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Which of the following compounds would be the major product of the E1 reaction of 3-bromide 4-ethyle heptane?
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.
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.
In the reaction below, the rate of methane and bromine reactions decreases by adding HBr. Which of the following reactions comes from this step?
Since HBr can break, and produce free radicals, giving H. to CH.3 reduces the rate of reaction.
Since HBr can break, and produce free radicals, giving H. to CH.3 reduces the rate of reaction.
Which of the molecule pairs shown below has a DIASTEREOMERIC relationship?
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).
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).
This is an acid compound. Acids are soluble in polar solvents. Among the above solvents, only Tetrahydrofuran is a non-polar solvent.
This is an acid compound. Acids are soluble in polar solvents. Among the above solvents, only Tetrahydrofuran is a non-polar solvent.
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–
Rank the following amines from the strongest to the weakest base, while the amines are in aqueous solution.
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.
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.
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.
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).
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).
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.
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.