Question:
The decreasing order of hydride affinity for following carbocations is:
Choose the correct answer from the options given below:
The decreasing order of hydride affinity for following carbocations is:
Choose the correct answer from the options given below:
Choose the correct answer from the options given below:
Updated On: Mar 21, 2025
- C, A, D, B
- A, C, D, B
- C, A, B, D
- A, C, B, D
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Analyzing the stability and hydride affinity of the given carbocations.
Carbocation A is stabilized through conjugation with a double bond.
Carbocation B is stabilized by conjugation with three phenyl rings, making it the most stable and thus possessing the highest hydride affinity.
Carbocation D is the least stable due to the absence of resonance and inductive effects.
Therefore, the decreasing order of hydride affinity is: C, A, B, D.
Carbocation A is stabilized through conjugation with a double bond.
Carbocation B is stabilized by conjugation with three phenyl rings, making it the most stable and thus possessing the highest hydride affinity.
Carbocation D is the least stable due to the absence of resonance and inductive effects.
Therefore, the decreasing order of hydride affinity is: C, A, B, D.
Was this answer helpful?
0
0
Top Questions on Structure of atom
- Isotopes of an element have a different number of
- UPCATET - 2025
- Physics
- Structure of atom
- The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom is –
- UPCATET - 2025
- Physics
- Structure of atom
- An element 'M' has 25% of the electrons filled in the third shell as in the second shell. The element 'M' is :
- CBSE Class X - 2025
- Science
- Structure of atom
- Which of the following is correct about the stability of half-filled and completely filled subshells?
(i) Relatively small shielding
(ii) Larger coulombic repulsion energy
(iii) Smaller exchange energy
(iv) Smaller coulombic repulsion energy
(v) Larger exchange energy- KEAM - 2025
- Chemistry
- Structure of atom
- Consider the last electron of an element having atomic number $9$ and choose the correct option.
- JEE Main - 2025
- Chemistry
- Structure of atom
View More Questions
Questions Asked in JEE Main exam
- Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is:
- JEE Main - 2025
- Arithmetic Progression
- Let \( f(x) = \frac{x - 5}{x^2 - 3x + 2} \). If the range of \( f(x) \) is \( (-\infty, \alpha) \cup (\beta, \infty) \), then \( \alpha^2 + \beta^2 \) equals to.
- JEE Main - 2025
- Functions
- CrCl\(_3\).xNH\(_3\) can exist as a complex. 0.1 molal aqueous solution of this complex shows a depression in freezing point of 0.558°C. Assuming 100\% ionization of this complex and coordination number of Cr is 6, the complex will be:
- JEE Main - 2025
- Colligative Properties
- Suppose that the number of terms in an A.P. is \( 2k, k \in \mathbb{N} \). If the sum of all odd terms of the A.P. is 40, the sum of all even terms is 55, and the last term of the A.P. exceeds the first term by 27, then \( k \) is equal to:
- JEE Main - 2025
- Arithmetic Progression
- If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:
- JEE Main - 2025
- Complex numbers
View More Questions