Answer:
The answer is Option a, that is "−9kJmole,5kJmole".
Explanation:
Please find the complete question in the attached file.
In the question, it uses the catalyst inside a process, which does not modify the process eigenvalues, however, it decreases the active energy with an enthalpy of -9kJmole, and also the power for activating decreases around 13 to 5 kJ mole, that's why the choice a is correct.
Answer:
See figures 1 and 2
Explanation:
For this question, we can analyze each compound:
<u>a. (4Z,3S)-2,4,7-trimethyl-3-amino-4-octene</u>
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In this case, we have to start with a <u>carbon chain of 8 carbons</u>. Then in carbons 2, 4 and, 7, we have to put three methyl groups (). In carbon 4 we have to put an amino group (
). When we put the amino group in carbon 4 we will have a <u>chiral carbon</u>. So, we have to indicate the orientation in the bond of the amino group. If we want an "S" configuration the atoms must have a <u>counterclockwise orientation</u>. So, the amino group must have a dashed bond. Finally, in carbon 4 we have to put a double bond, the configuration of this double bond is <u>"Z"</u>. So, the most important groups must be on the <u>same side</u>, in this case.
See figure 1 for further explanations
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<u>b. (3S,6R)-6-(2,4-dinitrophenyl)-3-amino-1-heptyne</u>
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In this case, we have to start with a <u>chain of 7 carbons</u>. In carbon 3 we have to put an amino group (). When we put this group in carbons we will have a chiral carbon is the name we have an <u>"S" configuration</u> for this carbon. So, we have to use the dashed bond in order to have a <u>counterclockwise orientation</u> for all the atoms bonded on carbon 3. In carbon 6 we have to put a big group "2,4-dinitrophenyl". In this group, we have a bond with the main carbon chain in carbon 6 (of benzene), in this same benzene ring we have to put nitro groups (
) in carbons 2 and 4. Additionally, carbon 6 is a chiral carbon, and the configuration of this carbon is "R". Therefore we have to use a wedge bond for all the "2,4-dinitrophenyl" in order to have a clockwise orientation for all the atoms bonded on carbon 6. Finally, we have to put a triple bond between carbons 2 and 3.
See figure 2 to further explanations
