Q.(a) Explain why :

Showing the 12 most recent of 52 on this concept.

1. CBSE 2026Set A1 markMCQ
   Q.When vapours of an alcohol are passed over hot reduced copper, it gives an alkene. The alcohol is

   (a) Primary

   (b) Secondary

   (c) Tertiary

   (d) None of these

   ›Reveal solutionSolution

   Over hot reduced copper (573 K), a primary alcohol gives an aldehyde, a secondary gives a ketone, and a tertiary gives an alkene.

   When alcohol vapours are passed over hot reduced copper the behaviour depends on the class of alcohol:

   • Primary alcohol -> dehydrogenation -> aldehyde
   • Secondary alcohol -> dehydrogenation -> ketone …

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2. CBSE 2026Set ANNUAL1 mark
   Q.Write the name of product obtained when vapour of ethyl alcohol are passed over heated Copper at 573 K.

   ›Reveal solutionSolution

   Passing alcohol vapours over heated copper catalyses either dehydrogenation (for 1° and 2° alcohols) or dehydration (for 3° alcohols), depending on alcohol type.

   …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Dehydration of tertiary alcohols with copper at 573 K gives:

   (a) Aldehyde

   (b) Ketone

   (c) Alkene

   (d) None of these

   ›Reveal solutionSolution

   Passing alcohol vapours over heated copper at 573 K is a classification test: 1° alcohols → aldehydes, 2° alcohols → ketones, but 3° alcohols (no α-H on the carbinol carbon available for dehydrogenation) undergo dehydration to give an alkene.

   When vapours of an alcohol are passed over copper catalyst at 573 K:

   • Primary alcohols are dehydrogenated (lose $H_2$) to aldehydes: $RCH_2OH \xrightarrow{Cu, 573K} RCHO + H_2$
   • Secondary alcohols are dehydrogenated to ketones: $R_2CHOH \xrightarrow{Cu, 573K} R_2C=O + H_2$ …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.When vapour's of a compound X are passed over heated copper, the major product obtained is the acetone. The compound X is:

   (a) n-Propyl alcohol

   (b) Iso-propyl alcohol

   (c) Acetaldehyde

   (d) Propane

   ›Reveal solutionSolution

   Vapours passed over heated copper dehydrogenate 2° alcohols to ketones; since the product is acetone (a ketone), X must be a secondary alcohol — isopropyl alcohol.

   Over heated copper (573 K), alcohols are catalytically dehydrogenated based on their class:

   • 1° alcohol → aldehyde
   • 2° alcohol → ketone
   • 3° alcohol → alkene (dehydration) …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.The most suitable reagent for the conversion of $RCH_2OH \rightarrow RCHO$ is

   (a) $KMnO_4$

   (b) $K_2Cr_2O_7$

   (c) $LiAlH_4$

   (d) PCC (Pyridinium Chlorochromate)

   ›Reveal solutionSolution

   Selective oxidation of a 1° alcohol to an aldehyde (without over-oxidation to the acid) requires an anhydrous, mild oxidant — PCC — rather than a strong aqueous oxidant.

   Why $KMnO_4$ and $K_2Cr_2O_7$ (a, b) fail: these are strong oxidising agents used in aqueous, typically acidified medium. The initially formed aldehyde reacts with water to form a geminal diol (aldehyde hydrate), $RCH(OH)_2$, which is itself readily oxidised further by these strong oxidants to the carboxylic acid, $RCOOH$. So the reaction cannot be stopped cleanly at the aldehyde stage.

   Why $LiAlH_4$ (c) fails: this is a powerful reducing agent (it reduces esters, acids and other carbonyls down to alcohols) — the wrong direction entirely for an oxidation.

   …

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6. CBSE 2026Set ANNUAL1 mark
   Q.Complete the following reaction: aniline (benzene ring with an $-NH_2$ substituent) $+ Br_2(aq) \rightarrow$ ?

   ›Reveal solutionSolution

   The $-NH_2$ group is a powerful activating, ortho/para-directing group, so aniline reacts instantly with bromine water at all three activated ring positions to give 2,4,6-tribromoaniline as a white precipitate.

   The lone pair on the amino nitrogen delocalises into the aromatic ring by resonance, strongly raising electron density especially at the ortho ($2,6$) and para ($4$) positions. This makes those three positions so reactive toward electrophiles that no Lewis-acid catalyst is required (unlike ordinary benzene bromination, which needs $FeBr_3$), and substitution does not stop after one bromination — it proceeds at all three activated sites simultaneously: …

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7. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: In addition of bromine in CCl4 to an alkene resulting in disappearance of reddish brown colour of bromine constitutes, an important method for the detection of ______ in a molecule.

   ›Reveal solutionSolution

   Decolourisation of bromine in CCl4 detects unsaturation (C=C double bond).

   An alkene readily adds bromine across its carbon-carbon double bond to form a colourless dibromide:

   C=C + Br2 -> Br-C-C-Br

   …

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8. CBSE 2026Set ANNUAL1 mark
   Q.Write True or False: Bromo, iodo and polychloro derivatives of hydrocarbons are heavier than water.

   ›Reveal solutionSolution

   True - these halogen derivatives are denser than water.

   The heavy halogen atoms (Br, I) and multiple chlorine atoms greatly increase the molar mass and density of the molecule. As a result, bromo, iodo and polychloro derivatives of hydrocarbons (e.g. bromoform, iodoform, chloroform, …

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9. CBSE 2026Set SEM31 markMCQ
   Q.The reagent which can be used for the following transformation is: phenol (C6H5OH) -> salicylaldehyde (2-hydroxybenzaldehyde, OH and CHO on adjacent ring carbons)

   (a) i) CHCl3, NaOH, 60-80 C ii) dil. HCl

   (b) i) CO2, NaOH, 120-140 C ii) dil. HCl

   (c) i) CCl4, NaOH, 60-80 C ii) dil. HCl

   (d) i) HCHO, NaOH ii) dil. HCl

   ›Reveal solutionSolution

   Phenol + CHCl3 + NaOH (60-80 C) then acidification gives 2-hydroxybenzaldehyde (salicylaldehyde) by the Reimer-Tiemann reaction. Correct option (a).

   In the Reimer-Tiemann reaction, chloroform (CHCl3) with aqueous NaOH generates dichlorocarbene (:CCl2), the electrophile. It attacks the phenoxide ring, chiefly at the ortho position; subsequent hydrolysis on acidification (dil. HCl) converts the -CHCl2 group into -CHO, introducing an aldehyde group ortho to -OH.

   Product: salicylaldehyde (2-hydroxybenzaldehyde).

   • CO2/NaOH (option b) is the Kolbe reaction, giving salicylic acid (-COOH), not the aldehyde. …

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10. CBSE 2025Set 56/5/11 markMCQ
    Q.$CH_3CH_2OH$ can be converted to $CH_3CHO$ by : (A) catalytic hydrogenation (B) treatment with $LiAlH_4$ (C) treatment with PCC (D) treatment with $KMnO_4$

    ›Reveal solutionSolution

    The key idea is that converting ethanol ($CH_3CH_2OH$) to ethanal ($CH_3CHO$) is a controlled oxidation of a primary alcohol to an aldehyde. The correct reagent is PCC (pyridinium chlorochromate), which stops at the aldehyde stage without over-oxidizing to a carboxylic acid.

    This question tests your understanding of alcohol oxidation — a core reaction in organic chemistry. Ethanol is a primary alcohol. To get an aldehyde, you need to oxidize it partially. The challenge is that many strong oxidizers will push the reaction all the way to the carboxylic acid (acetic acid, $CH_3COOH$). So the trick is choosing a reagent that is mild enough to stop at the aldehyde.

    Let’s examine each option.

    1. Option (A): Catalytic hydrogenation

       Hydrogenation ($H_2$ with a metal catalyst like Pd, Pt, or Ni) is a reduction process. It adds hydrogen across double or triple bonds. Ethanol has no multiple bonds to reduce — it’s already saturated. This would do nothing. So this is wrong.

    2. Option (B): Treatment with $LiAlH_4$

       Lithium aluminium hydride is a powerful reducing agent. It reduces carbonyl compounds (aldehydes, ketones, acids, esters) to alcohols. Using it on ethanol would be pointless — ethanol is already an alcohol. It cannot oxidize anything. So this is also wrong.

    3. Option (C): Treatment with PCC

       PCC (pyridinium chlorochromate, $C_5H_5NH^+CrO_3Cl^-$) is a mild oxidizing agent specifically designed for the conversion of primary alcohols to aldehydes. It works in anhydrous conditions (typically in dichloromethane) and stops cleanly at the aldehyde stage.

       The reaction:

    $CH_3CH_2OH \xrightarrow{PCC} CH_3CHO$

    This is the textbook method. So this is correct.

    4. Option (D): Treatment with $KMnO_4$ …

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11. CBSE 2025Set 56/6/11 markMCQ
    Q.Which one of the following amines gives an alcohol on reaction with $HNO_2$ ? (A) $C_6H_5NH_2$ (aniline) (B) $C_2H_5NH_2$ (C) $(C_2H_5)_2NH$ (D) $(C_2H_5)_3N$

    ›Reveal solutionSolution

    The key idea is that primary aliphatic amines react with nitrous acid ($HNO_2$) to give alcohols via a diazonium intermediate that decomposes. Among the options, only $C_2H_5NH_2$ (ethylamine) is a primary aliphatic amine, so it yields ethanol. The correct option is (B).

    The reaction of an amine with nitrous acid ($HNO_2$) is a classic test to distinguish between primary, secondary, and tertiary amines. Nitrous acid is unstable and is prepared in situ by reacting sodium nitrite ($NaNO_2$) with a mineral acid like HCl or $H_2SO_4$. The outcome depends entirely on the class of the amine.

    For primary aliphatic amines (like ethylamine), the reaction proceeds through an unstable alkyldiazonium salt. This salt spontaneously decomposes to give a carbocation, which then reacts with water to form an alcohol. This is the only case where an alcohol is the major product.

    For primary aromatic amines (like aniline), the diazonium salt formed is stable at low temperatures (0–5°C) and does not give an alcohol with water — it gives phenol only upon heating or under specific conditions. At room temperature, aniline reacts with $HNO_2$ to give a diazonium salt that can couple or decompose to other products, but not ethanol.

    For secondary amines (like diethylamine), the reaction yields a yellow, oily N-nitrosamine — no alcohol is formed.

    For tertiary amines (like triethylamine), the reaction gives a nitrosamine salt or simply dissolves, again no alcohol.

    So the only amine that reliably gives an alcohol under standard conditions is a primary aliphatic amine.

    Let’s check each option:

    1. Option (A): $C_6H_5NH_2$ (aniline) — This is a primary aromatic amine. With $HNO_2$ at 0–5°C, it forms a stable benzenediazonium salt. This salt does not decompose to give an alcohol at low temperature; it requires heating with water to yield phenol. Under the usual conditions of the reaction (room temperature or slightly above), aniline gives a diazonium salt that may undergo coupling or other reactions, but not an alcohol. So this is not the answer.

    2. Option (B): $C_2H_5NH_2$ (ethylamine) — This is a primary aliphatic amine. The reaction with $HNO_2$ proceeds as:

    $C_2H_5NH_2 + HNO_2 \rightarrow [C_2H_5N_2^+] \xrightarrow{H_2O} C_2H_5OH + N_2 + H^+$

    The intermediate ethyldiazonium ion is unstable and immediately loses $N_2$ to form an ethyl carbocation, which then reacts with water to give ethanol. This is the classic case where an alcohol is produced. So this is the correct option.

    3. Option (C): $(C_2H_5)_2NH$ (diethylamine) — This is a secondary amine. With $HNO_2$, it forms a yellow, oily N-nitrosamine: …

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12. CBSE 2025Set ANNUAL1 mark
    Q.Aniline does not undergo Friedel-Crafts reaction. Give reason.

    ›Reveal solutionSolution

    The catalyst itself reacts with aniline's basic amino group, deactivating the ring before any substitution can occur.

    Friedel–Crafts reactions (alkylation/acylation) require the Lewis acid catalyst $AlCl_3$. Aniline's $-NH_2$ group is strongly basic (it has a lone pair on nitrogen), so it readily reacts with $AlCl_3$ to form a salt/complex ($C_6H_5\overset{+}{N}H_2-AlCl_3^-$).

    …

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