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CG PET Aldehydes, Ketones and Carboxylic Acids — practice questions

Explore 51 free Aldehydes, Ketones and Carboxylic Acids questions in CG PET 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 51.

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 →Hydrocarbons are organic compounds composed exclusively of:Read the answer and worked solution →In the presence of organic peroxides, addition of HBr to propene gives predominantly:Read the answer and worked solution →Wurtz reaction is used to prepare: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 →Carbonyl carbon hybridization: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 →Split the carbonyl family in twoRead the answer and worked solution →Rank four acids by strengthRead the answer and worked solution →Match each scientist to their contributionRead the answer and worked solution →Ionic or covalent?Read the answer and worked solution →Complete the driver of bond formationRead the answer and worked solution →Single, double and triple bondsRead the answer and worked solution →True or false: Lewis structuresRead 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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