Chapter 15.6, Problem 13ATS

Interpretation Introduction

Interpretation:

The given Chrysanthemic acid exhibit the IR spectrum of five signals above $\text{1500}{\text{cm}}^{\text{-1}}$ and the source for each of these signals should be identified.

Concept introduction

IR spectrum can be divided into two main regions,

1. 1. Diagnostic region (wavenumber above $\text{1500}{\text{cm}}^{\text{-1}}$)
2. 2. Finger print region(wavenumber below $\text{1500}{\text{cm}}^{\text{-1}}$)

The diagnostic region consists of signals from the double bonds, triple bonds and the $\text{X}\text{-}\text{H}$ type (O-H, C-H or N-H) bonds.

Finger print region contains signals from the single bonds only.

Reciprocal of wavelength is called wavenumber or Number of waves in one centimeter is called wavenumber

Unit of wave number is reciprocal of Centimeter.

According to Hooke’s law the factors affecting wave numbers are

1. Bond strength

2. Reduced mass

So, in the diagnostic region on comparing double bonds, triple bonds and the $\text{X}\text{-}\text{H}$ type (O-H, C-H or N-H) bond on the basis Hooke’s law double bond appears at wave number 1600-1850 ${\text{cm}}^{\text{-1}}$ and the $\text{X}\text{-}\text{H}$ type bonds appears at the left side of the spectrum (above 2700 ${\text{cm}}^{\text{-1}}$). Triple bonds appears in between these two bonds(2100-2300 ${\text{cm}}^{\text{-1}}$)

To find: the double bonds in the given Chrysanthemic acid that responsible for the signals above 1500 $c{m}^{-1}$.