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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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sveticcg [70]

Correct answer: has a completely filled outermost shell

Atoms of the element with complete outermost shells are stable. So, in order to attain stability the atom either loses electrons or gains electrons to completely fill the outermost shell. The stable electronic configuration for the s and p-block elements is exhibited by the noble gases or the group 8 elements. All the unstable atoms try to attain the electronic configuration of the nearest noble gas with completely filled outermost shell.

5 0
3 years ago
200.0 mL of 3.85 M HCl is added to 100.0 mL of 4.6 M barium hydroxide. The reaction goes to completion. What is the concentratio
Ede4ka [16]

Answer:

2.387 mol/L

Explanation:

The reaction that takes place is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

First we <u>calculate how many moles of each reagent were added</u>:

  • HCl ⇒ 200.0 mL * 3.85 M = 203.85 mmol HCl
  • Ba(OH)₂ ⇒ 100.0 mL * 4.6 M = 460 mmol Ba(OH)₂

460 mmol of Ba(OH)₂ would react completely with (2*460) 920 mmol of HCl. There are not as many mmoles of HCl so Ba(OH)₂ will remain in excess.

Now we <u>calculate how many moles of Ba(OH)₂ reacted</u>, by c<em>onverting the total number of HCl moles to Ba(OH)₂ moles</em>:

  • 203.85 mmol HCl * \frac{1mmolBa(OH)_{2}}{2mmolHCl}= 101.925 mmol Ba(OH)₂

This means the remaining Ba(OH)₂ is:

  • 460 mmol - 101.925 mmol = 358.075 mmoles Ba(OH)₂

There are two OH⁻ moles per Ba(OH)₂ mol:

  • OH⁻ moles = 2 * 358.075 = 716.15 mmol OH⁻

Finally we <u>divide the number of OH⁻ moles by the </u><u><em>total</em></u><u> volume</u> (100 mL + 200 mL):

  • 716.15 mmol OH⁻ / 300.0 mL = 2.387 M

So the answer is 2.387 mol/L

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Answer:

A) (3.2g)

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n200080 [17]

Answer:

Explanation:

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