A reaction was set up to convert an alcohol to a carbonyl. Samples of the reaction mixture were checked by infrared spectrometry every couple of minutes. Use the infrared spectra provided to determine if the reaction mixture contains: pure alcohol, a mixture of alcohol and carbonyl, or pure carbonyl. The reaction mixture contains pure carbonyl. pure alcohol. mixture of alcohol and carbonyl. The reaction mixture contains mixture of alcohol and carbonyl. pure carbonyl. pure alcohol.

First spectrum = contains only alcohol Compound.

Second spectrum= Contains both alcohol and carbonyl compounds.

Explanation:

The concept known as "spectroscopy'' is an important aspect of Chemistry. It is used in the identification of chemical compounds. Some of the techniques used in spectroscopy is through the use of NMR, IR spectroscopy and UV-Vis spectroscopy.

This question is about IR spectroscopy. Here each functional group has its own unique frequency of absorption. For instance the frequency of absorption of Alkane, C- H( stretch) is 2850 - 3000, alkyne, C-H( stretch) is 3300, amine, N-H(bending) is 1600 and so on.

For the characteristic frequency of absorption found in OH(stretch) is 3300 - 3700. This can be seen in the spectrum given. Thus, OH bond is present.

For carbonyl group, C= O the characteristic absorption frequency is 1670-1820. This frequency can not be found in the first spectrum. But checking the second spectrum a frequency of 1700 can be seen.

Thus, the C=O bond is absent on the first spectrum but present in the second spectrum.

In the second spectrum too, the characteristic frequency of absorption of 3300 is found. Thus, in the second spectrum the compound have both OH and C=O.

answerto predict;

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o