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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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Which fundamental force is primarily responsible for the attraction between protons and electrons?
Inessa [10]
The electromagnetic force is responsible for the electrostatic attraction and repulsion between charged particles and since the protons have positive charge and electrons have negative charge so the electromagnetic force will be the fundamental force primarily responsible for the attraction between electrons and protons
7 0
3 years ago
Given that 2S (s)+3O2 (g)→2SO3 (g)2SO2 (g)+O2 (g)→2SO3 (g) has an enthalpy change of −790.4 kJ has an enthalpy change of −198.2
abruzzese [7]

Answer:

The heat of formation of SO2 is -296.1 kJ

Explanation:

<u>Step 1:</u> Data given

2S (s)+3O2 (g)→2SO3 (g)     ΔH = -790.4 kJ  

2SO2 (g)+O2 (g)→2SO3 (g)  ΔH = -198.2 kJ

<u>Step 2</u>: Calculate the heat of formation of SO2

2 S(s) + 3 O2(g) --> 2 SO3(g) ΔH = -790.4 kJ  

S(s) + 3/2 O2(g) → SO3(g)    ΔH = -395.2 kJ

2SO2 (g)+O2 (g)→2SO3 (g)  ΔH = -198.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g)   ΔH = 99.1 kJ

----------------------------------------------------------------

S(s) + 3/2 O2(g) → SO3(g)    ΔH = -395.2 kJ

SO3(g) → SO2(g) + 1/2 O2(g)   ΔH = 99.1 kJ

-------------------------------------------------------------------

S (s)+O2 (g)→SO2 (g)

ΔHrxn = (-790.4 /2) kJ + (198.2/2) kJ

ΔHrxn = -395.2 kJ + 99.1 kJ = 296.1 kJ

The heat of formation of SO2 is -296.1 kJ

4 0
3 years ago
Question 11 (5 points)
LenKa [72]

Answer:

False

Explanation:

On the left side of the equation (Li + O2), there is 1 Li atom and 2 O atoms.

but on thw right side of the equation (Li2O,) there is 2 Li atoms and 1 O atom

8 0
3 years ago
t physiological pH, the carboxylic acid group of an amino acid will be ________, while the amino group will be ________, yieldin
astraxan [27]

Answer:

At physiological pH,the carboxylic acid group of an amino acid will be deprotonated while the amino group will be protonated,yielding the zwitter ion form.

Explanation:

Deprotonated means removal of protons in an acid base reaction and protonated means addition of protons in an acid base reaction.

Both protonated and de protonated reaction takes place in catalytic acid bade reaction by changing either it's mass or it's charge.

During formation of zwitter ion, the carboxylic acid will be deprotonated by donating the H+ ion while the amino acid is protonated by taking the H+ ion.

R-CH-COOH                               R-CH-COO-

   I                                ⇒                 I

  NH₂                                              NH₃               (Zwitter ion)

5 0
3 years ago
How do rivers affect watersheds?
beks73 [17]

Answer:

Explanation:

the last one I read it

5 0
2 years ago
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