MP Board Class 12 (MPBSE) Aldehydes, Ketones and Carboxylic Acids — practice questions

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

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._

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._