Question

5. Examine the IR below and classify the compound. 3000 asoo 00 2000 1000 1500 C Alcohol A. B. Aldehyde C. Amine D. Ketone E. Acid 6. Examine the IR below and classify the compound. 3500 3000 2500 2000 1500 1000 cn 500 A. Alcohol B. Aldehyde C. Amine D. Ketone E. Acid 9 PE 0c091 s4 e 62

Answer 1

0.3 Solution :- In this IR Spectra the Compound is known as (c. Amine Summary :- the most Characteristic band in Amines is due to the N-H bond stretch, and it appear as a weak to medium, this band is Positioned at the left end of the spectrum, in the range about 3200 - 3600 Cmt

Solution :- the IR known Sepectra as Compound Carboxylic is Acid. E Summary :- In IR Spectroscopy the Carboxylic acids Show a strong, wide band for the O-H Stretch. Unlike the O-H stretch band Observed in alcohols , The Carboxylic acid 0-H Stretch Appears as a very Broad band in the region 3300 - 2500 Cmt bad Centered at about 3000 Cmt this is in the same region as the C-H Stretching bands of both alkyl and aromatics groups. the Carbonyl Stretch C= 0 of a Carboxylic acid appears as an intende band from 1760 - 1690 cm? Ic-0 Stretch from 1320 1210 Cnt

Solution :- the as Compound is known D. ketone. 2 Presence of C o ~ 1700 Cmt Summary :- IR Now , Let's look at some actual output from IR Spectroscopy experiments. Below in the IR Spectrum. Iust the key absorption beat in this Spectrum is that from the Carbonyl double bond, at 1716 cm-t.) yola will notice that there are many additional beaks in this spectrum in the Longer - wavelength 400 - 1400 Cmt region. this part of the Spectrum is called the fingerprint region. H

0.6 Solution :- the Compound & known 2 Cor Amine !! as Sumaray : NH Stretch 3400 - 3 250 Cnt. N-H bend ( Primary Amines only) from 1650 - 1580 Cmt - GN Stretch [Aromatic Amines] from 1335-1250 Cmt.