Exams › JEE Advanced › Chemistry
Among the marked alpha-hydrogen atoms in the following compounds, which compound has the marked hydrogen with the lowest pKa (most acidic)?
- H2C-COOH
- H2C-C(=O)-CH3
- Ph-CH(H)-C(=O)-CH3
- Ph-CH(H)-C(=O)-OH
Correct answer: H2C-COOH
Solution
The compound H2C-COOH can be interpreted as malonic acid or a diacid where the alpha carbon is flanked by two -COOH groups, providing maximum resonance and inductive stabilization of the carbanion. This gives the lowest pKa among the listed options. The phenyl group in options C and D also stabilizes, but two strong EW groups in option A are more effective.
Related JEE Advanced Chemistry questions
- In its interactions with potassium periodate (KIO4) and hydroxylamine (NH2OH), hydrogen peroxide functions respectively as a
- Among the two isomers of hydroxybenzoic acid, the ortho isomer is a stronger acid compared to its para counterpart. Which of the following best explains this observation?
- In which of the following reactions is hydrogen gas produced?
- How many of the following compounds are classified as saline (ionic) hydrides? NaH, MgH2, CaH2, VH0.56, Fe3H, NH3
- In how many of the following processes is H2 gas released as a product? (i) Na + H2O --> (ii) KH + H2O --> (iii) CO + H2 (steam, Fe2O3/Cr2O3 catalyst) --> (iv) Electrolysis of fused (molten) NaH (v) Electrolysis of acidic water
- The wavelength of the series limit (longest wavelength limit, i.e., the convergence limit) of the Lyman series of hydrogen is x angstrom. Calculate the wave number (in angstrom⁻¹) of the third line of the Balmer series of He+ ion.
⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →