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Airida [17]
3 years ago
14

Alkenes can be converted to alcohols by reaction with mercuric acetate to form a ?-hydroxyalkylmercury(II) acetate compound, a r

eaction called oxymercuration. Subsequent reduction with NaBH4 reduces the C?Hg bond to a C?H bond, forming the alkyl alcohol, a reaction called demercuration. Draw the structures of the Hg-containing compound(s) and the final alcohol product(s) formed in the following reaction sequence, omitting, by products. If applicable, draw hydrogen at a chirality center and indicate stereochemistry via wedge-and-dash bonds.
Neutral produst (s) of oxymercuration. Include stereoisomers, if any

image from custom entry tool HG(OOCCH3)2

----------------->

H2O,THF

Chemistry
2 answers:
Pie3 years ago
8 0

Answer:

Alcohol will be end product in any case.

Explanation:

Alkenes are the class of organic compounds which are termed as unsaturated organic compounds. These compounds contain a single or multiple double bond depending upon the carbon chain length. The simplest alkene among many others is called Ethene with the molecular formula C2H4. Its molecular structure is H2C=CH2. The double bond present in any alkene serve as a nucleophilic component, meaning it will attack any element with partial positive to positive charge.

The process of introducing mercuric acetate to any alkene is termed as oxymercuration. This is a three step process. Please see the reference image.

As you can see there,

in first step, the double bond of an alkene attacks the mercury element which carries the positive, thereby dispositioning the acetate bond, and forming a cyclic mercury complex.

In second step, water acts as a nucleophile and attack on electrophilic (electron loving) carbon, which moves the bond towards mercury retaining mercury stability. In this step, oxygen of water molecule carries positive charge because of donating lone pair of electrons to the alkene carbon.

In third step, the acetate ion which left the mercury in first step, being a nucleophile attack the hydrogen of water molecule, donating the bonded pair to oxygen and restoring neutral charge over oxygen. It forms acetic acid which reconverts to mercuric acetate by subsequent reaction and therefore restores the catalyst.

If the first step, the oxymercurated compound is treated with Sodium Borohydride (NaBH4), the demercuration occurs leaving hydrogen at the same place where mercury was previously attached. It is termed as Oxymercuration-Reduction Reaction.It is a popular laboratory technique to achieve alkene hydration with Markovnikov selectivity while avoiding carbocation intermediates and thus the rearrangement which can lead to complex product mixtures.

Goshia [24]3 years ago
3 0

Answer:

The simplified mechanism and products are on the picture.

Explanation:

If we have the symmetrical alkene the addition of mercury and OH group is not regioselective but when we have more donors for one of Carbons in alkene then the OH group will go there.

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Determine which elements or compounds are products in the<br> follow reaction equation:
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Answer: D is the answer since it is the product of this equation

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4 0
3 years ago
A gas has a volume of 340.0 mL at 45.90 oC. What is the new temperature of the gas, in Kelvin, if the volume increased to 550.0
d1i1m1o1n [39]

Answer:

Solution for A gas has a volume of 340.0 mL at 45.90 degree celsius. What is the new temperature of the gas, in kelvin, if the volume increased to 550.0 mL.

Missing: oC. ‎| Must include: oC.

Explanation:

6 0
2 years ago
4Na + O2 2Na2O
beks73 [17]
4
N
a
+
O
2
→
2
N
a
2
O
.
By the stoichiometry of this reaction if 5 mol natrium react, then 2.5 mol
N
a
2
O
should result.
Explanation:
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61.98

g
⋅
m
o
l
−
1
. If
5

m
o
l
natrium react, then
5
2

m
o
l
×
61.98

g
⋅
m
o
l
−
1

=

154.95

g
natrium oxide should result.
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5 0
2 years ago
Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that
iragen [17]

Answer:

108.43 grams KNO₃

Explanation:

To solve this problem we use the formula:

  • ΔT = Kf * b * i

Where

  • ΔT is the temperature difference (14.5 K)
  • Kf is the cryoscopic constant (1.86 K·m⁻¹)
  • b is the molality of the solution (moles KNO₃ per kg of water)
  • and<em> i</em> is the van't Hoff factor (2 for KNO₃)

We <u>solve for b</u>:

  • 14.5 K = 1.86 K·m⁻¹ * b * 2
  • b = 3.90 m

Using the given volume of water and its density (aprx. 1 g/mL) we <u>calculate the necessary moles of KNO₃</u>:

  • 275 mL water ≅ 275 g water
  • 275 g /1000 = 0.275 kg
  • moles KNO₃ = molality * kg water = 3.90 * 0.275
  • moles KNO₃ = 1.0725 moles KNO₃

Finally we <u>convert KNO₃ moles to grams</u>, using its molecular weight:

  • 1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃
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2 years ago
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