Free Organic Chemistry Exam
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This is the Free Organic Chemistry Exam. Please read the following tips before beginning:
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You will have 10 minutes to answer 10 questions.
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- Acid-base Chemistry 0%
- Aromatics and Bonding 0%
- Chemical and Physical Properties 0%
- Individual and Combination Reactions 0%
- Mechanism 0%
- Nomenclature 0%
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Question 1 of 10
1. Question
Which of the following compounds would be the major product of the E1 reaction of 3-bromide 4-ethyle heptane?
Correct
The E1 reaction will produce the most thermodynamically stable alkene as the major product. The first step in the E1 mechanism (shown below) is when the leaving group levels, creating a carbocation, now one of hydrogen Ha will be removed to form an alkene. Removal of the labeled proton results in the least substituted and least thermodynamically stable alkenes and are more thermodynamically stable. The E alkene has less steric hindrance and is, therefore, more stable and the major product.
Incorrect
The E1 reaction will produce the most thermodynamically stable alkene as the major product. The first step in the E1 mechanism (shown below) is when the leaving group levels, creating a carbocation, now one of hydrogen Ha will be removed to form an alkene. Removal of the labeled proton results in the least substituted and least thermodynamically stable alkenes and are more thermodynamically stable. The E alkene has less steric hindrance and is, therefore, more stable and the major product.
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Question 2 of 10
2. Question
In the reaction below, the rate of methane and bromine reactions decreases by adding HBr. Which of the following reactions comes from this step?
Correct
Since HBr can break, and produce free radicals, giving H. to CH.3 reduces the rate of reaction.
Incorrect
Since HBr can break, and produce free radicals, giving H. to CH.3 reduces the rate of reaction.
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Question 3 of 10
3. Question
Which of the molecule pairs shown below has a DIASTEREOMERIC relationship?
Correct
Compounds in C, D, and E are enantiomeric (non-superimposable mirror image). Convince yourself of this by rotating one of the compounds about a vertical axis passing through the cyclohexane center until the compounds are a mirror image. Compounds in A are identical (convince yourself of this by passing priority and determining whether R or S stereochemistry is present, both molecules are in the R configuration. Compounds in B ARE DIASTEREOMERIC, i.e. stereoisomers that are not enantiomeric (are not mirror image of each other).
Incorrect
Compounds in C, D, and E are enantiomeric (non-superimposable mirror image). Convince yourself of this by rotating one of the compounds about a vertical axis passing through the cyclohexane center until the compounds are a mirror image. Compounds in A are identical (convince yourself of this by passing priority and determining whether R or S stereochemistry is present, both molecules are in the R configuration. Compounds in B ARE DIASTEREOMERIC, i.e. stereoisomers that are not enantiomeric (are not mirror image of each other).
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Question 4 of 10
4. Question
Which solvent would NOT best dissolve the compound below?
Correct
This is an acid compound. Acids are soluble in polar solvents. Among the above solvents, only Tetrahydrofuran is a non-polar solvent.
Incorrect
This is an acid compound. Acids are soluble in polar solvents. Among the above solvents, only Tetrahydrofuran is a non-polar solvent.
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Question 5 of 10
5. Question
Predict the major product of the following reaction.
Correct
The reaction of an alkyl halide in the presence of a Lewis acid such as AlCl3 will undergo an acid-base reaction forming a carbocation and AlCl4–
Incorrect
The reaction of an alkyl halide in the presence of a Lewis acid such as AlCl3 will undergo an acid-base reaction forming a carbocation and AlCl4–
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Question 6 of 10
6. Question
Which reagent(s) would best accomplish the following transformation?
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Incorrect
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Question 7 of 10
7. Question
What is the major organic product of the following reaction?
Correct
Incorrect
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Question 8 of 10
8. Question
Rank the following amines from the strongest to the weakest base, while the amines are in aqueous solution.
Correct
Because alkyl groups donate electrons to the more electronegative nitrogen. The inductive effect makes the electron density on the alkylamine’s nitrogen greater than the nitrogen of ammonium. Correspondingly, primary, secondary, and tertiary alkyl amines are more basic than ammonia.
With an alkyl amine, the lone pair electron is localized on the nitrogen. However, the lone pair electron on an amide is delocalized between the nitrogen and the oxygen through resonance. This makes amides much less basic compared to alkylamines.
When a nitrogen atom is incorporated directly into an aromatic ring, its basicity depends on the bonding context. In a pyridine ring, for example, the nitrogen lone pair occupies an sp2-hybrid orbital, and is not part of the aromatic sextet – it is essentially imine nitrogen. Its electron pair is available for forming a bond to a proton, and thus the pyridine nitrogen atom is somewhat basic.
Incorrect
Because alkyl groups donate electrons to the more electronegative nitrogen. The inductive effect makes the electron density on the alkylamine’s nitrogen greater than the nitrogen of ammonium. Correspondingly, primary, secondary, and tertiary alkyl amines are more basic than ammonia.
With an alkyl amine, the lone pair electron is localized on the nitrogen. However, the lone pair electron on an amide is delocalized between the nitrogen and the oxygen through resonance. This makes amides much less basic compared to alkylamines.
When a nitrogen atom is incorporated directly into an aromatic ring, its basicity depends on the bonding context. In a pyridine ring, for example, the nitrogen lone pair occupies an sp2-hybrid orbital, and is not part of the aromatic sextet – it is essentially imine nitrogen. Its electron pair is available for forming a bond to a proton, and thus the pyridine nitrogen atom is somewhat basic.
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Question 9 of 10
9. Question
Urea-formaldehyde is commonly used as a polar solvent in organic reactions. Its Lewis structure is shown below. Rank the indicated bond angles from smallest to largest angle.
Correct
The central atom in an sp2 hybridized and is therefore predicted to have a bound angle of 120o having a trigonal planar geometry. The central atom in c is sp3 hybridized and therefore assume a tetrahedral geometry with a bond angle of 109.5o. The central atom in b is sp3 hybridized as well, but the lone pair on N exerts greater repulsion to the bonding electron on N causing the bonds to move away from the lone pair. This has an effect of decreasing the angle from that predicted of an sp that predicted an sp3 hybridized atom. The bond angle will, therefore, be less than 109.5o (˜ 107o). The correct order therefore is:
b (˜ 107o) < c (109.5o) < a (120o).
Incorrect
The central atom in an sp2 hybridized and is therefore predicted to have a bound angle of 120o having a trigonal planar geometry. The central atom in c is sp3 hybridized and therefore assume a tetrahedral geometry with a bond angle of 109.5o. The central atom in b is sp3 hybridized as well, but the lone pair on N exerts greater repulsion to the bonding electron on N causing the bonds to move away from the lone pair. This has an effect of decreasing the angle from that predicted of an sp that predicted an sp3 hybridized atom. The bond angle will, therefore, be less than 109.5o (˜ 107o). The correct order therefore is:
b (˜ 107o) < c (109.5o) < a (120o).
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Question 10 of 10
10. Question
What is the IUPAC name for the following compound?
Correct
First, the IUPAC name of this compound will be based on the longest carbon chain. This chain is seven carbons in length, as labeled below, giving the compound a designation of hyptyne. Next, the substitution is listed in alphabetical order (ignoring the prefixes di, tri, etc). this means that the fluoro-group should be listed first, finally methyl groups.
Incorrect
First, the IUPAC name of this compound will be based on the longest carbon chain. This chain is seven carbons in length, as labeled below, giving the compound a designation of hyptyne. Next, the substitution is listed in alphabetical order (ignoring the prefixes di, tri, etc). this means that the fluoro-group should be listed first, finally methyl groups.