Tentatively identify the lower boiling solvent from the temperature of the first
ID: 543804 • Letter: T
Question
Tentatively identify the lower boiling solvent from the temperature of the first plateau and the ketone or alcohol from the temperature of the second plateau (the approximate boiling point of each component.) Examine the IR of the higher-boiling fraction and determine whether it is an alcohol or a ketone from the characteristic O-H or C=O stretch. Identify which peak in the IR is from the –OH group of the alcohol or the C=O peak from the ketone. Draw the structure of the compound.
Analyst Date 3361 Thursday, October 12, 2017 3:28 PM 100 364899cm-1 95 2935.66cm-1 85 2886.39cm-1 1467.05 80- 1340 703.70cm-1 3337.04cm-1 28147c 1378.89cm-1 1275.9 110788_1 1 1604cm-1 816.64pm-1 65- 29 70.58cm-1 60- 1128.05cm-1 50 45- 950.46cm-1 38- 4000 2000 1500 1000 3500 3000 2500 650 cm-1 3361283 Sample 283 By 3361 Date Thursday, October 12 2017 ugher bonting purntExplanation / Answer
This compound is an alcohol.
The boiling point of alcohols is higher due to intermolecular hydrogen bonding. During vaporization, first, we have to break the intermolecular hydrogen bonding. For this purpose, we have to supply more energy. Thus boiling points of alcohols are generally higher.
Absence of C=O, stretching around 1600 cm-1, clearly says that this compound is not a ketone
Important peaks are
The broad peak at 3307 is due to O-H stretching
The peaks coming around 2900 is due to C-H stretching.
All other peaks are coming in the fingerprint region. That is below 1500 cm-1
Another important peak corresponding toC-O peak around 1100 cm-1.
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