The ascending acidity order of the following H atoms is

Solution

Chemistry ma’am · 4 steps

Pull off each H, then ask how happily that carbon holds the lone pair left behind.

1. Look at the carbanion. Removing an H as H⁺ leaves a lone pair on carbon. The more stable that carbanion, the more acidic the H.
2. Hybridisation first. More s character holds the lone pair closer to the nucleus, so the carbon behaves as more electronegative. The order is sp (50% s) > sp² (33% s) > sp³ (25% s), which puts A above B, and B above both C and D.
3. Then alkyl groups. C and D are both sp³, so look at the substituents. Methyl groups push electrons in (+I effect), which piles more charge onto a carbanion and destabilises it. The tertiary carbanion from C has three methyls, while the primary one from D has only one, so C is the least acidic.

   H	Carbon	Carbanion formed	Acidity rank
   A	sp, ethyne	HC≡C⁻, lone pair in sp orbital	most acidic
   B	sp², ethene	H₂C=CH⁻, lone pair in sp² orbital	second
   D	sp³, primary	CH₃CH₂⁻, one methyl pushing in	third
   C	sp³, tertiary	(CH₃)₃C⁻, three methyls pushing in	least acidic

4. So. In ascending order the acidity is C < D < B < A, which is option (d).

The idea to keep. For C–H acidity, rank by s character first (sp > sp² > sp³). Among sp³ carbons, every extra alkyl group destabilises the carbanion, so 1° > 2° > 3°, the reverse of the carbocation order.

Answer from the NTA final answer key. Solution by PrepWiser.