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ira [324]
3 years ago
10

Why is only one diastereomer formed in this reaction? Relate your answer to the mechanism you drew. b) If you used cis-stilbene

as your reactant, what would your products be?

Chemistry
1 answer:
crimeas [40]3 years ago
4 0

Answer:

1. Diastereomers have different physical properties (unlike most aspects of enantiomers) and often different chemical reactivity. ... Many conformational isomers are diastereomers as well. Diastereoselectivity is the preference for the formation of one or more than one diastereomer over the other in an organic reaction.

2. The result is a trans dibromide, as shown in the equation below

Explanation:

Diastereomers (sometimes called diastereoisomers) are a type of a stereoisomer.[1] Diasteoreomers are defined as non-mirror image non-identical stereoisomers. Hence, they occur when two or more stereoisomers of a compound have different configurations at one or more (but not all) of the equivalent (related) stereocenters and are not mirror images of each other.[2] When two diastereoisomers differ from each other at only one stereocenter they are epimers. Each stereocenter gives rise to two different configurations and thus typically increases the number of stereoisomers by a factor of two.

2. the addition of bromine to the trans and

cis isomers of 1,2-diphenylethene, more commonly known as trans- and cis-stilbene.

H

H

H H

trans-stilbene cis-stilbene

m.p. 122-124°C b.p. 82-84°C

density 0.970 g/mL density 1.011 g/mL

M.W. 180.25 g/mol M.W. 180.25 g/mol

In both cases, the nucleophilic double bond undergoes an electrophilic addition reaction

by the bromine reagent which proceeds via a cyclic bromonium ion. The addition of

bromine begins at one side of the double bond (either side is equally likely, but only one

option is drawn) and is followed by attack of bromide ion on the bromonium ion (again,

attack could occur at either carbon since the ion is symmetric, but only one option is

drawn). The result is a trans dibromide, as shown in the equation below:

Since the cis and trans isomers of stilbene have different geometries, it follows

that upon reaction with bromine they give rise to stereoisomeric bromonium ions and,

eventually, products that differ only by their stereochemistry.

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68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi
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Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

  • 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:

  • 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
  • 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>

Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:

  • 0.528 mol Mg(NO₃)₂ * \frac{1molMg(OH)_2}{1molMg(NO_3)_2} = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

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A student was given a piece of metal with a mass of 85.0 g. She placed it
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Answer:

22mL

Explanation:

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Which example illustrates a chemical change
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Two moles of magnesium (Mg) and five moles of oxygen (O2) are placed in a reaction vessel. When magnesium is ignited, it reacts
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Answer:

\boxed{\text{Mg is the limiting reactant}}

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble the data in one place.

           2Mg + O₂ ⟶ 2MgO

n/mol:    2       5

Calculate the moles of MgO we can obtain from each reactant.

From Mg:  

The molar ratio of MgO:Mg is 2:2

\text{Moles of MgO} = \text{2 mol Mg} \times \dfrac{\text{2 mol MgO}}{\text{2 mol Mg}} = \text{2 mol MgO}

From O₂:  

The molar ratio of MgO:O₂ is 2:1.

\text{Moles of MgO} = \text{5 mol O}_{2} \times \dfrac{\text{2 mol MgO}}{\text{1 mol O}_{2}} = \text{10 mol MgO}\\\\\boxed{\textbf{Mg is the limiting reactant}} \text{ because it gives the smaller amount of MgO}

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