Chemistry, 27.06.2019 07:20 diven8765

The following iupac name is incorrect. explain why it is incorrect and give the correct iupac name. 2,2−dimethyl−4−ethylheptane a. the compound name is alphabetized incorrectly. b. the compound has an incorrectly labeled parent chain. c. the compound has incorrectly labeled substituents. d. the compound name is spelled incorrectly.

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Answer from: famouzgal

Choice a. The compound name is alphabetized incorrectly.

The correct IUPAC name of this compound should be:

4-ethyl-2,2-dimethylheptane.

Explanation:

Draw the carbon backbone structure for "2,2-dimethyl-4-ethylheptane":

$\begin{array}{ccccccccccccc}& &\text{C}\\& & |\\\text{C}&-&\text{C}&-&\text{C}&-&\text{C}&-&\text{C}&-&\text{C}&-&\text{C}\\&&|&&&&|\\&&\text{C}&&&&\text{C}\\&&&&&&|\\&&&&&&\text{C} \end{array}$ .

The seven-carbon backbone is indeed the longest chain in this compound.

$\begin{array}{ccccccccccccc}& &\text{C}\\& & |\\\text{C}^1&-&\text{C}^2&-&\text{C}^3&-&\text{C}^4&-&\text{C}^5&-&\text{C}^6&-&\text{C}^7\\&&|&&&&|\\&&\text{C}&&&&\text{C}\\&&&&&&|\\&&&&&&\text{C} \end{array}$ .

Numbering from the end with the two methyl groups indeed minimizes the number on each group. However, the alphabetical order of the groups is not correct. There are two kinds of substitute groups in this compound:

Two methyl groups, andOne ethyl group.

The name for them will be

2,2-dimethyl, and4-ethyl.

Ignore the multiplier prefixes (di-, tri-, etc.) while alphabetizing.

The ethyl group shall be placed before the two methyl groups. The name of the compound shall be:

4-ethyl-2,2-dimethylheptane.

Answer from: Quest

answerlinnaeus;

Answer from: Quest

Answerthis solution will not form a precipitateexplanationsolubility equilbrium: pbf₂(s) ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is $k_{sp} = [\text{pb}^{2+}][\text{f}^- ]^2 = 4 \times 10^{-8}$ (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as $q = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}}$ then, we can interpret q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product: $\begin{aligned} q & = [\text{pb}^{2+}]_{\text{initial}} \cdot [\text{f}^- ]^2_{\text{initial}} \\ & = (1.9 \times 10^{-3} ) (3.8\times 10^{3})^2 \\ & = 2.7 \times 10^{-8} \end{aligned}$ since q = 2.7 × 10⁻⁸ < ksp = 4 × 10⁻⁸, a precipitate will not form.

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