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Ierofanga [76]
3 years ago
11

Two stereoisomers are obtained from the reaction of cyclopentene oxide with dimethylamine. the r,r-isomer is used in the manufac

ture of eclanamine, an antidepressant. what other isomer is obtained?

Chemistry
1 answer:
-Dominant- [34]3 years ago
5 0
I have attached an image that shows the reaction and each isomer formed.

The oxygen of the epoxide adds syn to the alkene, which means the oxygen is on one face of the molecule. Therefore, the epoxide has 2 chiral centers. The dimethyl amine is a nucleophile and it attack a carbon bound to oxygen in an sn2 fashion. This causes the epoxide ring to open, and after a proton transfer, the molecule now has a dimethyl amino group and an adjacent hydroxy group in an anti-relationship. The amine is able to attack either of the two carbons bound to the oxygen in the epoxide and this leads to two isomers. The isomers formed are a pair of enantiomers with the stereochemistry (<em>R,R</em>) and (<em>S,S</em>). Therefore, the second isomer that the question asks for is the (<em>S,S</em>) product shown.

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If a substance has a high hydronium concentration, the substance has a ____ pH.
aalyn [17]

Answer: low pH

Hydronium is H^+ and acids give hydronium ions to water solution.

PH = -log H^+ concentration. The higher the concentration, the lower pH value

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2 years ago
Rank and explain how the freezing point of 0.100 m solutions of the following ionic electrolytes compare. List from lowest freez
Elden [556K]

Answer:

The solutions are listed from the lowest freezing point to highest freezing point.

Mg₃(PO₄)₂ -  AlBr₃ - BeBr₂ -  KBr

Explanation:

Colligative property of freezing point, is the freezing point depression.

Formula is:

ΔT = Kf . m . i

Where ΔT = T° freezing pure solvent - T° freezing solution

Kf is the cryoscopic constant

m is molality and i is the Van't Hoff factor (number of ions, dissolved in solution)

In this case, T° freezing pure solvent is 0° because it's water, so the Kf is 1.86 °C/m, and m is 0.1 molal.

The i modifies the T° freezing solution. The four salts, are ionic compounds, so for each case:

BeBr₂  → Be²⁺  +  2Br⁻   i = 3

AlBr₃ →  Al³⁺  + 3Br⁻   i = 4

Mg₃(PO₄)₂  →  3Mg²⁺  +  2PO₄⁻³    i = 5

KBr →  K⁺  +  Br⁻   i = 2

Solution of Mg₃(PO₄)₂ will have the lowest temperature of freezing and the solution of KBr, the highest (always lower, than 0°).

8 0
3 years ago
On a roller coaster, where is maximum kinetic energy I’ll mark the brainiest.
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its the pink one I think

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4 0
2 years ago
Read 2 more answers
Using complete subshell notation (not abbreviations, 1s 22s 22p 6 , and so forth), predict the electron configuration of each of
dybincka [34]

<u>Answer:</u> The electronic configuration of the elements are written below.

<u>Explanation:</u>

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

For the given options:

  • <u>Option a:</u>  Carbon (C)

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is 1s^22s^22p^2

  • <u>Option b:</u>  Phosphorus (P)

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is 1s^22s^22p^63s^23p^3

  • <u>Option c:</u>  Vanadium (V)

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is 1s^22s^22p^63s^23p^64s^23d^3

  • <u>Option d:</u>  Antimony (Sb)

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3

  • <u>Option e:</u>  Samarium (Sm)

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6

Hence, the electronic configuration of the elements are written above.

4 0
3 years ago
Gold is currently trading at very high price. Suppose that gold is selling for around $1860/ounce. How
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Answer:

The answer is "3.81041978"

Explanation:

\to 1 \ OZ= 28.349523125 \ grams\\\\

              =28.349523125\times {1000} \ miligrams\\\\= 28349. 5231  \ miligrams\\

In 1860 = 28349.5231 \ miligrams\\

\to In \$ \ 1 = \frac{28349.5231}{1860} \ \ miligrams\\

             = 15.2416791 \ miligrams

\ In \  1 \ quarter =  \$ \ 0.25

\to \$ \ 0.25 =  15.2416791  \times 0.25 \  miligrams\\

              = 3.81041978

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3 years ago
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