Q.Suggest a reagent for the following conversion. The starting material is a secondary allylic alcohol, pent-3-en-2-ol (CH3-CH(OH)-CH=CH-CH3), and the product is the corresponding alpha,beta-unsaturated ketone, pent-3-en-2-one (CH3-CO-CH=CH-CH3); the carbon-carbon double bond is retained unchanged and only the -CH(OH)- group is oxidised to a >C=O group.

Showing the 12 most recent of 39 on this concept.

1. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The compound which is not isomeric with ethoxyethane? (A) Propylmethylether (B) Butan-2-ol (C) Butanone (D) 2-Methylpropan-1-ol

   ›Reveal solutionSolution

   Isomers share the same molecular formula. Ethoxyethane (diethyl ether) is $\ce{C4H10O}$; butanone is $\ce{C4H8O}$, so it is not an isomer. The correct option is (C).

   Concept & Intuition

   Isomerism requires identical molecular formulas. Ethoxyethane (commonly called diethyl ether) has the structure $\ce{CH3CH2-O-CH2CH3}$, giving it the formula $\ce{C4H10O}$. Any compound that is isomeric with it must also be $\ce{C4H10O}$. The trick is to check each option’s molecular formula — if it differs in the number of hydrogens (or oxygens), it cannot be an isomer.

   Step-by-step reasoning

   1. Determine the molecular formula of ethoxyethane

      Ethoxyethane = two ethyl groups linked by an oxygen: $\ce{CH3CH2-O-CH2CH3}$.

      Count atoms: 4 carbons, 10 hydrogens, 1 oxygen → $\ce{C4H10O}$.

   2. Check each option’s formula

      • (A) Propylmethylether: $\ce{CH3CH2CH2-O-CH3}$ → also $\ce{C4H10O}$. Isomer.
      • (B) Butan-2-ol: $\ce{CH3CH(OH)CH2CH3}$ → $\ce{C4H10O}$. Isomer (alcohol isomer of ether).
      • (C) Butanone: $\ce{CH3COCH2CH3}$ → $\ce{C4H8O}$ (a ketone). This has two fewer hydrogens than $\ce{C4H10O}$, so it is not an isomer.
      • (D) 2-Methylpropan-1-ol: $\ce{(CH3)2CHCH2OH}$ → $\ce{C4H10O}$. Isomer.

   3. Identify the odd one out …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The IUPAC name of the end product Z is (CH$_3$)$_3$C–OH $\xrightarrow{\text{Cu} \ 573\text{K}}$ W $\xrightarrow{\text{(i) O}_3}$ X + Y $\xrightarrow{\text{(ii) Zn | H}_2\text{O}}$ $\xrightarrow{\text{dil. NaOH}}$ Z (A) 4-Hydroxybutan-2-one (B) 3-Hydroxybutan-2-one (C) 4-Hydroxybutanal (D) But-3-en-2-one

   ›Reveal solutionSolution

   The reaction sequence starts with dehydration of tert-butanol to isobutylene, ozonolysis gives acetone and formaldehyde, and an aldol condensation yields 4-hydroxybutan-2-one. The correct option is (A).

   Concept & Intuition

   This problem tests your ability to follow a multi-step organic reaction sequence, recognizing each transformation. The key is to identify the starting alcohol, predict the product of dehydration (since Cu at 573 K is a classic condition for alcohol dehydration to an alkene), then apply ozonolysis to cleave the alkene, and finally recognize that the last step (dilute NaOH) is an aldol condensation between the two carbonyl compounds formed. The pitfall is forgetting that the alkene from a tertiary alcohol is highly substituted, and that ozonolysis with reductive workup gives two carbonyl fragments.

   Step-by-step reasoning

   1. Identify the starting material and first step The starting compound is (CH₃)₃C–OH, tert-butanol. Heating it with copper at 573 K is a standard method for dehydrating alcohols to alkenes. For a tertiary alcohol, elimination occurs readily to give the most substituted alkene:

   $(CH_3)_3C-OH \xrightarrow{Cu,\,573\,K} (CH_3)_2C=CH_2 + H_2O$

   So W is isobutylene (2-methylpropene).

   2. Ozonolysis of the alkene Ozone (O₃) adds across the double bond, forming an ozonide. Reductive workup with Zn/H₂O cleaves the ozonide to give two carbonyl compounds. For an unsymmetrical alkene like isobutylene:

   $(CH_3)_2C=CH_2 \xrightarrow{(i)\,O_3,\, (ii)\,Zn/H_2O} (CH_3)_2C=O + HCHO$

   Thus X = acetone (propanone) and Y = formaldehyde (methanal).

   3. The final step: aldol condensation The mixture of X and Y is treated with dilute NaOH. This is a classic crossed aldol reaction. Formaldehyde has no α-hydrogens, so it acts only as the electrophile. Acetone has α-hydrogens and can form an enolate. The enolate of acetone attacks the carbonyl carbon of formaldehyde: $$$$

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Observe the following reaction C$_8$H$_8$O $\xrightarrow{\text{Y}}$ Z Aromatic compound X gives both iodoform and 2,4-DNP tests. Product Z liberates CO$_2$ with NaHCO$_3$ solution. Identify the set(s) in which Y is correctly represented from the following I. Zn – Hg | HCl; KMnO$_4$ | OH$^-$, H$_3$O$^+$ II. KMnO$_4$ | OH$^-$; H$_3$O$^+$ III. LiAlH$_4$, H$_2$O (A) I, II only (B) I, III only (C) II, III only (D) I only

   ›Reveal solutionSolution

   X ($\text{C}_8\text{H}_8\text{O}$, positive iodoform + 2,4-DNP) is acetophenone $\text{C}_6\text{H}_5\text{COCH}_3$; Z (liberates $\text{CO}_2$ with $\text{NaHCO}_3$) is benzoic acid. Routes I and II both give benzoic acid; III only reduces. Answer (A).

   Identify X. $\text{C}_8\text{H}_8\text{O}$ has 5 degrees of unsaturation (aromatic ring = 4, C=O = 1). A positive 2,4-DNP test means a carbonyl; a positive iodoform test means a $\text{CH}_3\text{CO}-$ group. The aromatic compound fitting both is acetophenone, $\text{C}_6\text{H}_5\text{COCH}_3$.

   Identify Z. Z liberates $\text{CO}_2$ with $\text{NaHCO}_3$, so Z is a carboxylic acid - here benzoic acid, $\text{C}_6\text{H}_5\text{COOH}$.

   Evaluate the reagent sets Y.

   • I. Zn-Hg/HCl; then $\text{KMnO}_4/\text{OH}^-,\ \text{H}_3\text{O}^+$. Clemmensen reduction converts $-\text{COCH}_3$ to $-\text{CH}_2\text{CH}_3$ (ethylbenzene); alkaline $\text{KMnO}_4$ then oxidises the whole side chain to $-\text{COOH}$, giving benzoic acid. Works. …

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4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Observe the following reaction sequence $\mathrm{C_6H_5\overset{+}{N_2}\ X^-} \xrightarrow{\ X\ } Y \xrightarrow{\ Z\ } \text{1,4-benzoquinone (cyclohexa-2,5-diene-1,4-dione)}$ From the given list of reagents, identify the correct sequence of X and Z for the above reaction sequence

   $\mathrm{H_2SO_4}$	$\mathrm{HNO_3}$	$\mathrm{C_2H_5OH}$	$\mathrm{Na_2Cr_2O_7/H^+}$	$\mathrm{H_2O/283K}$
   I	II	III	IV	V

   The correct answer is (A) V & II (B) III & IV (C) III & I (D) V & IV

   ›Reveal solutionSolution

   The diazonium salt is hydrolysed by water at 283 K (V) to phenol, and phenol is oxidised by $\mathrm{Na_2Cr_2O_7/H^+}$ (IV) to 1,4-benzoquinone. So X & Z = V & IV — option (D).

   The concept: read a sequence from both ends

   When a scheme has two unknown reagents and a named final product, the fastest route is retrosynthesis — ask what the product is characteristically made from.

   1,4-Benzoquinone ($\mathrm{cyclohexa}$-2,5-diene-1,4-dione) has carbonyls at the 1- and 4-positions and is not aromatic. It is the classic oxidation product of phenol (or of hydroquinone/aniline). That single fact fixes the intermediate: $\mathrm{Y} =$ phenol, and $\mathrm{Z}$ must be a strong oxidising agent.

   Going forwards, a diazonium ion loses $\mathrm{N_2}$ — an outstanding leaving group — and what replaces it depends entirely on the reagent:

   • $\mathrm{H_2O}$, warm (283 K) $\rightarrow$ phenol
   • $\mathrm{C_2H_5OH}$ or $\mathrm{H_3PO_2}$ $\rightarrow$ benzene (reduction; the $\mathrm{N_2^+}$ is simply replaced by H)
   • $\mathrm{CuCl/HCl}$, $\mathrm{CuBr/HBr}$ $\rightarrow$ halobenzene (Sandmeyer)

   Step-by-step

   1. Fix Y from the product. 1,4-benzoquinone comes from oxidation of phenol:

   $\mathrm{C_6H_5OH} \xrightarrow{\ \mathrm{Na_2Cr_2O_7/H^+}\ } \text{1,4-benzoquinone}$

   So $\mathrm{Z} = \mathrm{Na_2Cr_2O_7/H^+} =$ IV. ($\mathrm{HNO_3}$ would nitrate; $\mathrm{H_2SO_4}$ would sulphonate — neither builds the dione.)

   2. Fix X. To get phenol from $\mathrm{C_6H_5N_2^+X^-}$: …

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5. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.In acid medium, H$_2$O$_2$ reacts with aqueous KMnO$_4$ to form Mn$^{2+}$, H$_2$O and X. In basic medium, H$_2$O$_2$ reacts with aqueous KMnO$_4$ to form MnO$_2$, H$_2$O, OH$^-$ and Y. What are X and Y respectively? (A) O$_2$, H$_2$ (B) H$_2$, H$_2$ (C) O$_2$, O$_2$ (D) H$_2$, O$_2$

   ›Reveal solutionSolution

   The key is that H₂O₂ acts as a reducing agent in both acidic and basic media when reacting with KMnO₄. In acid, MnO₄⁻ is reduced to Mn²⁺ and H₂O₂ is oxidised to O₂ (X = O₂). In base, MnO₄⁻ is reduced to MnO₂ and H₂O₂ is again oxidised to O₂ (Y = O₂). So both X and Y are O₂ — option (C).

   The problem is about the redox behaviour of hydrogen peroxide with potassium permanganate under different pH conditions. Many students memorise that KMnO₄ is a strong oxidising agent, but they forget that H₂O₂ can act as either an oxidant or a reductant depending on the partner. Here, H₂O₂ is the reducing agent — it gets oxidised itself. The product of that oxidation is always oxygen gas (O₂), regardless of whether the medium is acidic or basic. The only thing that changes is the fate of MnO₄⁻.

   Let’s walk through each case.

   1. In acid medium: The half-reaction for permanganate in acid is:

   $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

   H₂O₂, when oxidised, gives:

   $\text{H}_2\text{O}_2 \rightarrow \text{O}_2 + 2\text{H}^+ + 2e^-$

   Balancing the electrons (multiply the H₂O₂ half by 5 and the MnO₄⁻ half by 2) and adding gives the net reaction:

   $2\text{MnO}_4^- + 5\text{H}_2\text{O}_2 + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{O}_2 + 8\text{H}_2\text{O}$

   So X is clearly O₂.

   2. In basic medium: Here permanganate is reduced to MnO₂ (manganese dioxide, a brown precipitate). The half-reaction in base is:

   $\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-$

   H₂O₂ again gets oxidised to O₂:

   $\text{H}_2\text{O}_2 + 2\text{OH}^- \rightarrow \text{O}_2 + 2\text{H}_2\text{O} + 2e^-$

   (Notice: in base, the oxidation of H₂O₂ consumes OH⁻ and produces water — but the key product is still O₂.) …

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The increasing order of boiling points of the following is

   [!FORMULA] $\begin{array}{|c|c|c|c|} \hline \text{CH}_3\text{–O–CH}_3 & \text{CH}_3\text{CHO} & \text{CH}_3\text{CH}_2\text{CH}_3 & \text{CH}_3\text{CH}_2\text{OH} \\ \text{I} & \text{II} & \text{III} & \text{IV} \\ \hline \end{array}$

   (A) I < III < II < IV (B) III < I < II < IV (C) I < IV < III < II (D) III < I < IV < II

   ›Reveal solutionSolution

   Boiling point depends on intermolecular forces: stronger forces (hydrogen bonding > dipole-dipole > London dispersion) give higher boiling points. The order is propane (III) < dimethyl ether (I) < acetaldehyde (II) < ethanol (IV), so the correct option is (B).

   Concept & Intuition

   Boiling point is the temperature at which a liquid’s vapor pressure equals atmospheric pressure. To boil, molecules must overcome the attractive forces holding them together in the liquid. The stronger these intermolecular forces, the more energy (higher temperature) needed. The key forces here, in order of increasing strength, are:

   • London dispersion forces (present in all molecules, increase with molecular size/shape)
   • Dipole-dipole interactions (present in polar molecules)
   • Hydrogen bonding (a special, very strong dipole-dipole interaction when H is bonded to N, O, or F)

   We have four small molecules of similar molar mass (~44–46 g/mol), so dispersion forces are comparable. The deciding factor is the type of polarity and hydrogen bonding.

   Step-by-step reasoning

   1. Identify the molecules and their key features

      • I: CH₃–O–CH₃ (dimethyl ether) — polar C–O bonds, but no O–H bond; only dipole-dipole and dispersion.
      • II: CH₃CHO (acetaldehyde) — polar C=O bond, strong dipole-dipole, but no O–H or N–H; no hydrogen bonding as a donor.
      • III: CH₃CH₂CH₃ (propane) — nonpolar; only weak London dispersion forces.
      • IV: CH₃CH₂OH (ethanol) — has an O–H group; can form hydrogen bonds (both donor and acceptor). This is the strongest intermolecular force among the four.

   2. Rank by intermolecular force strength

      • Weakest: Propane (III) — only dispersion.
      • Next: Dimethyl ether (I) — dispersion + dipole-dipole.
      • Next: Acetaldehyde (II) — dispersion + stronger dipole-dipole (due to C=O, which is more polar than C–O).
      • Strongest: Ethanol (IV) — dispersion + dipole-dipole + hydrogen bonding.

      So the boiling point order should be:

      $\text{III} < \text{I} < \text{II} < \text{IV}$ …

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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.An alkene X on ozonolysis gives a mixture of simplest ketone (Y) and 3-Pentanone. The IUPAC name of the alkene X is (A) 2,3-Dimethylbut-2-ene (B) 3-Ethyl-4-methylpent-3-ene (C) 3-Ethyl-2-methylpent-2-ene (D) 2-Methyl-3-ethylpent-2-ene

   ›Reveal solutionSolution

   Ozonolysis cleaves the C=C bond to give two carbonyl compounds; the simplest ketone is acetone (propanone), and 3‑pentanone is a symmetrical ketone. Reversing the cleavage shows the alkene must be 3‑ethyl‑2‑methylpent‑2‑ene, which is option (C).

   Concept & Intuition

   Ozonolysis of an alkene breaks the double bond and inserts oxygen atoms, turning each doubly‑bonded carbon into a carbonyl group. If the alkene is unsymmetrical, you get two different carbonyl compounds. Here the products are the “simplest ketone” (acetone, CH₃COCH₃) and 3‑pentanone (CH₃CH₂COCH₂CH₃). To find the original alkene, we “glue” the two carbonyl fragments back together at the carbon that was the carbonyl carbon in each product. That means the alkene’s double bond was between the carbon that came from acetone’s carbonyl and the carbon that came from 3‑pentanone’s carbonyl.

   Step‑by‑step reasoning

   1. Identify the two carbonyl products

      • The simplest ketone is acetone: (CH₃)₂C=O. Its carbonyl carbon is the central carbon, which was one of the alkene’s doubly‑bonded carbons.
      • 3‑Pentanone is CH₃CH₂–CO–CH₂CH₃. Its carbonyl carbon is the middle carbon, which was the other doubly‑bonded carbon of the alkene.

   2. Reconstruct the alkene by joining the two carbonyl carbons

      Remove the oxygen from each carbonyl and connect the two carbons with a double bond.

      • From acetone: the carbon has two methyl groups attached.
      • From 3‑pentanone: the carbon has an ethyl group on each side. So the alkene’s double bond is between

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

   That gives the structure:

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

   3. Name the alkene systematically The longest continuous carbon chain that includes the double bond:
      • The left side has a central carbon with two methyls; the right side has two ethyls.
      • The longest chain is actually 5 carbons: count from the leftmost methyl through the double bond to the end of an ethyl group.
      • Number so that the double bond gets the lowest locant:

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

    This is **3‑ethyl‑2‑methylpent‑2‑ene**.  
    (Double bond between C2 and C3; methyl on C2, ethyl on C3.)
   

   4. Match with the options

   • (A) 2,3‑Dimethylbut‑2‑ene → gives acetone + acetone (not 3‑pentanone). …

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8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Metal X obtained from sphalerite ore can be purified by which of the following methods? (A) Distillation (B) Poling (C) Zone refining (D) Vapour phase refining

   ›Reveal solutionSolution

   Sphalerite is the chief ore of zinc (Zn), and the metal obtained from it is purified by distillation because zinc has a low boiling point (907 °C) and can be vaporised and condensed, leaving impurities behind. The correct option is (A).

   ---

   The key idea here is that the purification method for a metal depends heavily on its physical properties, especially its boiling point relative to its impurities. Zinc, obtained from sphalerite (ZnS), has a relatively low boiling point (907 °C) compared to many common impurities like cadmium, lead, or iron. This makes distillation an ideal method: you heat the impure zinc above its boiling point, collect the zinc vapour, and condense it back to pure metal. The impurities, having much higher boiling points, remain as a solid residue.

   Let’s walk through why the other options don’t fit:

   1. Distillation (Option A) – This works for metals with low boiling points. Zinc’s boiling point (907 °C) is low enough that it can be vaporised without melting the furnace. Cadmium, a common impurity in zinc, has an even lower boiling point (767 °C) and can be removed first by careful temperature control. This is the standard industrial method for purifying zinc.

   2. Poling (Option B) – Poling is used for metals like copper or tin that contain dissolved oxides. A green wood pole is stirred into the molten metal; the hydrocarbons release gases that reduce the oxides. Zinc does not typically have oxide impurities that require this treatment, and poling is not suited for a metal that vaporises easily.

   3. Zone refining (Option C) – This method relies on differences in solubility of impurities in the solid vs. liquid state. It is used for extremely high-purity metals like silicon or germanium (for semiconductors). Zinc is not typically purified this way because distillation is far cheaper and simpler for achieving the required purity. …

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Which of the following can undergo Hell-Volhard-Zelinsky reaction? (A) $\mathrm{C_6H_5-CH_2CHO}$ (phenylacetaldehyde) (B) $\mathrm{C_6H_5-CO_2H}$ (benzoic acid) (C) $\mathrm{C_6H_5-CH_2CO_2H}$ (phenylacetic acid) (D) $\mathrm{C_6H_5-CH_2-CO-CH_3}$ (phenylacetone)

   ›Reveal solutionSolution

   HVZ needs a carboxylic acid with at least one $\alpha$-hydrogen. Benzoic acid has no $\alpha$-H, and the aldehyde and ketone are not acids at all — leaving phenylacetic acid, option (C).

   The concept first

   The Hell–Volhard–Zelinsky (HVZ) reaction is the standard way to put a halogen on the carbon next to a $-\mathrm{COOH}$ group:

   $\mathrm{R\text{-}CH_2\text{-}COOH}\ \xrightarrow[\text{(ii) }\mathrm{H_2O}]{\text{(i) }\mathrm{Cl_2\ or\ Br_2\ /\ red\ P}}\ \mathrm{R\text{-}CHX\text{-}COOH}$

   Why the red phosphorus? A carboxylic acid enolises very badly. Red P + $\mathrm{X_2}$ generates $\mathrm{PX_3}$, which turns a small amount of the acid into its acyl halide $\mathrm{RCH_2COX}$. That species enolises readily, and the enol is attacked by $\mathrm{X_2}$ at the $\alpha$-carbon; hydrolysis then hands back the $\alpha$-halo acid.

   So HVZ has two non-negotiable requirements:

   1. the substrate is a carboxylic acid, and
   2. it possesses an $\alpha$-hydrogen (no $\alpha$-H $\Rightarrow$ no enol $\Rightarrow$ no reaction).

   Step-by-step through the options

   (A) $\mathrm{C_6H_5CH_2CHO}$, phenylacetaldehyde. It has $\alpha$-hydrogens, but it is an aldehyde. Its $\alpha$-halogenation is ordinary acid/base-catalysed halogenation — not HVZ. ✗ …

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10. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.A sample of water contains Mg(HCO$_3$)$_2$ and Ca(HCO$_3$)$_2$. On boiling this water, these hydrogen carbonates are removed as precipitates. The precipitates are (A) Mg(OH)$_2$, Ca(OH)$_2$ (B) MgCO$_3$, Ca(OH)$_2$ (C) MgCO$_3$, CaCO$_3$ (D) Mg(OH)$_2$, CaCO$_3$

    ›Reveal solutionSolution

    On boiling, magnesium hydrogen carbonate decomposes to magnesium hydroxide (not carbonate) because Mg(OH)$_2$ is less soluble than MgCO$_3$, while calcium hydrogen carbonate gives calcium carbonate. The precipitates are Mg(OH)$_2$ and CaCO$_3$, so the correct option is (D).

    The key here is understanding how the hydrogen carbonates of magnesium and calcium behave when heated. Both Mg(HCO$_3$)$_2$ and Ca(HCO$_3$)$_2$ are soluble in cold water, but boiling drives off carbon dioxide and water, leaving behind insoluble products. However, the exact product depends on the relative solubilities of the possible carbonates and hydroxides — and this is where magnesium and calcium differ.

    For calcium, CaCO$_3$ is far less soluble than Ca(OH)$_2$, so the precipitate is calcium carbonate. For magnesium, the situation is reversed: Mg(OH)$_2$ is much less soluble than MgCO$_3$, so the precipitate is magnesium hydroxide. Let’s walk through the chemistry step by step.

    1. The decomposition reaction for a hydrogen carbonate When any hydrogen carbonate is heated, it decomposes to the carbonate, water, and carbon dioxide:

    $M(\text{HCO}_3)_2 \xrightarrow{\Delta} M\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$

    This is the general pattern. If this were the whole story, both Mg and Ca would give their carbonates. But magnesium carbonate itself can further react with water under these conditions.

    2. Why magnesium hydroxide forms instead MgCO$_3$ is sparingly soluble, but Mg(OH)$_2$ is even less soluble — its solubility product is about $5.6 \times 10^{-12}$, compared to $6.8 \times 10^{-6}$ for MgCO$_3$. In the hot aqueous environment, any MgCO$_3$ that forms can undergo hydrolysis:

    $\text{MgCO}_3 + \text{H}_2\text{O} \rightarrow \text{Mg(OH)}_2 + \text{CO}_2$

    The driving force is the precipitation of the much less soluble hydroxide. So the net reaction for magnesium hydrogen carbonate on boiling is:

    $\text{Mg(HCO}_3)_2 \xrightarrow{\Delta} \text{Mg(OH)}_2 + 2\text{CO}_2$

    (Water is also produced, but it’s part of the medium.)

    3. Calcium stays as carbonate …

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11. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
    Q.An isomer of $C_7H_{16}$ is X. This has five primary, one tertiary and one quaternary carbon. What is X? (A) 3-Ethylpentane (B) 3,3-Dimethylpentane (C) 2,2,3-Trimethylbutane (D) 2,4-Dimethylpentane

    ›Reveal solutionSolution

    The key is to count the types of carbons (primary, tertiary, quaternary) in each isomer. Only 2,2,3-trimethylbutane has exactly five primary, one tertiary, and one quaternary carbon — so the answer is (C).

    The question gives you a molecular formula $C_7H_{16}$ (a heptane isomer) and tells you the carbon classification: five primary carbons (bonded to only one other carbon), one tertiary carbon (bonded to three other carbons), and one quaternary carbon (bonded to four other carbons). That’s a total of seven carbons, which matches the formula. Your job is to find which of the four options fits this exact pattern.

    Let’s walk through each option and count the carbon types.

    1. Option (A): 3-Ethylpentane

       Draw the structure: a pentane chain with an ethyl group on carbon 3.

       • The chain carbons: C1, C2, C3, C4, C5. C3 is bonded to the ethyl group, so it’s tertiary (connected to three carbons: C2, C4, and the ethyl carbon).
       • The ethyl group has two carbons: one attached to C3 (that’s a secondary carbon, bonded to two carbons) and a terminal CH₃ (primary).
       • Count: primary carbons = the two terminal CH₃ on the main chain (C1 and C5) plus the CH₃ on the ethyl = 3 primary. Tertiary = 1 (C3). Quaternary = 0. That’s only 3 primary, not 5. So (A) is out.

    2. Option (B): 3,3-Dimethylpentane

       Structure: pentane chain with two methyl groups on carbon 3.

       • C3 is bonded to C2, C4, and two methyl groups — that’s four bonds to carbon, so it’s quaternary.
       • The two methyl groups on C3 are primary. The terminal CH₃ on C1 and C5 are primary. That gives 4 primary carbons.
       • C2 and C4 are secondary (each bonded to two carbons). No tertiary carbon.
       • Count: primary = 4, tertiary = 0, quaternary = 1. Not matching (needs 5 primary and 1 tertiary). So (B) is out.

    3. Option (C): 2,2,3-Trimethylbutane

       Structure: a butane chain (4 carbons) with three methyl substituents: two on carbon 2 and one on carbon 3.

       • Let’s label the main chain: C1, C2, C3, C4.
         • C1: bonded only to C2 and three H’s — primary.
         • C2: bonded to C1, C3, and two methyl groups — that’s four carbon bonds, so quaternary.
         • C3: bonded to C2, C4, and one methyl group — three carbon bonds, so tertiary.
         • C4: bonded only to C3 and three H’s — primary. …

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12. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
    Q.Which of the following is an example of antifertility drug? (A) Bithionol (B) Sucralose (C) Novestrol (D) Terpineol

    ›Reveal solutionSolution

    Antifertility drugs are chemical substances designed to prevent pregnancy. Among the given choices, Novestrol is a synthetic steroid hormone used as an antifertility drug, typically found in oral contraceptives. The correct option is (C).

    Antifertility drugs, also known as contraceptives, are chemical substances used to prevent conception. Their primary role is in family planning, allowing individuals to control the timing and spacing of pregnancies. These drugs typically work by interfering with the female reproductive cycle, most commonly by inhibiting ovulation (the release of an egg from the ovary) or by altering the uterine environment to prevent the implantation of a fertilized egg.

    Most antifertility drugs are synthetic derivatives of natural hormones like estrogen and progesterone. By mimicking or modulating the effects of these hormones, they disrupt the normal hormonal cascade required for pregnancy.

    Let's examine each option:

    1. Understanding Antifertility Drugs:

       Antifertility drugs are substances that prevent pregnancy. They achieve this primarily by interfering with the hormonal regulation of the female reproductive cycle. The most common types are oral contraceptives, which contain synthetic estrogen and/or progestin derivatives. These synthetic hormones inhibit the release of gonadotropins (FSH and LH) from the pituitary gland, thereby preventing ovulation. They can also thicken cervical mucus, making it difficult for sperm to reach the egg, and alter the uterine lining to prevent implantation.

    2. Analyzing Option (A) Bithionol:

       Bithionol is an antiseptic. It is commonly added to soaps to reduce the growth of microorganisms on the skin. It has no role in preventing pregnancy.

    3. Analyzing Option (B) Sucralose:

       Sucralose is an artificial sweetener. It is a calorie-free sugar substitute derived from sucrose. It is used in various food and beverage products and has no medicinal or antifertility properties. …

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