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

Showing the 12 most recent of 33 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.What are X and Y respectively in the following set of reactions? $\mathrm{C_6H_5CN} \xrightarrow{X} A$ , $\mathrm{C_6H_5CN} \xrightarrow{Y} B$ ; $A, B \xrightarrow{H^+} \mathrm{C_6H_5-C(CH_3)=N-CH_2C_6H_5}$ (A) $H_2/cat$ ; $(CH_3)_2Cd$ (B) $Na/C_2H_5OH$ ; $CH_3MgBr, H_2O$ (C) $DIBAL-H, H_2O$ ; $CH_3MgBr, H_2O$ (D) $LiAlH_4, H_2O$ ; $(CH_3)_2Cd$

   ›Reveal solutionSolution

   The final imine comes from condensing acetophenone with benzylamine. Benzonitrile is reduced by Na/C₂H₅OH to give benzylamine (A), and reacted with CH₃MgBr then hydrolysed to give acetophenone (B); acid catalyses their condensation to the imine.

   Concept and Intuition

   Nitriles ($\mathrm{R-C\equiv N}$) are versatile precursors to both amines and ketones depending on the reagent used:

   • Full reduction of a nitrile (e.g. dissolving-metal reduction with sodium in ethanol, or catalytic $\mathrm{H_2}$/LiAlH₄) adds four hydrogens across the triple bond, converting $-\mathrm{C\equiv N}$ directly into a primary amine, $-\mathrm{CH_2NH_2}$.
   • Grignard addition to a nitrile gives a metallated imine salt, $\mathrm{R-C(=NMgBr)-R'}$, which on aqueous acidic hydrolysis ($\mathrm{H_2O}/\mathrm{H_3O^+}$) is hydrolysed straight through to the ketone $\mathrm{R-CO-R'}$ (releasing ammonia).

   A ketone and a primary amine then condense under acid catalysis (loss of water) to form an imine ($\mathrm{C=N-R}$) — exactly the target product here.

   Step-by-Step Solution

   1. Target imine: $\mathrm{C_6H_5-C(CH_3)=N-CH_2C_6H_5}$. Break the C=N bond conceptually: the $\mathrm{C_6H_5-C(CH_3)=}$ portion comes from a ketone, $\mathrm{C_6H_5-CO-CH_3}$ (acetophenone); the $\mathrm{=N-CH_2C_6H_5}$ portion comes from a primary amine, $\mathrm{C_6H_5CH_2-NH_2}$ (benzylamine).
   2. Route to A (benzylamine) from $\mathrm{C_6H_5CN}$: full reduction of the nitrile using sodium and ethanol (a dissolving-metal / nascent-hydrogen reduction) directly gives $\mathrm{C_6H_5CH_2NH_2}$.
   3. Route to B (acetophenone) from $\mathrm{C_6H_5CN}$: react with the Grignard reagent $\mathrm{CH_3MgBr}$, then hydrolyse with $\mathrm{H_2O}$ (aqueous acid workup) to give $\mathrm{C_6H_5COCH_3}$.
   4. A (amine) + B (ketone) condense under $\mathrm{H^+}$ catalysis, losing water, to give the target imine.
   5. So X (→A) = $\mathrm{Na/C_2H_5OH}$ and Y (→B) = $\mathrm{CH_3MgBr, H_2O}$ — matching option (B).

   Common Mistakes

   • Using DIBAL-H (a partial reducing agent) in place of a full-reduction reagent — DIBAL-H on a nitrile with aqueous workup stops at the aldehyde stage, not the amine, so it cannot supply the $-\mathrm{CH_2NH_2}$ needed here.
   • Trying to use $(\mathrm{CH_3})_2\mathrm{Cd}$ on the nitrile directly — dimethylcadmium reacts characteristically with acid chlorides to give ketones, not with nitriles.

   ✓Final answer

   The correct option is (B) — $Na/C_2H_5OH$ ; $CH_3MgBr, H_2O$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Which of the following are the correct statements about D-glucose? I. It forms oxime with hydroxyl amine II. It forms addition product with $NaHSO_3$ III. It forms cyanohydrin with $HCN$ IV. It forms saccharic acid with bromine water The correct answer is (A) II & III only (B) I & III only (C) II & IV only (D) III & IV only

   ›Reveal solutionSolution

   Glucose's structural-elucidation reactions: it does form an oxime and a cyanohydrin (confirming the aldehyde group exists in equilibrium with the ring form), but it does NOT form a bisulphite addition product, and bromine water oxidises it only to gluconic acid, not saccharic acid.

   Concept and Intuition

   Even though glucose predominantly exists as a cyclic hemiacetal, a small equilibrium amount of the open-chain aldehyde form is always present. This small amount is enough to react with small, reactive nucleophiles like $HCN$ and $NH_2OH$ (whose reactions pull the equilibrium forward), giving cyanohydrin and oxime respectively. But bulkier/other characteristic aldehyde tests (like the bisulphite addition with $NaHSO_3$, or Schiff's test) fail for glucose — this mismatch between "should behave like an aldehyde" and "doesn't give all aldehyde tests" was itself key historical evidence for glucose's cyclic hemiacetal structure.

   Step-by-Step Solution

   1. Statement I: Glucose + $NH_2OH \rightarrow$ glucose oxime (via the open-chain $-CHO$). This reaction is well-documented and used as evidence for the carbonyl group. True.
   2. Statement II: Glucose does not give the $NaHSO_3$ addition product — a well-known NCERT fact distinguishing glucose from typical aldehydes despite having a carbonyl group in its open form. False.
   3. Statement III: Glucose + HCN $\rightarrow$ glucose cyanohydrin (nucleophilic addition at the carbonyl carbon), again evidence for the $-CHO$ group. True.
   4. Statement IV: Bromine water (a mild oxidant) oxidises only the aldehyde end of glucose to a carboxylic acid, giving gluconic acid (monocarboxylic acid), not saccharic acid. Saccharic acid (glucaric acid, dicarboxylic) requires oxidation of BOTH the $-CHO$ and the terminal $-CH_2OH$, which needs the stronger oxidant dilute $HNO_3$. False as stated (wrong reagent–product pairing).
   5. So only statements I and III are correct.

   Common Mistakes

   • Assuming any reagent that normally reacts with aldehydes will react with glucose (some diagnostic reactions like $NaHSO_3$ addition and Schiff's test famously fail).
   • Confusing gluconic acid (from bromine water) with saccharic acid (from dilute $HNO_3$) — different oxidants oxidise different ends of the glucose molecule.

   ✓Final answer

   The correct option is (B) — I & III only.

   ANSWER: B

3. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.What are B and C respectively in the given sequence of reactions? Bromocyclohexane $\xrightarrow[\text{ether}]{Mg\ |\ dry}$ A; A $\xrightarrow{CH_3CH_2OH}$ B; A $\xrightarrow[(ii)\ H_3O^+]{(i)\ HCHO}$ C (A) Cyclohexanol , Cyclohexylmethanol (B) Cyclohexane , Cyclohexylmethanol (C) Cyclohexane , Ethylcyclohexane (D) $C_2H_6$ , Ethoxycyclohexane (cyclohexane ring bearing an $-OCH_2CH_3$ substituent)

   ›Reveal solutionSolution

   Tests Grignard reagent formation and its two contrasting fates: quenching by an acidic O-H gives the alkane; addition to formaldehyde gives a one-carbon-homologated primary alcohol.

   Concept and Intuition

   A Grignard reagent (R-MgX) is a strong carbanion-like nucleophile and an extremely strong base. It reacts instantly with any acidic proton (even a weak one like the O-H of an alcohol) faster than it can do anything else — this simply protonates the carbanion and regenerates R-H. Only when there is no acidic proton available (e.g. with a carbonyl compound like HCHO) does the Grignard get the chance to act as a nucleophile and add across the C=O bond, building a new C-C bond and a new alcohol after the acidic workup.

   Step-by-Step Solution

   1. Bromocyclohexane + Mg in dry ether → cyclohexylmagnesium bromide (A), by oxidative insertion of Mg into the C-Br bond.
   2. A + $CH_3CH_2OH$: the O-H proton of ethanol quenches the Grignard immediately (acid-base reaction), giving cyclohexane (B) + $CH_3CH_2OMgBr$. No new C-C bond forms.
   3. A + (i) HCHO (ii) $H_3O^+$: the nucleophilic cyclohexyl carbon attacks the electrophilic carbonyl carbon of formaldehyde, forming an alkoxide; aqueous acid workup protonates it to the primary alcohol cyclohexylmethanol (C), i.e. $C_6H_{11}-CH_2OH$.

   Common Mistakes

   • Forgetting that alcohols (even though "just" weak acids) destroy Grignard reagents instantly — many students expect an ether or addition product instead.
   • Miscounting the extra carbon contributed by HCHO and naming C as cyclohexanol instead of cyclohexylmethanol.

   ✓Final answer

   The correct option is (B) — Cyclohexane, Cyclohexylmethanol.

   ANSWER: B

4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.Consider the following sequence of reactions Isopropyl benzene $\xrightarrow{\mathrm{O_2}} X \xrightarrow[\mathrm{H_2O}]{\mathrm{H^+}}$ phenol $+ Z$ The incorrect statement about z is (A) z gives yellow precipitate of $\mathrm{CHI_3}$ with NaOH + $\mathrm{I_2}$ solution (B) z gives isopropyl alcohol on reduction with $\mathrm{H_2}$ in the presence of Pd catalyst (C) z on reaction with $\mathrm{CH_3MgBr}$ followed by hydrolysis gives 2° alcohol (D) z does not give positive test with Fehling's reagent

   ›Reveal solutionSolution

   Tests the cumene-to-phenol process and acetone's reactions; the answer is (C), since a ketone + Grignard reagent gives a tertiary (not secondary) alcohol.

   Concept and Intuition

   The industrial cumene process converts isopropylbenzene to phenol in two steps: air oxidation to cumene hydroperoxide (X), followed by acid-catalysed hydrolytic rearrangement to give phenol and acetone (Z) as co-product:

   $\text{Cumene} \xrightarrow{O_2} \text{Cumene hydroperoxide (X)} \xrightarrow[H_2O]{H^+} \text{Phenol} + \text{Acetone (Z)}$

   So Z = acetone, $(CH_3)_2C=O$, a ketone. To find the incorrect statement, we check each of acetone's known reactions:

   • Iodoform test: acetone has a $CH_3-CO-$ group, so it gives a positive iodoform test (yellow $CHI_3$ precipitate) with NaOH/I2 — true.
   • Catalytic hydrogenation: ketones are reduced to secondary alcohols by H2/Pd; acetone → isopropyl (2°) alcohol — true.
   • Grignard addition: a ketone reacting with a Grignard reagent, upon hydrolysis, gives a tertiary alcohol (since the ketone carbon already bears two alkyl/aryl groups, and the Grignard's R group adds a third) — acetone + CH3MgBr → 2-methylpropan-2-ol (tert-butanol), a tertiary alcohol, NOT secondary. This makes the given statement false.
   • Fehling's test: ketones (other than special reducing sugars) generally do not give a positive Fehling's test since it requires an oxidisable aldehyde-type group; acetone gives a negative result — true.

   Step-by-Step Solution

   1. Deduce Z = acetone from the cumene process.
   2. Check (A): acetone + iodoform test → positive (true statement, so not the answer).
   3. Check (B): acetone + H2/Pd → isopropyl alcohol (true statement, so not the answer).
   4. Check (C): acetone (a ketone, R2C=O with two alkyl groups) + Grignard reagent → after hydrolysis, the product alcohol carbon bears three alkyl/aryl groups → tertiary alcohol, not secondary. This statement is FALSE — this is the incorrect statement asked for.
   5. Check (D): acetone does not reduce Fehling's solution (true statement, so not the answer).

   Common Mistakes

   • Forgetting the general rule that aldehydes + Grignard → secondary alcohols, while ketones + Grignard → tertiary alcohols (except formaldehyde + Grignard → primary alcohol) — mixing these up leads to picking the wrong option here.

   ✓Final answer

   The correct option is (C) — z on reaction with $CH_3MgBr$ followed by hydrolysis gives 2° alcohol.

   ANSWER: C

5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The correct statement about the product of the following reaction is $\mathrm{CH_3CHO} \xrightarrow[\text{(ii) } H_2O]{\text{(i) } C_2H_5MgBr} \text{product}$ (A) It undergoes dehydration with 20% $\mathrm{H_3PO_4}$ at 358 K (B) It gives ketone on oxidation with $\mathrm{CrO_3}$ (C) It does not give positive iodoform test (D) It is a vinylic alcohol

   ›Reveal solutionSolution

   Tests identifying the Grignard-addition product (a secondary alcohol) and its correct chemical behaviour on oxidation, dehydration, and the iodoform test.

   Concept and Intuition

   An aldehyde plus a Grignard reagent adds the alkyl group to the carbonyl carbon; after aqueous workup this gives a SECONDARY alcohol (aldehyde + RMgX → 2° alcohol; ketone + RMgX → 3° alcohol). Once the product's structure is nailed down, each option can be checked against that structure's real chemistry.

   Step-by-Step Solution

   1. $\mathrm{CH_3CHO} + \mathrm{C_2H_5MgBr} \rightarrow \mathrm{CH_3-CH(OMgBr)-C_2H_5}$; hydrolysis with $\mathrm{H_2O}$ gives $\mathrm{CH_3-CH(OH)-CH_2CH_3}$ = butan-2-ol, a secondary alcohol.
   2. Secondary alcohols are oxidised by chromium(VI) reagents such as $\mathrm{CrO_3}$ to ketones (here, butan-2-one, $\mathrm{CH_3COCH_2CH_3}$) — this statement is straightforwardly true.
   3. Checking (C): butan-2-ol has the fragment $\mathrm{CH_3-CH(OH)}-$ (a methyl group directly on the carbinol carbon) — this is exactly the structural requirement for a POSITIVE iodoform test, so the claim that it does "not" give one is false.
   4. Checking (D): the molecule is a fully saturated secondary alcohol, not an enol/vinylic alcohol — false.
   5. So the only statement that is factually correct about the product is (B).

   Common Mistakes

   • Missing that butan-2-ol, despite being a "simple" secondary alcohol, still satisfies the methyl-carbinol pattern needed for a positive iodoform test.
   • Assuming any alcohol with an –OH not on a terminal carbon is somehow atypical for CrO₃ oxidation; the 1°/2°/3° rule (2° → ketone) applies cleanly here.

   ✓Final answer

   The correct option is (B) — it gives a ketone on oxidation with $\mathrm{CrO_3}$.

   ANSWER: B

6. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.Glucose on reaction with HCN forms a compound 'A'. Acid hydrolysis of A gives B. The molecular formula of B is (A) $\mathrm{C_7H_{14}O_8}$ (B) $\mathrm{C_6H_{14}O_7}$ (C) $\mathrm{C_7H_{14}O_6}$ (D) $\mathrm{C_7H_{12}O_5}$

   ›Reveal solutionSolution

   Glucose's cyanohydrin (from HCN addition) hydrolyses its nitrile group to a carboxylic acid, giving B with molecular formula $C_7H_{14}O_8$.

   Concept and Intuition

   This is the classic 'glucose chain-length extension' sequence used historically (Kiliani–Fischer style logic) to probe glucose's structure: HCN adds across the free aldehyde group to form a cyanohydrin (adding one carbon), and subsequent acid hydrolysis converts the newly added nitrile into a carboxylic acid group, adding two oxygens (from two water molecules) while releasing ammonia.

   Step-by-Step Solution

   1. Glucose (open-chain aldose): $C_6H_{12}O_6$, with a free $-CHO$ group.
   2. Reaction with HCN: nucleophilic addition of HCN across the aldehyde carbonyl adds the entire HCN unit to the molecule, forming cyanohydrin A: $A = C_6H_{12}O_6 + HCN = C_7H_{13}NO_6$.
   3. Acid hydrolysis of the nitrile group in A: $R-C\equiv N + 2H_2O \to R-COOH + NH_3$.
   4. Applying atom balance: $C_7H_{13}NO_6 + 2H_2O \to B + NH_3$.
   5. Left side atoms: $C_7\,H_{13+4}\,N_1\,O_{6+2} = C_7H_{17}NO_8$.
   6. Subtracting $NH_3$ (the byproduct): $B = C_7H_{17}NO_8 - NH_3 = C_7H_{14}O_8$.

   Common Mistakes

   • Forgetting that nitrile hydrolysis to a carboxylic acid consumes 2 water molecules (not 1) and releases NH₃, not just adding one oxygen.
   • Miscounting hydrogens when the ammonia byproduct is subtracted.

   ✓Final answer

   The correct option is (A) — $\mathrm{C_7H_{14}O_8}$.

   ANSWER: A

7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.An alkyl bromide X($\mathrm{C_5H_{11}Br}$) undergoes hydrolysis in a two step mechanism. X is converted to Grignard reagent and then reacted with $\mathrm{CO_2}$ in dry ether followed by acidification gave Y. What is Y? (A) $\mathrm{(CH_3)_2CH{-}CH_2{-}CH_2{-}COOH}$ (skeletal structure: a branched chain with a methyl branch, then two $\mathrm{CH_2}$ groups, ending in $\mathrm{COOH}$) (B) $\mathrm{CH_3{-}CH_2{-}CH_2{-}CH_2{-}CH_2{-}COOH}$ (skeletal structure: an unbranched chain ending in $\mathrm{COOH}$) (C) $\mathrm{CH_3{-}CH_2{-}CH(CH_3){-}CH_2{-}COOH}$ (skeletal structure: a chain with a methyl branch nearer the $\mathrm{COOH}$ end) (D) $\mathrm{(CH_3)_3C{-}CH_2{-}COOH}$ (skeletal structure: a gem-dimethyl branched carbon adjacent to $\mathrm{CH_2COOH}$)

   ›Reveal solutionSolution

   This tests the difference between a rearranging $S_N1$ hydrolysis and a rearrangement-free Grignard carboxylation of the same neopentyl-type halide. Answer: $\mathrm{(CH_3)_3C{-}CH_2{-}COOH}$ (D).

   Concept and Intuition

   The phrase "hydrolysis in a two-step mechanism" is the giveaway for identifying $X$: among the $\mathrm{C_5H_{11}Br}$ isomers, neopentyl bromide, $\mathrm{(CH_3)_3C{-}CH_2{-}Br}$, is the standard example of a primary halide that is nevertheless forced into an $S_N1$ (two-step: ionisation, then nucleophilic capture) pathway on hydrolysis, because the bulky tert-butyl group next to the leaving carbon blocks backside $S_N2$ attack. Once ionised, the resulting primary carbocation is so unstable that it instantly rearranges by a 1,2-methyl shift to a tertiary carbocation, so hydrolysis of $X$ actually gives a rearranged tertiary alcohol.

   But the question does not hydrolyse $X$ directly — it first converts $X$ to its Grignard reagent. Formation of a Grignard reagent (insertion of Mg metal into the C–Br bond) is a concerted/radical-pair process at the original carbon and does not proceed through a free carbocation, so no rearrangement occurs; the alkyl skeleton of the Grignard reagent is identical to that of $X$. Reaction of a Grignard reagent with $\mathrm{CO_2}$ (dry ether) forms a magnesium carboxylate at that same carbon, and acidification liberates the carboxylic acid — effectively inserting one new carbon ($\mathrm{COOH}$) onto the unrearranged skeleton.

   Step-by-Step Solution

   1. Identify $X$: $\mathrm{C_5H_{11}Br}$ whose hydrolysis is a two-step (carbocation, rearranging) process $\Rightarrow$ neopentyl bromide, $\mathrm{(CH_3)_3C{-}CH_2{-}Br}$.
   2. Form the Grignard reagent: $\mathrm{(CH_3)_3C{-}CH_2{-}Br} + \mathrm{Mg} \xrightarrow{\text{dry ether}} \mathrm{(CH_3)_3C{-}CH_2{-}MgBr}$ — no rearrangement, skeleton preserved.
   3. React with dry-ice $\mathrm{CO_2}$: $\mathrm{(CH_3)_3C{-}CH_2{-}MgBr} + \mathrm{CO_2} \to \mathrm{(CH_3)_3C{-}CH_2{-}COOMgBr}$.
   4. Acidify: $\mathrm{(CH_3)_3C{-}CH_2{-}COOMgBr} + \mathrm{H_3O^+} \to \mathrm{(CH_3)_3C{-}CH_2{-}COOH}$ ($Y$), i.e. 3,3-dimethylbutanoic acid.
   5. This exactly matches option (D)'s gem-dimethyl carbon adjacent to $\mathrm{CH_2COOH}$.

   Common Mistakes

   • Assuming that because hydrolysis of $X$ would rearrange, the Grignard route must also give a rearranged product — it does not, since Grignard formation avoids the carbocation altogether.
   • Picking a straight-chain or differently-branched acid (options A, B, C) that would correspond to a different $\mathrm{C_5H_{11}Br}$ isomer, ignoring the two-step-hydrolysis clue that pins down neopentyl bromide specifically.

   ✓Final answer

   The correct option is (D) — $\mathrm{(CH_3)_3C{-}CH_2{-}COOH}$.

   ANSWER: D

8. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.What is A in the following reaction? $\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH_2{-}CN} \xrightarrow[\text{2) }\mathrm{H_2O}]{\text{1) } \mathrm{AlH(i{-}Bu)_2}} (A)$ (A) $\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH_2{-}CH_2{-}NH_2}$ (B) $\mathrm{CH_3{-}CH_2{-}CH_2{-}CH_2{-}CH_2{-}CH_2{-}NH_2}$ (C) $\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH_2{-}CHO}$ (D) $\mathrm{CH_3{-}CH_2{-}CH_2{-}CH_2{-}CH_2{-}CHO}$

   ›Reveal solutionSolution

   DIBAL-H stops nitrile reduction at the aldehyde stage (via an imine intermediate hydrolysed on work-up) and leaves alkenes untouched. Answer: (C).

   Concept and Intuition

   DIBAL-H, $\mathrm{AlH(i{-}Bu)_2}$, delivers a single hydride. When it attacks a nitrile ($\mathrm{R{-}C{\equiv}N}$), the hydride adds once to give a stable metalated imine (an aluminium–nitrogen chelate) that does not collapse further at low temperature because the aluminium coordinates and "locks" the intermediate. Aqueous hydrolysis of this imine/enamine–aluminium complex during work-up then releases the corresponding aldehyde, $\mathrm{R{-}CHO}$. This is the standard method for stopping a nitrile reduction one oxidation level short of the amine that full reduction (e.g. with $\mathrm{LiAlH_4}$ or catalytic hydrogenation) would give. DIBAL-H is also a mild, chemoselective hydride source that does not reduce ordinary isolated alkene double bonds, so the $\mathrm{CH_3{-}CH{=}CH{-}}$ portion of the substrate is unaffected.

   Step-by-Step Solution

   1. Substrate: $\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH_2{-}C{\equiv}N}$.
   2. Step 1, $\mathrm{AlH(i{-}Bu)_2}$ (1 equivalent, controlled conditions): hydride adds once to the nitrile carbon, forming the imine–aluminium adduct $\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH_2{-}CH{=}NAl(i{-}Bu)_2}$ (conceptually); the alkene is untouched throughout.
   3. Step 2, $\mathrm{H_2O}$ work-up: hydrolyses the imine/aluminium complex to the aldehyde, releasing $\mathrm{NH_3}$ (as ammonium salts) and giving $\mathrm{CH_3{-}CH{=}CH{-}CH_2{-}CH_2{-}CHO}$.
   4. This matches option (C) exactly — same carbon count, double bond retained, terminal group is $\mathrm{{-}CHO}$ not $\mathrm{{-}CH_2NH_2}$.

   Common Mistakes

   • Assuming any hydride reduction of a nitrile goes all the way to the primary amine (true for $\mathrm{LiAlH_4}$/$\mathrm{H_2}$/catalyst, but not for controlled DIBAL-H).
   • Forgetting DIBAL-H leaves the isolated alkene alone and mistakenly reducing/hydrating the double bond as well.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

9. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.What are X and Y in the following reaction sequence? $\mathrm{C_6H_5N_2^+Cl^-} \xrightarrow{X} \mathrm{C_6H_5CN} \xrightarrow{\text{(i) }\mathrm{CH_3MgBr}\ \text{(ii) }\mathrm{H_2O}} Y$ (A) KCN; $\mathrm{C_6H_5COCH_3}$ (B) KCN; $\mathrm{C_6H_5C(OH)(CH_3)_2}$ (C) CuCN|KCN; $\mathrm{C_6H_5CH(OH)CH_3}$ (D) CuCN|KCN; $\mathrm{C_6H_5COCH_3}$

   ›Reveal solutionSolution

   Diazonium → nitrile is a Sandmeyer reaction (CuCN|KCN); a Grignard reagent adding to a nitrile and then hydrolysing gives a ketone, not an alcohol — so $Y=\mathrm{C_6H_5COCH_3}$.

   Concept and Intuition

   Sandmeyer-type reactions replace the $-\mathrm{N_2^+}$ group of a diazonium salt with a nucleophile delivered via a copper(I) catalyst; with cyanide (as CuCN, generated from CuCN|KCN) this installs $-\mathrm{CN}$ directly onto the aromatic ring, retaining the ring intact — a route to aryl nitriles that plain $\mathrm{KCN}$ alone (without Cu(I)) does not reliably achieve on an aryl diazonium.

   A nitrile's carbon is only singly electrophilic toward a Grignard: one equivalent of $\mathrm{RMgX}$ adds across the $\mathrm{C\equiv N}$ triple bond to give a metalated imine (an imino-magnesium salt). This does not react further with a second equivalent of Grignard (unlike an ester, which can go on to a tertiary alcohol) because the intermediate is stable to the reaction conditions until workup. Aqueous acidic hydrolysis of that imine salt then gives a ketone, not an alcohol.

   Step-by-Step Solution

   1. $\mathrm{C_6H_5N_2^+Cl^-} \xrightarrow{\text{CuCN|KCN}} \mathrm{C_6H_5CN}$ — Sandmeyer reaction; $X = \text{CuCN|KCN}$.
   2. $\mathrm{C_6H_5CN} \xrightarrow{\mathrm{CH_3MgBr}} \mathrm{C_6H_5C(=NMgBr)CH_3}$ (Grignard adds once across the nitrile).
   3. Aqueous workup hydrolyses the imine-magnesium salt to the ketone:

   $Y = \mathrm{C_6H_5COCH_3} \quad (\text{acetophenone})$

   Common Mistakes

   • Using plain $\mathrm{KCN}$ without $\mathrm{Cu}$ and expecting the same clean aryl nitrile (Sandmeyer specifically needs the Cu(I) catalytic cycle).
   • Assuming Grignard + nitrile gives a tertiary alcohol like it would with an ester or two Grignard equivalents on a ketone — it stops at the ketone stage after hydrolysis.

   ✓Final answer

   The correct option is (D) — CuCN|KCN; $\mathrm{C_6H_5COCH_3}$.

   ANSWER: D

10. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.What are X and Y respectively in the following reaction sequence? (g = వాయువు, dil = విలీన) $\mathrm{4\text{-}O_2N\text{-}C_6H_4\text{-}CHO \xrightarrow[HCl(g)]{C_2H_5OH} X \xrightarrow{dil.\ HCl} Y}$ (A) $X = \mathrm{4\text{-}O_2NC_6H_4CH(OC_2H_5)_2}$ (diethyl acetal) ; $Y = \mathrm{4\text{-}O_2NC_6H_4CHO}$ (the aldehyde) (B) $X = \mathrm{4\text{-}O_2NC_6H_4CH(OH)(OC_2H_5)}$ (hemiacetal, OH and $OC_2H_5$ on the same carbon) ; $Y = \mathrm{4\text{-}O_2NC_6H_4CH(OC_2H_5)_2}$ (diethyl acetal) (C) $X = \mathrm{4\text{-}O_2NC_6H_4CH(OC_2H_5)_2}$ (diethyl acetal) ; $Y = \mathrm{4\text{-}O_2NC_6H_4CH(OH)(OC_2H_5)}$ (hemiacetal) (D) $X = \mathrm{4\text{-}O_2NC_6H_4CH(OH)(OC_2H_5)}$ (hemiacetal) ; $Y = \mathrm{4\text{-}O_2NC_6H_4CH_2OH}$ (the benzylic alcohol)

    ›Reveal solutionSolution

    This tests acetal formation (protection) and its acid hydrolysis (deprotection) of an aldehyde. The answer is (A): X is the diethyl acetal, and Y is the regenerated aldehyde.

    Concept and Intuition

    Aldehydes react with excess alcohol under anhydrous acidic conditions (e.g. dry HCl gas) to form acetals, $\mathrm{RCH(OR')_2}$, via a hemiacetal intermediate that is not isolated under these forcing (excess alcohol, anhydrous acid, often with removal of water) conditions — the reaction proceeds essentially straight through to the acetal. Acetals are a classic protecting group for the carbonyl: they are stable under basic and neutral conditions but are cleaved by aqueous acid, regenerating the original aldehyde plus the alcohol (acetal hydrolysis is simply the reverse of acetal formation, driven by the presence of water).

    Step-by-Step Solution

    1. Step 1 (aldehyde $\to$ X): 4-nitrobenzaldehyde reacts with excess ethanol in the presence of dry HCl gas. Under these anhydrous, acid-catalyzed, excess-alcohol conditions, the aldehyde is converted all the way to its diethyl acetal: $X = \mathrm{4\text{-}O_2NC_6H_4CH(OC_2H_5)_2}$.
    2. Step 2 (X $\to$ Y): treating the acetal X with dilute (aqueous) HCl reverses acetal formation — water attacks and the acetal hydrolyzes back to the free aldehyde, releasing 2 equivalents of ethanol: $Y = \mathrm{4\text{-}O_2NC_6H_4CHO}$, i.e., the original 4-nitrobenzaldehyde is regenerated.
    3. So overall this sequence is simply 'protect the aldehyde as its acetal, then deprotect it back' — X is the acetal, Y is the aldehyde again.
    4. This matches option (A) exactly, ruling out (B)/(C) (which reverse the order or insert a hemiacetal where an isolable acetal/aldehyde is expected) and (D) (which wrongly reduces the aldehyde to a benzylic alcohol — no reducing agent is present in either step).

    Common Mistakes

    • Assuming the hemiacetal is the isolated product of step 1 — under anhydrous conditions with excess alcohol and HCl gas as catalyst, the reaction proceeds to the full acetal, not the hemiacetal.
    • Thinking dilute HCl (aqueous) would do anything other than hydrolyze the acetal back to the carbonyl compound — there is no reducing or other functional-group-changing agent present.

    ✓Final answer

    The correct option is (A) — $X = \mathrm{4\text{-}O_2NC_6H_4CH(OC_2H_5)_2}$ (diethyl acetal); $Y = \mathrm{4\text{-}O_2NC_6H_4CHO}$ (the aldehyde).

    ANSWER: A

11. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.An alcohol, X ($C_5H_{12}O$) in the presence of Cu/573K gives Y ($C_5H_{10}$). The reactants required for the preparation of X are (A) acetone, $\mathrm{CH_3COCH_3}$ (drawn as a skeletal structure with the carbonyl O at the bottom), and $\mathrm{C_2H_5MgBr}$ (B) $\mathrm{HCHO}$, $(CH_3)_3CMgBr$ (C) [FIGURE] (a five-carbon ketone: an ethyl group and a branched, isopropyl-type group attached to a central carbonyl carbon), and $\mathrm{C_2H_5MgBr}$ (D) acetone, $\mathrm{CH_3COCH_3}$ (drawn as a skeletal structure with the carbonyl O at the top), and $(CH_3)_2CHMgBr$

    ›Reveal solutionSolution

    X loses $H_2O$ (not $H_2$) over Cu/573K to give $C_5H_{10}$, which is the signature of a tertiary alcohol undergoing dehydration; only acetone + $C_2H_5MgBr$ builds a tertiary $C_5H_{12}O$ alcohol. Answer: (A).

    Concept and Intuition

    Over copper at 573 K, alcohols react differently depending on their class:

    • 1° alcohols dehydrogenate (lose $H_2$) to give aldehydes.
    • 2° alcohols dehydrogenate (lose $H_2$) to give ketones.
    • 3° alcohols have no H on the carbinol carbon to remove for dehydrogenation, so instead they undergo dehydration (lose $H_2O$) to give alkenes.

    Here, X ($C_5H_{12}O$) gives Y ($C_5H_{10}$). Checking the atom balance: $C_5H_{12}O - H_2O = C_5H_{10}$ — mass balance fits loss of water, confirming X must be a tertiary alcohol.

    A tertiary alcohol is made by adding a Grignard reagent to a ketone (not an aldehyde, which gives a secondary alcohol): $R_2C=O + R'MgX \to R_2C(OH)R'$. We need the resulting alcohol to have exactly 5 carbons.

    Step-by-Step Solution

    1. Determine X's alcohol class from the reaction Cu/573K: since $Y = C_5H_{10}$ (loss of $H_2O$ from X, not $H_2$), X must be a tertiary alcohol giving an alkene by dehydration.
    2. Test option (A): acetone $(CH_3)_2C=O$ (3 carbons) + $C_2H_5MgBr$ (2 carbons) → $(CH_3)_2C(OH)C_2H_5$ = 2-methyl-2-butanol, a tertiary alcohol with 5 carbons, formula $C_5H_{12}O$. ✓ Matches on both carbon count and alcohol class.
    3. Test option (D): acetone (3C) + $(CH_3)_2CHMgBr$ (isopropyl, 3C) → $(CH_3)_2C(OH)CH(CH_3)_2$ = 2,3-dimethyl-2-butanol, which has 6 carbons ($C_6H_{14}O$) — too many carbons, rejected.
    4. Test option (B): $HCHO$ + $(CH_3)_3CMgBr$ → $(CH_3)_3C-CH_2OH$ (neopentyl alcohol), which is a primary alcohol ($C_5H_{12}O$ by formula) — but a 1° alcohol would dehydrogenate to an aldehyde (losing $H_2$, giving $C_5H_{10}O$), not match $Y=C_5H_{10}$. Rejected.
    5. Test option (C): the drawn 5/6-carbon ketone + $C_2H_5MgBr$ adds too many total carbons (well beyond 5). Rejected.
    6. Only (A) gives a tertiary $C_5H_{12}O$ alcohol consistent with dehydration to $C_5H_{10}$.

    Common Mistakes

    • Forgetting that Cu/573K behaves differently for 1°/2°/3° alcohols — treating all alcohols as simply "dehydrogenating" would wrongly point to options that give aldehydes/ketones instead of alkenes.
    • Not checking total carbon count carefully in the Grignard addition — options (C) and (D) fail purely on carbon-count grounds.

    ✓Final answer

    The correct option is (A) — acetone, $CH_3COCH_3$, and $C_2H_5MgBr$.

    ANSWER: A

12. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.What are X and Y respectively in the following reaction sequence? [FIGURE] (benzaldehyde $\xrightarrow{X}$ Oxime $\xrightarrow{Y}$ benzonitrile) (A) NH$_2$NH$_2$, C$_6$H$_5$SO$_2$Cl / Pyridine (B) NH$_2$NH$_2$, (CH$_3$CO)$_2$O (C) NH$_2$OH, C$_6$H$_5$SO$_2$Cl / Pyridine (D) NH$_2$OH, (CH$_3$CO)$_2$O

    ›Reveal solutionSolution

    Aldehyde $\to$ oxime needs hydroxylamine; oxime $\to$ nitrile needs a dehydrating agent, classically acetic anhydride — X, Y = $NH_2OH$, $(CH_3CO)_2O$, option (D).

    Concept and Intuition

    This is the standard two-step laboratory route from an aldehyde to the one-carbon-longer nitrile (useful because nitriles hydrolyse further to carboxylic acids, or reduce to amines):

    1. Oxime formation: aldehydes and ketones condense with hydroxylamine ($NH_2OH$) to form oximes ($C=N-OH$), releasing water. This is a standard carbonyl condensation, analogous to hydrazone/semicarbazone formation.
    2. Dehydration of the oxime to a nitrile: removing a molecule of water from the oxime ($-CH=N-OH \to -C\equiv N + H_2O$) requires a dehydrating agent. The reagent commonly cited in NCERT-level organic chemistry for this step is acetic anhydride, $(CH_3CO)_2O$ (it also works with reagents like $P_2O_5$, $SOCl_2$, or tosyl chloride/pyridine in more advanced contexts, but the textbook-standard answer is acetic anhydride).

    Step-by-Step Solution

    1. Benzaldehyde ($C_6H_5CHO$) reacts with $NH_2OH$ to give benzaldoxime, $C_6H_5CH=N-OH$: this identifies X = $NH_2OH$.
    2. The oxime is dehydrated to benzonitrile, $C_6H_5C\equiv N$, using acetic anhydride, $(CH_3CO)_2O$, as the standard reagent: this identifies Y = $(CH_3CO)_2O$.
    3. Matching X = $NH_2OH$ and Y = $(CH_3CO)_2O$ to the options gives option (D).

    Common Mistakes

    • Swapping which step needs hydroxylamine vs which needs the dehydrating agent.
    • Confusing hydrazine ($NH_2NH_2$, used for hydrazones/Wolff-Kishner reduction) with hydroxylamine ($NH_2OH$, used for oximes) — they look similar but serve different purposes.

    ✓Final answer

    The correct option is (D) — X = $NH_2OH$, Y = $(CH_3CO)_2O$.

    ANSWER: D