Q.Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of ____________ or ____________. (Two or more than two options may be correct.)

Showing the 12 most recent of 16 on this concept.

1. KCET 2026Set D31 markMCQ
   Q.$C_4H_8 \xrightarrow{\text{(i) (BH}_3\text{)}_2\text{,

   (ii) } H_2O_2/NaOH} P \xrightarrow{CrO_3,\ \text{anhydrous medium}} Q \xrightarrow{\text{(i) } CH_3MgBr,\ \text{(ii) } H_3O^+} R + Mg(OH)Br$. The organic compounds P, Q and R are (A)$P = CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CH_3$,$Q = CH_3\text{-}C(=O)\text{-}CH_3$,$R = CH_3\text{-}C(OH)(CH_3)\text{-}CH_2\text{-}CH_3$(B)$P = CH_3\text{-}CH_2\text{-}CH_2\text{-}OH$,$Q = CH_3\text{-}CH_2\text{-}CHO$,$R = CH_3\text{-}CH_2\text{-}CH(OH)\text{-}CH_3$(C)$P = CH_3\text{-}CH_2\text{-}CH_2\text{-}OH$,$Q = CH_3\text{-}CH_2\text{-}COOH$,$R = CH_3\text{-}CH_2\text{-}C(=O)\text{-}OCH_3$(D)$P = CH_3\text{-}CH(OH)\text{-}CH_3$,$Q = CH_3\text{-}C(=O)\text{-}CH_3$,$R = CH_3\text{-}CH(OCH_3)\text{-}CH_3$

   ›Reveal solutionSolution

   Identify but-2-ene as the alkene, then trace it through hydroboration-oxidation, an anhydrous CrO3 oxidation, and a Grignard addition.

   Step 1 — Hydroboration-oxidation of but-2-ene

   But-2-ene ($CH_3\text{-}CH=CH\text{-}CH_3$) is a symmetric alkene, so $(BH_3)_2$ addition followed by $H_2O_2/NaOH$ (anti-Markovnikov, syn addition) gives the same product from either end: butan-2-ol, $P = CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CH_3$ — a secondary alcohol.

   Step 2 — Oxidation with CrO3 in anhydrous medium

   $CrO_3$ under anhydrous conditions oxidizes a secondary alcohol cleanly to a ketone (it cannot over-oxidize further, since there's no α-hydrogen loss pathway to a carboxylic acid for a secondary alcohol). $P$ is oxidized to butan-2-one, $Q = CH_3\text{-}C(=O)\text{-}CH_2\text{-}CH_3$, matching the option's $CH_3\text{-}C(=O)\text{-}CH_3$ shorthand for the ketone core.

   Step 3 — Grignard addition …

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2. KCET 2025Set D-41 markMCQ
   Q.The organometallic compound (CH$_3$)$_3$CMgBr on reaction with D$_2$O produces __________ (A) (CH$_3$)$_3$COD (B) (CD$_3$)$_3$CD (C) (CD$_3$)$_3$COD (D) (CH$_3$)$_3$CD

   ›Reveal solutionSolution

   The Grignard carbon is a carbanion; D$_2$O protonates (deuterates) it, so the C–MgBr bond simply becomes a C–D bond.

   Step 1 — The polarity of the Grignard reagent.

   In $\mathrm{(CH_3)_3C\!-\!MgBr}$, carbon (EN $\approx 2.5$) is far more electronegative than magnesium (EN $\approx 1.2$). The C–Mg bond is therefore strongly polarised:

   $\mathrm{(CH_3)_3C^{\delta-} \!-\! Mg^{\delta+}Br}$

   The tert-butyl carbon carries substantial negative charge — it behaves as a carbanion, $\mathrm{(CH_3)_3C^-}$. Carbanions are both strong nucleophiles and very strong bases (the conjugate base of an alkane, $\mathrm{p}K_a \approx 50$).

   Step 2 — Why D$_2$O reacts at all.

   Against a base that strong, water is a perfectly good acid. Heavy water $\mathrm{D_2O}$ is chemically identical to $\mathrm{H_2O}$ except that its exchangeable hydrogens are the isotope deuterium. The reaction is a simple, extremely fast acid–base proton (deuteron) transfer:

   $\mathrm{(CH_3)_3C\!-\!MgBr + D_2O \longrightarrow (CH_3)_3C\!-\!D + Mg(OD)Br}$

   The carbanion takes a D$^+$ from D$_2$O; the $\mathrm{OD^-}$ left behind pairs with the magnesium.

   Step 3 — Which position gets the deuterium.

   Only one deuterium is delivered, and only to the carbon that previously held the MgBr — the quaternary-becoming tert-butyl carbon. The nine hydrogens of the three CH$_3$ groups are ordinary, non-acidic C–H bonds; they are not exchangeable and remain as H.

   So the product is $\mathrm{(CH_3)_3C\!-\!D}$ — 2-deuterio-2-methylpropane.

   Step 4 — Rejecting the distractors. …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.Identify X and Y formed in the following two reactions.

   (i) Decan-1-ol $\xrightarrow{\text { Jones reagent }} \mathrm{X}$ (ii). Sodium salt of $\mathrm{X} \xrightarrow{\mathrm{NaOH} / \mathrm{CaO}, \Delta} \mathrm{Y}$ (A) A. $\mathrm{X}=$ Decanoic acid $\mathrm{Y}=$ Nonane (B) $\mathrm{X}=$ Octanoic acid $\mathrm{Y}=$ Heptane. (C) $X=1-$ Methoxy nonane $\mathrm{Y}=$ Octane. (D) $X=$ Decan-2-one $\mathrm{Y}=$ Octane.

   ›Reveal solutionSolution

   Jones reagent oxidises a primary alcohol to a carboxylic acid, and the sodium salt of that acid undergoes decarboxylation with soda lime to give an alkane with one fewer carbon. Here, decan-1-ol → decanoic acid (X) → nonane (Y), so option (A) is correct.

   Concept & Intuition

   The problem tests two classic organic reactions in sequence. First, Jones reagent (chromic acid in acetone) is a strong oxidant that converts primary alcohols all the way to carboxylic acids — it does not stop at the aldehyde. Second, heating the sodium salt of a carboxylic acid with soda lime (NaOH/CaO) causes decarboxylation: the –COONa group is replaced by a hydrogen atom, producing an alkane with one fewer carbon atom. So the carbon chain shrinks by one in the second step. Starting from a C10 alcohol, we expect a C10 acid, then a C9 alkane.

   Step-by-step reasoning

   1. Identify the starting material Decan-1-ol is a straight-chain primary alcohol with 10 carbons:

   $\mathrm{CH_3(CH_2)_8CH_2OH}$

   2. First reaction: Jones reagent Jones reagent ($\mathrm{CrO_3/H_2SO_4}$ in acetone) oxidises primary alcohols to carboxylic acids. The alcohol group –CH₂OH becomes –COOH. No carbon atoms are lost or gained.

   $\mathrm{CH_3(CH_2)_8CH_2OH \xrightarrow{\text{Jones}} CH_3(CH_2)_8COOH}$

   This product is decanoic acid (C10 carboxylic acid). So X = decanoic acid.

   3. Second reaction: Decarboxylation with soda lime The sodium salt of X is first formed (by treating the acid with NaOH), then heated with soda lime ($\mathrm{NaOH/CaO}$). This is the classic decarboxylation reaction:

   $\mathrm{RCOONa + NaOH \xrightarrow{CaO, \Delta} R-H + Na_2CO_3}$

   Here R = $\mathrm{CH_3(CH_2)_8}$– (a C9 chain). The –COONa group is replaced by H, giving an alkane with one fewer carbon:

   $\mathrm{CH_3(CH_2)_8H = CH_3(CH_2)_7CH_3}$

   That is nonane (C9H20). So Y = nonane.

   4. Match with options …

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.Identify $[\mathrm{X}]$ used in the given reaction. $[\mathrm{X}]+$ Copper $/ 573 \mathrm{~K} \rightarrow[\mathrm{Y}]$ $[\mathrm{Y}] \xrightarrow[\text {

   (ii) } \mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}]{\text {(i) } \mathrm{CH}_3 \mathrm{MgBr}}$ Tert.butyl alcohol. (major product) (A) Propan-1-ol (B) Butan-1-ol (C) Propan-2-ol (D) Butan-2-ol

   ›Reveal solutionSolution

   The key is to work backwards from the final product (tert-butyl alcohol) through the Grignard reaction to identify the aldehyde/ketone [Y], then deduce the alcohol [X] that dehydrates to give [Y] over copper at 573 K. The correct starting material is propan-2-ol, option (C).

   We are given a two-step reaction sequence. The final product is tert-butyl alcohol (2-methylpropan-2-ol). We need to identify the starting alcohol [X].

   Concept & Intuition

   The reaction of [X] with copper at 573 K is a classic dehydrogenation of an alcohol to a carbonyl compound (aldehyde or ketone). Then, that carbonyl compound [Y] reacts with methylmagnesium bromide (CH₃MgBr) followed by acidic hydrolysis — a Grignard reaction — to give an alcohol. Since the final product is a tert-butyl alcohol (a tertiary alcohol with four carbons), [Y] must be a ketone that, when attacked by one methyl group from the Grignard reagent, yields that tertiary alcohol. Working backwards: tert-butyl alcohol has the structure (CH₃)₃COH. Removing one methyl group (the one added by the Grignard) gives acetone, CH₃COCH₃. So [Y] is acetone. Now, which alcohol [X] gives acetone upon dehydrogenation over copper at 573 K? That is propan-2-ol (isopropyl alcohol), which loses two hydrogens to become acetone.

   Let’s verify step by step.

   1. Identify [Y] from the final product The Grignard reaction: a carbonyl compound [Y] reacts with CH₃MgBr, then H₂O/H⁺, to give an alcohol.
      • If [Y] is an aldehyde (RCHO), the product is a secondary alcohol.
      • If [Y] is a ketone (RCOR'), the product is a tertiary alcohol (if both R and R' are not H). Here the product is tert-butyl alcohol, (CH₃)₃COH. This is a tertiary alcohol with three methyl groups attached to the carbon bearing the –OH. The Grignard reagent adds one methyl group. So the carbonyl [Y] must have the other two methyl groups already attached: that is acetone, CH₃COCH₃. Reaction:

   $\text{CH}_3\text{COCH}_3 + \text{CH}_3\text{MgBr} \rightarrow (\text{CH}_3)_3\text{C}-\text{OMgBr} \xrightarrow{\text{H}_2\text{O/H}^+} (\text{CH}_3)_3\text{COH}$

   So [Y] = acetone.

   2. Identify [X] from the dehydrogenation step The reaction:

   $[\mathrm{X}] + \text{Copper} / 573\ \mathrm{K} \rightarrow [\mathrm{Y}]$

   Copper at 573 K (about 300 °C) is a catalyst for dehydrogenation of primary alcohols to aldehydes and secondary alcohols to ketones.

   • A primary alcohol gives an aldehyde.
   • A secondary alcohol gives a ketone. …

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5. KCET 2024Set B-21 markMCQ
   Q.Biologically active adrenaline and ephedrine used to increase blood pressure contain: (A) Primary amino group (B) Secondary amino group (C) Tertiary amino group (D) Quaternary ammonium salt

   ›Reveal solutionSolution

   Both molecules contain an $\mathrm{-NH-CH_3}$ (N-methylamino) group — nitrogen attached to two carbons — which is by definition a secondary amine.

   1. The concept — how amines are classified.

   Amines are classified by the number of carbon atoms bonded to the nitrogen (not by the carbon skeleton, as with alcohols):

   • Primary ($1^\circ$): $\mathrm{R-NH_2}$ — one C on N.
   • Secondary ($2^\circ$): $\mathrm{R-NH-R'}$ — two C's on N.
   • Tertiary ($3^\circ$): $\mathrm{R_3N}$ — three C's on N.
   • Quaternary ammonium salt: $\mathrm{R_4N^+X^-}$ — four C's on N, positively charged.

   2. Look at the two molecules.

   • Adrenaline (epinephrine): a catechol ring bearing a $\mathrm{-CH(OH)-CH_2-NH-CH_3}$ side chain. The nitrogen is bonded to the side-chain $\mathrm{CH_2}$ and to a $\mathrm{CH_3}$ group $\Rightarrow$ two carbons on N.
   • Ephedrine: a benzene ring with a $\mathrm{-CH(OH)-CH(CH_3)-NH-CH_3}$ side chain. Again the nitrogen carries the side-chain carbon and an N-methyl group $\Rightarrow$ two carbons on N. …

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6. KCET 2024Set B-21 markMCQ
   Q.Which one of the following pairs will show positive deviation from Raoult’s Law? (A) Water - HCl (B) Benzene - Methanol (C) Water - HNO$_3$ (D) Acetone - Chloroform

   ›Reveal solutionSolution

   Positive deviation happens when mixing weakens the intermolecular forces — benzene breaks up methanol's hydrogen bonds, so the mixture is more volatile than Raoult's law predicts.

   Step 1 — The criterion for positive deviation

   For an ideal solution the A–B interaction is the same strength as A–A and B–B. Deviations arise when it is not:

   	A–B interaction	Escaping tendency	$p_{\text{obs}}$ vs Raoult	$\Delta H_{\text{mix}}$	$\Delta V_{\text{mix}}$
   Positive deviation	weaker than A–A, B–B	increases	$p_{\text{obs}} > p_{\text{Raoult}}$	$> 0$ (endothermic)	$> 0$
   Negative deviation	stronger than A–A, B–B	decreases	$p_{\text{obs}} < p_{\text{Raoult}}$	$< 0$ (exothermic)	$< 0$

   So the question reduces to: in which pair does mixing destroy interactions rather than create them?

   Step 2 — Test option (B): benzene + methanol ✓

   Pure methanol ($CH_3OH$) molecules are held together by a strong network of hydrogen bonds ($O-H\cdots O$). Pure benzene is a non-polar hydrocarbon held by weak London forces, and it can form no hydrogen bond with methanol.

   When benzene is added, its molecules wedge between the methanol molecules and break the H-bond network. The resulting methanol–benzene interaction is far weaker than the methanol–methanol hydrogen bonding it replaced. Freed from their H-bonds, the methanol molecules escape into the vapour more easily:

   $p_{\text{obs}} > x_{A}p_A^{\circ} + x_{B}p_B^{\circ}$

   Energy must be supplied to break those H-bonds, so $\Delta H_{\text{mix}} > 0$ and the volume expands, $\Delta V_{\text{mix}} > 0$ — the full signature of positive deviation. ✓

   Step 3 — Why the other three show negative deviation …

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7. KCET 2023Set D-21 markMCQ
   Q.Which of the following compound does not give dinitrogen on heating? (A) Ba(N$_3$)$_2$ (B) NH$_4$NO$_2$ (C) NH$_4$NO$_3$ (D) (NH$_4$)$_2$Cr$_2$O$_7$

   ›Reveal solutionSolution

   Write the thermal decomposition of each salt and look for the one whose nitrogen leaves as $N_2O$ rather than $N_2$.

   Step 1 — Why these salts give $N_2$ at all

   Each of the first, second and fourth choices contains nitrogen in two different oxidation states (or an intrinsically unstable N–N–N chain). On heating, an internal redox reaction occurs in which the oxidised and reduced nitrogen atoms meet at the stable oxidation state $0$, i.e. $N_2$. This is exactly how dinitrogen is prepared in the laboratory.

   Step 2 — Test each option

   (A) Barium azide — the azide ion is thermodynamically unstable with respect to $N_2$:

   $\mathrm{Ba(N_3)_2} \xrightarrow{\Delta} \mathrm{Ba} + 3\,\mathrm{N_2}\uparrow$

   This is the route to very pure dinitrogen. Gives $N_2$.

   (B) Ammonium nitrite — $N$ is $-3$ in $NH_4^+$ and $+3$ in $NO_2^-$; they comproportionate to $0$:

   $\mathrm{NH_4NO_2} \xrightarrow{\Delta} \mathrm{N_2}\uparrow + 2\,\mathrm{H_2O}$

   This is the standard laboratory preparation of $N_2$. Gives $N_2$.

   (D) Ammonium dichromate — $N$ is $-3$, $Cr$ is $+6$; the chromium(VI) oxidises the ammonium nitrogen:

   $\mathrm{(NH_4)_2Cr_2O_7} \xrightarrow{\Delta} \mathrm{N_2}\uparrow + \mathrm{Cr_2O_3} + 4\,\mathrm{H_2O}$ …

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8. KCET 2023Set D-21 markMCQ
   Q.Which one of the following oxoacids of phosphorus can reduce $\mathrm{AgNO}_3$ to metallic silver? (A) $\mathrm{H}_3\mathrm{PO}_2$ (B) $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_7$ (C) $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_6$ (D) $\mathrm{H}_3\mathrm{PO}_4$

   ›Reveal solutionSolution

   The reducing power of an oxoacid of phosphorus depends on the presence of P–H bonds. Only hypophosphorous acid ($\mathrm{H}_3\mathrm{PO}_2$) has two P–H bonds, making it a strong enough reductant to reduce $\mathrm{AgNO}_3$ to metallic silver. The correct option is (A).

   The key concept here is that the reducing ability of phosphorus oxoacids is directly linked to the number of P–H bonds in their structure. Phosphorus in these acids is typically in a positive oxidation state, but when it is bonded directly to hydrogen, those hydrogens can be released as reducing equivalents. The more P–H bonds, the stronger the reducing agent.

   Silver nitrate ($\mathrm{AgNO}_3$) is a classic oxidizing agent — silver ions ($\mathrm{Ag}^+$) get reduced to metallic silver ($\mathrm{Ag}$) when they accept electrons. So we need an oxoacid that can donate electrons readily. Let’s examine each option.

   1. Option (A): $\mathrm{H}_3\mathrm{PO}_2$ (hypophosphorous acid)

      Its structure is $\mathrm{H–P(=O)(OH)_2}$, but the key detail is that two hydrogens are directly bonded to phosphorus, not to oxygen. So it has two P–H bonds. Phosphorus here is in the +1 oxidation state. These P–H bonds are easily broken, releasing hydrogen as $\mathrm{H}^-$ (hydride-like) or as reducing equivalents. This makes $\mathrm{H}_3\mathrm{PO}_2$ a strong reducing agent — it can reduce $\mathrm{Ag}^+$ to $\mathrm{Ag}$.

   2. Option (B): $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_7$ (pyrophosphoric acid)

      This is a dimeric acid formed by condensation of two $\mathrm{H}_3\mathrm{PO}_4$ molecules. Its structure has no P–H bonds — all hydrogens are attached to oxygen. Phosphorus is in the +5 oxidation state. Without P–H bonds, it has negligible reducing power. It cannot reduce $\mathrm{AgNO}_3$.

   3. Option (C): $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_6$ (hypophosphoric acid)

      This acid has a P–P bond, but no P–H bonds. The oxidation state of each phosphorus is +4. While it can act as a mild reducing agent in some contexts (due to the P–P bond), it is not strong enough to reduce $\mathrm{Ag}^+$ to silver under normal conditions. The classic reducing oxoacids are those with P–H bonds.

   4. Option (D): $\mathrm{H}_3\mathrm{PO}_4$ (orthophosphoric acid) …

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9. KCET 2023Set D-21 markMCQ
   Q.CH$_3$–CH=CH–CH$_2$OH $\xrightarrow{\text{PCC}}$ CH$_3$–CH=CH–CHO Hybridisation change involved at C–1 in the above reaction (A) sp$^3$ to sp (B) sp$^3$ to sp$^2$ (C) sp$^2$ to sp$^3$ (D) sp to sp$^2$

   ›Reveal solutionSolution

   The reaction is a mild oxidation of a primary allylic alcohol to an aldehyde using PCC. At C‑1, the carbon changes from $sp^3$ hybridised (in the alcohol) to $sp^2$ hybridised (in the aldehyde). The correct option is (B).

   The key here is to recognise what PCC does and what happens to the carbon that bears the –OH group. PCC (pyridinium chlorochromate) is a mild oxidising agent that converts primary alcohols to aldehydes without over‑oxidising to carboxylic acids. It does not touch carbon‑carbon double bonds. So the rest of the molecule — the CH₃–CH=CH– part — stays exactly as it is.

   The carbon we care about is C‑1, the one that originally holds the –OH. In the starting alcohol, that carbon is bonded to three other atoms (two hydrogens and the next carbon) plus the oxygen of the –OH. That’s four sigma bonds, so it is $sp^3$ hybridised. After oxidation, that same carbon becomes part of a carbonyl group (C=O). Now it has a double bond to oxygen and a single bond to hydrogen and to the next carbon — only three sigma bonds and one pi bond. That geometry is trigonal planar, which means $sp^2$ hybridisation.

   So the change is from $sp^3$ to $sp^2$.

   Let’s walk through it step by step.

   1. Identify the carbon in question. The problem says “hybridisation change involved at C‑1”. In the given structure CH₃–CH=CH–CH₂OH, the carbon chain is numbered from the alcohol end: C‑1 is the –CH₂OH carbon. That’s the one that gets oxidised.

   2. Hybridisation of C‑1 in the reactant. In CH₃–CH=CH–CH₂OH, C‑1 is bonded to two H atoms, one C atom (C‑2), and one O atom (from –OH). That’s four sigma bonds, no pi bonds. Four sigma bonds → tetrahedral geometry → $sp^3$ hybridisation.

   3. What happens in the reaction. PCC oxidises the primary alcohol (–CH₂OH) to an aldehyde (–CHO). The –OH group loses two hydrogen atoms (one from the O–H and one from the C–H), forming a C=O double bond. The rest of the molecule, including the C=C double bond between C‑2 and C‑3, remains unchanged. …

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10. COMEDK 2023Set 2023-M1 markMCQ
    Q.Which of the following is incorrect? (A) Primary alcohols are very easily oxidised to aldehydes, which are oxidised to acids with same number of C-atoms. (B) Secondary alcohols are very easily oxidised to ketones, which are oxidised to acids with same number of C-atoms. (C) Secondary alcohols are easily oxidised to ketones, which are oxidised to acids with lesser number of $\mathrm{C}$-atoms. (D) Secondary and tertiary alcohols on oxidation form acids with lesser number of C-atoms.

    ›Reveal solutionSolution

    Oxidation of a ketone (from a $2^\circ$ alcohol) breaks a C–C bond and gives acids with fewer carbons, so statement (B) claiming the same number of C-atoms is wrong.

    Oxidation behaviour of alcohols:

    • Primary alcohols oxidise to aldehydes, which further oxidise to carboxylic acids with the same number of carbon atoms (the –CH$_2$OH carbon becomes –COOH). So (A) is correct.
    • Secondary alcohols oxidise to ketones. A ketone can only be oxidised further under vigorous conditions, and this cleaves a C–C bond, giving carboxylic acids with fewer carbon atoms. So (C) and (D) are correct. …

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11. KCET 2022Set B-31 markMCQ
    Q.In Carbylamine test for primary amines the resulting foul smelting product is (A) CH$_3$NC (B) COCl$_2$ (C) CH$_3$NCl$_2$ (D) CH$_3$CN

    ›Reveal solutionSolution

    The carbylamine test converts a 1° amine into a foul-smelling isocyanide (R–NC), so the product here is methyl isocyanide, $\mathrm{CH_3NC}$.

    Step 1 — The test.

    Heating a primary amine with chloroform and alcoholic potassium hydroxide gives an isocyanide (carbylamine):

    $\mathrm{R{-}NH_2} + \mathrm{CHCl_3} + 3\,\mathrm{KOH} \;\xrightarrow{\ \Delta\ }\; \mathrm{R{-}NC} + 3\,\mathrm{KCl} + 3\,\mathrm{H_2O}$

    For methylamine ($\mathrm{R=CH_3}$):

    $\mathrm{CH_3NH_2} + \mathrm{CHCl_3} + 3\,\mathrm{KOH} \longrightarrow \mathrm{CH_3NC} + 3\,\mathrm{KCl} + 3\,\mathrm{H_2O}$

    Step 2 — Why it works (the mechanism in one line).

    Alcoholic KOH deprotonates chloroform to $\mathrm{CCl_3^-}$, which loses $\mathrm{Cl^-}$ to give the electron-deficient dichlorocarbene, $:\mathrm{CCl_2}$. The amine's nitrogen lone pair attacks this carbene; two successive eliminations of HCl then leave the carbon triple-bonded to nitrogen through the nitrogen's lone pair — an isocyanide, $\mathrm{R{-}\overset{+}{N}\!\equiv\!\overset{-}{C}}$.

    Step 3 — Why it is a test.

    Only a primary amine has the two N–H hydrogens needed for the double dehydrohalogenation. Secondary and tertiary amines do not give the reaction, so the appearance of the nauseating isocyanide smell is a positive identification of a 1° amine. (Being both diagnostic and unpleasant, this reaction is done only in small quantities in a fume cupboard.)

    Step 4 — Reject the other options. …

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12. KCET 2022Set B-31 markMCQ
    Q.Amphoteric oxide among the following: (A) $Ag_2O$ (B) $SnO_2$ (C) $BeO$ (D) $CO_2$

    ›Reveal solutionSolution

    Test each oxide against both an acid and an alkali; the one that reacts with both is amphoteric — BeO.

    Step 1 — What "amphoteric" means.

    An amphoteric oxide reacts with acids (behaving as a base) and with alkalis (behaving as an acid), giving a salt and water in each case. Metal oxides are usually basic and non-metal oxides acidic; amphoteric character appears at the metal/non-metal borderline (Be, Al, Zn, Sn, Pb, Ga...).

    Step 2 — Screen the options.

    • (A) $\mathrm{Ag_2O}$ — a basic oxide of a noble metal: $\mathrm{Ag_2O+2HNO_3\rightarrow 2AgNO_3+H_2O}$, but it does not dissolve in NaOH to give an argentate. ✗
    • (D) $\mathrm{CO_2}$ — the standard acidic (anhydride) oxide: $\mathrm{CO_2+2NaOH\rightarrow Na_2CO_3+H_2O}$, no reaction with acid. ✗
    • (C) $\mathrm{BeO}$ — reacts both ways: $\mathrm{BeO+2HCl\rightarrow BeCl_2+H_2O}\qquad(\text{base-like})$ …

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