4AB › Chemistry PYQ › CSIR-NET › December 2013 › Q81Q81 · CSIR-NET Chemistry, December 2013 Paper: CSIR-NET December 2013 · Subject: Organic Chemistry · Chapter: Organic Spectroscopy · Topic: NMR 1H · Marks: 2 · Difficulty: Medium
Methyl 4-oxopentanoate exhibited signals at $\delta$ 208, 172, 51, 37, 32 and 27 ppm in its ${ }^{13} \mathrm{C}$ NMR spectrum. The signals due to the methoxy, $\mathrm{C}_{1}, \mathrm{C}_{4}$ and $\mathrm{C}_{5}$ carbons are
(a) OMe -32; $\mathrm{C}_{1}-208; \mathrm{C}_{4}-172; \mathrm{C}_{5}-51$
(b) OMe-51; $\mathrm{C}_{1}-208; \mathrm{C}_{4}-172; \mathrm{C}_{5}-32$
(c) OMe-32; $\mathrm{C}_{1}-172; \mathrm{C}_{4}-208; \mathrm{C}_{5}-51$
(d) OMe-51; $\mathrm{C}_{1}-172, \mathrm{C}_{4}-208; \mathrm{C}_{5}-32$
Answer Answer: D ✓ checked by 4AB · confidence high
Explanation CH₃O–C(O)–CH₂CH₂–C(O)–CH₃: OMe 51, ester C-1 172, ketone C-4 208, C-5 (acetyl CH₃) 30–32; the CH₂ carbons are 37 and 27.
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