ORGANIC CHEMISTRY HELP!!!!! IR Spectroscopy.....SOS Carbon dioxide (co_2) has tw
ID: 913181 • Letter: O
Question
ORGANIC CHEMISTRY HELP!!!!! IR Spectroscopy.....SOS Carbon dioxide (co_2) has two types of bond stretches: symmetrical and unsymmetrical. Classify each one of these stretches as IR active or IR inactive and explain your choice. How could you see IR spectroscopy to differentiate between the two isomers: 1-butyne and 2-butyne? Which bond is stronger: the C=0 of an ester (1735 cm^-1) or the C=0 bond of a ketene (1715 cm^-1). Explain the answer. For each compound below, approximate the most important IR absorptions that you would expect to see. 5. For each group of IR frequencies listed below, suggest the functional group that is present.Explanation / Answer
1. Carbon dioxide has two types of bond stretches : symmetrical stretch does not result in change of initial dipole and thus is IR inactive.
unsymmetrical strech results in change of dipole from intial state of zero dipole and thus is IR active.
2. 1-butyne is terminal alkyne which shows an IR peak at around 3300 cm-1 specific for terminal alkyneC-H stretching. This peak is absent in case of internal alkyne as 2-butyne. Thus IR can be used to distinguish between the two alkynes easily.
3. C=O bond of an ester is stronger than C=O of a ketone. The IR frequency is directly related to energy of the bond by equation,
E = hv(wavenumber)
h being Planck's constant
Thus, greater wavenumber value in IR of bond suggests greater energy of bond. In case of ester, the electronegative O pulls e- from the C=O bond through the C=O oxygen, thus giving more double bond character to the C=O bond. Which results in greater strength of bond. This is however not the case in ketone, where, the R groups are electron donating and thus, reduces the overall double bond character of C=O bond, thus resulting in low bond strength for C=O.
4. In case of first molecule from left to right.
molecule 1 : would show two peaks at arpund 2750 cm-1 and 27850 cm-1 for C-H of aldehyde. Another peak at around 1740 cm-1 for C=O stretch.
molecule 2 : A peak at around 3300 cm-1 for N-H stretch of amide. Another peak at 1650 cm-1 for C=O.
molecule 3 : A broad peak at around 3200-3600 cm-1 for O-H stretch. Another peak at around 1620-1680 cm-1 for C=C stretch of alkene.
molecule 4 : Terminal alkyne would show strong sharp peak at around 330 cm-1 for alkyneC-H stretch.
5. Identification of functional group based on IR frequency,
a) This is a ketone. 1734 cm- is typical of C=O stretch.
b) This is an alcohol. 3400 cm-1 broad peak is for O-H stretch of alcohol.
c) This is an alkene. 3050 cm-1 is for =C-H stretch and 1650 cm- is for C=C stretch.
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