Interpret the peaks with the Lewis Structures drawn in the spaces provided. A su

Question

Interpret the peaks with the Lewis Structures drawn in the spaces provided.

A suggested approach for spectroscopy problems:

Calculate the degree of unsaturation to limit the number of possible structures.

Look at the IR absorption bands > 1500 cm-1 to determine what functional groups are likely in the compound. Remember that these functional groups must be consistent with the degree of unsaturation.

Look at the NMR to determine the connectivity of the compound.

Draw some possible structures and see if they "work" with the IR, NMR, and unsaturation.

Draw the Lewis Structures of the one that fits all the spectra in the box provided.

Degree of Unsaturation A calculation of the degree of unsaturation is a good way to start a spectroscopy problem. It tells you how many rings and double bonds are in the molecule; thus, you know if you should look for a carbonyl or a carbon-carbon double bond, or a ring or an aromatic ring. The degree of unsaturation can be calculated readily from the molecular formula of all compounds containing carbon, hydrogen, oxygen, nitrogen, sulfur, or the halogens, by applying the following rules: Rule 1: Replace all halogens in the molecular formula by hydrogens. Rule 2: Omit oxygens and sulfurs. Rule 3: For each nitrogen, omit the nitrogen and one hydrogen. Applications of these rules reduces the molecular formula in question to the molecular formula of the hydrocarbon which has the same degree of unsaturation. The degree of unsaturation of a hydrocarbon is easily deduced if one remembers that a saturated hydrocarbon has the formula CrH +2 Thus for the formula C H 2n (2n 2) m 1 E n Applying these rules to the molecular formula CBHRNOBr. Rule 1, replace halogens with hydrogens: CH NO Rule 2, omit oxygens C HON Rule 3, omit the nitrogen and one hydrogen: CaHR Therefore, 1- 5 This means that the molecule has five rings and/or pi bonds (or any combination of the two). A degree of unsaturation greater than or equal to 4 doesn't demand but should suggest the possibility of an aromatic (benzene) ring.

Explanation / Answer

molecule is simple CH3-CH2-CH2-CH2-C(O)H

ANALYSIS

CH3-CH2-CH2-CH2-C(O)H==10PPM SINGLET

CH3-CH2-CH2-CH2-C(O)H==2.5 TRIPLET

CH3-CH2-CH2-CH2-C(O)H==1.5 MULTI

CH3-CH2-CH2-CH2-C(O)H==1.2 MULTI

CH3-CH2-CH2-CH2-C(O)H==0.9 TRIPLET

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