$\mathrm{Na}\left[\left(\eta^{5}-\mathrm{C}_{5}\mathrm{H}_{5}\right)\mathrm{Fe}(\mathrm{CO})_{2}\right]$ reacts with $\mathrm{Br}_{2}$ to give A. Reaction of A with $\mathrm{LiAlH}_{4}$ results in B. The proton NMR spectrum of B consists of two singlets of relative intensity $5:1$. Compounds A and B respectively, are
(a)$\left(\eta^{5}-\mathrm{C}_{5} \mathrm{H}_{5}\right) \mathrm{Fe}(\mathrm{CO})_{2} \mathrm{Br}$, and $\left(\eta^{5}-\mathrm{C}_{5} \mathrm{H}_{5}\right) \mathrm{Fe}(\mathrm{CO})_{2} \mathrm{H}$
(b)$\left(\eta^{4}-\mathrm{C}_{5} \mathrm{H}_{5}\right) \mathrm{Fe}(\mathrm{CO})_{2} \mathrm{Br}_{2}$ and $\left(\eta^{4}-\mathrm{C}_{5} \mathrm{H}_{5}\right) \mathrm{Fe}(\mathrm{CO})_{2} \mathrm{HBr}$
(c)$\left(\eta^{5}-\mathrm{C}_{5} \mathrm{H}_{5}\right) \mathrm{Fe}(\mathrm{CO})_{2} \mathrm{Br}$ and $\left(\eta^{4}-\mathrm{C}_{5} \mathrm{H}_{5}\right) \mathrm{Fe}(\mathrm{CO})_{2}(\mathrm{H})_{2}$
(d)$\left(\eta^{5}-\mathrm{C}_{5} \mathrm{H}_{5}\right) \mathrm{Fe}(\mathrm{CO})_{2} \mathrm{Br}$ and $\left(\eta^{5}-\mathrm{C}_{5} \mathrm{H}_{5}\right) \mathrm{Fe}(\mathrm{CO})_{2} \mathrm{HBr}$
Answer
Answer: A ✓ checked by 4AB · confidence high
Explanation
Br₂ cleaves Na[CpFe(CO)₂] to CpFe(CO)₂Br (A). LiAlH₄ gives the hydride CpFe(CO)₂H (B), whose Cp : H singlets are 5 : 1 (a).