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Vesna [10]
3 years ago
10

Draw the structure corresponding to the followingname:_________.

Chemistry
1 answer:
natita [175]3 years ago
7 0

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>

<u />

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 (CH_3). In carbon 4 we have to put an amino group (NH_2). 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

<u />

<u>b. (3S,6R)-6-(2,4-dinitrophenyl)-3-amino-1-heptyne</u>

<u />

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 (NH_2). 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 (NH_2) 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

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goldfiish [28.3K]

Answer:

Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g &times; (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g &times; (1 mole/ 32 g) = 1.53 moles O2 B.

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield

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Together, you would have "-18h^2 + 81h + 4h - 18". Combine like terms (81h and 4h), and you'll get the answer of "-18h^2 + 85h - 18".

I hope this helps!!
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