Chapter 13, Problem 13B.2E

(a)

Interpretation Introduction

Interpretation:

For protons in acetaldehyde, the difference in local magnetic field betwenn the two regions has to be calculated at $\text{1}\text{.2}\text{T}$.

Explanation of Solution

Given information,

  $\begin{array}{l}\text{Acetaldehyde}{\text{CH}}_{\text{3}}\text{CHO}\\ \delta \text{C}{H}_{\text{3}}\text{CHO}=2.20\\ \delta {\text{CH}}_{\text{3}}\text{C}H\text{O}=9.80\\ \text{Applied}\text{field}\text{=}\text{1}\text{.2}\text{T}\end{array}$

First calculate the resonance frequency for proton,

$\begin{array}{l}\text{ν}\text{=}\frac{{\text{γ}}_{\text{N}}{\text{B}}_{\text{0}}}{\text{2π}}\\ \\ \text{=}\frac{\text{2}\text{.6752}\text{×}{\text{10}}^{\text{8}}{\text{T}}^{\text{-1}}{\text{s}}^{\text{-1}}\text{×}\text{1}\text{.2}\text{T}}{\text{2}\text{×}\text{3}\text{.14}}\text{=}\text{0}\text{.511}\text{×}{\text{10}}^{\text{8}}{\text{s}}^{\text{-1}}\\ \text{=}\text{51}\text{.1}\text{MHz}\left(\text{1}\text{MHz}\text{=}{\text{10}}^{\text{6}}\text{Hz}\right)\end{array}$

Now calculate the local magnetic field difference between the two protons,

$\begin{array}{l}\text{δ}\text{=}\frac{\text{ν}\text{-}{\text{ν}}_{\text{0}}}{{\text{ν}}_{\text{0}}}\times {10}^6\\ \\ \text{ν}\text{-}{\text{ν}}_{\text{0}}\text{=}{\text{δν}}_{\text{0}}\text{×}{\text{10}}^{\text{-6}}\text{=}\left(\text{δCHO}\text{-}{\text{δCH}}_{\text{3}}\right){\text{ν}}_{\text{0}}\\ \\ \text{=}\left(\text{9}\text{.80}\text{-}\text{2}\text{.20}\right)\text{51}\text{.1}\text{Hz}\text{=}\text{388}\text{Hz}\end{array}$

Therefore, The difference between the local magnetic field of ${\text{-CH}}_{\text{3}}$ and $\text{-CHO}$ group in $\text{1}\text{.2 T}$ is $\text{388}\text{Hz}$.

(b)

Interpretation Introduction

Interpretation:

For protons in acetaldehyde, the difference in local magnetic field betwenn the two regions has to be calculated at $5.0\text{T}$.

Explanation of Solution

Given information,

  $\begin{array}{l}\text{Acetaldehyde}{\text{CH}}_{\text{3}}\text{CHO}\\ \delta \text{C}{H}_{\text{3}}\text{CHO}=2.20\\ \delta {\text{CH}}_{\text{3}}\text{C}H\text{O}=9.80\\ \text{Applied}\text{field}\text{=}\text{5}\text{.0}\text{T}\end{array}$

First calculate the resonance frequency for proton,

$\begin{array}{l}\text{ν}\text{=}\frac{{\text{γ}}_{\text{N}}{\text{B}}_{\text{0}}}{\text{2π}}\\ \\ \text{=}\frac{\text{2}\text{.6752}\text{×}{\text{10}}^{\text{8}}{\text{T}}^{\text{-1}}{\text{s}}^{\text{-1}}\text{×}\text{5}\text{.0}\text{T}}{\text{2}\text{×}\text{3}\text{.14}}\text{=}\text{2}\text{.13}\text{×}{\text{10}}^{\text{8}}{\text{s}}^{\text{-1}}\\ \text{=}\text{213}\text{MHz}\left(\text{1}\text{MHz}\text{=}{\text{10}}^{\text{6}}\text{Hz}\right)\end{array}$

Now calculate the local magnetic field difference between the two protons,

$\begin{array}{l}\text{δ}\text{=}\frac{\text{ν}\text{-}{\text{ν}}_{\text{0}}}{{\text{ν}}_{\text{0}}}\times {10}^6\\ \\ \text{ν}\text{-}{\text{ν}}_{\text{0}}\text{=}{\text{δν}}_{\text{0}}\times {10}^{-6}\text{=}\left(\text{δCHO}\text{-}{\text{δCH}}_{\text{3}}\right){\text{ν}}_{\text{0}}\\ \\ \text{=}\left(\text{9}\text{.80}\text{-}\text{2}\text{.20}\right)\text{213}\text{Hz}\text{=}\text{1619}\text{Hz}\end{array}$

Therefore, The difference between the local magnetic field of ${\text{-CH}}_{\text{3}}$ and $\text{-CHO}$ group in $5.0\text{T}$ is $\text{1619}\text{Hz}$.