What is the correct order of acidity of the protons marked A-D in the given compounds?
The key to determining acidity is analyzing the stability of the conjugate base. Factors like resonance, electronegativity, and hybridization play a crucial role. Carboxylic acids are generally much more acidic than carbon acids, which are more acidic than terminal alkynes. Alkynes with a negative charge adjacent to the triple bond are highly unstable.
\(H_C > H_D > H_A > H_B\)
HB > HA > HD > HC
HA > HB > HC > HD
\(H_C > H_A > H_D > H_B\)
The acidity of a proton depends on the stability of the conjugate base formed after its removal. The more stable the conjugate base, the more acidic the proton.
Step 1: Acidity of \(H_C\)
\(H_C\) is the most acidic proton. Removal of \(H_C\) results in a carboxylate anion, which is highly stabilized by resonance.\\[6pt]
Step 2: Acidity of \(H_D\)
\(H_D\) is the second most acidic proton. Its removal forms a carbanion that is stabilized by resonance with the benzene ring.\\[6pt]
Step 3: Acidity of \(H_A\) vs \(H_B\)
\(H_A\) is more acidic than \(H_B\). The conjugate base formed after the removal of \(H_A\) is stabilized by resonance with the triple bond, resulting in a negative charge being distributed over two carbon atoms. However, the conjugate base formed after removal of \(H_B\) results in the negative charge adjacent to the triple bond, with no resonance stabilization. This is highly unstable due to the electron-withdrawing nature of the sp-hybridized carbon in the alkyne.
Step 4: Overall Acidity Order
Therefore, the correct order of acidity is:
\[H_C > H_D > H_A > H_B.\]
Conclusion: Option \((2)\) is correct.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,