Given: The reactivity of aldehydes and ketones in nucleophilic addition reactions is influenced by the electron-withdrawing or electron-donating effects of substituents attached to the aromatic ring or carbonyl group.
- Electron-withdrawing groups (such as \( -NO_2 \)) attached to the aromatic ring increase the electrophilicity of the carbonyl carbon, thus increasing reactivity towards nucleophiles. - Electron-donating groups (such as \( -CH_3 \)) decrease the electrophilicity of the carbonyl carbon, decreasing reactivity towards nucleophiles.
- Benzaldehyde has no electron-donating or electron-withdrawing groups attached to the benzene ring. - It has moderate reactivity towards nucleophiles.
- The \( -CH_3 \) group is an electron-donating group, which decreases the electrophilicity of the carbonyl carbon. - Therefore, acetophenone has lower reactivity compared to benzaldehyde.
- The \( -CH_3 \) group is an electron-donating group, but less so than in acetophenone. - It lowers the electrophilicity of the carbonyl group but is more reactive than acetophenone.
- The \( -NO_2 \) group is an electron-withdrawing group, which significantly increases the electrophilicity of the carbonyl carbon. - This makes p-nitrobenzaldehyde highly reactive towards nucleophiles.
Based on the electron-donating and electron-withdrawing effects of the substituents, the reactivity order is: \[ \text{Acetophenone} < \text{p-Tolualdehyde} < \text{Benzaldehyde} < \text{p-Nitrobenzaldehyde}. \]
The correct answer is \( \boxed{(3)} \), which corresponds to **p-Tolualdehyde**.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.