I\'m having trouble with these three home work questions. If possible could you

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I'm having trouble with these three home work questions. If possible could you explain how to do these as well? Clear steps would be highly appreciated. Thank You!

1. Draw a structure for a compound that meets the following description: An optically active compound, C5H10O2, with broad strong IR absorption from 2600 to 3100 cm-1 and strong IR absorption at 1715 cm-1.

2.Draw a structure for a ketone that exhibits a molecular ion at M+ = 86 and that produces a fragment at m/z = 57.

3.Halogenated compounds are particularly easy to identify by their mass spectra because chlorine and bromine occur naturally as mixtures of two abundant isotopes.

Chlorine occurs as 35Cl (75.8%) and 37Cl (24.2%);
Bromine occurs as 79Br (50.7%) and 81Br (49.3%);
Boron compounds also stand out owing to the two isotopes 10B (19.9%) and 11B (80.1%).
For the compound Bromobenzene, C6H5Br:

At what masses do the molecular ions occur?

(List in order of increasing mass separated by commas, e.g. 120,122.)

What are the percentages of each molecular ion?

(List to nearest 1% in order of increasing mass separated by commas, e.g. 55,45.)

Explanation / Answer

(1) consider the information "broad strong IR absorption from 2600 to 3100 cm-1". This simply means strong,broad O-H stretching that is usually found in carboxylic acid.

Again consider "strong IR absorption at 1715 cm-1". This represents keto C=O group.

So, overall this must be a carboxylic acid.

Now, it has formula of C5H10O2. So, 5 carbon with a carboxylic acid group.

So, the probable structures are:

(a) CH3-CH2-CH2-CH2-COOH (pentanoic acid)

(b) CH3-CH(CH3)-CH2-COOH (3-methylbutanoic acid)

(c) CH3-CH2-CH(CH3)-COOH (2-methylbutanoic acid)

Now, the problem says, the compound is optically active. So, among these three, only (c) is optically acitive (has a chiral center).

So, the compound is 2-methylbutanoic acid.

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(2) The ketone exhibits a molecular ion M+ = 86. That means, if we remove the electron fro a ketone, then it will give M+. Now, since the mass of electron is negligible, so the keton will have molar mass of 86 g/mol.

Now, the general formula of ketone is CnH2nO. So, the molar mass will be:

(12*n + 2n*1 + 16) Here atomic mass of C is 12, of H is 1 and O is 16.

So,

12*n + 2n*1 + 16 = 86

12n + 2n + 16 =86

14n = 70

n =70/14 = 5

So, the ketone formula is C5H10O.

Now, let us consider the mass of fragment parts. probable fragment is m/z =57. So, the other part will be

m/z = 86-57 =29

Now, consider the fraction -CH2CH3. This has 2 carbons and 5 hydrogens. So, mass: 2(12) + 5(1) =29 (m/z)

So, the rest part will be CH3CH2C=O . It has 3 carbons, 5 hydrogens and 1 oxygen. So, 3(12) + 5(1) + 16 = 57 (m/z)

So, the ketone is CH3-CH2-CO-CH2-CH3 or (CH3CH2)2C=O. Name: 3-pentanone.

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(c) Consider 79Br (50.7%). The atomic masses are: C =12, H = 1 and Br = 79.

So, m/z of C6H579Br+ will be:

12*6+1*5+79 = 156

Percent of each molecular ion will be same as percent of isotope, so it will be 50.7% or almost 51%

Consider 81Br (49.3%). The atomic masses are: C =12, H = 1 and Br = 81.

So, m/z of C6H581Br+ will be:

12*6+1*5+81 = 158

Percent of each molecular ion will be same as percent of isotope, so it will be 49.3% or almost 49%

So, masses at which the molecular ions occur (156,158)

Percentage of each molecular ion (51,49)

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