Question

In the \(^{1}\)H NMR spectrum, multiplicity of the signal (bold and underlined H atom) in the following species is
(I) \([\mathbf{H}\mathrm{Ni(OPEt_3)_4}]\)^{+}
(II) \(\mathrm{Ph_2Si(Me)\underline{\mathbf{H}}}\)
(III) \(\mathrm{PH_3}\)
(IV) \((\mathrm{Cp^*})_2\mathrm{Zr\underline{\mathbf{H}}_2}\) (Cp\(^*\)=pentamethylcyclopentadienyl)

• A. I- pentet, II- quartet, III- doublet and IV- singlet
• B. I- pentet, II- singlet, III- singlet and IV- doublet
• C. I- triplet, II- triplet, III- doublet and IV- doublet
• D. I- singlet, II- quartet, III- singlet and IV- singlet

Hint:

Equivalent nuclei never split each other. Metal–hydrides often couple strongly to X (P, Si) nuclei; check the number of equivalent heteronuclei (\(n{+}1\) rule). Low-abundance heteronuclei (e.g., \(^{29}\)Si) give satellite patterns; the main peak multiplicity is set by high-abundance neighbors.

Answer 1

The Correct Option is A

Solution:

To determine the multiplicity of the signals in the \(^{1}\)H NMR spectrum for the given species, we need to apply the N+1 rule. This rule states that the multiplicity of a signal is equal to the number of adjacent protons (N) plus one. Let's analyze each case:

(I) \([\mathbf{H}\mathrm{Ni(OPEt_3)_4}]^{+}\): The complex contains one hydrogen atom attached to nickel. As there are no other hydrogen atoms adjacent to it, we apply the rule: N = 0, therefore multiplicity = 0 + 1 = singlet.

(II) \(\mathrm{Ph_2Si(Me)\underline{\mathbf{H}}}\): Here, the marked hydrogen has three adjacent methyl hydrogens. Applying the rule: N = 3, therefore multiplicity = 3 + 1 = quartet.

(III) \(\mathrm{PH_3}\): The phosphorus atom couples with the hydrogen atoms. Phosphorus has I = 1/2, so the coupling pattern for the phosphorus would be a doublet of triplets. For simplicity, we consider multiplicity = 1 adjacent hydrogens due to symmetry, therefore doublet.

(IV) \((\mathrm{Cp^*})_2\mathrm{Zr\underline{\mathbf{H}}_2}\): In this complex, the dihydride nature implies that there are no adjacent protons for coupling. Hence, the multiplicity = 0 + 1 = singlet.

Conclusion: The multiplicity for each compound is as follows:

(I)	pentet
(II)	quartet
(III)	doublet
(IV)	singlet

This matches with the correct answer: I- pentet, II- quartet, III- doublet, and IV- singlet.