



To determine which compound has a highly acidic hydrogen, we need to evaluate the structure of the compounds presented in the options. A hydrogen atom is considered acidic if it can be easily removed as a proton (H+) from a molecule. This typically occurs in certain functional groups known for having acidic hydrogens, such as carboxylic acids, phenols, or enols.
Among common compounds, acidic hydrogens are typically found in:
In the context of this question, the options are given in the form of images, and one of them has been indicated as the correct answer. To confirm why the correct option contains a highly acidic hydrogen, let's consider the typical examples:
1. **Carboxylic Acids**: Hydrogen attached to the -COOH group is highly acidic.
2. **Phenols**: Hydrogen in the -OH group when attached directly to an aromatic ring is moderately acidic.
3. **Enoles**: In keto-enol tautomerism, the hydrogen in the enol form can also be acidic.
The correct answer image provided:
This particular image corresponds to a structure that potentially contains a group like the carboxylic acid or a similar functional group with an acidic hydrogen. As confirmed by selecting this image as the correct answer, the molecule in question likely contains such a functional group.
In conclusion, the chosen image suggests the presence of a compound or group with significantly acidic hydrogen. This reasoning ensures the understanding that only certain functional groups exhibit high acidity, which fits the correct answer.
In structure (4), the methylene group (CH2) between two carbonyl groups (C=O) is highly acidic due to the resonance stabilization of the conjugate base. When the hydrogen is removed, the negative charge on the carbon is delocalized between the two carbonyl groups, making the conjugate base more stable. This increased resonance makes the hydrogen highly acidic.
The correct increasing order of stability of the complexes based on \( \Delta \) value is:
Match List-I with List-II: List-I
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $