Using the following spectra and other information provided, identify the compound. Assign pertinent peaks in the spectra. Label the peaks on the spectrum and place the structure of the compound in the box on the right side of the spectrum - Identify correct compound - label proton spectrum - label carbon spectrum
Analyzing Infrared Spectra
The electromagnetic radiation or frequency is classified into radio-waves, micro-waves, infrared, visible, ultraviolet, X-rays and gamma rays. The infrared spectra emission refers to the portion between the visible and the microwave areas of electromagnetic spectrum. This spectral area is usually divided into three parts, near infrared (14,290 – 4000 cm-1), mid infrared (4000 – 400 cm-1), and far infrared (700 – 200 cm-1), respectively. The number set is the number of the wave (cm-1).
IR Spectrum Of Cyclohexanone
It is the analysis of the structure of cyclohexaone using IR data interpretation.
IR Spectrum Of Anisole
Interpretation of anisole using IR spectrum obtained from IR analysis.
IR Spectroscopy
Infrared (IR) or vibrational spectroscopy is a method used for analyzing the particle's vibratory transformations. This is one of the very popular spectroscopic approaches employed by inorganic as well as organic laboratories because it is helpful in evaluating and distinguishing the frameworks of the molecules. The infra-red spectroscopy process or procedure is carried out using a tool called an infrared spectrometer to obtain an infrared spectral (or spectrophotometer).
Using the following spectra and other information provided, identify the compound. Assign pertinent peaks in the spectra.
Label the peaks on the spectrum and place the structure of the compound in the box on the right side of the spectrum
- Identify correct compound
- label proton spectrum
- label carbon spectrum
The molecule formula of the compound is C11H13NO. The double bond equivalence(DBE) of the compound cab ebe calculated by using the formula.
DBE = x + 1 - ( y - z) /2
where, x = number of carbon atoms
y = number of hydrogen atoms and
z = number of nitrgen atoms
Here, x = 11, y = 13 and z = 1
therefore, by piting this values in the above equation becomes,
DBE = 11 + 1 - (13 - 1) / 2
DBE = 12 - 12/2
DBE = 12-6
DBE = 6
Double bond equivalence 6 represents molecule is highly unsaturated. From the 1H-NMR and 13C-NMR spectra it is assumed that it may be contain the benzene ring.
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