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AP Intermediate 2nd Year Aldehydes, Ketones and Carboxylic Acids — practice questions

Explore 54 free Aldehydes, Ketones and Carboxylic Acids questions in AP Intermediate 2nd Year 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 54.

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 →The carbonyl carbon in aldehydes and ketones is:Read the answer and worked solution →Glutaric acid is the common name for:Read the answer and worked solution →A reaction of an aldehyde with ammonia followed by a primary amine R-NH₂ gives:Read the answer and worked solution →**Clemmensen reduction** converts C=O of a ketone to CH₂ using: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 →As per your syllabus, nitration, the actual ELECTROPHILE that attacks benzene in HNO₃ + H₂SO₄ mixture is which species?Read the answer and worked solution →As per your syllabus, halogenation, the role of anhydrous AlCl₃ (or FeCl₃) in benzene chlorination is which?Read the answer and worked solution →As per your syllabus, Friedel-Crafts ALKYLATION of benzene uses which reagent system?Read the answer and worked solution →As per your syllabus, Friedel-Crafts ACYLATION of benzene uses an acyl halide + AlCl₃ to generate which electrophile?Read the answer and worked solution →As per your syllabus, which of the following groups is a META-directing substituent on benzene?Read the answer and worked solution →As per your syllabus, halogens (-Cl, -Br) on benzene are which kind of directors despite their deactivating -I effect?Read the answer and worked solution →The general formula for alkanes is:Read the answer and worked solution →Tollens' reagent gives a silver mirror with:Read the answer and worked solution →The IUPAC name of the simplest aldehyde, $\mathrm{HCHO}$, isRead the answer and worked solution →The carbonyl carbon in an aldehyde or ketone isRead the answer and worked solution →Which carboxylic acid has the highest $K_a$ (i.e. is the most acidic)?Read 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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