Chemical Shifts and Coupling Constants in NMR Spectroscopy

Introduction:
Nuclear Magnetic Resonance (NMR) spectroscopy is a powerful analytical technique used to determine the molecular structure and composition of organic compounds. It provides valuable information about the chemical environment and interactions of atoms within a molecule. In NMR spectra, chemical shifts and coupling constants are two important parameters that help in the interpretation of the spectra.

Chemical Shift:
Chemical shift is a measure of the resonance frequency of a nucleus in a magnetic field, relative to a reference compound. It is expressed in parts per million (ppm) and provides information about the electronic environment surrounding the nucleus. The chemical shift depends on various factors such as electronegativity, hybridization, and neighboring atoms.

Coupling Constant:
Coupling constant, denoted as J, is a measure of the splitting or multiplicity of a signal in an NMR spectrum. It represents the degree of magnetic interaction between two or more nuclei in a molecule. Coupling constants are measured in Hertz (Hz) and can provide information about the connectivity and stereochemistry of the molecule.

Interpretation of the Problem:
In the given problem, we have a doublet signal for a proton in an NMR spectrum with chemical shifts of 4.08 and 4.06 ppm. We need to determine the coupling constant in Hz.

Solution:
To determine the coupling constant, we first need to understand the splitting pattern of a doublet. In a doublet, the signal is split into two peaks of equal intensity due to the coupling with one neighboring proton. The coupling constant (J) is equal to the difference in resonance frequencies between these two peaks.

Calculation:
Given chemical shifts: 4.08 ppm and 4.06 ppm
Frequency of the NMR spectrometer: 400 MHz

The difference in resonance frequencies (Δv) can be calculated using the formula:

Δv = (Δδ) * ν
where Δv is the difference in resonance frequencies, Δδ is the difference in chemical shifts, and ν is the frequency of the NMR spectrometer.

Δv = (4.08 ppm - 4.06 ppm) * 400 MHz
Δv = 0.02 ppm * 400 MHz
Δv = 8 Hz

Therefore, the coupling constant (J) is equal to 8 Hz.

Answer:
The coupling constant in Hz is 8.

Explanation:
The chemical shifts of 4.08 and 4.06 ppm indicate the presence of a doublet signal in the NMR spectrum. The difference in chemical shifts between the two peaks is 0.02 ppm. By multiplying this value with the frequency of the NMR spectrometer (400 MHz), we can calculate the difference in resonance frequencies. This difference corresponds to the coupling constant, which in this case is 8 Hz. The coupling constant provides information about the interaction between the proton of interest and its neighboring proton, helping to determine the connectivity and stereochemistry of the molecule.