Q.Which of the following compounds is most reactive towards nucleophilic addition reactions?

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The correct statement about an $S_N1$ reaction is (A) It is favoured by polar aprotic solvent (B) It follows second order kinetics (C) It involves racemisation (D) It involves Walden inversion

   ›Reveal solutionSolution

   The key idea is that an $S_N1$ reaction proceeds through a planar carbocation intermediate, which allows attack from either face, leading to racemisation. The correct statement is that it involves racemisation.

   The concept behind $S_N1$ reactions is all about the stability of the intermediate. Unlike $S_N2$ reactions, which happen in one smooth step with backside attack, $S_N1$ reactions are stepwise. The first and slowest step is the departure of the leaving group, forming a carbocation. This carbocation is sp²-hybridised and therefore planar — a flat, trigonal structure. Because it’s planar, the nucleophile can attack from either the top or the bottom with equal probability. If the starting material was a single enantiomer (chiral), this equal attack from both sides gives a 50:50 mixture of the two enantiomers — a racemic mixture. That’s why racemisation is the hallmark of $S_N1$ reactions.

   Now let’s examine each option carefully:

   1. Option (A): “It is favoured by polar aprotic solvent”

      Polar aprotic solvents (like acetone or DMF) are great for $S_N2$ reactions because they don’t solvate the nucleophile too tightly, leaving it “naked” and reactive. But $S_N1$ reactions need to stabilise the carbocation intermediate, and polar protic solvents (like water or ethanol) do that better by solvating both the carbocation and the leaving group. So $S_N1$ is actually favoured by polar protic solvents, not aprotic. This statement is false.

   2. Option (B): “It follows second order kinetics”

      The rate of an $S_N1$ reaction depends only on the concentration of the substrate (the alkyl halide), because the slow step is the dissociation of the leaving group. The nucleophile doesn’t appear in the rate law. So the reaction is first order, not second order. This statement is false.

   3. Option (C): “It involves racemisation”

      As explained, the planar carbocation intermediate leads to attack from both sides with equal probability. If the starting material is optically active, the product will be a racemic mixture (or at least partially racemised, depending on the substrate and conditions). This is a defining feature of $S_N1$ reactions. This statement is true.

   4. Option (D): “It involves Walden inversion”

      Walden inversion is the inversion of configuration that occurs in $S_N2$ reactions due to backside attack. In $S_N1$, because attack can occur from either side, you get a mixture of retention and inversion, not a clean inversion. So this is false.

   Watch out

   A common mistake is to confuse the solvent preference: $S_N1$ loves polar protic solvents (which stabilise ions), while $S_N2$ loves polar aprotic solvents (which leave nucleophiles unencumbered).

   Tip

   A quick memory aid: “$S_N1$ — one step slow, one flat carbocation, one racemic product.” The “1” in $S_N1$ reminds you it’s unimolecular (first order) and leads to racemisation.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Which of the following will be the product of Hell-Volhard-Zelinsky reaction? (A) R–CH$_2$OH (B) $\begin{aligned} &\text{R–CH–COOH} \\ &\quad\,\,\,\,\,| \\ &\quad\,\,\,\,\,\text{Cl} \end{aligned}$ (C) $\begin{aligned} &\text{R–C–NH}_2 \\ &\quad\,\,\,\,\,|| \\ &\quad\,\,\,\,\,\text{O} \end{aligned}$ (D) $\begin{aligned} &\text{R–C–Cl} \\ &\quad\,\,\,\,\,|| \\ &\quad\,\,\,\,\,\text{O} \end{aligned}$

   ›Reveal solutionSolution

   The Hell-Volhard-Zelinsky (HVZ) reaction replaces the α‑hydrogen of a carboxylic acid with a halogen (usually bromine or chlorine) in the presence of a catalytic amount of PBr₃, giving an α‑halo acid. The product is an α‑chloro carboxylic acid, which matches option (B).

   The Hell-Volhard-Zelinsky reaction is a classic method for selectively halogenating the carbon atom adjacent to the carboxyl group (the α‑position) of a carboxylic acid. The key insight is that without a catalyst, carboxylic acids do not easily undergo electrophilic substitution at the α‑carbon because the carboxyl group is deactivating. The HVZ reaction uses a small amount of PBr₃ (or red phosphorus + Br₂) to first form an acyl bromide, which is much more reactive toward enolization. Once the enol forms, bromination (or chlorination) occurs at the α‑position, and subsequent hydrolysis gives the α‑halo acid.

   Let’s walk through the reasoning step by step.

   1. Identify the reaction type.

      The HVZ reaction specifically converts a carboxylic acid (R–CH₂–COOH) into an α‑halo carboxylic acid (R–CHX–COOH, where X = Cl, Br, or I). The halogen ends up on the carbon next to the –COOH group.

   2. Examine each option.

      • (A) R–CH₂OH is an alcohol — no carboxyl group, no halogen. Not the HVZ product.
      • (B) R–CH(Cl)–COOH is an α‑chloro carboxylic acid. This matches exactly what HVZ produces.
      • (C) R–CO–NH₂ is an amide. HVZ does not introduce nitrogen.
      • (D) R–CO–Cl is an acyl chloride. While an acyl chloride is an intermediate in the HVZ mechanism, the final product after aqueous workup is the α‑halo acid, not the acyl chloride.

   3. Confirm the mechanism (briefly).

      • PBr₃ converts the carboxylic acid to an acyl bromide.
      • The acyl bromide enolizes more readily, and the enol attacks Br₂ (or Cl₂) at the α‑position.
      • Hydrolysis of the resulting α‑bromoacyl bromide yields the α‑halo acid. Thus, the final isolated product is always an α‑halo carboxylic acid.

   4. Eliminate distractors.

      Option (D) is a common trap: students sometimes remember the acyl bromide intermediate and forget the hydrolysis step. But the question asks for the product of the reaction, not an intermediate.

   Watch out

   A frequent mistake is to pick the acyl chloride (D) because it appears during the mechanism. Remember: the HVZ reaction ends with aqueous workup, which converts the acyl halide back to the carboxylic acid — but now with a halogen at the α‑position.

   Tip

   A quick memory aid: HVZ = “Halogenate the Vicinal carbon Zero steps away from the carbonyl” — that is, the α‑carbon. The product is always an α‑halo acid.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.In the following reaction sequence, what are X and Y respectively? $\mathrm{C_6H_5{-}COOEt} \xrightarrow{X} \mathrm{A} \xrightarrow[\text{(ii) } Y]{\text{(i) } \mathrm{NH_2OH}} \mathrm{C_6H_5{-}CN}$ (A) $\mathrm{H_2\,|\,Pd;\ (CH_3CO)_2O}$ (B) $\mathrm{H_2\,|\,Pd;\ C_6H_5SO_2Cl}$, Pyridine (C)

   (i) DIBAL-H

   (ii) $\mathrm{H_2O}$; $\mathrm{(CH_3CO)_2O}$ (D)

   (i) DIBAL-H

   (ii) $\mathrm{H_2O}$; $\mathrm{C_6H_5SO_2Cl}$

   ›Reveal solutionSolution

   Since $\mathrm{NH_2OH}$ must react with an aldehyde, $A$ = benzaldehyde — which means $X$ is the partial reducing agent DIBAL-H (then $\mathrm{H_2O}$). $Y$ then dehydrates the aldoxime to benzonitrile: $(\mathrm{CH_3CO})_2\mathrm{O}$. Option (C).

   The concept first

   The cleanest way through any reagent-identification question is to work backwards from the product and ask: what functional group must have been sitting there just before?

   (a) The last step tells you what $A$ is. Hydroxylamine, $\mathrm{NH_2OH}$, is an ammonia derivative: it condenses with a carbonyl compound to give an oxime:

   $\mathrm{R{-}CHO} + \mathrm{NH_2OH} \longrightarrow \mathrm{R{-}CH{=}N{-}OH}\ (\text{aldoxime}) + \mathrm{H_2O}$

   An aldoxime has the formula $\mathrm{R{-}CH{=}N{-}OH}$; take one molecule of water out of it and you are left with $\mathrm{R{-}C\equiv N}$ — a nitrile. That is the standard aldehyde $\to$ nitrile route:

   $\mathrm{C_6H_5CHO} \xrightarrow{\mathrm{NH_2OH}} \mathrm{C_6H_5CH{=}N{-}OH} \xrightarrow{-\mathrm{H_2O}} \mathrm{C_6H_5C\equiv N}$

   A ketoxime could not do this (no H on the carbon), so $A$ must be an aldehyde — benzaldehyde.

   (b) The first step is therefore a controlled reduction. We need $\mathrm{C_6H_5COOEt} \to \mathrm{C_6H_5CHO}$: the ester must be reduced only as far as the aldehyde and no further. This is exactly the job of DIBAL-H (di-isobutylaluminium hydride) at low temperature — it delivers a single hydride, giving a tetrahedral intermediate that survives until aqueous work-up releases the aldehyde. Catalytic $\mathrm{H_2/Pd}$ is not a reagent that converts an ester into an aldehyde at all (Rosenmund's controlled hydrogenation works on an acyl chloride, not an ester, and needs the poisoned catalyst).

   (c) $Y$ is a dehydrating agent. Converting the aldoxime to the nitrile is a loss of water, so $Y$ must dehydrate: acetic anhydride $(\mathrm{CH_3CO})_2\mathrm{O}$ is the standard reagent (it mops up the water as acetic acid).

   Step-by-step

   1. Target: $\mathrm{C_6H_5{-}C\equiv N}$ (benzonitrile), reached from $A$ by (i) $\mathrm{NH_2OH}$ then (ii) $Y$.
   2. Deduce $A$. $\mathrm{NH_2OH}$ makes an oxime; only an aldoxime can be dehydrated to a nitrile. Hence $A$ is the aldehyde $\mathrm{C_6H_5CHO}$ (benzaldehyde).
   3. Deduce $X$ (ethyl benzoate $\to$ benzaldehyde). This is a partial reduction of an ester, which is precisely what DIBAL-H does:

   $\mathrm{C_6H_5COOC_2H_5} \xrightarrow[\text{(ii) } \mathrm{H_2O}]{\text{(i) DIBAL-H},\ -78\,^\circ\mathrm{C}} \mathrm{C_6H_5CHO}$

   $\Rightarrow X =$ (i) DIBAL-H (ii) $\mathrm{H_2O}$. This immediately eliminates both $\mathrm{H_2|Pd}$ options.

   4. Form the oxime:

   $\mathrm{C_6H_5CHO} + \mathrm{NH_2OH} \longrightarrow \mathrm{C_6H_5CH{=}N{-}OH} + \mathrm{H_2O}$

   5. Deduce $Y$ — it must remove $\mathrm{H_2O}$ from the aldoxime:

   $\mathrm{C_6H_5CH{=}N{-}OH} \xrightarrow{(\mathrm{CH_3CO})_2\mathrm{O}} \mathrm{C_6H_5C\equiv N} + \mathrm{H_2O}$

   Acetic anhydride is the classical dehydrating agent for this conversion.

   6. Assemble: $X$ = DIBAL-H / $\mathrm{H_2O}$; $Y$ = $(\mathrm{CH_3CO})_2\mathrm{O}$.

   ✓Final answer

   $X$ is DIBAL-H followed by water (giving benzaldehyde) and $Y$ is acetic anhydride (dehydrating the aldoxime to benzonitrile), so the correct option is (C).

   ANSWER: C

4. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.Acrolein (X) is one of the chemicals formed when O$_3$ and NO$_2$ react with unburnt hydrocarbons present in the polluted air. The structure of 'X' is (A) $\mathrm{CH_3-CH=CH_2}$ (B) $\mathrm{CH_2=CH-CHO}$ (C) $\mathrm{CH_2=CH-CN}$ (D) $\mathrm{CH_3CO(OO)NO_2}$

   ›Reveal solutionSolution

   Acrolein is the simplest unsaturated aldehyde, formed from incomplete combustion or atmospheric reactions of hydrocarbons; its structure is $\mathrm{CH_2=CH-CHO}$, so the correct option is (B).

   Acrolein is a well-known compound in atmospheric chemistry and organic chemistry. The key here is to recognize that "acrolein" is the common name for propenal — an aldehyde with a carbon‑carbon double bond. The question describes it as forming when ozone ($\mathrm{O_3}$) and nitrogen dioxide ($\mathrm{NO_2}$) react with unburnt hydrocarbons in polluted air. This is a classic photochemical smog reaction, and acrolein is a major irritant produced. So we need the structure that matches the name "acrolein."

   Let’s examine each option:

   1. Option (A): $\mathrm{CH_3-CH=CH_2}$

      This is propene, an alkene. It has no oxygen atom, so it cannot be an aldehyde or any oxygen-containing compound. Acrolein must contain oxygen (the “-oin” ending hints at an aldehyde or alcohol derivative). So this is incorrect.

   2. Option (B): $\mathrm{CH_2=CH-CHO}$

      This is propenal: a three-carbon chain with a double bond between C1 and C2, and an aldehyde group ($-\mathrm{CHO}$) at the end. The common name for propenal is indeed acrolein. It is formed in smog from the oxidation of hydrocarbons. This matches perfectly.

   3. Option (C): $\mathrm{CH_2=CH-CN}$

      This is acrylonitrile, a nitrile (contains a cyano group, $-\mathrm{CN}$). It has no aldehyde group and is not called acrolein. So this is wrong.

   4. Option (D): $\mathrm{CH_3CO(OO)NO_2}$

      This is peroxyacetyl nitrate (PAN), a well-known component of photochemical smog. PAN is an irritant, but its structure is completely different from acrolein — it contains a peroxy and nitrate group. The name “acrolein” does not refer to PAN.

   Tip

   A quick memory aid: “acrolein” sounds like “acrid” (sharp, irritating smell) + “olein” (oil/alkene). It’s the simplest unsaturated aldehyde — just remember double bond + aldehyde on a three-carbon chain.

   Thus, by matching the name to the structure, we see that only option (B) is acrolein.

   Watch out

   A common mistake is confusing acrolein with acrylonitrile (option C) because both have a $\mathrm{CH_2=CH-}$ group. But acrolein ends with -CHO (aldehyde), not -CN (nitrile). Always check the functional group.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

5. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.What is the IUPAC name of the product Y in the given reaction sequence?

   [!FORMULA] $\begin{array}{c} \text{CHO} \\ | \\ (\text{CHOH})_4 \xrightarrow{\text{HCN}} X \xrightarrow{\text{H}^+/\text{H}_2\text{O}} Y \\ | \\ \text{CH}_2\text{OH} \end{array}$

   (A) 2,3,4,5,6,7-hexahydroxyheptanoic acid (B) 2,3,4,5,6-pentahydroxyhexanoic acid (C) 3,4,5-trihydroxyheptanoic acid (D) 3,4,5-trihydroxyhexanoic acid

   ›Reveal solutionSolution

   The reaction sequence involves extending an aldohexose by one carbon via cyanohydrin formation, followed by hydrolysis of the nitrile to a carboxylic acid, resulting in a 7-carbon polyhydroxy carboxylic acid named 2,3,4,5,6,7-hexahydroxyheptanoic acid.

   Concept and Intuition: Chain Lengthening of Sugars

   This problem demonstrates a classic method for increasing the carbon chain length of aldoses (sugars containing an aldehyde group), often referred to as the Kiliani-Fischer synthesis. The core idea is to convert the aldehyde group into a nitrile, which then extends the carbon backbone by one carbon. This new nitrile group can subsequently be transformed into other functional groups, such as a carboxylic acid, as seen in this problem.

   Why this approach works:

   1. Aldehyde Reactivity: Aldehydes are electrophilic at their carbonyl carbon. They readily react with nucleophiles like cyanide (CN⁻).
   2. Cyanohydrin Formation: The addition of HCN to an aldehyde forms a cyanohydrin, which contains both a hydroxyl group and a nitrile group on the same carbon. This reaction effectively adds a new carbon atom to the chain.
   3. Nitrile Hydrolysis: Nitriles (-C≡N) are versatile functional groups that can be hydrolyzed under acidic or basic conditions to yield carboxylic acids (-COOH). This transformation is robust and allows for the conversion of the newly introduced carbon into a carboxylic acid functionality.

   By combining these two steps, we can systematically lengthen the carbon chain of an aldose and introduce a carboxylic acid group at the new terminal end.

   ---

   Let's break down the reaction sequence step by step:

   1. Identify the Starting Material

   The given structure is:

   $\begin{array}{c} \text{CHO} \\ | \\ (\text{CHOH})_4 \\ | \\ \text{CH}_2\text{OH} \end{array}$

   This represents an aldohexose. Let's count the carbons:

   • One aldehyde carbon (CHO)
   • Four secondary alcohol carbons (CHOH)
   • One primary alcohol carbon (CH₂OH) Total carbons = $1 + 4 + 1 = 6$ carbons. So, the starting material is a 6-carbon sugar (an aldohexose), like glucose, mannose, or galactose. For the purpose of naming the final product's carbon skeleton, the specific stereochemistry of the hydroxyl groups is not required.

   2. Step 1: Reaction with HCN to form X

   The first step is the reaction of the aldohexose with hydrogen cyanide (HCN). This is a cyanohydrin formation reaction.

   • The aldehyde group (CHO) is the reactive site.
   • The cyanide ion (CN⁻) acts as a nucleophile, attacking the electrophilic carbonyl carbon.
   • The carbonyl oxygen is protonated to form a hydroxyl group.
   • This reaction adds one carbon atom to the chain.

   The transformation is:

   $\text{R-CHO} + \text{HCN} \longrightarrow \text{R-CH(OH)-CN}$

   Applying this to our aldohexose:

   Original aldohexose: $\text{CHO}-(\text{CHOH})_4-\text{CH}_2\text{OH}$ (6 carbons)

   Product X (cyanohydrin): $\text{CN}-\text{CH(OH)}-(\text{CHOH})_4-\text{CH}_2\text{OH}$

   • Carbon Count: The chain has now been extended by one carbon, so product X has $6 + 1 = 7$ carbons.
   • Functional Groups: Product X contains one nitrile group (-CN) and six hydroxyl groups (-OH). The carbon that was originally the aldehyde carbon (C1) is now a new chiral center (C2 in the new numbering scheme) bearing a hydroxyl group.

   Cyanohydrin Formation:

   $\text{R-CHO} + \text{HCN} \rightleftharpoons \text{R-CH(OH)-CN}$

   This reaction is reversible and typically catalyzed by a base.

   3. Step 2: Hydrolysis of X with H⁺/H₂O to form Y

   The second step involves the hydrolysis of product X (the cyanohydrin) under acidic conditions ($\text{H}^+/\text{H}_2\text{O}$).

   • Nitriles (-CN) are readily hydrolyzed to carboxylic acids (-COOH) in the presence of acid (or base) and water.
   • The hydroxyl groups in the molecule are stable under these conditions and remain unchanged.

   The transformation is:

   $\text{R'-CN} \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{R'-COOH}$

   Applying this to product X:

   Product X: $\text{CN}-\text{CH(OH)}-(\text{CHOH})_4-\text{CH}_2\text{OH}$

   Product Y: $\text{COOH}-\text{CH(OH)}-(\text{CHOH})_4-\text{CH}_2\text{OH}$

   • Carbon Count: The carbon chain length remains 7 carbons.
   • Functional Groups: Product Y contains one carboxylic acid group (-COOH) and six hydroxyl groups (-OH).

   Tip

   Nitrile Hydrolysis: Nitriles are versatile precursors. Acidic hydrolysis yields carboxylic acids, while basic hydrolysis also yields carboxylic acids (as carboxylate salts). Reduction with LiAlH₄ yields primary amines.

   4. IUPAC Naming of Product Y

   Now, let's determine the IUPAC name for product Y:

   $\text{COOH}-\text{CH(OH)}-\text{CH(OH)}-\text{CH(OH)}-\text{CH(OH)}-\text{CH(OH)}-\text{CH}_2\text{OH}$

   1. Identify the longest carbon chain containing the principal functional group: The principal functional group is the carboxylic acid (-COOH). The chain containing this group has 7 carbons.
   2. Determine the parent name: A 7-carbon chain is "heptane". Since it's a carboxylic acid, the suffix is "-oic acid". So, the parent name is heptanoic acid.
   3. Number the carbon chain: The carbon of the carboxylic acid group is always designated as C1.
      • C1: -COOH
      • C2: -CH(OH)-
      • C3: -CH(OH)-
      • C4: -CH(OH)-
      • C5: -CH(OH)-
      • C6: -CH(OH)-
      • C7: -CH₂OH
   4. Identify and locate substituents: There are six hydroxyl (-OH) groups.
      • One at C2
      • One at C3
      • One at C4
      • One at C5
      • One at C6
      • One at C7 Therefore, we have "hexahydroxy" at positions 2, 3, 4, 5, 6, 7.
   5. Assemble the full IUPAC name: Combine the substituent names and positions with the parent name. 2,3,4,5,6,7-hexahydroxyheptanoic acid

   5. Compare with Options

   Let's check our derived name against the given options:

   (A) 2,3,4,5,6,7-hexahydroxyheptanoic acid - Matches our result.

   (B) 2,3,4,5,6-pentahydroxyhexanoic acid - Incorrect carbon count (hexanoic acid implies 6 carbons, but Y has 7) and incorrect number of hydroxyls.

   (C) 3,4,5-trihydroxyheptanoic acid - Incorrect number and positions of hydroxyls.

   (D) 3,4,5-trihydroxyhexanoic acid - Incorrect carbon count and number/positions of hydroxyls.

   Therefore, option (A) is the correct IUPAC name for product Y.

   ---

   ✓Final answer

   The IUPAC name of product Y is 2,3,4,5,6,7-hexahydroxyheptanoic acid. The correct option is (A).

   ANSWER: A

6. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
   Q.The hybridisation of carbon atoms from left to right in the given compound are respectively $\mathrm{H_2C=C=CH-C\equiv N}$ (A) $\mathrm{sp^3, sp, sp^2, sp}$ (B) $\mathrm{sp^2, sp^2, sp^2, sp}$ (C) $\mathrm{sp, sp^2, sp, sp^2}$ (D) $\mathrm{sp^2, sp, sp^2, sp}$

   ›Reveal solutionSolution

   The hybridisation of a carbon atom is determined by its steric number (number of sigma bonds + number of lone pairs). For the given compound, the hybridisations from left to right are $\mathrm{sp^2, sp, sp^2, sp}$.

   The hybridisation of a central atom in a molecule is a concept that describes the mixing of atomic orbitals to form new hybrid orbitals, which are then used to form sigma bonds and accommodate lone pairs. The type of hybridisation (e.g., $\mathrm{sp}$, $\mathrm{sp^2}$, $\mathrm{sp^3}$) is directly related to the number of electron domains (sigma bonds and lone pairs) around the atom, often called the steric number.

   Here's how to determine the hybridisation for each carbon atom:

   1. Understand the Steric Number Rule: The steric number for an atom is calculated as:

   $\text{Steric Number} = (\text{Number of sigma bonds}) + (\text{Number of lone pairs})$

   Once the steric number is known, the hybridisation can be determined:
   *   Steric Number = 2 $\implies$ $\mathrm{sp}$ hybridisation
   *   Steric Number = 3 $\implies$ $\mathrm{sp^2}$ hybridisation
   *   Steric Number = 4 $\implies$ $\mathrm{sp^3}$ hybridisation
   
   > [!IMPORTANT]
   > Remember that a single bond contains one sigma bond. A double bond contains one sigma bond and one pi bond. A triple bond contains one sigma bond and two pi bonds. Only the sigma bonds contribute to the steric number for hybridisation. Lone pairs also contribute to the steric number. Carbon atoms in stable organic compounds typically do not have lone pairs.
   

   2. Draw the Lewis Structure (or interpret the condensed formula):

   The given compound is $\mathrm{H_2C=C=CH-C\equiv N}$. Let's number the carbon atoms from left to right for clarity:

   $\overset{1}{\mathrm{H_2C}} = \overset{2}{\mathrm{C}} = \overset{3}{\mathrm{CH}} - \overset{4}{\mathrm{C}} \equiv \mathrm{N}$

   3. Determine Hybridisation for Carbon 1 ($\mathrm{H_2C=}$):

      • This carbon is bonded to two hydrogen atoms and double-bonded to Carbon 2.
      • Number of sigma bonds: 2 (to H) + 1 (to C2, from the double bond) = 3
      • Number of lone pairs: 0 (carbon typically forms 4 bonds and has no lone pairs)
      • Steric Number = 3 + 0 = 3
      • Hybridisation = $\mathrm{sp^2}$

   4. Determine Hybridisation for Carbon 2 ($\mathrm{=C=}$):

      • This carbon is double-bonded to Carbon 1 and double-bonded to Carbon 3.
      • Number of sigma bonds: 1 (to C1, from the double bond) + 1 (to C3, from the double bond) = 2
      • Number of lone pairs: 0
      • Steric Number = 2 + 0 = 2
      • Hybridisation = $\mathrm{sp}$

   5. Determine Hybridisation for Carbon 3 ($\mathrm{=CH-}$):

      • This carbon is double-bonded to Carbon 2, single-bonded to one hydrogen atom, and single-bonded to Carbon 4.
      • Number of sigma bonds: 1 (to C2, from the double bond) + 1 (to H) + 1 (to C4) = 3
      • Number of lone pairs: 0
      • Steric Number = 3 + 0 = 3
      • Hybridisation = $\mathrm{sp^2}$

   6. Determine Hybridisation for Carbon 4 ($\mathrm{-C\equiv N}$):

      • This carbon is single-bonded to Carbon 3 and triple-bonded to the nitrogen atom.
      • Number of sigma bonds: 1 (to C3) + 1 (to N, from the triple bond) = 2
      • Number of lone pairs: 0
      • Steric Number = 2 + 0 = 2
      • Hybridisation = $\mathrm{sp}$

   Combining these results, the hybridisations of the carbon atoms from left to right are $\mathrm{sp^2, sp, sp^2, sp}$.

   ✓Final answer

   The hybridisation of carbon atoms from left to right in the given compound are respectively $\boxed{\mathrm{sp^2, sp, sp^2, sp}}$.

7. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.A primary alcohol was reacted with pyridinium chlorochromate (PCC), which resulted in a product P. The product P on treatment with ammoniacal silver nitrate solution produces (A) Anhydride of carboxylic acid (B) Aldehyde (C) Amide (D) Carboxylate anion

   ›Reveal solutionSolution

   PCC oxidises a primary alcohol to an aldehyde (P). An aldehyde reacts with ammoniacal silver nitrate (Tollens’ reagent) to give the carboxylate anion. The correct option is (D).

   The key here is to track the oxidation state of carbon through two sequential reactions. PCC is a mild oxidising agent — it stops at the aldehyde stage for primary alcohols, unlike stronger oxidants like $\ce{K2Cr2O7}$ or $\ce{KMnO4}$ which would push all the way to the carboxylic acid. So product P is an aldehyde.

   Now, ammoniacal silver nitrate is Tollens’ reagent. It contains the diamminesilver(I) ion, $\ce{[Ag(NH3)2]+}$, in a basic medium. This reagent is famous for the silver mirror test — it oxidises an aldehyde to a carboxylate anion while reducing silver(I) to metallic silver. The reaction happens in basic solution, so the immediate product is the carboxylate salt, not the free acid.

   Let’s walk through it step by step.

   1. First reaction: primary alcohol + PCC PCC ($\ce{C5H5NH+CrO3Cl-}$) is a selective oxidant. It converts a primary alcohol ($\ce{R-CH2OH}$) to an aldehyde ($\ce{R-CHO}$) without over-oxidising.

   $\ce{R-CH2OH + PCC -> R-CHO}$

   So product P is an aldehyde.

   2. Second reaction: aldehyde + Tollens’ reagent Tollens’ reagent is prepared by adding aqueous ammonia to silver nitrate until the initially formed $\ce{Ag2O}$ precipitate just dissolves. The active species is $\ce{[Ag(NH3)2]+OH-}$. The aldehyde is oxidised to a carboxylate ion, and silver(I) is reduced to metallic silver (the mirror).

   $\ce{R-CHO + 2[Ag(NH3)2]+ + 2OH- -> R-COO- + 2Ag(s) + 4NH3 + H2O}$

   Notice the product is $\ce{R-COO-}$, the carboxylate anion — not the free carboxylic acid, because the medium is basic.

   3. What about the other options?
      • Anhydride of carboxylic acid: requires two carboxyl groups to couple, not formed here.
      • Aldehyde: that’s P, not the final product.
      • Amide: would need an amine and a carboxylic acid derivative under dehydrating conditions — not what Tollens’ does.
      • Carboxylate anion: exactly what we get.

   Watch out

   A common mistake is to think Tollens’ reagent gives the free carboxylic acid. But the reaction is run in basic ammonia solution, so the product is the carboxylate salt. If you acidify later, you’d get the acid — but the question asks for the product on treatment with ammoniacal silver nitrate, which is the carboxylate anion.

   Tip

   Remember the mnemonic: PCC stops at aldehyde; Tollens’ gives the silver mirror and the carboxylate salt. This pair of reactions is a classic way to distinguish aldehydes from ketones (ketones don’t react with Tollens’).

   ✓Final answer

   The product is the carboxylate anion, so the correct option is (D).

8. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.Ammonia is a Lewis base because it is (A) Electron pair donor (B) Electron pair acceptor (C) Proton donor (D) Proton acceptor

   ›Reveal solutionSolution

   Ammonia donates its lone pair of electrons to form a coordinate bond, making it a Lewis base. The correct answer is (A) Electron pair donor.

   The Lewis theory of acids and bases is broader than the older Brønsted–Lowry theory. A Lewis base is any species that can donate an electron pair, while a Lewis acid is any species that can accept an electron pair. This definition focuses entirely on electron pair transfer, not on protons.

   Ammonia ($NH_3$) has a nitrogen atom with five valence electrons. Three of these are used in covalent bonds with hydrogen atoms, leaving one lone pair of electrons on the nitrogen. This lone pair is available to be shared with an electron-deficient species — for example, when ammonia reacts with a proton ($H^+$) to form the ammonium ion ($NH_4^+$), the lone pair forms a coordinate bond with the proton.

   1. Identify the key feature of ammonia. The nitrogen atom in $NH_3$ has a lone pair of electrons. This is the structural feature that determines its chemical behaviour as a base.

   2. Apply the Lewis definition. A Lewis base is defined as an electron pair donor. Since ammonia can donate its lone pair, it qualifies as a Lewis base.

   3. Check the other options for clarity.

      • (B) Electron pair acceptor — This is the definition of a Lewis acid, not a base. Ammonia does not accept an electron pair; it donates one.
      • (C) Proton donor — This is the Brønsted–Lowry definition of an acid. Ammonia does not donate a proton; it accepts one (forming $NH_4^+$).
      • (D) Proton acceptor — This is the Brønsted–Lowry definition of a base. While ammonia is indeed a Brønsted–Lowry base (it accepts a proton), the question specifically asks about the Lewis definition, which is based on electron pairs, not protons.

   Watch out

   A common mistake is to pick (D) Proton acceptor because ammonia is famously a base in the Brønsted–Lowry sense. But the question explicitly asks for the Lewis definition. Always read the theory being referenced — Lewis bases are about electron pair donation, not proton acceptance.

   4. Confirm the correct choice. Since the Lewis definition centres on electron pair donation, and ammonia donates its lone pair, the correct answer is that ammonia is a Lewis base because it is an electron pair donor.

   ✓Final answer

   The correct option is (A) Electron pair donor.

9. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.The process in which colloids, when subjected to DC electric field move towards an electrode is (A) Brownian movement (B) Tyndall effect (C) Peptization (D) Electrophoresis

   ›Reveal solutionSolution

   The key idea is that the directed motion of colloidal particles under an applied DC electric field is called electrophoresis. The correct option is (D).

   Colloids are mixtures where tiny particles (1–1000 nm) are dispersed in a continuous medium. These particles often carry a surface charge — either positive or negative — due to adsorption of ions from the solution. When you apply a direct current (DC) electric field across such a colloidal system, the charged particles experience a force and begin to migrate toward the electrode of opposite charge. This phenomenon is not random; it is a directed, field-driven movement.

   The question asks for the specific name of this process. Let’s examine each option to see why only one fits.

   1. Brownian movement is the random, zigzag motion of colloidal particles caused by collisions with solvent molecules. It has nothing to do with an electric field — it’s purely thermal and statistical. So (A) is wrong.

   2. Tyndall effect is the scattering of light by colloidal particles, making a beam of light visible through the colloid. Again, no electric field is involved — it’s an optical property. So (B) is wrong.

   3. Peptization is the process of converting a freshly precipitated substance back into a colloidal state by adding an electrolyte. It’s a chemical dispersion method, not an electric-field-driven motion. So (C) is wrong.

   4. Electrophoresis is exactly the migration of charged colloidal particles under an applied electric field. The particles move toward the oppositely charged electrode — positive particles go to the cathode, negative ones to the anode. This is a standard technique used to separate or characterize colloids.

   Watch out

   A common mistake is to confuse electrophoresis with electro-osmosis, which is the movement of the liquid medium relative to a stationary charged surface under an electric field. Here, the particles themselves move — that’s electrophoresis.

   Tip

   In electrophoresis, if the colloidal particles are negatively charged (common for many metal sulfides and clays), they move toward the positive electrode (anode). This directional movement is the key experimental observation that distinguishes it from random Brownian motion.

   ✓Final answer

   The correct option is (D) Electrophoresis.

10. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.The major product in the following transformation is A cyclohexene ring bearing a $-\mathrm{CH_2CH_2CHO}$ side chain on one alkene carbon, a $-\mathrm{CH_3}$ group on the other alkene carbon, and a $-\mathrm{CO_2Me}$ group on the adjacent ring carbon $\xrightarrow{\mathrm{NaBH_4}}$ (A) [FIGURE] The ring double bond reduced, the side chain converted to $-\mathrm{CH_2CH_2CH_2OH}$ and the $-\mathrm{CO_2Me}$ converted to $-\mathrm{CH_2OH}$ (B) [FIGURE] The ring double bond retained, the side chain converted to $-\mathrm{CH_2CH_2CH_2OH}$ and the $-\mathrm{CO_2Me}$ converted to $-\mathrm{CH_2OH}$ (C) [FIGURE] The ring double bond reduced, the side chain converted to $-\mathrm{CH_2CH_2CH_2OH}$ and the $-\mathrm{CO_2Me}$ group retained (D) [FIGURE] The ring double bond retained, the side chain converted to $-\mathrm{CH_2CH_2CH_2OH}$ and the $-\mathrm{CO_2Me}$ group retained

    ›Reveal solutionSolution

    Sodium borohydride reduces the side-chain aldehyde to a primary alcohol and nothing else — the ring double bond and the methyl ester are untouched. Option (D).

    The concept: the reactivity window of $\mathrm{NaBH_4}$

    Hydride reagents are graded by how nucleophilic the hydride is:

    Reagent	Reduces
    $\mathrm{NaBH_4}$ (mild)	aldehydes, ketones, acid chlorides
    $\mathrm{LiAlH_4}$ (powerful)	aldehydes, ketones, esters, acids, amides, nitriles
    $\mathrm{H_2}$/Pd	alkenes, alkynes

    The $\mathrm{B-H}$ bond is far less polar than $\mathrm{Al-H}$, so $\mathrm{BH_4^{-}}$ is a weak hydride donor and can attack only the most electrophilic carbonyls. An ester carbonyl is deactivated by resonance donation from its $-\mathrm{OR}$ group and is inert to $\mathrm{NaBH_4}$. An isolated alkene is a nucleophile, not an electrophile — a hydride will never attack it.

    This chemoselectivity is what makes $\mathrm{NaBH_4}$ so valuable in synthesis, and it is exactly what the question probes.

    Step 1 — Identify every functional group

    1. Ring $\mathrm{C=C}$ (trisubstituted alkene, not conjugated with the ester).
    2. Side-chain $-\mathrm{CH_2CH_2CHO}$ (an aldehyde).
    3. $-\mathrm{CO_2Me}$ (a methyl ester).

    Step 2 — Apply $\mathrm{NaBH_4}$

    • Aldehyde: reduced.

    $\mathrm{-CH_2CH_2CHO} \xrightarrow{\mathrm{NaBH_4}} \mathrm{-CH_2CH_2CH_2O^{-}} \xrightarrow{\text{work-up}} \mathrm{-CH_2CH_2CH_2OH}$

    • Ester: untouched (too weakly electrophilic).
    • Alkene: untouched (no hydride adds to a $\mathrm{C=C}$).

    Step 3 — Read off the product

    A cyclohexene ring with its double bond still present, the $\mathrm{CH_3}$ still present, the $-\mathrm{CO_2Me}$ still present, and the side chain now ending in $-\mathrm{CH_2OH}$.

    Step 4 — Eliminate the distractors

    • (A) reduces everything — would need $\mathrm{LiAlH_4}$ plus hydrogenation. ✗
    • (B) reduces the ester too — needs $\mathrm{LiAlH_4}$. ✗
    • (C) hydrogenates the alkene — needs $\mathrm{H_2}$/catalyst. ✗
    • (D) only the aldehyde reduced. ✓

    ✓Final answer

    The product keeps the ring double bond and the $-\mathrm{CO_2Me}$ group, with the side chain converted to $-\mathrm{CH_2CH_2CH_2OH}$.

    ANSWER: D

11. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.The major product of the following reactions is Bromocyclohexane $\xrightarrow{\text{(i) } \mathrm{Mg};\ \text{(ii) } \mathrm{CO_2};\ \text{(iii) } \mathrm{H_3O^+};\ \text{(iv) } \mathrm{SOCl_2};\ \text{(v) } (\mathrm{CH_3})_2\mathrm{Cd}}$ (A) [FIGURE] Cyclohexane ring bearing a $-\mathrm{CH_2CH_2OH}$ group (B) [FIGURE] Cyclohexane ring bearing a $-\mathrm{CO-CH_3}$ group (1-cyclohexylethan-1-one) (C) [FIGURE] Cyclohexane ring bearing a $-\mathrm{CH(OH)CH_3}$ group (D) [FIGURE] Cyclohexane ring bearing a $-\mathrm{C(OH)(CH_3)_2}$ group

    ›Reveal solutionSolution

    Grignard $\to$ $\mathrm{CO_2}$ gives cyclohexanecarboxylic acid; $\mathrm{SOCl_2}$ gives the acyl chloride; dimethylcadmium then delivers just one methyl to give the ketone $\mathrm{C_6H_{11}COCH_3}$ — option (B).

    The concept: a three-move synthesis of a ketone

    Each reagent has one clean job:

    1. Grignard formation

    $\mathrm{C_6H_{11}Br} \xrightarrow{\mathrm{Mg},\ \text{dry ether}} \mathrm{C_6H_{11}MgBr}$

    The carbon that carried $\mathrm{Br}$ is now nucleophilic — a complete polarity reversal.

    2. Carboxylation with $\mathrm{CO_2}$

    The Grignard attacks the electrophilic carbon of $\mathrm{CO_2}$ (usually dry ice):

    $\mathrm{C_6H_{11}MgBr} + \mathrm{CO_2} \rightarrow \mathrm{C_6H_{11}COOMgBr} \xrightarrow{\mathrm{H_3O^{+}}} \mathrm{C_6H_{11}COOH}$

    This is the standard way of adding exactly one carbon as a $-\mathrm{COOH}$.

    3. Making the acid chloride

    $\mathrm{C_6H_{11}COOH} + \mathrm{SOCl_2} \rightarrow \mathrm{C_6H_{11}COCl} + \mathrm{SO_2}\uparrow + \mathrm{HCl}\uparrow$

    4. The cadmium step — the point of the question

    $\mathrm{R_2Cd}$ has a much more covalent $\mathrm{C-M}$ bond than a Grignard. It is nucleophilic enough to attack an acid chloride, but not a ketone:

    $2\,\mathrm{C_6H_{11}COCl} + (\mathrm{CH_3})_2\mathrm{Cd} \longrightarrow 2\,\mathrm{C_6H_{11}CO\!-\!CH_3} + \mathrm{CdCl_2}$

    So the reaction stops at the ketone. Had we used $\mathrm{CH_3MgBr}$ instead, a second methyl would add and we would end up with the tertiary alcohol of option (D) — which is exactly the trap.

    Final product

    1-Cyclohexylethan-1-one: a cyclohexane ring bearing $-\mathrm{COCH_3}$.

    Eliminating the rest

    • (A) $-\mathrm{CH_2CH_2OH}$ — no step here makes that chain. ✗
    • (C) $-\mathrm{CH(OH)CH_3}$ — would need reduction of the ketone. ✗
    • (D) $-\mathrm{C(OH)(CH_3)_2}$ — needs a Grignard (double addition), not $\mathrm{R_2Cd}$. ✗

    ✓Final answer

    The major product is cyclohexyl methyl ketone, $\mathrm{C_6H_{11}COCH_3}$.

    ANSWER: B

12. TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
    Q.The correct statement about the following chemical reaction is [FIGURE] (A) It is an SN$^2$ reaction with inversion of configuration of the reactant (B) It is an SN$^2$ reaction with retention configuration of the reactant (C) It is an SN$^1$ reaction with retention of configuration of the reactant (D) It is an SN$^1$ reaction with racemisation

    ›Reveal solutionSolution

    The Br sits on a primary $-\mathrm{CH_2}-$ group, so $\mathrm{OH^-}$ attacks by $S_N2$; but the chiral carbon is the adjacent one and none of its bonds break, so the configuration is retained. The answer is (B).

    The concept first — inversion is a statement about the attacked carbon. Students memorise "$S_N2$ = inversion" and then apply it to the whole molecule. That is the trap. In $S_N2$ the nucleophile attacks the carbon bearing the leaving group from the side opposite the leaving group; the three other bonds on that carbon flip through like an umbrella (Walden inversion). If the molecule's stereocentre is somewhere else, nothing at all happens to it — its four bonds are untouched, so its spatial arrangement (and its R/S label) is preserved.

    Step 1 — Identify the substrate. From the drawing, one carbon carries four different groups: $\mathrm{CH_2Br}$ (bold wedge), $\mathrm{H}$ (dashed), $\mathrm{Ph}$ and $\mathrm{CH_3}$. So the compound is

    $\mathrm{Ph{-}\overset{*}{C}H(CH_3){-}CH_2Br}$

    where $*$ marks the stereocentre.

    Step 2 — Locate the leaving group. The bromine is on the $-\mathrm{CH_2}-$ carbon. That carbon carries two hydrogens, one Br and one carbon — it is a primary carbon and it is not a stereocentre.

    Step 3 — Decide the mechanism. A primary carbon cannot form a stable carbocation, so an $S_N1$ path is out. With aqueous/alcoholic NaOH the strong nucleophile $\mathrm{OH^-}$ attacks directly:

    $\mathrm{HO^- + Ph{-}CH(CH_3){-}CH_2Br \longrightarrow Ph{-}CH(CH_3){-}CH_2OH + Br^-}$

    This is a one-step, bimolecular $S_N2$ reaction (rate $= k[\mathrm{RBr}][\mathrm{OH^-}]$).

    Step 4 — What happens to the stereocentre? Inversion occurs at the $-\mathrm{CH_2Br}$ carbon, but that carbon has two identical hydrogens, so "inversion" there is invisible — it produces no stereochemical change. The stereocentre $\mathrm{Ph{-}CH(CH_3)-}$ never has a bond broken. Its configuration is therefore retained, and the product is optically active with the same spatial arrangement as the reactant.

    Step 5 — Test the distractors. (A) says inversion — wrong, the chiral carbon isn't the reacting carbon. (C) and (D) assume $S_N1$ — impossible for a primary halide (and $S_N1$ would in any case demand a carbocation at the $\mathrm{CH_2}$ carbon).

    ✓Final answer

    The reaction is $S_N2$ at the primary carbon, and because the stereocentre is not attacked, the configuration of the reactant is retained.

    ANSWER: B