Write the pKa order of the following acids.
The pKa order depends on the stability of the conjugate base. More stable conjugate base means stronger acid (lower pKa).
Acid (x) loses a proton to form a conjugate base stabilized by two equivalent resonance structures and an electron-withdrawing nitro group. Acid (y) forms a conjugate base with resonance, but the nitro group is meta, offering less stabilization. Acid (z) forms a conjugate base with no resonance stabilization.
Therefore, acid strength is x > y > z, meaning pKa order is x < y < z. The correct option is the one stating x > y > z.
Final Answer: x > y > z
The pKa value indicates the acidity strength of a compound. A lower pKa means stronger acid (more tendency to donate a proton), while a higher pKa means weaker acid. For carboxylic acids, electron-withdrawing groups (EWG) increase acidity (lower pKa) by stabilizing the conjugate base, while electron-donating groups (EDG) decrease acidity (higher pKa) by destabilizing the conjugate base.
Identify the substituents on the carboxylic acid group:
Compare the electron-withdrawing/donating effects:
Strongest acid (lowest pKa) to weakest acid (highest pKa): y > z > x
Therefore, the pKa order is: y < z < x (meaning pKay < pKaz < pKax)
The correct order is: y > z > x (which corresponds to option: y > x > z is incorrect; the correct order should be y > z > x)
Among the given options, x > z > y is closest to the reverse of correct order, but none match exactly. The correct relationship is pKay < pKaz < pKax.
Acidity constant:
pKa definition:
Lower pKa = stronger acid
Electron-withdrawing groups stabilize the conjugate base, increasing acidity
Electron-donating groups destabilize the conjugate base, decreasing acidity
Arrange the following compounds in order of decreasing acidity :
The total number of basic groups in the following form of lysine is
Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below:
An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only organic product through the following sequence of reactions, in which Q is an intermediate organic compound.
P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2/H2O. On heating, P forms the cyclic anhydride.
Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U.
