Q.896 mL vapour of a hydrocarbon 'A' having carbon 87.80% and hydrogen 12.19% weighs 3.28g at STP. Hydrogenation of 'A' gives 2-methylpentane. Also 'A' on hydration in the presence of H2SO4 and HgSO4 gives a ketone 'B' having molecular formula C6H12O. The ketone 'B' gives a positive iodoform test. Find the structure of 'A' and give the reactions involved.

Showing the 12 most recent of 14 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.The incorrect statement regarding 'C' formed in the given sequence of reactions is

   [!FORMULA] $\ce{CH2O} \xrightarrow{\text{(i) CH3MgBr}} \xrightarrow{\text{(ii) H2O}} \text{A} \xrightarrow{\text{PCC}} \text{B} \xrightarrow{\text{dil.NaOH}} \text{C}$

   (A) It is $\beta$-Hydroxybutyraldehyde (B) It has no asymmetric carbons (C) The IUPAC name of it is 3-Hydroxybutanal (D) This on heating gives $\alpha,\beta$-unsaturated aldehyde

   ›Reveal solutionSolution

   The sequence converts formaldehyde into 3-hydroxybutanal (aldol product), which has one chiral carbon, so the statement claiming it has no asymmetric carbons is false; the incorrect statement is (B).

   Concept & Intuition

   This problem tests your ability to trace a multi-step organic reaction sequence and spot a structural claim that contradicts the actual product. The key is to recognize each transformation:

   • Grignard addition to formaldehyde gives a primary alcohol.
   • PCC oxidation turns that alcohol into an aldehyde.
   • Dilute NaOH then triggers an aldol condensation between two identical aldehyde molecules, yielding a β-hydroxy aldehyde. Once you draw the final product, you can evaluate each statement about its structure, chirality, and reactivity.

   Step-by-step reasoning

   1. First step: Grignard reaction with formaldehyde Formaldehyde ($\ce{H2C=O}$) reacts with methylmagnesium bromide ($\ce{CH3MgBr}$). The Grignard reagent attacks the carbonyl carbon, and after aqueous workup, you get a primary alcohol:

   $\ce{CH2O + CH3MgBr ->[H2O] CH3CH2OH}$

   So A is ethanol ($\ce{CH3CH2OH}$).

   2. Second step: PCC oxidation Pyridinium chlorochromate (PCC) oxidizes primary alcohols to aldehydes without overoxidation to carboxylic acids.

   $\ce{CH3CH2OH ->[PCC] CH3CHO}$

   Thus B is acetaldehyde ($\ce{CH3CHO}$).

   3. Third step: Aldol condensation with dilute NaOH Dilute NaOH catalyzes an aldol reaction between two molecules of acetaldehyde. One molecule forms an enolate, which attacks the carbonyl of another, giving a β-hydroxy aldehyde:

   $\ce{2 CH3CHO ->[dil.NaOH] CH3CH(OH)CH2CHO}$

   This product is 3-hydroxybutanal (common name: β-hydroxybutyraldehyde). So C is $\ce{CH3CH(OH)CH2CHO}$.

   4. Evaluate each statement about C
      • (A) It is β-Hydroxybutyraldehyde — True. The hydroxyl is on the β-carbon relative to the aldehyde group.
      • (B) It has no asymmetric carbons — False. The carbon bearing the –OH group (C3) is attached to four different groups: H, OH, CH3, and CH2CHO. This is a chiral (asymmetric) carbon. …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.Which of the following does not undergo disproportionation reaction in the presence of Conc. NaOH? (A) HCHO (B) C_6H_5CHO (C) (CH_3)_3CCHO (D) CH_3CH_2CHO

   ›Reveal solutionSolution

   Disproportionation (Cannizzaro reaction) requires an aldehyde with no $\alpha$-hydrogen. Among the given options, only $(CH_3)_3CCHO$ has no $\alpha$-hydrogen, so it does undergo the reaction — meaning the one that does not undergo disproportionation is the one that has $\alpha$-hydrogens. That is CH$_3$CH$_2$CHO.

   The question is about the Cannizzaro reaction — a disproportionation of aldehydes in concentrated base. In this reaction, one molecule of aldehyde is reduced to an alcohol and another is oxidized to a carboxylic acid (which appears as its salt in strong base). The key condition: the aldehyde must have no $\alpha$-hydrogen atoms (i.e., no hydrogen on the carbon next to the $-\text{CHO}$ group). Why? Because if $\alpha$-hydrogens are present, the base will instead deprotonate that $\alpha$-carbon, leading to an enolate, and the reaction takes a different path (aldol condensation). So the Cannizzaro reaction is the only disproportionation path available to aldehydes that cannot form enolates.

   Let’s examine each option.

   1. HCHO (formaldehyde) — The carbon attached to the $-\text{CHO}$ group is simply a hydrogen atom. There is no $\alpha$-carbon at all, so certainly no $\alpha$-hydrogen. Formaldehyde readily undergoes the Cannizzaro reaction with conc. NaOH to give methanol and sodium formate. So it does undergo disproportionation.

   2. C$_6$H$_5$CHO (benzaldehyde) — The $\alpha$-carbon is the one directly attached to the $-\text{CHO}$ group; here that carbon is part of the aromatic ring. Aromatic ring carbons do not have $\alpha$-hydrogens (the hydrogen is on the ring, not on the $\alpha$-carbon). So benzaldehyde has no $\alpha$-hydrogen and undergoes the Cannizzaro reaction to give benzyl alcohol and sodium benzoate. It does undergo disproportionation.

   3. (CH$_3$)$_3$CCHO (pivalaldehyde, trimethylacetaldehyde) — The $\alpha$-carbon is the quaternary carbon bearing three methyl groups. That carbon has no hydrogen atoms at all — it is fully substituted. So there are no $\alpha$-hydrogens. Pivalaldehyde undergoes the Cannizzaro reaction to give neopentyl alcohol and the salt of pivalic acid. It does undergo disproportionation. …

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3. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.What are X and Y in the following reaction sequence? $\mathrm{C_5H_{12}O} \xrightarrow{\text{Cu, 573 K}} \mathrm{C_5H_{10}} \xrightarrow{\text{(i) O}_3 \text{

   (ii) Zn + H}_2\mathrm{O}} X + Y$ (A) $\begin{array}{c} \mathrm{O} \\ \vert \\ \mathrm{CH_3CH_2CH_2CCH_3} \end{array} + \mathrm{CH_3CHO}$ (B) $\begin{array}{c} \mathrm{CH_3CH_2CH_2CHO} \end{array} + \mathrm{HCHO}$ (C) $\begin{array}{c} \mathrm{CH_3CH_2CO} \\ \vert \\ \mathrm{H} \end{array} + \mathrm{CH_3CHO}$ (D) $\begin{array}{c} \mathrm{O} \\ \vert \\ \mathrm{CH_3CH_2CH_2CCH_3} \end{array} + \mathrm{HCO_2H}$

   ›Reveal solutionSolution

   The starting alcohol is 3‑methylbutan‑2‑ol; dehydration gives 3‑methylbut‑1‑ene, which upon ozonolysis yields acetone and acetaldehyde — matching option (A).

   Concept & Intuition

   We are given a three‑step sequence: an alcohol C₅H₁₂O → alkene C₅H₁₀ → ozonolysis products X + Y.

   • The first step (Cu, 573 K) is dehydrogenation of a secondary alcohol to a ketone, but here the product is an alkene (C₅H₁₀), not a ketone. That means the alcohol must be tertiary (cannot be dehydrogenated) or the conditions cause dehydration instead. In fact, hot copper can catalyse dehydration of alcohols to alkenes, especially for tertiary alcohols.
   • The second step is ozonolysis (O₃ then Zn/H₂O), which cleaves the alkene double bond to give carbonyl compounds (aldehydes and/or ketones).
   • We need to deduce the structure of the original alcohol so that the alkene formed gives the observed fragments.

   Step‑by‑step reasoning

   1. Identify the alcohol’s skeleton from the alkene formula

      The alkene is C₅H₁₀. The alcohol C₅H₁₂O must have the same carbon skeleton. Possible skeletons for C₅ alkenes:

      • pent‑1‑ene
      • pent‑2‑ene
      • 2‑methylbut‑1‑ene
      • 2‑methylbut‑2‑ene
      • 3‑methylbut‑1‑ene The alcohol that dehydrates to a given alkene must have the –OH on a carbon adjacent to the double bond.

   2. Work backwards from the ozonolysis products

      Ozonolysis of an alkene R₁R₂C=CR₃R₄ gives two carbonyl compounds:

   $\text{R}_1\text{R}_2\text{C=O} \quad\text{and}\quad \text{R}_3\text{R}_4\text{C=O}$

   Look at the options:

   • (A) gives acetone (CH₃COCH₃) + acetaldehyde (CH₃CHO)
   • (B) gives butanal (CH₃CH₂CH₂CHO) + formaldehyde (HCHO)
   • (C) gives propanal (CH₃CH₂CHO) + acetaldehyde
   • (D) gives acetone + formic acid (HCO₂H)

   Only (A) and (D) contain a ketone (acetone). Since ozonolysis with reductive work‑up (Zn/H₂O) gives aldehydes and ketones, not carboxylic acids, (D) is ruled out (formic acid would require oxidative work‑up). So the alkene must yield acetone + acetaldehyde.

   3. Deduce the alkene structure Acetone = (CH₃)₂C=O, acetaldehyde = CH₃CHO. The alkene double bond must be between the carbons that become the carbonyl groups.
      • The carbon that becomes the carbonyl of acetone is a disubstituted carbon (two methyls).
      • The carbon that becomes the aldehyde is a monosubstituted carbon (one H, one CH₃). Hence the alkene is:

   $(\text{CH}_3)_2\text{C}=\text{CHCH}_3$

   That is 2‑methylbut‑2‑ene.

   4. Find the alcohol that dehydrates to 2‑methylbut‑2‑ene Dehydration of an alcohol removes –OH and an adjacent H to form the double bond. …

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4. TG EAPCET 2025Set ap-2025-04-30-FN1 markMCQ
   Q.Alkene ($\mathrm{C_3H_6}$) reacts with X to give $\mathrm{C_3H_8O}$, which reacts with Y to give Z. Z forms 2,4-dinitrophenylhydrazone and gives yellow ppt with NaOH|$I_2$. What are X and Y? (A) $\mathrm{BH_3, H_2O_2/OH^-; CrO_3}$ (B) $\mathrm{BH_3, H_2O_2/OH^-; PCC}$ (C) $\mathrm{H_2O/H^+; CrO_3}$ (D) $\mathrm{H_2O/H^+; KMnO_4|OH^-, \Delta}$

   ›Reveal solutionSolution

   The alkene propene undergoes hydration to give an alcohol, which is then oxidised to a methyl ketone — the key is that the final product gives a positive iodoform test. The correct reagents are $\mathrm{H_2O/H^+}$ followed by $\mathrm{CrO_3}$, which is option (C).

   The problem gives you a sequence: an alkene $\mathrm{C_3H_6}$ (propene) reacts with X to give $\mathrm{C_3H_8O}$ (an alcohol), which then reacts with Y to give Z. Z forms a 2,4-dinitrophenylhydrazone (so it's a carbonyl compound — aldehyde or ketone) and gives a yellow precipitate with NaOH/$\mathrm{I_2}$ (the iodoform test). The iodoform test is positive only for methyl ketones ($\mathrm{R-CO-CH_3}$) or ethanol/ secondary alcohols that can be oxidised to such ketones. Since Z is already a carbonyl, it must be a methyl ketone.

   The alkene is propene, $\mathrm{CH_3-CH=CH_2}$. The first step (X) must add the elements of water across the double bond to give an alcohol $\mathrm{C_3H_8O}$. There are two possible alcohols: propan-1-ol ($\mathrm{CH_3CH_2CH_2OH}$) and propan-2-ol ($\mathrm{CH_3CH(OH)CH_3}$). Only propan-2-ol can be oxidised to a methyl ketone (propanone, $\mathrm{CH_3COCH_3}$), which gives the iodoform test. So X must be a reagent that gives Markovnikov addition of water — that is, $\mathrm{H_2O/H^+}$ (acid-catalysed hydration). The alternative, hydroboration-oxidation ($\mathrm{BH_3, H_2O_2/OH^-}$), gives anti-Markovnikov addition, producing propan-1-ol, which would oxidise to an aldehyde (propanal) — and propanal does NOT give the iodoform test. …

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5. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.A hydrocarbon has 85.7% of carbon by weight. 56 g (2 moles) of hydrocarbon was completely burnt in oxygen and obtained CO$_2$ and H$_2$O. What is the weight (in g) of CO$_2$ formed? (C = 12 u, H = 1 u, O = 16 u) (A) 176 (B) 88 (C) 132 (D) 352

   ›Reveal solutionSolution

   The hydrocarbon has molar mass $28\text{ g/mol}$ with $2$ carbons per molecule (C$_2$H$_4$); $2$ mol gives $4$ mol CO$_2$ $= 176\text{ g}$.

   Molar mass of the hydrocarbon $= \dfrac{56\text{ g}}{2\text{ mol}} = 28\text{ g/mol}$.

   Mass of carbon per mole $= 85.7\% \times 28 = 24\text{ g}$, so number of C atoms $= \dfrac{24}{12} = 2$.

   Hydrogen per mole $= 28 - 24 = 4\text{ g} \Rightarrow 4$ H atoms. The hydrocarbon is $\text{C}_2\text{H}_4$. …

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6. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.An alkene X ($\mathrm{C_4H_8}$) exhibits geometrical isomerism. Oxidation of A with $\mathrm{KMnO_4}$ $|$ $\mathrm{H^+}$ gave Y. On heating sodium salt of Y with a mixture of NaOH and CaO gave Z. What is Z? (A) $\mathrm{CH_3CH_3}$ (B) $\mathrm{CH_3CH_2CH_3}$ (C) $\mathrm{CH_3CH_2CH_2CH_3}$ (D) $\mathrm{CH_4}$

   ›Reveal solutionSolution

   The alkene X is but-2-ene (the only C₄H�8 alkene that shows geometrical isomerism). Its oxidative cleavage gives two molecules of acetic acid (Y). The sodium salt of Y undergoes decarboxylation to yield methane (Z). So the correct option is (D).

   Concept & Intuition

   The problem tests two key reactions:

   1. Oxidative cleavage of alkenes with hot acidic KMnO₄ – this breaks the double bond and turns each carbon of the double bond into a carboxylic acid (or ketone if the carbon is disubstituted).
   2. Decarboxylation of sodium carboxylates (the soda-lime reaction) – heating the sodium salt of a carboxylic acid with NaOH/CaO removes CO₂ and leaves an alkane with one fewer carbon.

   We need to identify which C₄H₈ alkene shows geometrical isomerism, then follow the reaction sequence.

   Step-by-step reasoning

   1. Identify the alkene X

      The molecular formula C₄H₈ corresponds to several alkenes: but-1-ene, but-2-ene, and 2-methylpropene.

      Geometrical isomerism (cis/trans) requires that each carbon of the double bond has two different substituents.

      • But-1-ene (CH₂=CH–CH₂–CH₃): one carbon has two H’s → no geometrical isomers.
      • 2-Methylpropene (CH₂=C(CH₃)₂): one carbon has two H’s, the other has two CH₃’s → no geometrical isomers.
      • But-2-ene (CH₃–CH=CH–CH₃): each double-bond carbon has one H and one CH₃ → can exist as cis and trans. Hence X must be but-2-ene.

   2. Oxidation of X with KMnO₄ / H⁺

      Hot acidic KMnO₄ cleaves the double bond completely. For but-2-ene:

   $\text{CH}_3\text{–CH=CH–CH}_3 \xrightarrow{\text{KMnO}_4, \text{H}^+, \Delta} 2\,\text{CH}_3\text{COOH}$

   Each half becomes a carboxylic acid. So Y is acetic acid (CH₃COOH).

   3. Formation of the sodium salt

      The sodium salt of Y is sodium acetate: CH₃COONa.

   4. Decarboxylation with soda lime …

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7. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.A hydrocarbon containing C and H has 92.3% of C. When 52 g of hydrocarbon is completely burnt in oxygen, x moles of water and y moles of CO$_2$ were formed. The liberated water is sufficient to liberate one mole of H$_2$ when reacted with sodium metal. What is the weight (in g) of O$_2$ consumed? (C = 12u; H = 1u; O = 16u) (A) 80 (B) 160 (C) 240 (D) 320

   ›Reveal solutionSolution

   The key is to first find the empirical formula from the carbon percentage, then use the water-liberation clue to determine the molecular formula and the moles of hydrocarbon burned. From the combustion reaction, we calculate the moles of O₂ consumed and convert to grams, yielding 160 g.

   Concept and Intuition

   We have a hydrocarbon (only C and H) with a known mass percentage of carbon. That lets us find the simplest whole-number ratio of C to H — the empirical formula. But we need the actual molecular formula to know how much oxygen is consumed per mole of hydrocarbon. The extra clue: the water produced from burning 52 g of the hydrocarbon is enough to liberate exactly 1 mole of H₂ when reacted with sodium. Sodium reacts with water:

   $2\text{Na} + 2\text{H}_2\text{O} \rightarrow 2\text{NaOH} + \text{H}_2$

   So 2 moles of water give 1 mole of H₂. Therefore, the water produced must be 2 moles. That tells us exactly how many moles of hydrocarbon were burned, and from there we can deduce the molecular formula and the oxygen consumed.

   ---

   Step-by-step solution

   1. Find the empirical formula from the percentage composition

      • 92.3% C means 92.3 g C per 100 g hydrocarbon, and the rest is H: 7.7 g H.
      • Moles of C = $\frac{92.3}{12} \approx 7.692$
      • Moles of H = $\frac{7.7}{1} = 7.7$
      • Ratio C : H = $7.692 : 7.7 \approx 1 : 1$ (very close, tiny rounding).
      • So the empirical formula is CH.

   2. Use the water clue to find the molecular formula

      • From the reaction with sodium: 2 H₂O → 1 H₂.
      • Liberating 1 mole of H₂ requires 2 moles of H₂O.
      • Therefore, burning 52 g of hydrocarbon produces 2 moles of water.
      • In a combustion reaction:

   $\text{C}_n\text{H}_m + \left(n + \frac{m}{4}\right) \text{O}_2 \rightarrow n \text{CO}_2 + \frac{m}{2} \text{H}_2\text{O}$

    So moles of H₂O = $ \frac{m}{2} \times $ (moles of hydrocarbon).  
   

   • Let the molecular formula be (CH)ₖ, so m = k.
   • Then moles of H₂O = $\frac{k}{2} \times$ (moles of hydrocarbon) = 2.
   • Also, the mass of hydrocarbon burned is 52 g. Molar mass of (CH)ₖ = (12+1)k = 13k g/mol.
   • Moles of hydrocarbon = $\frac{52}{13k} = \frac{4}{k}$.
   • Plug into water equation:

   $\frac{k}{2} \times \frac{4}{k} = 2$

    This simplifies to $ 2 = 2 $, which is always true — so any k works? That can’t be right. Wait — we need to check: the water produced is exactly 2 moles, but that doesn’t fix k uniquely unless we also know the CO₂ produced. Let’s use the CO₂ clue.
   

   3. Use the CO₂ produced to find k

   • From combustion, moles of CO₂ = $n \times$ (moles of hydrocarbon) = $k \times \frac{4}{k} = 4$ moles.
   • So y = 4 moles of CO₂.
   • Now we have: from 52 g of hydrocarbon, we get 4 moles CO₂ and 2 moles H₂O. …

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8. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.1-Propanol can be distinguished from 2-propanol by which test? (A) 2,4-DNP test (B) Tollens' test (C) Lucas test (D) Fehling's test

   ›Reveal solutionSolution

   The Lucas test distinguishes 1‑propanol from 2‑propanol because 2‑propanol (a secondary alcohol) reacts quickly with Lucas reagent to form a cloudy alkyl chloride, while 1‑propanol (a primary alcohol) does not react at room temperature. The correct option is (C).

   The key concept here is the relative reactivity of alcohols toward nucleophilic substitution — specifically, the ease with which an alcohol can be converted to an alkyl halide. The Lucas test (ZnCl₂ in concentrated HCl) exploits the fact that tertiary alcohols react instantly, secondary alcohols react within minutes (forming a cloudy emulsion), and primary alcohols do not react appreciably at room temperature. Since 1‑propanol is primary and 2‑propanol is secondary, only 2‑propanol gives a positive Lucas test under standard conditions.

   Let’s walk through the reasoning step by step:

   1. Understand what each test detects

      • 2,4‑DNP test (2,4‑dinitrophenylhydrazine) reacts with carbonyl groups (aldehydes and ketones) to form a yellow/orange precipitate. Neither 1‑propanol nor 2‑propanol has a carbonyl group, so both would give a negative result.
      • Tollens’ test (ammoniacal silver nitrate) oxidizes aldehydes to carboxylic acids, depositing a silver mirror. Alcohols are not aldehydes, so neither propanol reacts.
      • Fehling’s test (copper(II) complex in alkaline solution) also detects aldehydes (and some α‑hydroxy ketones). Again, no reaction with simple alcohols.
      • Lucas test specifically distinguishes primary, secondary, and tertiary alcohols based on the rate of formation of an alkyl chloride.

   2. Focus on the Lucas test mechanism

      The Lucas reagent (ZnCl₂ in concentrated HCl) protonates the alcohol’s –OH group, turning it into a better leaving group (water). A carbocation intermediate then forms. The stability of this carbocation determines the reaction rate:

      • Tertiary carbocations are very stable → immediate reaction.
      • Secondary carbocations are moderately stable → reaction in 5–10 minutes (cloudiness appears).
      • Primary carbocations are highly unstable → no reaction at room temperature.

   3. Apply to the two propanols

      • 1‑Propanol (CH₃CH₂CH₂OH): If it lost water, it would form a primary carbocation (CH₃CH₂CH₂⁺), which is too unstable to form under these mild conditions. Hence, no visible change. …

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9. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.Identify A and B from the following reactions I) $\mathrm{CH_3-CH=CH-CH_3} \xrightarrow{\text{(i) O}_3} \xrightarrow{\text{(ii) Zn-H}_2\text{O}} 2X$ II) $2X \xrightarrow{\text{1. NaOH(dil)}} \xrightarrow{\text{2. A}} Z \xrightarrow{\text{(i) O}_3} \xrightarrow{\text{(ii) Zn-H}_2\text{O}} A+B$ (A) $\mathrm{CH_3-CHO+HCHO}$ (B) $\begin{array}{c} \mathrm{CH_3CHO+} \\ \quad \mathrm{CHO} \\ \quad \mathrm{CHO} \end{array}$ (C) $\mathrm{CH_3-CH_2-CHO+CH_2O}$ (D) $\begin{array}{c} \mathrm{CH_3-\overset{\displaystyle O}{\displaystyle \vert}C-CH_3+CHO} \\ \qquad\qquad \mathrm{CHO} \end{array}$

   ›Reveal solutionSolution

   Ozonolysis of but-2-ene gives $2$ molecules of acetaldehyde ($X$). Aldol condensation of $X$ then dehydration gives but-2-enal (crotonaldehyde), $Z$. Ozonolysis of $Z$ cleaves its double bond to give acetaldehyde and glyoxal, so $A+B=\text{CH}_3\text{CHO}+\text{OHC-CHO}$.

   Reaction I — identifying X. But-2-ene, $\text{CH}_3\text{-CH=CH-CH}_3$, on reductive ozonolysis (O$_3$; Zn/H$_2$O) cleaves the C=C bond to give two identical carbonyls:

   $\text{CH}_3\text{-CH=CH-CH}_3 \longrightarrow 2\,\text{CH}_3\text{CHO}.$

   So $X=\text{acetaldehyde }(\text{CH}_3\text{CHO})$.

   Reaction II — building Z. Two molecules of acetaldehyde undergo base-catalysed aldol addition with dilute NaOH to give $3$-hydroxybutanal, which on warming eliminates water (aldol condensation) to give the $\alpha,\beta$-unsaturated aldehyde but-2-enal (crotonaldehyde): …

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10. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.A solid has a structure in which ‘W’ atoms are located at the corners of a cubic lattice, oxygen atoms at the edge centre and Na atom at the body centre. The formula of the compound is (A) NaWO$_2$ (B) Na$_2$WO$_3$ (C) NaWO$_3$ (D) NaWO$_4$

    ›Reveal solutionSolution

    To find the formula of a compound from its crystal structure, we determine the effective number of each type of atom within a single unit cell by considering their positions and fractional contributions. For this compound, the effective number of Na, W, and O atoms are 1, 1, and 3 respectively, leading to the formula $\boxed{\text{NaWO}_3}$.

    When atoms arrange themselves in a crystal lattice, they form a repeating unit called a unit cell. The chemical formula of the compound is determined by the net number of each type of atom present within one such unit cell. Atoms located at different positions within the unit cell (like corners, edges, faces, or the body center) are shared by a different number of adjacent unit cells, and thus contribute only a fraction of themselves to any single unit cell.

    Here's how we account for these contributions in a cubic lattice:

    • An atom at a corner is shared by 8 unit cells, so its contribution to one unit cell is $1/8$.
    • An atom at an edge center is shared by 4 unit cells, so its contribution to one unit cell is $1/4$.
    • An atom at a face center is shared by 2 unit cells, so its contribution to one unit cell is $1/2$.
    • An atom at the body center is entirely within one unit cell, so its contribution is $1$.

    Let's apply this to find the effective number of each atom in the given compound:

    1. Calculate the effective number of 'W' atoms:

       'W' atoms are located at the corners of the cubic lattice. A cubic unit cell has 8 corners.

       Each corner atom contributes $1/8$ to the unit cell.

       Effective number of W atoms = (Number of corners) $\times$ (Contribution per corner atom)

       Effective number of W atoms = $8 \times \frac{1}{8} = 1$

    2. Calculate the effective number of 'O' atoms:

       Oxygen atoms are located at the edge centers. A cubic unit cell has 12 edges. …

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11. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
    Q.An iodo-organic compound, when fused with sodium metal, followed by acidification with nitric acid and then treated with silver nitrate gives (A) A white precipitate is formed, which is soluble in $\mathrm{NH_4OH}$ solution (B) A yellow precipitate is formed, which is insoluble in $\mathrm{NH_4OH}$ solution (C) A red precipitate is formed, which is insoluble in $\mathrm{NH_4OH}$ solution (D) A yellow precipitate is formed, which is soluble in $\mathrm{NH_4OH}$ solution

    ›Reveal solutionSolution

    The Lassaigne test for iodine produces a yellow precipitate of AgI that is insoluble in NH₄OH, so the correct choice is (B).

    The question tests the Lassaigne test (sodium fusion test) for halogens in organic compounds. When an organic compound containing iodine is fused with sodium, the iodine is converted to sodium iodide (NaI). After acidification with nitric acid (to destroy any cyanide or sulfide ions that might interfere), adding silver nitrate (AgNO₃) gives a silver halide precipitate. The key is identifying the color and solubility of silver iodide (AgI) in ammonium hydroxide.

    1. Fusion with sodium – The iodo-organic compound is heated with sodium metal. This breaks the C–I bond, forming NaI:

    $\text{R–I} + 2\text{Na} \rightarrow \text{NaI} + \text{R–Na}$

    2. Acidification with nitric acid – After dissolving the fused mass in water, dilute HNO₃ is added. This serves two purposes: it neutralizes any excess NaOH and decomposes any NaCN or Na₂S (from nitrogen or sulfur in the compound) so they don’t form false precipitates later.

    3. Addition of silver nitrate – AgNO₃ reacts with iodide ions:

    $\text{Ag}^+ + \text{I}^- \rightarrow \text{AgI} \downarrow$

    Silver iodide is pale yellow (often described simply as “yellow” in textbooks). …

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12. TG EAPCET 2021Set ap-2021-08-10-FN1 markMCQ
    Q.An iodo-organic compound, when fused with sodium metal, followed by acidification with nitric acid and then treated with silver nitrate gives (A) A white precipitate is formed, which is soluble in $\mathrm{NH_4OH}$ solution (B) A yellow precipitate is formed, which is insoluble in $\mathrm{NH_4OH}$ solution (C) A red precipitate is formed, which is insoluble in $\mathrm{NH_4OH}$ solution (D) A yellow precipitate is formed, which is soluble in $\mathrm{NH_4OH}$ solution

    ›Reveal solutionSolution

    The test detects iodine in an organic compound via sodium fusion, forming NaI; acidification with HNO₃ and addition of AgNO₃ yields AgI, a yellow precipitate insoluble in NH₄OH. The correct option is (B).

    Concept & Intuition

    When an organic compound containing iodine is fused with sodium metal, the iodine is converted into sodium iodide (NaI). This is the Lassaigne’s test for halogens. After acidification with nitric acid (to destroy any cyanide or sulfide ions that might interfere), silver nitrate is added. Silver iodide (AgI) is a pale yellow solid that is insoluble in aqueous ammonia (NH₄OH), unlike silver chloride (white, soluble) and silver bromide (pale yellow, sparingly soluble). This difference in solubility in ammonia is the key to identifying the halogen.

    Step-by-step reasoning

    1. Sodium fusion The iodo-organic compound is heated with sodium metal. The reaction converts covalently bonded iodine into ionic iodide:

    $\text{R–I} + 2\text{Na} \rightarrow \text{NaI} + \text{R–Na}$

    The sodium iodide dissolves in water when the fused mass is extracted.

    2. Acidification with nitric acid

       Nitric acid is added to the aqueous extract. This serves two purposes:

       • It neutralizes any leftover sodium hydroxide.
       • It decomposes sodium cyanide (if nitrogen was present) and sodium sulfide (if sulfur was present) as HCN and H₂S gases, preventing them from forming precipitates with silver nitrate later. The iodide ion remains unaffected.

    3. Addition of silver nitrate

       Silver nitrate reacts with iodide ions to form silver iodide:

    $\text{AgNO}_3 + \text{NaI} \rightarrow \text{AgI} \downarrow + \text{NaNO}_3$

    Silver iodide is a pale yellow precipitate.

    4. Solubility in ammonium hydroxide …

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