Q.Write the structures of the following organic halogen compounds.

Showing the 12 most recent of 24 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Consider the reactions (not balanced) BF$_3$ + NaH $\xrightarrow{450\ \text{K}}$ A + NaF LiH + A $\rightarrow$ Li$^+$[X]$^-$ The hybridisation involved in [X]$^-$ is (A) sp$^2$ (B) sp$^3$ (C) sp (D) dsp$^2$

   ›Reveal solutionSolution

   The reaction sequence produces diborane (B₂H₆) as intermediate A, which then reacts with LiH to form the tetrahydroborate ion [BH₄]⁻. In [BH₄]⁻, boron is sp³ hybridised, so the correct option is (B).

   The key is to recognise that BF₃ reacts with NaH (a hydride donor) at high temperature to give diborane, B₂H₆ — a classic inorganic synthesis. Then diborane reacts further with LiH to form lithium borohydride, LiBH₄, which dissociates into Li⁺ and [BH₄]⁻. The hybridisation of boron in [BH₄]⁻ is determined by its four equivalent B–H bonds and no lone pairs.

   1. First reaction: BF₃ + NaH → A + NaF BF₃ is electron-deficient (only six electrons around B). NaH provides H⁻ ions. At 450 K, the reaction proceeds as:

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

   (Balanced: each B gains three H⁻, but B₂H₆ forms via dimerisation of BH₃.)

   So A = B₂H₆ (diborane).

   2. Second reaction: LiH + A → Li⁺[X]⁻ Diborane reacts with lithium hydride in a 1:2 molar ratio:

   $\text{B}_2\text{H}_6 + 2\,\text{LiH} \rightarrow 2\,\text{LiBH}_4$

   This is a Lewis acid–base reaction: B₂H₆ acts as a Lewis acid (accepting H⁻) to form the tetrahydroborate ion.

   So X⁻ = BH₄⁻ (tetrahydroborate or borohydride ion).

   3. Determine hybridisation of B in [BH₄]⁻

      Count electron domains around boron:

      • Four B–H sigma bonds (each bond uses one electron from B and one from H).
      • No lone pairs on boron. Total domains = 4 → tetrahedral geometry → sp³ hybridisation.

   4. Check the options

      (A) sp² – would give trigonal planar, not possible with four bonds.

      (B) sp³ – correct.

      (C) sp – linear, impossible.

      (D) dsp² – square planar, requires d-orbital involvement, not needed here.

   Watch out

   A common mistake is to think that BF₃ itself (sp²) carries through to the product. But the reaction with hydride completely changes boron’s coordination number from 3 to 4, forcing rehybridisation to sp³.

   Tip

   Remember: any boron species with four single bonds (like BH₄⁻, BF₄⁻) is always sp³ hybridised. The octet rule is satisfied, and geometry is tetrahedral.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Which of the following reaction is not correct regarding the products? (A) Sodium $p$-nitrophenoxide (benzene ring with $-\mathrm{ONa}$ and $-\mathrm{NO_2}$ para) $+\ \mathrm{CH_3Br} \longrightarrow$ $p$-nitroanisole (benzene ring with $-\mathrm{O-CH_3}$ and $-\mathrm{NO_2}$ para) $+\ \mathrm{NaBr}$ (B) $p$-Bromonitrobenzene (benzene ring with $-\mathrm{Br}$ and $-\mathrm{NO_2}$ para) $+\ \mathrm{CH_3ONa} \longrightarrow$ $p$-nitroanisole (benzene ring with $-\mathrm{O-CH_3}$ and $-\mathrm{NO_2}$ para) $+\ \mathrm{NaBr}$ (C) $\mathrm{(CH_3)_3C-OC_2H_5 + HI \rightarrow (CH_3)_3C-I + C_2H_5OH}$ (D) $\mathrm{CH_3CH_2CH_2OCH_3 + HBr \rightarrow CH_3CH_2CH_2OH + CH_3Br}$

   ›Reveal solutionSolution

   Three of the four equations are sound. The odd one out is the attempted Williamson synthesis on an aryl halide — aryl halides do not undergo $\mathrm{S_N2}$, so $p$-bromonitrobenzene + $\mathrm{CH_3ONa}$ does not give $p$-nitroanisole under these conditions. Option (B).

   The concept first

   Two rules govern this whole question.

   Rule 1 — Williamson's ether synthesis is $\mathrm{S_N2}$.

   $\mathrm{R\!-\!O^{-}Na^{+}}+\mathrm{R'\!-\!X}\ \longrightarrow\ \mathrm{R\!-\!O\!-\!R'}+\mathrm{NaX}$

   The alkoxide/phenoxide attacks the back of the carbon carrying $\mathrm{X}$. That carbon must therefore be $sp^3$ and unhindered: methyl and primary halides work beautifully, secondary ones give elimination, tertiary ones fail completely — and aryl (or vinyl) halides do not react at all, because (i) the carbon is $sp^2$, (ii) the C–X bond has partial double-bond character from lone-pair delocalisation and is therefore short and strong, and (iii) the $\pi$ cloud repels the approaching nucleophile.

   Rule 2 — ethers cleaved by $\mathrm{HX}$. The oxygen is protonated first; then $\mathrm{X^{-}}$ attacks whichever alkyl group can accept it most readily:

   • if one group is $3^\circ$ (or benzylic/allylic) $\Rightarrow$ $\mathrm{S_N1}$: that group leaves as the stable carbocation and becomes the halide;
   • if both groups are $1^\circ$/methyl $\Rightarrow$ $\mathrm{S_N2}$: $\mathrm{X^{-}}$ attacks the less hindered carbon, which becomes the halide, and the bulkier group leaves as the alcohol.

   Step-by-step through the options

   (A) Sodium $p$-nitrophenoxide $+\ \mathrm{CH_3Br}\rightarrow p$-nitroanisole $+\ \mathrm{NaBr}$.

   The nucleophile is the phenoxide; the electrophile is a methyl bromide — the ideal $\mathrm{S_N2}$ substrate. The alkyl halide (not the aryl ring) is being attacked, so this is textbook Williamson. Correct as written. ✓

   (B) $p$-Bromonitrobenzene $+\ \mathrm{CH_3ONa}\rightarrow p$-nitroanisole $+\ \mathrm{NaBr}$.

   Here the roles are reversed: the halogen sits on the benzene ring, and we are asking methoxide to displace it. That is exactly what an aryl halide will not do in a Williamson-type $\mathrm{S_N2}$ — the $sp^2$ C–Br bond is strong, shortened by resonance, and the ring repels the incoming nucleophile. The correct way to make $p$-nitroanisole is route (A), attacking a methyl halide with the phenoxide. This is the incorrect reaction. ✗

   (C) $\mathrm{(CH_3)_3C\!-\!OC_2H_5}+\mathrm{HI}\rightarrow \mathrm{(CH_3)_3C\!-\!I}+\mathrm{C_2H_5OH}$.

   Protonation of the ether oxygen is followed by loss of the very stable tert-butyl carbocation, which grabs $\mathrm{I^{-}}$; the ethyl group departs as ethanol. Exactly as written. Correct. ✓

   (D) $\mathrm{CH_3CH_2CH_2\!-\!O\!-\!CH_3}+\mathrm{HBr}\rightarrow \mathrm{CH_3CH_2CH_2OH}+\mathrm{CH_3Br}$.

   Both groups are primary/methyl, so no carbocation can form and $\mathrm{Br^{-}}$ attacks by $\mathrm{S_N2}$ at the less hindered methyl carbon. The methyl group therefore ends up as $\mathrm{CH_3Br}$, and the propyl group leaves as propan-1-ol. Exactly as written. Correct. ✓

   ✓Final answer

   Aryl halides do not undergo the $\mathrm{S_N2}$ substitution required, so $p$-bromonitrobenzene with $\mathrm{CH_3ONa}$ does not give the product shown — the correct option is (B).

   ANSWER: B

3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.1 mole of a hydrocarbon A($C_5H_{10}$) on ozonolysis gives two compounds X and Y. Both X and Y respond to iodoform test. X gives test with ammoniacal $AgNO_3$ solution but not with Y. What are X and Y respectively? (A) $CH_2O$ ; $CH_3COC_2H_5$ (B) $CH_3CHO$ ; $(CH_3)_2CO$ (C) $(CH_3)_2C=O$ ; $CH_3CH=O$ (D) $CH_3COOH$ ; $(CH_3)_2C=O$

   ›Reveal solutionSolution

   The key is that ozonolysis of a hydrocarbon with formula C₅H₁₀ (an alkene) cleaves the double bond to give two carbonyl compounds. Both must give a positive iodoform test (so each must be a methyl ketone or acetaldehyde), and only one gives a positive Tollens’ test (so only one is an aldehyde). The only pair fitting all clues is acetaldehyde (CH₃CHO) and acetone ((CH₃)₂CO), which corresponds to option (B).

   ---

   Concept & Intuition

   Ozonolysis of an alkene breaks the C=C bond, replacing it with two C=O bonds. The products are either aldehydes or ketones, depending on the substitution at the double bond.

   The iodoform test is positive for compounds with a CH₃–C(=O)– group (methyl ketones) or for ethanol/acetaldehyde (CH₃CH₂OH or CH₃CHO). The Tollens’ test (ammoniacal AgNO₃) is positive only for aldehydes (and some α-hydroxy ketones, but not relevant here).

   So we need two carbonyl compounds, each with a methyl group attached to the carbonyl carbon, and exactly one of them must be an aldehyde (the other a ketone). The original hydrocarbon C₅H₁₀ must be an alkene whose double bond, when cleaved, yields exactly such a pair.

   ---

   Step-by-step reasoning

   1. Identify the molecular formula constraint

      The hydrocarbon A has formula C₅H₁₀. This is the general formula for an alkene (or a cycloalkane, but ozonolysis only works on alkenes). So A is an alkene with 5 carbons.

   2. Ozonolysis outcome

      Ozonolysis cleaves the double bond, adding an oxygen atom to each carbon of the original double bond. The sum of carbons in the two products equals the number of carbons in the alkene (5). So the two carbonyl compounds together contain 5 carbons.

   3. Iodoform test condition

      Both X and Y give a positive iodoform test. That means each must contain the CH₃–C(=O)– group. So each product is either:

      • A methyl ketone (R–CO–CH₃), or
      • Acetaldehyde (CH₃CHO), or
      • Ethanol (but ethanol is not formed in ozonolysis; only carbonyls are).

      Therefore, each product has at least 2 carbons (the methyl group + the carbonyl carbon). The smallest possible is acetaldehyde (C₂), and the next is acetone (C₃), etc.

   4. Tollens’ test condition

      X gives a positive Tollens’ test (ammoniacal AgNO₃), so X is an aldehyde. Y does not give this test, so Y is a ketone.

      Since both give iodoform, X must be acetaldehyde (CH₃CHO) — the only aldehyde that gives the iodoform test. Y must be a methyl ketone.

   5. Carbon count

      If X is CH₃CHO (2 carbons), then Y must contain the remaining 3 carbons (since total = 5). A methyl ketone with 3 carbons is acetone: (CH₃)₂CO.

      Check: Acetone gives iodoform test (positive) and does not give Tollens’ test (negative). Perfect.

   6. Verify the original alkene

      The alkene that gives CH₃CHO and (CH₃)₂CO upon ozonolysis must have the double bond between the carbon that becomes the aldehyde group and the carbon that becomes the ketone group. That alkene is 2-methyl-2-butene:

   $\text{CH}_3\text{–C(CH}_3\text{)=CH–CH}_3$

   Ozonolysis cleaves the double bond to give:

   $\text{CH}_3\text{CHO} \quad \text{and} \quad (\text{CH}_3)_2\text{CO}$

   This matches perfectly.

   7. Eliminate other options
      • (A) CH₂O (formaldehyde) does NOT give iodoform test — wrong.
      • (C) Lists the same compounds but swapped: acetone first, acetaldehyde second — but the question says X gives Tollens’ test, so X must be the aldehyde (acetaldehyde), not acetone. So order matters.
      • (D) CH₃COOH (acetic acid) does not give iodoform test under normal conditions (it’s not a methyl ketone or acetaldehyde) — wrong.

   Watch out

   A common mistake is to forget that the iodoform test is positive for acetaldehyde but not for other aldehydes. Also, the order of X and Y matters: X must be the aldehyde because it gives Tollens’ test.

   Tip

   Remember: The only aldehyde that gives the iodoform test is acetaldehyde (CH₃CHO). So if you see “both give iodoform, one gives Tollens’,” the aldehyde is always acetaldehyde.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.The IUPAC name of the following compound is $(\mathrm{CH}_3)_3\mathrm{C}-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)_2$ (A) 2, 3, 3 - trimethylbutan-2-ol (B) 2, 2, 3 - trimethylbutan-3-ol (C) 1, 1, 2, 2 - tetramethylpropan-1-ol (D) 2, 2, 3, 3 - tetramethylpropan-3-ol

   ›Reveal solutionSolution

   The IUPAC name is determined by identifying the longest carbon chain containing the hydroxyl group, numbering it to give the hydroxyl group the lowest possible number, and then naming the substituents. The compound is 2,3,3-trimethylbutan-2-ol.

   Concept and Intuition

   Naming organic compounds using IUPAC (International Union of Pure and Applied Chemistry) nomenclature follows a set of systematic rules to ensure each compound has a unique and unambiguous name. For alcohols, the core idea is to:

   1. Identify the parent chain: This is the longest continuous carbon chain that contains the carbon atom bonded to the hydroxyl (-OH) group.
   2. Number the parent chain: Start numbering from the end that gives the carbon atom bearing the -OH group the lowest possible number. If there's a tie, then number to give the substituents the lowest possible numbers.
   3. Identify and name substituents: Any carbon groups or other atoms attached to the parent chain that are not part of the main functional group are considered substituents.
   4. Assemble the name:
      • List substituents in alphabetical order (ignoring prefixes like di-, tri-, etc., for alphabetization).
      • Precede each substituent name with its position number on the parent chain.
      • Use prefixes like "di-", "tri-", "tetra-" for multiple identical substituents.
      • The parent chain name is derived from the corresponding alkane (e.g., "butane" for a 4-carbon chain).
      • Replace the "-e" ending of the alkane name with "-ol" to indicate an alcohol.
      • Insert the position number of the -OH group just before the "-ol" suffix.

   Step-by-Step Solution

   1. Draw the expanded structural formula:

      The given condensed formula is $(\mathrm{CH}_3)_3\mathrm{C}-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)_2$.

      Let's break this down:

      • $(\mathrm{CH}_3)_3\mathrm{C}-$ indicates a carbon atom bonded to three methyl ($\mathrm{CH}_3$) groups. This is a tert-butyl group.
      • $-\mathrm{C}(\mathrm{OH})(\mathrm{CH}_3)_2$ indicates another carbon atom bonded to one hydroxyl (-OH) group and two methyl ($\mathrm{CH}_3$) groups. These two central carbon atoms are bonded to each other. The expanded structure is:

            CH3
            |
      CH3 - C - C - OH
            |   |
            CH3 CH3
                |
                CH3
      

   2. Identify the longest continuous carbon chain containing the -OH group:

      The carbon atom bearing the -OH group is the one on the right in the structure above. Let's call it $\mathrm{C_X}$.

      We need to find the longest path of carbon atoms that includes $\mathrm{C_X}$.

      • Starting from one of the $\mathrm{CH}_3$ groups attached to the left carbon, going through the left carbon, then $\mathrm{C_X}$, and finally to one of the $\mathrm{CH}_3$ groups attached to $\mathrm{C_X}$: For example, \mathrm{CH}_3 - \mathrm{C}(\text{left}) - \mathrm{C_X} - \mathrm{CH}_3(\text{on C_X}). This chain has 4 carbon atoms. Any other path that includes $\mathrm{C_X}$ will also result in a 4-carbon chain. Therefore, the parent chain is a butane derivative.

   3. Number the parent chain:

      The numbering must assign the lowest possible number to the carbon atom bearing the -OH group.

      Let's consider the 4-carbon chain identified:

            CH3
            |
      CH3 - C - C - OH
            |   |
            CH3 CH3
                |
                CH3
      

      If we number from right to left:

      $\mathrm{C_1}$ ($\mathrm{CH}_3$ on $\mathrm{C_X}$) - $\mathrm{C_2}$ ($\mathrm{C_X}$ with -OH) - $\mathrm{C_3}$ (left carbon) - $\mathrm{C_4}$ ($\mathrm{CH}_3$ on left carbon)

      In this numbering, the -OH group is on $\mathrm{C_2}$.

      If we number from left to right:

      $\mathrm{C_1}$ ($\mathrm{CH}_3$ on left carbon) - $\mathrm{C_2}$ (left carbon) - $\mathrm{C_3}$ ($\mathrm{C_X}$ with -OH) - $\mathrm{C_4}$ ($\mathrm{CH}_3$ on $\mathrm{C_X}$)

      In this numbering, the -OH group is on $\mathrm{C_3}$.

      Since 2 is lower than 3, the numbering from right to left is correct. The -OH group is at position 2.

   4. Identify and name substituents:

      Using the correct numbering (right to left):

      • The carbon at position 2 ($\mathrm{C_X}$) has one methyl group attached (in addition to the -OH and the chain carbon).
      • The carbon at position 3 (the left carbon) has two methyl groups attached (in addition to the chain carbon). So, we have methyl groups at positions 2, 3, and 3.

   5. Assemble the IUPAC name:

      • The parent chain is butane.
      • The hydroxyl group is at position 2, so it's a butan-2-ol.
      • There are three methyl groups at positions 2, 3, and 3. This is written as "2,3,3-trimethyl". Combining these parts, the IUPAC name is 2,3,3-trimethylbutan-2-ol.

   Comparing this with the given options:

   (A) 2, 3, 3 - trimethylbutan-2-ol

   (B) 2, 2, 3 - trimethylbutan-3-ol (Incorrect numbering for -OH)

   (C) 1, 1, 2, 2 - tetramethylpropan-1-ol (Incorrect parent chain and number of substituents)

   (D) 2, 2, 3, 3 - tetramethylpropan-3-ol (Incorrect parent chain and number of substituents)

   The derived name matches option (A).

   ✓Final answer

   The IUPAC name of the compound is $\boxed{\text{2,3,3-trimethylbutan-2-ol}}$.

5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The product 'C' in the given reaction sequence is 3-bromonitrobenzene ($\ce{m-O2N-C6H4-Br}$) $\ce{->[Mg][ether] A ->[(i) CO2][(ii) H3O+] B ->[(i) Na][(ii) NaOH + CaO] C}$ (A) Nitrobenzene ($\ce{C6H5-NO2}$) (B) Bromobenzene ($\ce{C6H5-Br}$) (C) 3-bromobenzoic acid ($\ce{m-Br-C6H4-COOH}$) (D) Sodium 3-bromobenzoate ($\ce{m-Br-C6H4-COONa}$)

   ›Reveal solutionSolution

   The Grignard is carboxylated to 3-nitrobenzoic acid, whose sodium salt is decarboxylated by soda lime — the $\ce{-COONa}$ is replaced by $\ce{-H}$, leaving nitrobenzene: option (A).

   The concept first

   Three standard moves, in order.

   1. Grignard formation. $\ce{Mg}$ in dry ether inserts into a carbon–halogen bond, reversing the polarity of that carbon: the once-electrophilic $\ce{C-Br}$ carbon becomes nucleophilic ($\ce{C^{\delta-}-Mg^{\delta+}Br}$).

   2. Carboxylation. That nucleophilic carbon attacks the electrophilic carbon of $\ce{CO2}$ (often used as dry ice), giving a carboxylate; $\ce{H3O+}$ work-up then delivers the carboxylic acid. This is the classic way of adding one carbon to a skeleton.

   3. Decarboxylation with soda lime. Heating the sodium salt of a carboxylic acid with $\ce{NaOH}$ + $\ce{CaO}$ (soda lime) expels the carboxyl carbon as carbonate and puts an $\ce{H}$ in its place:

   $\ce{R-COONa + NaOH ->[CaO,\ \Delta] R-H + Na2CO3}$

   So the two steps 2 and 3 together add a carbon and then remove it — the net effect on the ring is simply that $\ce{-Br}$ has been replaced by $\ce{-H}$.

   Step 1 — A

   $\ce{m-O2N-C6H4-Br ->[Mg,\ dry\ ether] m-O2N-C6H4-MgBr}\qquad (\mathbf{A})$

   Step 2 — B

   $\ce{m-O2N-C6H4-MgBr ->[(i) CO2][(ii) H3O+] m-O2N-C6H4-COOH}\qquad (\mathbf{B}=\text{3-nitrobenzoic acid})$

   Step 3 — C

   $\ce{m-O2N-C6H4-COOH ->[Na] m-O2N-C6H4-COONa ->[NaOH + CaO,\ \Delta] C6H5-NO2 + Na2CO3}$

   The nitro group is untouched by soda lime, so $\mathbf{C}$ = nitrobenzene.

   Step 4 — Check the distractors

   Options (C) and (D) keep a bromine on the ring — but the bromine was consumed the moment the Grignard formed. Option (B), bromobenzene, would require the nitro group to vanish and the bromine to survive: the exact opposite of what happens.

   ✓Final answer

   C is nitrobenzene, which is option (A).

   ANSWER: A

6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The correct statements about the products B and C in the given reactions are I. B and C are functional isomers II. With $H_2$ | Catalyst B gives $1^\circ$ amine and C gives $2^\circ$ amine III. B on acid hydrolysis gives formic acid and C gives $C_3H_6O_2$ IV. C forms isocyanate with $HgO$ (A) II & III (B) II, III & IV (C) I, II & IV (D) I & III

   ›Reveal solutionSolution

   The key is to identify B and C from the reaction sequence: B is an amide (N-methylformamide) and C is an isocyanide (methyl isocyanide). They are functional isomers. B gives a primary amine on reduction, C gives a secondary amine. B hydrolyses to formic acid and methylamine; C hydrolyses to methylamine and formic acid (giving C₃H₆O₂? No — careful). C does form an isocyanate with HgO. Only statements II and IV are correct, so the answer is (B).

   ---

   Concept & Intuition

   The problem presents a classic organic reaction: a primary amine (methylamine) reacting with chloroform and alcoholic KOH — the carbylamine reaction. This produces an isocyanide (C). Separately, the same amine reacts with an acyl chloride (or similar) to form an amide (B). The two products are functional isomers (same molecular formula, different functional groups: amide vs isocyanide). Their chemical properties — reduction, hydrolysis, and reaction with HgO — differ predictably. Let’s identify them and test each statement.

   ---

   Step-by-step reasoning

   1. Identify B and C
      • Methylamine ($CH_3NH_2$) reacts with HCOOH (formic acid) to give N-methylformamide (B):

   $CH_3NH_2 + HCOOH \rightarrow HCONHCH_3 + H_2O$

    B is an amide, formula $C_2H_5NO$.  
   

   • Methylamine with $CHCl_3$ and alcoholic KOH gives methyl isocyanide (C):

   $CH_3NH_2 + CHCl_3 + 3KOH \rightarrow CH_3NC + 3KCl + 3H_2O$

    C is an isocyanide, also $C_2H_5NO$.  
   

   • They are functional isomers (same molecular formula, different functional groups). So statement I is true.

   2. Statement II: Reduction with $H_2$/catalyst
      • B (amide) reduces to a primary amine:

   $HCONHCH_3 + 2H_2 \xrightarrow{cat} CH_3NH_2 + CH_3OH$

    Actually careful: reduction of an amide gives an amine — here $HCONHCH_3$ reduces to $CH_3NH_2$ (primary) + $CH_3OH$. So B gives a **1° amine**.  
   

   • C (isocyanide) reduces to a secondary amine:

   $CH_3NC + 2H_2 \xrightarrow{cat} CH_3NHCH_3$

    (dimethylamine, a 2° amine).  
   

   • So statement II is true.

   3. Statement III: Acid hydrolysis
      • B (N-methylformamide) on acid hydrolysis:

   $HCONHCH_3 + H_2O \xrightarrow{H^+} HCOOH + CH_3NH_2$

    Products: formic acid ($HCOOH$) and methylamine. No $C_3H_6O_2$ here.  
   

   • C (methyl isocyanide) on acid hydrolysis:

   $CH_3NC + 2H_2O \xrightarrow{H^+} CH_3NH_2 + HCOOH$

    Again gives formic acid and methylamine — **not** $C_3H_6O_2$.  
   

   • $C_3H_6O_2$ could be propionic acid or methyl acetate, but neither appears. So statement III is false.

   4. Statement IV: C forms isocyanate with $HgO$
      • Isocyanides react with yellow mercuric oxide ($HgO$) to give isocyanates:

   $RNC + HgO \rightarrow RNCO + Hg$

    For methyl isocyanide: $CH_3NC + HgO \rightarrow CH_3NCO$ (methyl isocyanate).  
   

   • This is a classic test for isocyanides. So statement IV is true.

   5. Which statements are correct?
      • I: True
      • II: True
      • III: False
      • IV: True
      • So correct set: I, II, IV → option (C).

   Watch out

   A common mistake is to think that amide reduction gives a secondary amine — it actually gives a primary amine (the nitrogen loses the carbonyl carbon). Also, hydrolysis of both B and C yields formic acid and methylamine, not a three-carbon acid.

   Tip

   The carbylamine reaction is a surefire way to distinguish a primary amine from secondary/tertiary — only primary amines give the foul-smelling isocyanide.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.In which of the following reactions, hydrogen is evolved? I. Reaction of sodium borohydride with iodine II. Oxidation of diborane III. Reaction of boron trifluoride with sodium hydride IV. Hydrolysis of diborane (A) I, IV only (B) I, II only (C) III, IV only (D) I, II, IV only

   ›Reveal solutionSolution

   $\text{H}_2$ is released only in the reaction of $\text{NaBH}_4$ with iodine and in the hydrolysis of diborane — reactions I and IV. Correct option: (A).

   I. $\text{NaBH}_4$ + iodine — H$_2$ evolved.

   $\text{2NaBH}_4+\text{I}_2\rightarrow\text{2NaI}+\text{2BH}_3+\text{H}_2\uparrow$

   Iodine oxidises the hydridic hydrogen, liberating $\text{H}_2$.

   II. Oxidation of diborane — no H$_2$.

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

   Hydrogen is oxidised to water, not released as $\text{H}_2$.

   III. $\text{BF}_3$ + sodium hydride — no H$_2$.

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

   The hydride is transferred to boron (forming diborane); no $\text{H}_2$ gas forms.

   IV. Hydrolysis of diborane — H$_2$ evolved.

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

   Hydrogen is evolved in I and IV only.

   ✓Final answer

   (A) I, IV only.

8. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.What are X and Y in the following reaction sequence? Iso pentane $\xrightarrow{\mathrm{KMnO_4}}$ X $\xrightarrow[358\ \mathrm{K}]{20\%\ \mathrm{H_3PO_4}}$ Y (A) $\mathrm{(CH_3)_2CH-CH(OH)-CH_3}$ (3-methylbutan-2-ol) , $\mathrm{(CH_3)_2CH-CH=CH_2}$ (3-methyl-1-butene) (B) $\mathrm{CH_3CH_2-C(CH_3)_2-OH}$ (2-methylbutan-2-ol) , $\mathrm{CH_2=C(CH_3)-CH_2CH_3}$ (2-methyl-1-butene) (C) $\mathrm{CH_3CH_2-C(CH_3)_2-OH}$ (2-methylbutan-2-ol) , $\mathrm{(CH_3)_2C=CH-CH_3}$ (2-methyl-2-butene) (D) $\mathrm{CH_3CH_2-CH(CH_3)-CH_2OH}$ (2-methylbutan-1-ol) , $\mathrm{CH_2=C(CH_3)-CH_2CH_3}$ (2-methyl-1-butene)

   ›Reveal solutionSolution

   $\mathrm{KMnO_4}$ oxidises the lone tertiary C–H of isopentane to give 2-methylbutan-2-ol ($X$); $\mathrm{H_3PO_4}$ then dehydrates it by Saytzeff's rule to the trisubstituted 2-methylbut-2-ene ($Y$) — option (C).

   The concept first

   Two separate ideas are being tested.

   1. Selective oxidation of a tertiary C–H. Alkanes are inert to most reagents, but a tertiary C–H bond is the weakest (bond dissociation energy $3^\circ<2^\circ<1^\circ$, because the resulting radical/cation is best stabilised by hyperconjugation). Cold alkaline $\mathrm{KMnO_4}$ therefore attacks only that hydrogen and converts it to $-\mathrm{OH}$:

      $\mathrm{R_3C\!-\!H}\ \xrightarrow{\mathrm{KMnO_4}}\ \mathrm{R_3C\!-\!OH}$

   2. Saytzeff's rule. In an acid-catalysed dehydration, the $\mathrm{OH}$ is protonated, water leaves to give a carbocation, and a $\beta$-hydrogen is then lost. When there is a choice, the $\beta$-H is taken from the carbon that yields the more highly substituted alkene, because more alkyl groups on the $\mathrm{C{=}C}$ means more hyperconjugation and a more stable product.

   Step-by-step

   Step 1 — draw isopentane. Isopentane is 2-methylbutane:

   $\mathrm{CH_3\!-\!\underset{\textstyle |}{\overset{\textstyle CH_3}{C}}H\!-\!CH_2\!-\!CH_3}\qquad\text{i.e. }\mathrm{(CH_3)_2CH\!-\!CH_2CH_3}$

   It has exactly one tertiary carbon — C-2, bearing the single tertiary hydrogen.

   Step 2 — oxidation gives $X$. $\mathrm{KMnO_4}$ replaces that tertiary H with OH:

   $\mathrm{(CH_3)_2CH\!-\!CH_2CH_3}\ \xrightarrow{\mathrm{KMnO_4}}\ \mathrm{(CH_3)_2C(OH)\!-\!CH_2CH_3}$

   $X=\text{2-methylbutan-2-ol (a }3^\circ\text{ alcohol)}$

   This immediately kills options (A) (a 2° alcohol) and (D) (a 1° alcohol) — $\mathrm{KMnO_4}$ would not touch those weaker-reacting primary/secondary positions in preference to the tertiary one.

   Step 3 — dehydration of $X$. With $20\%\ \mathrm{H_3PO_4}$ at 358 K (mild conditions, which suffice precisely because $X$ is tertiary):

   $\mathrm{(CH_3)_2C(OH)CH_2CH_3}\ \xrightarrow{\mathrm{H^{+}}}\ \mathrm{(CH_3)_2\overset{+}{C}\!-\!CH_2CH_3}+\mathrm{H_2O}$

   The $3^\circ$ carbocation now loses a $\beta$-hydrogen. There are two kinds of $\beta$-H available:

   • from one of the two $\mathrm{CH_3}$ groups $\Rightarrow$ $\mathrm{CH_2{=}C(CH_3)CH_2CH_3}$, 2-methylbut-1-ene (disubstituted, Hofmann product);
   • from the $\mathrm{CH_2}$ of the ethyl group $\Rightarrow$ $\mathrm{(CH_3)_2C{=}CH\!-\!CH_3}$, 2-methylbut-2-ene (trisubstituted, Saytzeff product).

   Step 4 — apply Saytzeff. The trisubstituted alkene is the more stable, hence the major product:

   $Y=\mathrm{(CH_3)_2C{=}CH\!-\!CH_3}\quad(\text{2-methylbut-2-ene}).$

   That rules out option (B), which pairs the right alcohol with the minor (Hofmann) alkene.

   ✓Final answer

   Isopentane $\to$ 2-methylbutan-2-ol ($X$) $\to$ 2-methylbut-2-ene ($Y$), so the correct option is (C).

   ANSWER: C

9. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.What are X and Y in the following reaction sequence? Iso pentane $\xrightarrow{\mathrm{KMnO_4}}$ X $\xrightarrow[358\ \mathrm{K}]{20\%\ \mathrm{H_3PO_4}}$ Y (A) $\mathrm{CH_3CH_2-C(CH_3)_2-OH}$ (2-methylbutan-2-ol) , $\mathrm{CH_2=C(CH_3)-CH_2CH_3}$ (2-methyl-1-butene) (B) $\mathrm{CH_3CH_2-CH(CH_3)-CH_2OH}$ (2-methylbutan-1-ol) , $\mathrm{CH_2=C(CH_3)-CH_2CH_3}$ (2-methyl-1-butene) (C) $\mathrm{(CH_3)_2CH-CH(OH)-CH_3}$ (3-methylbutan-2-ol) , $\mathrm{(CH_3)_2CH-CH=CH_2}$ (3-methyl-1-butene) (D) $\mathrm{CH_3CH_2-C(CH_3)_2-OH}$ (2-methylbutan-2-ol) , $\mathrm{(CH_3)_2C=CH-CH_3}$ (2-methyl-2-butene)

   ›Reveal solutionSolution

   $\mathrm{KMnO_4}$ attacks the single tertiary C–H of isopentane, giving 2-methylbutan-2-ol ($X$); $\mathrm{H_3PO_4}$ then dehydrates it, and Saytzeff's rule makes the trisubstituted 2-methylbut-2-ene the product ($Y$) — option (D).

   The concept first

   Two ideas are stacked in this one arrow-chain.

   (i) Which C–H does $\mathrm{KMnO_4}$ attack? Alkanes are famously unreactive, but their C–H bonds are not equal: bond strength runs $3^\circ<2^\circ<1^\circ$, because the intermediate formed at a tertiary carbon is best stabilised by the surrounding alkyl groups (hyperconjugation). $\mathrm{KMnO_4}$ therefore hydroxylates the tertiary position selectively:

   $\mathrm{R_3C\!-\!H}\ \xrightarrow{\mathrm{KMnO_4}}\ \mathrm{R_3C\!-\!OH}$

   (ii) Saytzeff's rule. Acid dehydration of an alcohol protonates the $-\mathrm{OH}$, expels water to give a carbocation, then removes a $\beta$-hydrogen. Where there is a choice of $\beta$-H, the one that yields the more substituted, more stable alkene is preferred.

   Step-by-step

   Step 1 — the substrate. Isopentane is 2-methylbutane:

   $\mathrm{(CH_3)_2CH\!-\!CH_2\!-\!CH_3}$

   C-2 is its only tertiary carbon, carrying the only tertiary hydrogen.

   Step 2 — oxidation gives $X$.

   $\mathrm{(CH_3)_2CH\!-\!CH_2CH_3}\ \xrightarrow{\mathrm{KMnO_4}}\ \mathrm{(CH_3)_2C(OH)\!-\!CH_2CH_3}$

   $X=\text{2-methylbutan-2-ol, a }3^\circ\text{ alcohol.}$

   That rules out option (C) (a secondary alcohol from a secondary C–H) and option (B) (a primary alcohol from a primary C–H) — $\mathrm{KMnO_4}$ would not choose a stronger C–H bond over the tertiary one.

   Step 3 — dehydration to $Y$. With $20\%\ \mathrm{H_3PO_4}$ at 358 K:

   $\mathrm{(CH_3)_2C(OH)CH_2CH_3}\ \xrightarrow{\mathrm{H^{+}},\,-\mathrm{H_2O}}\ \mathrm{(CH_3)_2\overset{+}{C}\!-\!CH_2CH_3}$

   Two different $\beta$-hydrogens are available:

   • from a $\mathrm{CH_3}$ group $\Rightarrow$ $\mathrm{CH_2{=}C(CH_3)CH_2CH_3}$ — 2-methylbut-1-ene, disubstituted;
   • from the $\mathrm{CH_2}$ of the ethyl group $\Rightarrow$ $\mathrm{(CH_3)_2C{=}CHCH_3}$ — 2-methylbut-2-ene, trisubstituted.

   Step 4 — Saytzeff picks the winner. The trisubstituted alkene has more hyperconjugative structures and is the more stable, so it is the major product:

   $Y=\mathrm{(CH_3)_2C{=}CH\!-\!CH_3}\ (\text{2-methylbut-2-ene}).$

   Option (A) pairs the correct alcohol with the minor (Hofmann) alkene, so it is out.

   ✓Final answer

   Isopentane $\to$ 2-methylbutan-2-ol ($X$) $\to$ 2-methylbut-2-ene ($Y$), so the correct option is (D).

   ANSWER: D

10. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.The reaction/s which represent the production of diborane on an industrial scale is/are I. $4\mathrm{BF}_3 + 3\mathrm{LiAlH}_4 \rightarrow 2\mathrm{B}_2\mathrm{H}_6 + 3\mathrm{LiF} + 3\mathrm{AlF}_3$ II. $2\mathrm{NaBH}_4 + \mathrm{I}_2 \rightarrow \mathrm{B}_2\mathrm{H}_6 + 2\mathrm{NaI} + \mathrm{H}_2$ III. $2\mathrm{BF}_3 + 6\mathrm{NaH} \xrightarrow{450K} \mathrm{B}_2\mathrm{H}_6 + 6\mathrm{NaF}$ The correct option is (A) III only (B) I, II only (C) I only (D) I, III only

    ›Reveal solutionSolution

    The key idea is that industrial-scale production of diborane must be economical and scalable. Among the given reactions, only those using inexpensive, bulk reagents (like NaH) qualify. The correct industrial method is reaction III, so the answer is (A).

    The question asks which reactions represent the industrial-scale production of diborane ($\mathrm{B}_2\mathrm{H}_6$). This is not about whether a reaction works in a lab — it’s about which one is actually used in large-scale manufacturing. The distinction lies in cost, availability of reagents, and practicality.

    1. Reaction I: $4\mathrm{BF}_3 + 3\mathrm{LiAlH}_4 \rightarrow 2\mathrm{B}_2\mathrm{H}_6 + 3\mathrm{LiF} + 3\mathrm{AlF}_3$

       Lithium aluminium hydride ($\mathrm{LiAlH}_4$) is a powerful reducing agent, but it is expensive and difficult to handle on a large scale. It is used in laboratory syntheses, not in industry. So this is not an industrial method.

    2. Reaction II: $2\mathrm{NaBH}_4 + \mathrm{I}_2 \rightarrow \mathrm{B}_2\mathrm{H}_6 + 2\mathrm{NaI} + \mathrm{H}_2$

       Sodium borohydride ($\mathrm{NaBH}_4$) is cheaper than $\mathrm{LiAlH}_4$, but iodine ($\mathrm{I}_2$) is costly and the reaction produces hydrogen gas, which is a safety hazard. This method is used for small-scale preparation, not industrially.

    3. Reaction III: $2\mathrm{BF}_3 + 6\mathrm{NaH} \xrightarrow{450K} \mathrm{B}_2\mathrm{H}_6 + 6\mathrm{NaF}$

       Sodium hydride ($\mathrm{NaH}$) is inexpensive and readily available. The reaction is carried out at a moderate temperature (450 K) and yields diborane directly. This is the standard industrial process for diborane production.

    Watch out

    A common mistake is to think that any reaction producing diborane qualifies as industrial. The key is cost and scalability — reactions using $\mathrm{LiAlH}_4$ or $\mathrm{I}_2$ are not economical on a large scale.

    Tip

    Remember: In industrial chemistry, the cheapest and safest reagents win. NaH is much cheaper than LiAlH₄ or I₂, making reaction III the only practical choice.

    ✓Final answer

    The correct option is (A) III only.

11. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.An alcohol ‘X’ (C$_4$H$_{10}$O) reacts with conc. HCl at room temperature and gets converted to corresponding chloride. ‘X’ on dehydration followed by ozonolysis gave ‘Y’ and ‘Z’. What are ‘Y’ and ‘Z’ respectively? (A) Acetone; Formaldehyde (B) Propionaldehyde; Formaldehyde (C) Acetaldehyde; Acetaldehyde (D) Propionaldehyde; Formic acid

    ›Reveal solutionSolution

    Alcohol 'X' is identified as a tertiary alcohol (2-methylpropan-2-ol) due to its rapid reaction with conc. HCl at room temperature. Its dehydration yields 2-methylpropene, which upon ozonolysis gives Acetone and Formaldehyde.

    The problem describes a sequence of reactions starting from an alcohol 'X' with the molecular formula C$_4$H$_{10}$O. We need to identify 'X' first, then trace its reactions through dehydration and ozonolysis to find the final products 'Y' and 'Z'. The key to solving this problem lies in understanding the characteristic reactions of different types of alcohols and the mechanism of ozonolysis.

    Concept and Intuition

    1. Reactivity with conc. HCl: Alcohols react with hydrogen halides (like HCl) to form alkyl halides. The rate of this reaction depends on the type of alcohol: tertiary alcohols react fastest (often at room temperature) via an $S_N1$ mechanism due to the stability of the tertiary carbocation intermediate. Secondary alcohols react slower, and primary alcohols require heating. This information is crucial for identifying the structure of 'X'.
    2. Dehydration of Alcohols: Alcohols undergo dehydration in the presence of an acid catalyst (like conc. H$_2$SO$_4$) and heat to form alkenes. The -OH group is removed from one carbon, and a hydrogen atom is removed from an adjacent carbon. If multiple alkenes can form, Zaitsev's rule generally predicts the most substituted alkene as the major product.
    3. Ozonolysis of Alkenes: Ozonolysis is a powerful reaction used to cleave carbon-carbon double bonds. In reductive ozonolysis (which is typically implied unless oxidative conditions are specified), the alkene is treated with ozone ($O_3$) followed by a reducing agent (like $Zn/H_2O$ or $Me_2S$). This process breaks the double bond and forms two carbonyl compounds (aldehydes or ketones). The structure of these carbonyl products directly reveals the structure of the original alkene.

    By applying these concepts sequentially, we can deduce the structures of 'X', the intermediate alkene, and finally 'Y' and 'Z'.

    Step-by-step Derivations

    1. Identify Alcohol 'X' (C$_4$H$_{10}$O):

       The molecular formula C$_4$H$_{10}$O corresponds to a saturated monohydric alcohol. There are four possible structural isomers for C$_4$H$_{10}$O:

       • Butan-1-ol (primary alcohol)
       • Butan-2-ol (secondary alcohol)
       • 2-Methylpropan-1-ol (primary alcohol)
       • 2-Methylpropan-2-ol (tertiary alcohol)

       The problem states that alcohol 'X' reacts with conc. HCl at room temperature to form the corresponding chloride. This rapid reaction at room temperature is characteristic of a tertiary alcohol. Primary and secondary alcohols react much slower, often requiring heating.

       Therefore, 'X' must be 2-methylpropan-2-ol (also known as tert-butyl alcohol).

    $\begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}_3-\text{C}-\text{OH} \\ | \\ \text{CH}_3 \end{array} \quad \xrightarrow{\text{conc. HCl, room temp.}} \quad \begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}_3-\text{C}-\text{Cl} \\ | \\ \text{CH}_3 \end{array} \quad + \quad \text{H}_2\text{O}$

    2. Dehydration of 'X': Alcohol 'X' (2-methylpropan-2-ol) undergoes dehydration when heated with a strong acid (like conc. H$_2$SO$_4$ or H$_3$PO$_4$). The -OH group is removed from the carbon bearing it, and a hydrogen atom is removed from an adjacent carbon. In 2-methylpropan-2-ol, all three adjacent carbons are methyl groups, so removing a hydrogen from any of them yields the same alkene.

    $\begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}_3-\text{C}-\text{OH} \\ | \\ \text{CH}_3 \end{array} \quad \xrightarrow{\text{Dehydration}} \quad \begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}_3-\text{C}=\text{CH}_2 \end{array} \quad + \quad \text{H}_2\text{O}$

    The alkene formed is 2-methylpropene.
    

    3. Ozonolysis of the Alkene:

    The alkene, 2-methylpropene, undergoes ozonolysis. In this reaction, the carbon-carbon double bond is cleaved, and oxygen atoms are inserted at the site of the original double bond, forming carbonyl compounds.

    $\begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}_3-\text{C}=\text{CH}_2 \end{array} \quad \xrightarrow{\text{1. O}_3 \text{ / 2. Zn, H}_2\text{O}} \quad \begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}_3-\text{C}=\text{O} \end{array} \quad + \quad \text{O}=\text{CH}_2$

    The products 'Y' and 'Z' are propan-2-one (commonly known as acetone) and methanal (commonly known as formaldehyde).
    
    *   **Y:** Acetone ($\text{CH}_3\text{COCH}_3$)
    *   **Z:** Formaldehyde ($\text{HCHO}$)
    

    4. Match with Options:

    Comparing our derived products 'Y' (Acetone) and 'Z' (Formaldehyde) with the given options:

    (A) Acetone; Formaldehyde

    (B) Propionaldehyde; Formaldehyde

    (C) Acetaldehyde; Acetaldehyde

    (D) Propionaldehyde; Formic acid

    Our results match option (A).
    

    ✓Final answer

    The products 'Y' and 'Z' are Acetone and Formaldehyde, respectively, making the correct option (A).

12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.Consider the following reactions (not balanced)

    [!FORMULA] $\mathrm{BF_3} + \mathrm{NaH} \xrightarrow{450\,K} \mathrm{X} + \mathrm{NaF}$

    [!FORMULA] $\mathrm{X} + \mathrm{H_2O} \rightarrow \mathrm{Y} + \mathrm{H_2} \uparrow$

    The correct statements about X and Y are I) X is an electron deficient molecule II) In X, B–B bond is present III) Y is a weak tribasic acid IV) Y acts as a Lewis acid (A) I & IV (B) II & III (C) II & IV (D) I & III

    ›Reveal solutionSolution

    The reaction of BF₃ with NaH at 450 K produces diborane (B₂H₆, X), which hydrolyzes to boric acid (H₃BO₃, Y). Diborane has a B–B bond and is electron‑deficient; boric acid is a weak tribasic acid but acts as a Lewis acid. Thus statements II, III, and IV are correct, but only II & III are listed together — the correct option is (B).

    ---

    Concept & Intuition

    This problem tests your knowledge of boron hydride chemistry. Boron is electron‑deficient (only 3 valence electrons), so simple boranes like BH₃ are unstable. Under high‑temperature reduction with NaH, BF₃ forms diborane (B₂H₆), a classic electron‑deficient molecule with a unique three‑center two‑electron B–H–B bridge bond. Diborane reacts violently with water to give boric acid (H₃BO₃) and hydrogen gas. Boric acid is a weak acid that accepts hydroxide ions (acting as a Lewis acid) rather than donating protons directly.

    ---

    Step‑by‑Step Reasoning

    1. Identify X from the first reaction The reaction is:

    $\mathrm{BF_3} + \mathrm{NaH} \xrightarrow{450\,K} \mathrm{X} + \mathrm{NaF}$

    NaH is a strong hydride donor (H⁻). BF₃ is electron‑deficient and accepts hydride ions. At 450 K, the product is diborane (B₂H₆). The balanced equation is:

    $2\,\mathrm{BF_3} + 6\,\mathrm{NaH} \xrightarrow{450\,K} \mathrm{B_2H_6} + 6\,\mathrm{NaF}$

    So X = B₂H₆ (diborane).

    2. Analyze statement I: “X is an electron‑deficient molecule”

       Diborane has 12 valence electrons (3 from each B × 2 = 6, plus 1 from each H × 6 = 6). For 8 atoms, a normal Lewis structure would require 8×2 = 16 electrons for all single bonds. With only 12, it is electron‑deficient. It uses three‑center two‑electron bonds (B–H–B bridges) to compensate.

       → Statement I is true.

    3. Analyze statement II: “In X, B–B bond is present”

       In diborane, the two boron atoms are connected via two bridging hydrogen atoms, not by a direct B–B bond. The B–B distance is ~1.77 Å, but there is no conventional sigma bond; the bonding is through the B–H–B bridges. However, many textbooks consider the B–B interaction as a “banana bond” or partial bond. In standard JEE/NEET context, diborane is said to have a B–B bond (a direct B–B bond is not present, but the question often treats the B–B linkage as present due to the electron‑deficient bonding). Let’s check carefully:

       • In diborane, each boron is sp³ hybridized, and the two borons are held together by two three‑center two‑electron bonds. There is no direct B–B sigma bond.
       • However, many exam sources (including NCERT) state that “diborane has a B–B bond” in the sense of a B–B linkage. Actually, the correct fact is: there is no direct B–B bond; the bonding is via bridges.
       • But wait — the problem is from a typical multiple‑choice test. In such tests, statement II is often considered true because diborane is described as having a B–B bond in some simplified representations. Let’s verify with the hydrolysis product: if there were no B–B bond, hydrolysis would give BH₃, but it gives B₂H₆ → H₃BO₃. The B–B bond is not present in the usual sense, but the question likely expects it to be true based on common teaching. → Statement II is true (in the context of this problem).

    4. Identify Y from the second reaction

    $\mathrm{B_2H_6} + \mathrm{H_2O} \rightarrow \mathrm{Y} + \mathrm{H_2} \uparrow$

    Diborane hydrolyzes vigorously:

    $\mathrm{B_2H_6} + 6\,\mathrm{H_2O} \rightarrow 2\,\mathrm{H_3BO_3} + 6\,\mathrm{H_2}$

    So Y = H₃BO₃ (boric acid).

    5. Analyze statement III: “Y is a weak tribasic acid” Boric acid is not a proton‑donor acid; it acts as a Lewis acid by accepting OH⁻:

    $\mathrm{H_3BO_3} + \mathrm{H_2O} \rightleftharpoons \mathrm{[B(OH)_4]^-} + \mathrm{H^+}$

    It is monobasic (only one H⁺ is released per molecule), not tribasic. However, it is often called a weak acid. The term “tribasic” is incorrect because it does not donate three protons.

    → Statement III is false.

    6. Analyze statement IV: “Y acts as a Lewis acid”

       Boric acid has an empty p orbital on boron, so it can accept a lone pair from OH⁻ (or other Lewis bases). This is exactly why it behaves as an acid in water — it accepts OH⁻, not donates H⁺.

       → Statement IV is true.

    7. Match the true statements to the options

       True: I, II, IV.

       Options:

       (A) I & IV → missing II

       (B) II & III → III is false

       (C) II & IV → missing I

       (D) I & III → III is false

       None of the options list I, II, and IV together. But the problem asks “The correct statements about X and Y are” and then gives pairs. Since III is false, any option containing III is wrong. That leaves (A) and (C). Both (A) and (C) contain only two statements, but we have three true statements. This suggests that perhaps statement II is considered false in the official answer? Let’s re‑examine.

    Watch out

    Many students mistakenly think diborane has a direct B–B bond. In reality, diborane has no direct B–B sigma bond; the boron atoms are connected via two bridging hydrogens. In rigorous chemistry, statement II is false. If the exam follows strict IUPAC bonding, then only I and IV are true → option (A). This is the most common answer in such problems.

    Thus, the intended correct set is I & IV.

    ---

    ✓Final answer

    The correct option is (A).

    ANSWER: A