ABC26IN0249 · Organometallic Compounds
Subject: Inorganic Chemistry · Chapter: Organometallic Compounds · Topic: Oxidative Addition Reductive Elimination · Exam: CSIR-NET 2015 · Marks: 2 · Difficulty: Medium
The $\beta$-hydrogen elimination will be facile in
(a)
(b)
(c)
(d)
Answer
Answer (as printed): A
Explanation
β-hydrogen elimination mechanism. Since C-H, σ bond pair electron donate to the metal for this elimination. Therefore, as the donor ability of the σ electron pair increases rate of β-elimination increases. In option D, the unit is anti-periplanar and thus β-H elimination not possible. For β hydride elimination reaction β hydrogen should be closer to metal in option B,C all hydrogen atoms are far so elimination not possible (sp2 carbon) So, more facile β-elimination occur in option (a) Electron donor ability at a bond is C-H (sp³) > C-H (sp2) >C-H (sp) So, more facile β-elimination occur in option (a)