Q103 · CSIR-NET Chemistry, December 2019
Paper: CSIR-NET December 2019 · Subject: Organic Chemistry · Chapter: Organic Spectroscopy · Topic: Combined Structure Elucidation · Marks: 2 · Difficulty: Hard
The structure that exhibits the following spectral data is
\[ \begin{array}{l} \text{IR } (\nu_{\max}): 1740\,\mathrm{cm^{-1}} \\ {}^{1}\mathrm{H}\ \text{NMR}: \delta=0.9(\mathrm{t},3\mathrm{H}), 1.6(\mathrm{sext},2\mathrm{H}), 2.3(\mathrm{t},2\mathrm{H}), 4.6(\mathrm{d},2\mathrm{H}), 5.2(\mathrm{d},1\mathrm{H}), 5.4(\mathrm{d},1\mathrm{H}), 5.9(\mathrm{m},1\mathrm{H})\ \text{ppm} \\ \text{EI-MS } (m/z): 71\ (100\%) \end{array} \]
(a)
(b)
(c)
(d)
Answer
Answer: D ✓ checked by 4AB · confidence low
The source book printed A; on checking, D is correct — see the explanation.
Explanation
The 0.9 t / 1.6 sext / 2.3 t pattern is a butanoyl group (m/z 71 = C₃H₇CO⁺). OCH₂ at 4.6 (d) plus three vinyl H show an allyl ester, and ν 1740 cm⁻¹ an ester: allyl butanoate, closest to (d).
Study loop for Organic Spectroscopy