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JEE Advanced Aldehydes, Ketones and Carboxylic Acids — practice questions

Explore 122 free Aldehydes, Ketones and Carboxylic Acids questions in JEE Advanced Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Aldehydes, Ketones and Carboxylic Acids questions and exercises with answers

Showing questions 1–30 of 122.

The functional group $>\!\mathrm{C}\!=\!\mathrm{O}$ that is common to aldehydes, ketones, carboxylic acids and their derivatives is called:Read the answer and worked solution →Reduction of an aldehyde with $\mathrm{NaBH_4}$ or $\mathrm{LiAlH_4}$ produces:Read the answer and worked solution →Benzaldehyde is treated with concentrated $\mathrm{NaOH}$ (Cannizzaro reaction). The products are:Read the answer and worked solution →Functional group of aldehyde:Read the answer and worked solution →Functional group of ketone:Read the answer and worked solution →Functional group of carboxylic acid:Read the answer and worked solution →IUPAC name of CH₃CHO:Read the answer and worked solution →Fehling's test positive for:Read the answer and worked solution →Nucleophilic addition to carbonyl: nucleophile attacks:Read the answer and worked solution →Reduction of aldehyde RCHO by NaBH₄ gives:Read the answer and worked solution →For α,β-unsaturated ketones (enones), 1,4-addition (Michael) competes with 1,2-addition. Soft nucleophiles prefer:Read the answer and worked solution →Clemmensen reduction converts C=O to CH₂ using:Read the answer and worked solution →Esterification: RCOOH + R'OH + acid catalyst yields:Read the answer and worked solution →In Friedel-Crafts acylation, RCOCl + AlCl₃ + benzene gives:Read the answer and worked solution →Glycine has 2 amino acid forms in solution. Zwitterion:Read the answer and worked solution →The product of the Hunsdiecker reaction on $CH_3CH_2COOAg$ with $Br_2/CCl_4$ isRead the answer and worked solution →In **Dow's process**, chlorobenzene is converted to phenol by treatment withRead the answer and worked solution →The **Finkelstein reaction** uses which solvent for the conversion $RCl \to RI$?Read the answer and worked solution →Why does the Finkelstein reaction proceed efficiently in acetone but not in water?Read the answer and worked solution →A **Grignard reagent** has the general formulaRead the answer and worked solution →The reaction of $CH_3MgBr$ with formaldehyde ($HCHO$), followed by acidic workup, givesRead the answer and worked solution →$CH_3MgBr$ reacts with $CO_2$ followed by acidic workup. The product isRead the answer and worked solution →Which solvent is essential for preparing a Grignard reagent and why?Read the answer and worked solution →The Lewis-acid catalyst typically used in **Friedel–Crafts alkylation** isRead the answer and worked solution →Benzene treated with $CH_3COCl$ and anhydrous $AlCl_3$ givesRead the answer and worked solution →Friedel–Crafts alkylation of benzene with $1$-chloropropane and $AlCl_3$ gives, as the major product,Read the answer and worked solution →The **Swarts reaction** converts an alkyl chloride to an alkyl fluoride usingRead the answer and worked solution →**Williamson's synthesis** of ethers couples an alkoxide withRead the answer and worked solution →In **Wurtz's reaction**, two alkyl halides couple in the presence ofRead the answer and worked solution →The Wurtz reaction on $CH_3CH_2Br$ with sodium in dry ether gives mainlyRead the answer and worked solution →

Worked practice questions

The functional group $>\!\mathrm{C}\!=\!\mathrm{O}$ that is common to aldehydes, ketones, carboxylic acids and their derivatives is called:

  1. Hydroxyl group
  2. Carboxyl group
  3. Alkoxy group
  4. Carbonyl group
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Correct answer: D. Carbonyl group

$>\!\mathrm{C}\!=\!\mathrm{O}$ is the **carbonyl** group — a $C$ doubly bonded to $O$. The **carboxyl** group is $-\mathrm{COOH}$ (carbonyl + hydroxyl together — that's option B). The hydroxyl is $-\mathrm{OH}$ alone (alcohols, phenols). Alkoxy is $-\mathrm{OR}$ (ethers).

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Reduction of an aldehyde with $\mathrm{NaBH_4}$ or $\mathrm{LiAlH_4}$ produces:

  1. A carboxylic acid
  2. A secondary alcohol
  3. A primary alcohol
  4. An alkane
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Correct answer: C. A primary alcohol

Aldehydes ($\mathrm{R-CHO}$) are reduced to **primary** alcohols ($\mathrm{R-CH_2OH}$) because the carbonyl carbon ends up with just one $\mathrm{R}$ group on it. Ketones ($\mathrm{R_2C=O}$) reduce to *secondary* alcohols (option B) for the symmetric reason. Neither $\mathrm{NaBH_4}$ nor $\mathrm{LiAlH_4}$ goes all the way to alkane — that requires harsher conditions (Clemmensen, Wolff–Kishner).

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Benzaldehyde is treated with concentrated $\mathrm{NaOH}$ (Cannizzaro reaction). The products are:

  1. Salicylaldehyde and water
  2. Sodium benzoate and benzyl alcohol
  3. Benzoic acid only
  4. Benzyl alcohol only
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Correct answer: B. Sodium benzoate and benzyl alcohol

**Cannizzaro reaction** is the disproportionation of an aldehyde with **no $\alpha$-hydrogen** under concentrated alkali. One molecule of the aldehyde is oxidised, one is reduced. For benzaldehyde + NaOH (conc.): - One PhCHO is *oxidised* to PhCOO⁻Na⁺ (**sodium benzoate**). - One PhCHO is *reduced* to PhCH₂OH (**benzyl alcohol**). So the product mix is option C. Acidifying the workup converts the sodium benzoate to benzoic acid, but the raw Cannizzaro products are as written. **Why option A is wrong on its own.** Plain oxidation of benzaldehyde with stronger oxidants gives benzoic acid alone that's *not* Cannizzaro — Cannizzaro requires the simultaneous reduction half.

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