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 is the empirical formula of a compound containing 40% sulfur and 60% oxygen by mass?
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).

In the first configuration (1) as shown in the figure, four identical charges \( q_0 \) are kept at the corners A, B, C and D of square of side length \( a \). In the second configuration (2), the same charges are shifted to mid points C, E, H, and F of the square. If \( K = \frac{1}{4\pi \epsilon_0} \), the difference between the potential energies of configuration (2) and (1) is given by:
Given below are two statements:
Statement I:
will undergo alkaline hydrolysis at a faster rate than 
Statement II:
In
intramolecular substitution takes place first by involving lone pair of electrons on nitrogen.