Q55 · CSIR-NET Chemistry, June 2012

Paper: CSIR-NET June 2012 · Subject: Organic Chemistry · Chapter: Organic Spectroscopy · Topic: NMR 1H · Marks: 2 · Difficulty: Medium

In the $400 \mathrm{MHz},{ }^{1} \mathrm{H} \mathrm{NMR}$ spectrum of organic compound exhibited a doublet. The two lines of the doublet are at $\delta 2.35$ and 2.38 ppm . The coupling constant $(\mathrm{J})$ value is
(a)3 Hz
(b)6 Hz
(c)9 Hz
(d)12 Hz
Answer
Answer: D ✓ checked by 4AB · confidence high
Explanation
Δδ = 0.03 ppm × 400 MHz = 12 Hz.

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