Chemistry, 22.01.2020 20:31, sweatsierra

Draw the major organic product in the reaction scheme below. be sure to clearly show stereochemistry (if applicable).

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answered: holyturnley

see explanation below

Explanation:

You are missing the reaction scheme, but in picture 1, I found a question very similar to this, and after look into some other pages, I found the same scheme reaction, so I'm gonna work on this one, to show you how to solve it. Hopefully it will be the one you are asking.

According to the reaction scheme, in the first step we have NaNH2/NH3(l). This reactant is used to substract the most acidic hydrogen in the alkine there. In this case, it will substract the hydrogen from the carbon in the triple bond leaving something like this:

R: cyclopentane

R - C ≡ C (-)

Now, in the second step, this new product will experiment a SN2 reaction, and will attack to the CH3 - I forming another alkine as follow:

R - C ≡ C - CH3

Finally in the last step, Na in NH3 are reactants to promvove the hydrogenation of alkines. In this case, it will undergo hydrogenation in the triple bond and will form an alkene:

R - CH = CH - CH3

In picture 2, you have the reaction and mechanism.

Draw the major organic product in the reaction scheme below. be sure to clearly show stereochemistry

answered: Guest

the answer is:

line graphs show patterns in data with variables that change over time; whereas, circle graphs show data as parts out of a whole (such as out of 100%).

answered: Guest

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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Draw the major organic product in the reaction scheme below. be sure to clearly show stereochemistry...