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Indian Navy SSR Aldehydes, Ketones and Carboxylic Acids — practice questions

Explore 29 free Aldehydes, Ketones and Carboxylic Acids questions in Indian Navy SSR 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 1–29 of 29.

Benzaldehyde is treated with concentrated $\mathrm{NaOH}$ (Cannizzaro reaction). The products are: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 →As per your syllabus, Mechanism, the FIRST step of methane chlorination under sunlight is best described as which?Read the answer and worked solution →As per your syllabus, the carbon-carbon triple bond in ethyne (HC≡CH) consists of which σ/π combination?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 →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 →Complete the definition of resonanceRead the answer and worked solution →Tap every bond parameter the chapter namesRead the answer and worked solution →True or false: ionic solidsRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Bonded pair or lone pair?Read the answer and worked solution →Flip and pair: term with meaningRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Order the reasoning: why resonance is neededRead the answer and worked solution →Reading the definitions backRead the answer and worked solution →True or false: what bond parameters affectRead the answer and worked solution →Bond parameters in useRead the answer and worked solution →Which parameter answers this?Read the answer and worked solution →True or false: resonanceRead the answer and worked solution →Order the steps: draw a Lewis structureRead the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Assertion and reason: structureRead the answer and worked solution →Flip and pair: parameter with what it measuresRead the answer and worked solution →

Worked practice questions

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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In the presence of organic peroxides, addition of HBr to propene gives predominantly:

  1. 1-bromopropane (anti-Markovnikov)
  2. propan-1-ol
  3. 2-bromopropane (Markovnikov)
  4. propane
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Correct answer: A. 1-bromopropane (anti-Markovnikov)

Kharasch peroxide effect: HBr in the presence of organic peroxides (e.g. benzoyl peroxide) adds **anti-Markovnikov** via a free-radical mechanism. Br ends up on the **less** substituted carbon. Product: 1-bromopropane. Why: the radical chain involves a Br• radical adding to the alkene first. Br• prefers the less-substituted carbon, forming the more stable carbon radical. Then HBr donates H• to complete the addition. Notably, the peroxide effect is **specific to HBr**. HCl and HI don't show it.

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As per your syllabus, Mechanism, the FIRST step of methane chlorination under sunlight is best described as which?

  1. Heterolysis of C-H bond
  2. Markovnikov addition
  3. Homolysis of Cl-Cl bond
  4. Wurtz coupling
Show answer and solution

Correct answer: C. Homolysis of Cl-Cl bond

syllabus Mechanism (i): Cl-Cl homolysis under hν gives two Cl• free radicals. Free-radical chain mechanism.

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More Chemistry topics

Organic ChemistryAlkenesAlcohols Phenols EthersPolymersChemical Equilibriump-Block ElementsThe p-Block ElementsAtoms and Molecules