Step 1: Definition of acidic hydrogen
An acidic hydrogen is one that can be easily removed as a proton (\( \text{H}^+ \)), resulting in a stable conjugate base. The more stable the conjugate base, the more acidic the hydrogen.
Step 2: Analyze conjugate base stability
- In Option (B), the removal of the hydrogen leads to a carbanion which is resonance stabilized by conjugation with two adjacent double bonds. This forms an allylic anion extended over a conjugated cyclic system. - In Options (A), (C), and (D), although allylic positions are present, the resonance stabilization is less extensive compared to (B). Especially, (A) and (C) are sterically hindered or less effectively delocalized.
Step 3: Resonance in Option (B)
Removing the proton from the carbon adjacent to two double bonds gives a conjugate base that can delocalize the negative charge over a five-carbon conjugated system, similar to a cyclopentadienyl-like system — which is known for its aromatic stabilization when fully conjugated.
Conclusion: Option (B) has the most stable conjugate base, hence the most acidic hydrogen.
Final Answer: \( \boxed{\text{B}} \)
Given below are some nitrogen containing compounds:
Each of them is treated with HCl separately. 1.0 g of the most basic compound will consume ...... mg of HCl.
(Given Molar mass in g mol\(^{-1}\): C = 12, H = 1, O = 16, Cl = 35.5.)
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity):
In a scattering experiment, a particle of mass $ 2m $ collides with another particle of mass $ m $, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation $ \theta $ of the heavier particle, as shown in the figure, in radians is:
A conducting square loop initially lies in the $ XZ $ plane with its lower edge hinged along the $ X $-axis. Only in the region $ y \geq 0 $, there is a time dependent magnetic field pointing along the $ Z $-direction, $ \vec{B}(t) = B_0 (\cos \omega t) \hat{k} $, where $ B_0 $ is a constant. The magnetic field is zero everywhere else. At time $ t = 0 $, the loop starts rotating with constant angular speed $ \omega $ about the $ X $ axis in the clockwise direction as viewed from the $ +X $ axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. ($ V $) in the loop as a function of time:
Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $ D $ of a tube. The measured value of $ D $ is:
A conducting square loop of side $ L $, mass $ M $, and resistance $ R $ is moving in the $ XY $ plane with its edges parallel to the $ X $ and $ Y $ axes. The region $ y \geq 0 $ has a uniform magnetic field, $ \vec{B} = B_0 \hat{k} $. The magnetic field is zero everywhere else. At time $ t = 0 $, the loop starts to enter the magnetic field with an initial velocity $ v_0 \hat{j} \, \text{m/s} $, as shown in the figure. Considering the quantity $ K = \frac{B_0^2 L^2}{RM} $ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct: