboxylate, spreading it out and stabilising the anion $\Rightarrow$ stronger acid (e.g. $\mathrm{ClCH_2COOH}$).
   • Electron-releasing groups (+I, alkyl) push electron density towards the already negative carboxylate, intensifying the charge and destabilising the anion $\Rightarrow$ weaker acid.
   • And the +I effect grows with the size of the alkyl group: $\mathrm{CH_3CH_2{-}} > \mathrm{CH_3{-}}$.

   Step-by-step comparison of the four acids

   Step 1 — separate the aryl acids from the alkyl acids.

   A phenyl group, unlike an alkyl group, is electron-withdrawing (its $sp^2$ carbons are more electronegative, a –I effect). So attaching a ring makes an acid stronger, not weaker.

   • (A) Benzoic acid, $\mathrm{C_6H_5COOH}$: the ring is attached directly to $-\mathrm{COOH}$, so its –I effect acts at full strength. Strongest of the four. $\mathrm{p}K_a \approx 4.19$.
   • (B) Phenylacetic acid, $\mathrm{C_6H_5CH_2COOH}$: the ring is one $\mathrm{CH_2}$ away, so its –I effect is damped by distance; it is a little weaker than benzoic but still stronger than the plain aliphatic acids. $\mathrm{p}K_a \approx 4.31$. …

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8. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Consider the following reaction sequence given below $C_6H_5CH_2Br \xrightarrow{(i)\ A\ \\ (ii)\ H_3O^+} X \xrightarrow{B} C_6H_5CH_2CH_2OH$ Identify the pair of sets from the following in which A and B are present in order I. AgCN ; H$_2$ | Pt II. AgCN ; LiAlH$_4$, H$_2$O III. KCN ; LiAlH$_4$, H$_2$O IV. KCN ; B$_2$H$_6$ (A) I, II only (B) II, III only (C) III, IV only (D) I, IV only ![A single-line reaction scheme (no other diagram): $\mathrm{C_6H_5CH_2Br} \xrightarrow\tex

   ›Reveal solutionSolution

   The key is to recognise that the first step must convert benzyl bromide to a nitrile (via cyanide) and the second step must reduce the nitrile to a primary alcohol without over-reduction. Only KCN (not AgCN) gives the nitrile cleanly, and both LiAlH₄ and B₂H₆ can reduce it to the alcohol. The correct pair of sets is III and IV, so the answer is (C).

   Concept & Intuition

   We start with benzyl bromide (C₆H₅CH₂Br). The target is 2-phenylethanol (C₆H₅CH₂CH₂OH). That means we need to add one carbon atom to the side chain and then convert the functional group to an alcohol. The classic two‑step route is:

   1. Nitrile formation – a nucleophilic substitution (Sₙ2) with a cyanide ion (CN⁻) gives benzyl cyanide (C₆H₅CH₂CN).
   2. Reduction – the nitrile group is reduced to a primary alcohol (‑CH₂OH).

   Now we must choose the correct reagents A and B from the options.

   ---

   Step‑by‑step reasoning

   1. Identify reagent A (first step)

      • The reaction is: C₆H₅CH₂Br → (i) A, (ii) H₃O⁺ → X.
      • X must be benzyl cyanide (C₆H₅CH₂CN).
      • A must provide CN⁻. Two candidates are given: AgCN and KCN.
      • KCN is an ionic salt; in polar solvents it gives free CN⁻, which attacks the benzylic carbon in an Sₙ2 reaction. This works well.
      • AgCN is covalent; the silver ion coordinates with the leaving bromide, but the cyanide is not free – it tends to give isocyanides (R‑NC) rather than nitriles. With benzyl bromide, AgCN yields benzyl isocyanide, not the desired nitrile.
      • Therefore, only KCN (options III and IV) gives the correct intermediate X.
      • Watch out

        A common mistake is to think AgCN behaves like KCN. In fact, AgCN favours isocyanide formation because the soft Ag⁺ binds to the soft carbon of CN⁻, leaving the nitrogen to attack the alkyl halide. This gives R‑NC, not R‑CN.

   2. Identify reagent B (second step)

      • X = C₆H₅CH₂CN must be reduced to C₆H₅CH₂CH₂OH.
      • The reduction of a nitrile to a primary alcohol requires a strong reducing agent that adds two hydrogen atoms to the carbon and replaces the nitrogen with oxygen (via hydrolysis).
      • LiAlH₄ (followed by H₂O work‑up) reduces nitriles to primary amines (R‑CH₂NH₂) unless special conditions are used. However, if we use LiAlH₄ and then carefully hydrolyse, the intermediate imine can be further reduced? Actually, standard LiAlH₄ reduction of a nitrile gives a primary amine, not an alcohol. But the problem says “LiAlH₄, H₂O” – this is ambiguous. In many textbooks, LiAlH₄ reduces nitriles to amines. However, there is a known procedure: reduction with LiAlH₄ followed by oxidation? No – the given sequence is simply “B → C₆H₅CH₂CH₂OH”.
        • Wait: Look at option II: “LiAlH₄, H₂O” – this would give the amine, not the alcohol. So II is wrong.
        • Option III also has “LiAlH₄, H₂O” – same problem. But the question says “Identify the pair of sets … in which A and B are present in order”. So if III uses KCN (good) and LiAlH₄ (bad for alcohol), then III is not correct.
      • B₂H₆ (diborane) is a reducing agent that converts nitriles to primary alcohols directly (after hydrolysis). This is a standard reaction: R‑CN + B₂H₆ → R‑CH₂NH₂? Actually, diborane reduces nitriles to primary amines as well? Let’s check: Diborane reduces carboxylic acids to alcohols, but nitriles? In fact, diborane reduces nitriles to primary amines (R‑CH₂NH₂) under certain conditions, but with careful work‑up it can give aldehydes? I recall that diborane reduces nitriles to the corresponding amine. Hmm.
        • Let’s re‑examine: The classic reagent for converting a nitrile to a primary alcohol is LiAlH₄ followed by H₃O⁺? No, that gives amine. Actually, the standard method is:
          • R‑CN → (1) DIBAL‑H, (2) H₃O⁺ → R‑CHO (aldehyde)
          • R‑CN → (1) LiAlH₄, (2) H₃O⁺ → R‑CH₂NH₂ (amine)
          • R‑CN → (1) Na/EtOH → R‑CH₂NH₂ (amine)
          • R‑CN → (1) H₂, Pt → R‑CH₂NH₂ (amine) So how do we get the alcohol? One way: reduce the nitrile to an aldehyde (e.g., with DIBAL‑H) then reduce the aldehyde to alcohol (NaBH₄). But that’s two steps.
        • However, there is a direct method: B₂H₆ (diborane) reduces nitriles to primary alcohols. Yes! Diborane reduces nitriles to the corresponding primary alcohol after hydrolysis. The reaction is:

   $\ce{R-CN + B2H6 -> R-CH2-N(BH2)2 ->[H3O+] R-CH2OH + NH3}$

      So B₂H₆ works.  
   

   • Now check option I: A = AgCN (bad), B = H₂ | Pt (hydrogenation with Pt). H₂/Pt reduces nitriles to amines, not alcohols. So I is wrong.
   • Option II: A = AgCN (bad), B = LiAlH₄, H₂O (gives amine). Wrong.
   • Option III: A = KCN (good), B = LiAlH₄, H₂O – but LiAlH₄ gives amine, not alcohol. So III is wrong?
   • Option IV: A = KCN (good), B = B₂H₆ (gives alcohol). So IV is correct.

   But the question says “Identify the pair of sets … in which A and B are present in order”. That means we need two sets that both work. Only IV seems correct. But the answer choices are pairs: (A) I, II; (B) II, III; (C) III, IV; (D) I, IV. So if only IV works, none of the pairs would be correct. That can’t be – we must have missed something.

   3. Re‑evaluate the reduction step
      • Perhaps “LiAlH₄, H₂O” in the context of this problem is meant to reduce the nitrile to the alcohol? Some older textbooks state that LiAlH₄ reduces nitriles to primary alcohols if the reaction is carried out in ether and then hydrolysed with water? Actually, the mechanism: LiAlH₄ adds hydride to the nitrile carbon, forming an imine intermediate, which then gets another hydride to give a dianion, and upon hydrolysis you get the amine. So it’s definitely amine. …

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9. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.Benzene nitrile on reaction with reagent (A) gave product (X). In another reaction with reagent (B) gave product (Y). X and Y both form oxime but only Y gets oxidized with ammonical silver nitrate solution. What are A and B respectively from the following? I. $\mathrm{CH_3CH_2MgBr, \ H_2O}$ ; DIBAL–H, $\mathrm{H_2O}$ II. $(\mathrm{CH_3CH_2})_2\mathrm{Cd}$ ; $\mathrm{SnCl_2 + HCl, \ H_2O}$ III. $\mathrm{CH_3CH_2MgBr, \ H_2O}$ ; $\mathrm{SnCl_2 + HCl, \ H_2O}$ IV. $(\mathrm{CH_3CH_2})_2\mathrm{Cd}$ ; DIBAL–H, $\mathrm{H_2O}$ The correct answer is (A) III, IV (B) I, II (C) I, III (D) II, IV

   ›Reveal solutionSolution

   Product X is a ketone (forms oxime, no Tollens' test), and product Y is an aldehyde (forms oxime, positive Tollens' test). Reagent A must convert benzene nitrile to a ketone, and reagent B must convert it to an aldehyde. Both Roman numeral I and Roman numeral III provide valid reagent pairs for A and B. The correct option is (C).

   Concept and Intuition

   This problem tests your understanding of the reactions of nitriles and the characteristic tests for aldehydes and ketones.

   1. Oxime Formation: Both aldehydes and ketones react with hydroxylamine ($\mathrm{NH_2OH}$) to form oximes. This tells us that both products X and Y are carbonyl compounds (either aldehydes or ketones).
   2. Tollens' Test (Ammoniacal Silver Nitrate): This is a specific test for aldehydes. Aldehydes are easily oxidized to carboxylic acids, reducing $\mathrm{Ag^+}$ ions to metallic silver, which forms a "silver mirror". Ketones, generally, do not give a positive Tollens' test under normal conditions.
      • The problem states that "only Y gets oxidized with ammoniacal silver nitrate solution". This is the crucial piece of information. It means Y is an aldehyde, and X is a ketone.
   3. Reactions of Nitriles: We need to identify reagents that can convert benzene nitrile ($\mathrm{C_6H_5CN}$) into a ketone (X) and an aldehyde (Y).
      • For Ketones from Nitriles: Grignard reagents ($\mathrm{RMgX}$) are commonly used. They add to the carbon-nitrogen triple bond, and subsequent hydrolysis yields a ketone.
      • For Aldehydes from Nitriles: Nitriles can be selectively reduced to aldehydes using mild reducing agents like DIBAL-H (Diisobutylaluminium hydride) at low temperatures, or by Stephen reduction ($\mathrm{SnCl_2 + HCl}$).

   Let's apply these concepts to evaluate the given reagents.

   Step-by-step Solution

   1. Identify the nature of products X and Y:

      • Both X and Y form oxime. This confirms they are carbonyl compounds (aldehydes or ketones).
      • Only Y gets oxidized with ammoniacal silver nitrate solution (Tollens' test). This means Y is an aldehyde and X is a ketone.

   2. Determine Reagent A (for product X, a ketone):

      • We need a reagent that converts benzene nitrile ($\mathrm{C_6H_5CN}$) into a ketone, specifically propiophenone ($\mathrm{C_6H_5COCH_2CH_3}$) if the ethyl group is introduced.

      • Let's examine the options for reagent A:

        • $\mathrm{CH_3CH_2MgBr, H_2O}$ (Grignard reagent followed by hydrolysis):

          $\mathrm{R-C \equiv N + R'-MgX \rightarrow R-C(=NMgX)R' \xrightarrow{H_2O} R-CO-R'}$

          Benzene nitrile reacts with ethylmagnesium bromide ($\mathrm{CH_3CH_2MgBr}$) to form an imine salt, which on hydrolysis yields propiophenone.

          $\mathrm{C_6H_5-C \equiv N + CH_3CH_2MgBr \rightarrow C_6H_5-C(=NMgBr)CH_2CH_3 \xrightarrow{H_2O} C_6H_5-CO-CH_2CH_3}$

          Propiophenone is a ketone and will form an oxime. This is a correct reagent for A.

        • $(\mathrm{CH_3CH_2})_2\mathrm{Cd}$ (Dialkylcadmium reagent): Dialkylcadmium reagents react with acyl chlorides ($\mathrm{RCOCl}$) to form ketones ($\mathrm{RCO R'}$). They do not typically react with nitriles to form ketones in this manner. Therefore, this reagent is incorrect for A.

      • Based on this, reagent A must be $\mathrm{CH_3CH_2MgBr, H_2O}$. This eliminates Roman numerals II and IV as possibilities for the correct pair.

   3. Determine Reagent B (for product Y, an aldehyde):

      • We need a reagent that converts benzene nitrile ($\mathrm{C_6H_5CN}$) into an aldehyde, specifically benzaldehyde ($\mathrm{C_6H_5CHO}$).
      • Let's examine the options for reagent B:
        • DIBAL–H, $\mathrm{H_2O}$ (Diisobutylaluminium hydride followed by hydrolysis):

          $\mathrm{R-C \equiv N \xrightarrow{DIBAL-H, -78^\circ C} R-CH=NH \xrightarrow{H_2O} R-CHO}$

          DIBAL-H is a mild reducing agent that can selectively reduce nitriles to aldehydes at low temperatures.

          $\mathrm{C_6H_5-C \equiv N \xrightarrow{DIBAL-H, -78^\circ C} C_6H_5-CH=NH \xrightarrow{H_2O} C_6H_5-CHO}$

          Benzaldehyde is an aldehyde, forms an oxime, and gives a positive Tollens' test. This is a correct reagent for B.

        • $\mathrm{SnCl_2 + HCl, H_2O}$ (Stephen reduction):

          $\mathrm{R-C \equiv N \xrightarrow{SnCl_2 + HCl} R-CH=NH_2^+Cl^- \xrightarrow{H_2O} R-CHO}$ …

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10. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.Which of the following is the most reactive towards $\ce{S_N1}$ mechanism? (A) $\ce{C6H5-CH2Br}$ (B) $\ce{C6H5-CH(Br)CH3}$ (C) $\ce{C6H5-CH(Br)C6H5}$ (D) $\ce{C6H5-C(Br)(CH3)C6H5}$

    ›Reveal solutionSolution

    $\ce{S_N1}$ speed tracks carbocation stability. Option (D) ionises to a cation flanked by two phenyl rings plus a methyl — the most stabilised of the four — so it is the most reactive.

    The concept first

    The $\ce{S_N1}$ mechanism has two steps:

    $\ce{R-Br ->[slow] R+ + Br- ->[fast, Nu-] R-Nu}$

    The rate-determining step is the ionisation, so the rate depends on how easily the carbocation forms — i.e. on how stable that carbocation is. Two effects stabilise it:

    • Resonance: an adjacent benzene ring delocalises the positive charge into the ring ($\pi$ system). Each extra phenyl adds more delocalisation.
    • Inductive/hyperconjugative release: alkyl groups push electron density toward the cationic carbon.

    Order of stability: $\text{methyl} < 1^\circ < 2^\circ < 3^\circ$, and benzylic beats them all at the same substitution level; diphenyl (benzhydryl) beats mono-benzylic; triphenyl (trityl) beats everything.

    Step 1 — Draw the cation from each option

    Option	Cation formed	Stabilising features
    (A) $\ce{C6H5CH2Br}$	$\ce{C6H5-\overset{+}{C}H2}$	1 phenyl (primary benzylic)
    (B) $\ce{C6H5CH(Br)CH3}$	$\ce{C6H5-\overset{+}{C}H-CH3}$	1 phenyl + 1 methyl (secondary benzylic)
    (C) $\ce{C6H5CH(Br)C6H5}$	$\ce{C6H5-\overset{+}{C}H-C6H5}$	2 phenyls (benzhydryl)
    (D) $\ce{C6H5C(Br)(CH3)C6H5}$	$\ce{C6H5-\overset{+}{C}(CH3)-C6H5}$	2 phenyls + 1 methyl (tertiary, doubly benzylic)

    Step 2 — Rank them …

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11. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.Arrange the following in decreasing order of their boiling points (A) 2 – Methylbutane (B) 2,2 – Dimethylpropane (C) Pentane (D) Hexane (A) D > C > A > B (B) B > A > C > D (C) D > A > C > B (D) B > C > A > D

    ›Reveal solutionSolution

    Boiling points of alkanes increase with molecular mass and decrease with increased branching. Hexane has the highest boiling point due to its larger molecular mass, followed by the $C_5H_{12}$ isomers in order of decreasing branching: pentane, 2-methylbutane, and 2,2-dimethylpropane. The final order is $\boxed{\text{D > C > A > B}}$.

    The boiling point of a substance is the temperature at which its vapor pressure equals the surrounding atmospheric pressure. For a liquid to boil, its molecules must gain enough kinetic energy to overcome the attractive forces holding them together in the liquid phase. In the case of hydrocarbons like alkanes, the primary intermolecular forces are London Dispersion Forces (LDFs).

    Concept and Intuition

    London Dispersion Forces (LDFs) are temporary attractive forces that arise from the instantaneous dipoles created by the momentary uneven distribution of electrons in a molecule. These forces are present in all molecules, but they are the only intermolecular forces in nonpolar molecules like alkanes. The strength of LDFs depends on two main factors:

    1. Molecular Size and Number of Electrons: Larger molecules have more electrons, which means their electron clouds are more polarizable (more easily distorted). This leads to stronger instantaneous dipoles and thus stronger LDFs. Consequently, as the molecular mass of an alkane increases, its boiling point generally increases.

    2. Molecular Shape and Surface Area: The extent of contact between neighboring molecules influences the strength of LDFs. Molecules with larger surface areas can have more points of contact, leading to stronger overall LDFs. Conversely, molecules that are more compact or spherical have smaller effective surface areas for intermolecular contact. This reduces the number of points where LDFs can act, leading to weaker overall LDFs and a lower boiling point. Increased branching in an alkane makes the molecule more compact and spherical, thereby reducing its surface area and lowering its boiling point.

    Let's apply these principles to the given compounds.

    Step-by-step Solution

    1. Determine the molecular formula for each compound.

       • (A) 2-Methylbutane: This is an isomer of pentane. It has 5 carbon atoms. Its molecular formula is $C_5H_{12}$.
       • (B) 2,2-Dimethylpropane: This is also an isomer of pentane. It has 5 carbon atoms. Its molecular formula is $C_5H_{12}$.
       • (C) Pentane (n-pentane): This is a straight-chain alkane with 5 carbon atoms. Its molecular formula is $C_5H_{12}$.
       • (D) Hexane (n-hexane): This is a straight-chain alkane with 6 carbon atoms. Its molecular formula is $C_6H_{14}$.

    2. Compare compounds based on molecular mass.

       • Compounds (A), (B), and (C) are all isomers of pentane, having the molecular formula $C_5H_{12}$. They have the same molecular mass.
       • Compound (D), hexane, has the molecular formula $C_6H_{14}$. It has a higher molecular mass than the pentane isomers.
       • Since hexane has more carbon atoms and a higher molecular mass, it has a larger electron cloud, leading to stronger London Dispersion Forces compared to the $C_5H_{12}$ isomers.
       • Therefore, Hexane (D) will have the highest boiling point.

    3. Compare the $C_5H_{12}$ isomers based on branching.

       Now we need to arrange pentane, 2-methylbutane, and 2,2-dimethylpropane in decreasing order of their boiling points. All three have the same molecular formula ($C_5H_{12}$), so their boiling points will be determined by their molecular shapes and the extent of branching.

       • Pentane (C): This is a straight-chain alkane. It has the largest surface area for intermolecular contact among the three isomers. This allows for the most extensive London Dispersion Forces. …

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.Arrange the following in decreasing order of their boiling points (A) 2 – Methylbutane (B) 2,2 – Dimethylpropane (C) Pentane (D) Hexane (A) D > A > C > B (B) B > A > C > D (C) D > C > A > B (D) B > C > A > D

    ›Reveal solutionSolution

    Boiling points of alkanes increase with chain length and decrease with branching. The correct decreasing order is Hexane > Pentane > 2-Methylbutane > 2,2-Dimethylpropane, which corresponds to option (C).

    The boiling point of an alkane depends on the strength of intermolecular forces — specifically, London dispersion forces. These forces arise from temporary dipoles and increase with the surface area of the molecule. A longer, unbranched chain has a larger surface area for neighbouring molecules to "touch", so dispersion forces are stronger and more energy (higher temperature) is needed to overcome them. Branching reduces the surface area by making the molecule more compact, which weakens the dispersion forces and lowers the boiling point.

    So the key idea is simple: longer chain → higher boiling point; more branching → lower boiling point (for the same number of carbons).

    Let’s identify each compound:

    1. Hexane — 6 carbons, straight chain. Longest chain here, so the strongest dispersion forces. Highest boiling point.
    2. Pentane — 5 carbons, straight chain. Shorter than hexane, so lower boiling point than hexane, but higher than any branched C5 isomer.
    3. 2-Methylbutane — also 5 carbons, but with one methyl branch. Less surface area than pentane, so lower boiling point than pentane. …

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