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MP Board Class 12 (MPBSE) Aldehydes Ketones — practice questions

Explore 8 free Aldehydes Ketones questions in MP Board Class 12 (MPBSE) Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Aldehydes Ketones questions and exercises with answers

Showing questions 1–8 of 8.

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 →The reaction of formaldehyde with concentrated NaOH is calledRead the answer and worked solution →Split the carbonyl family in twoRead the answer and worked solution →Match each carbonyl testRead the answer and worked solution →Which one does NOT belong?Read the answer and worked solution →

Worked practice questions

Tollens' reagent gives a silver mirror with:

  1. Acetone, the classic ketone solvent on the shelf
  2. Methanol, an alcohol on the chemist shop shelf
  3. Acetaldehyde, an aldehyde that reduces Ag$^+$ to Ag
  4. Benzene, an aromatic hydrocarbon on the bench too
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Correct answer: C. Acetaldehyde, an aldehyde that reduces Ag$^+$ to Ag

Aldehydes reduce Tollens' Ag(NH$_3$)$_2^+$ to metallic silver (silver mirror). Acetone (ketone) does not.

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The IUPAC name of the simplest aldehyde, $\mathrm{HCHO}$, is

  1. methanal
  2. methanol
  3. methanoic acid
  4. ethanal
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Correct answer: A. methanal

1. Your syllabus: the IUPAC name of an aldehyde is formed by replacing the '-e' of the parent alkane with '-al'. 2. $\mathrm{HCHO}$ has ONE carbon atom (the parent alkane is methane). Adding the '-al' suffix gives METHANAL. 3. Common name: formaldehyde. 4. Distractors: methanol (option B) is $\mathrm{CH_3OH}$, an alcohol. Methanoic acid (option C) is $\mathrm{HCOOH}$, the carboxylic acid form. Ethanal (option D) is $\mathrm{CH_3CHO}$, the TWO-carbon aldehyde. _Source: syllabus Chemistry Part 2, Ch 8 "Aldehydes, Ketones and Carboxylic Acids", §8.1.1 (Nomenclature, IUPAC rules), p. 1–2._

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The carbonyl carbon in an aldehyde or ketone is

  1. $sp^3$ hybridised, with bond angles of $\sim 109.5^\circ$
  2. $sp$ hybridised, with bond angles of $180^\circ$
  3. $sp^2$ hybridised, with bond angles of $\sim 120^\circ$
  4. purely $p$-orbital, with no hybridisation
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Correct answer: C. $sp^2$ hybridised, with bond angles of $\sim 120^\circ$

1. Your syllabus: the carbonyl carbon has THREE atoms bonded to it (the doubly-bonded O and two other atoms — in formaldehyde, two H's; in acetaldehyde, one H and one C; in acetone, two C's). 2. Three sigma bonds around the central carbon → $sp^2$ HYBRIDISATION. Three sp$^2$ hybrids lie in a plane separated by $120^\circ$. 3. The fourth (unhybridised) $p$ orbital perpendicular to the plane forms the $\pi$ bond with the oxygen. 4. Hence the carbonyl group is PLANAR, with bond angles close to $120^\circ$. Real molecules show slight distortion (e.g. $\sim 121.7^\circ$ for $\angle\mathrm{HCO}$ in formaldehyde). 5. Option A is for tetrahedral $sp^3$ carbons (alkanes). Option B is the geometry of alkynes ($sp$). Option D is wrong — bonded atoms require hybridisation. _Source: syllabus Chemistry Part 2, Ch 8, §8.1.2 (Structure of the Carbonyl Group), p. 4._

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More Chemistry topics

Coordination CompoundsNamed ReactionsAlcohols Phenols EthersChemical KineticsSolid StateAlkenesAtomic combinationsAtoms and Elements