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never [62]
4 years ago
13

The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the con

figurations of the oxirane products, A and B are Product A: _______ Product B: _______ Will either of these two oxirane products rotate the plane of polarization of plane polarized light? _____

Chemistry
1 answer:
gtnhenbr [62]4 years ago
8 0

Answer:

Product A: cis; no

Product B: cis: no  

Explanation:

Two common methods of forming oxiranes from alkenes are:

  • Reaction with peroxyacids
  • Formation of a halohydrin followed by reaction with base

1. Reaction with peroxyacids

(a) Stereochemistry

The reaction with a peroxyacid is a syn addition, so the product has the same stereochemistry as the alkene.

The starting alkene is cis, so the product is <em>cis</em>-2,3-diethyloxirane.

(b) Configuration

The product is optically inactive because it has an internal plane of symmetry.

It will not rotate the plane of polarized light.

2. Halohydrin formation

(a) Stereochemistry

The halogenation of the alkene proceeds via a cyclic halonium ion.

The backside displacement of halide ion by alkoxide is also stereospecific, so a cis alkene gives a cis epoxide.

The product is <em>cis</em>-2,3-diethyloxirane.

(b) Configuration

The cyclic halonium ion has an internal plane of symmetry, as does the product (meso).

The oxirane will not rotate the plane of polarized light.

 

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What aqueous solution has the highest boiling point at standard pressure? A) 1.0 M KCl(aq) B) 1.0 M CaCl2(aq) C) 2.0 M KCl(aq) D
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The increase in the boiling point of a solvent is a colligative property.


That means that the increase in the boling point will be related to the number of particles (molecules or ions) present in the solution.


The higher the number of particles (molecules or ions) the higher the increase in the boiling point.


All the aqueous solutions presented are electrolytes, i.e. the solutes are ionic compounds.


Then, you have to compare the number of ions that you have in each solution.


A) 1.0 M KCl ---> 1.0 M K+     +      1.0 MCl-    = 2 moles of particles / liter


B) 1.0 M CaCl2 --> 1.0M Ca(2+)      +      1.0M * 2 Cl (-)    = 3 moles of particle / liter


C) 2.0M KCl ---> 2.0 M K+      +      2.0 M Cl-  = 4 moles of particle / liter


D) 2.0 M CaCl2 ----> 2.0 M Ca (2+)      + 2.0M * 2 Cl (-)  = 6 moles of particle / liter.


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Read 2 more answers
15.89 percent carbon, 21.18 percent oxygen, and 62.93 percent osmium. what is the empirical formula
otez555 [7]

Answer:

OsCO or COOs

Explanation:

Data given

Carbon = 15.89 %

Oxygen = 21.18 %

Osmium = 62.93%

Empirical formula = ?

Solution:

First find the masses of each component

Consider total compound is 100g

As we now

mass of element = % of component

So,

15.89 g of C     = 15.89 % Carbon

21.18 g of O      =   21.18 % Oxygen

62.93 g of Os  =   62.93% Osmium

Now convert the masses to moles

For Carbon

Molar mass of C = 12 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  15.89 g/ 12 g/mol

                  no. of mole =  1.3242

mole of C = 1.3242

For Oxygen

Molar mass of O = 16 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  21.18 g/ 16 g/mol

                  no. of mole =  

mole of O = 1.3238

For Os

Molar mass of Os = 190 g/mol

                  no. of mole = mass in g / molar mass

Put value in above formula

                  no. of mole =  62.93 g/ 190 g/mol

                  no. of mole =  

mole of Os = 1.3312

Now we have values in moles as below

C = 1.3242

O = 1.3238

Os = 1.3312

Divide the all values on the smalest values to get whole number ratio

C = 1.3242 /1.3238 = 1.0003

O = 1.3238 /1.3238 = 1

Os = 1.3312 /1.3238 = 1.0056

So all have round value 1 mole

C = 1

O = 1

Os = 1

So the empirical formula will be (OsCO) i.e. all 3 atoms in simplest small ratio

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4 years ago
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