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3 years ago

Propose a mechanism for the following reaction.

Chemistry

1 answer:

raketka [301]3 years ago

5 0

First, recognize that this is an elimination reaction in which hydroxide must leave and a double bond must form in its place. It is likely an E2 reaction. Here is an efficient mechanism:
1) Pre-reaction: Protonate the -OH to make it a good leaving group, water. H2SO4 or any strong H+ donor works. The water is positively charged but still connected to the compound.
2) E2: Use a sterically hindered base, such as tert-butoxide (tButO-) to abstract the hydrogen from the secondary carbon. [You want a sterically hindered base because a strong, non-sterically hindered base could also abstract a hydrogen from one of the two methyl groups on the tertiary carbon, and that leads to unwanted products, which is not efficient]. As the proton of hydrogen is abstracted, water leaves at the same time, creating an intermediate tertiary carbocation, and the 2 electrons in the C-H bond immediately are used to make a double bond towards the partial positive charge.
In the products we see the major product and water, as expected. Even though you have an intermediate, remember that an E2 mechanism technically happens in one step after -OH protonation.

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3. What noble gas would be part of the electron configuration notation for Mn?

shusha [124]

Answer:

Argon {Ar}

Explanation:

The noble gas used for a condensed electron configuration is the one before the element which you are configuring.

In this case, the element (Mn) is manganese

The noble gas that is before this element is Argon which is the row above it

so your configuration would be {Ar} 3d^5 4s^2

7 0

3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago

If you burn 50.6 g of hydrogen and produce 452 g of water, how much oxygen reacted?

Svetllana [295]

Acc. to Law of Conservation of Mass
Mass of reactants=Mass of Products
Let mass of Oxygen be x.
So,
50.6+x=452
x=452-50.6
=401.4 g

4 0

3 years ago

What is the main difference between electron configuration and orbital notation?

8_murik_8 [283]

Answer:

Orbital Notation is more specific on where exactly the electron is placed.

Explanation:

When writing an electron configuration for an atom, rather than writing out the occupation of each and every orbital specifically, you instead lump all the core electrons together and designate it with a symbol of the corresponding noble gas on the Periodic Table.

the arrangement of electrons in the orbitals of an atom or molecule

While Orbital Notation is a visual transformation of the electron configuration. It shows you where each specific electron is placed and what its "spin" is.

Glad I could help!

6 0

3 years ago

A. 20 moles of NH3 are needed to produce how many moles of H2O?

tiny-mole [99]

a. 30 moles of H₂O

b. 2.33 moles of N₂

<h3>Further explanation</h3>

Given

a. 20 moles of NH₃

b. 3.5 moles of O₂

Required

a. moles of H₂O

b. moles of N₂

Solution

Reaction

4NH₃+3O₂⇒2N₂+6H₂O

a. From the equation, mol ratio NH₃ : H₂O = 4 : 6, so mol H₂O :

=6/4 x mol NH₃

= 6/4 x 20 moles

= 30 moles

b. From the equation, mol ratio N₂ : O₂ = 2 : 3, so mol N₂ :

=2/3 x mol O₂

= 2/3 x 3.5 moles

= 2.33 moles

7 0

2 years ago