Question

High‐resolution mass spectral analysis of a pure sample of compound P reveals it to have a molecular formula of C₅H₁₀O₃. The ¹H NMR spectrum of P in CDCl₃has the following signals: A. 11.20ppm chemical shift singlet, broad with an integration of 32 B. 3.71ppm chemical shift triplet with an integration of 67 C. 3.53ppm chemical shift quartet with an integration of 64 D.a 2.64ppm shift triplet with an integration of 65 E. 1.21ppm chemcial shift triplet with an integration of 98. Draw out the lewis structure of compound P and draw an arrow pointing to one of the hydrogens that gives rise to the signal for each chemical shift.

Answer 1

Answer – Given, molecular formula, C₅H₁₀O₃ –

Degree of unsaturation = C₅H₁₂ - C₅H₁₀ = 2H = 1

11.20 ppm chemical shift singlet, broad with an integration of 32, means it is for carboxylic acid proton, so 32 integration = 1 H

3.71ppm chemical shift triplet with an integration of 67, means 2 H and it like –CH₂-CH­₂-

3.53 ppm chemical shift quartet with an integration of 64 means 2 H and it is like –CH₂-CH₃

2.64 ppm shift triplet with an integration of 65 means for 2 H and it is like -CH₂-CH­₂-

1.21 ppm chemcial shift triplet with an integration of 98 for 3H and it is like –CH₂-CH₃

So structure is –