Acidity of a compound is related to the stability of its conjugate base. The more stabilized the conjugate base, the stronger the acid.
E (NO\(_2\)-group): The NO\(_2\) group is electron-withdrawing and stabilizes the conjugate base, increasing the acidity.
C (Phenol): This has no electron-donating or electron-withdrawing group, so it has moderate acidity.
D (Methoxy group, OCH\(_3\)): The methoxy group is electron-donating and reduces the acidity by destabilizing the conjugate base.
A (Methanol): Alcohols generally have low acidity compared to phenols due to the absence of a conjugate base that can be stabilized by resonance.
B (Tertiary alcohol): The tertiary alcohol, due to steric hindrance and electron-donating alkyl groups, is the least acidic.
Thus, the correct order of acidity is: \[ \text{E} > \text{C} > \text{D} > \text{A} > \text{B} \]
The correct increasing order of stability of the complexes based on \( \Delta \) value is:
Match List-I with List-II: List-I
List I (Molecule) | List II (Number and types of bond/s between two carbon atoms) | ||
A. | ethane | I. | one σ-bond and two π-bonds |
B. | ethene | II. | two π-bonds |
C. | carbon molecule, C2 | III. | one σ-bonds |
D. | ethyne | IV. | one σ-bond and one π-bond |
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: