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Setler [38]
3 years ago
11

Oxidation of a dithiol such as 2,5-hexanedithiol forms a six-membered ring containing a disulfide group as part of the ring. Dra

w the structure of this cyclic disulfide (Hint: Draw the starting compound in line structure format first).
Chemistry
1 answer:
rosijanka [135]3 years ago
6 0

Answer:

See picture below

Explanation:

In general terms, oxidation of thiol are rather different than oxidations with alcohols.

A dithiol is a molecule that has two thiols in an organic compound. In the case of the 2,5 hexanedithiol, we have the thiols in the carbon 2 and carbon 5.

When oxidation of thiols occurs, and depending of the number of thiols in the molecule, the final compound will have a sulfide group.

This is because the oxydation makes that the hydrogen atom of the sulfur is substracted. So, in the case of the dithiols, we have that both hydrogen atoms are substracted and then, they are available to join. Therefore, As the thiols are in the same molecule, it will occur a self condensation reaction.

The picture below will show the final product.

Hope this helps

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2p is the correct representation for the sub-shell with n = 2 and l = 1.
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Explanation:

Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....

Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...

s = 1 orbital

p = 3 orbitals

d = 5 orbitals

f = 7 orbitals

For n = 4

l = 0 to (n-1) = 0 to 3 = (4s , 4p , 4d , 4f)

Number of subshells = 4

Number of orbitals =         1 + 3 + 5 + 7  = 16

The maximum number of electrons the n = 4 shell can contain:

Each orbital can holds upto two electrons, then 16 orbitals will have :

16\times 2=32

32 is the maximum number of electrons the n = 4 shell can contain

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3 years ago
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Which compound lauric acid or sucrose is more soluble in water?
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Now that Snape and Dumbledore has taught you the finer points of hydration calculations they have a slightly more challenging pr
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Answer:

The value of an integer x in the hydrate is 10.

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Moles of hydrated sodium carbonate = n

0.0366 M=\frac{n}{5.00 L}

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Molar mass of hydrated sodium carbonate = 106 g/mol+x18 g/mol

n=\frac{\text{mass of Compound}}{\text{molar mass of compound}}

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Solving for x, we get:

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Answer:

D) Chromatography

Explanation:

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