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Dominik [7]
3 years ago
13

Part b (r)–3-chloro-2-methylhexane will undergo a nucleophilic substitution reaction in the presence of sodium ethoxide and etha

nol. complete the mechanism for the sn2 reaction and draw the products of the reaction. draw all missing reactants and/or products in the appropriate boxes by placing atoms on the grid and connecting them with bonds and including charges where needed. indicate the mechanism by drawing the electron-flow arrows on the molecules. arrows should start on an atom or a bond and should end on an atom, bond, or where a new bond should be created.
Chemistry
1 answer:
arsen [322]3 years ago
8 0

Answer:

See answer below

Explanation:

To understand this, we need to make the reaction involved with the given configuration. Configuration R means that the substituents of the molecule, the priority order goes clockwise (From heaviest to lightest) so, if an Sn2 is ocurring in the reaction this means that the nucleophile, which in this case, is the sodium ethoxide will attack the molecule in the opposite side of the leaving group (which is the chlorine).

Also when this occurs, as Sn2 is a bimolecular reaction and is held in one step, it occurs an inversion in the configuration of the product. So, it's the innitial reactant was R, then the final product will be S.

The mechanism and product can be watched in the attached picture

Hope this helps.

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An element is a pure substance in which there are how many kinds of atoms?
Ostrovityanka [42]

Answer:

One

Explanation:

An element is a pure substance in which there are only one kind of atom. Elements are distinct substances that cannot be split up into simpler substances.

Such substances consists of only one kind of atom. There are over a hundred known elements to date.

Generally, as a pure substance, the composition of an element is definite and they are homogenous in all parts.

8 0
3 years ago
Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno
melamori03 [73]

1.062 mol/kg.

<em>Step 1</em>. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)

Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)

= 4.035 mmol KHP

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)

= 4.035 mmol NaOH

<em>Step 4</em>. Calculate the mass of the NaOH

Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)

= 161 mg NaOH

<em>Step 5</em>. Calculate the mass of the water

Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g

= 37.973 g

<em>Step 6</em>. Calculate the molal concentration of the NaOH

<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

3 0
3 years ago
If there are 40 mol of NBr3 and 48 mol of NaOH, what is the excess reactant?
Nata [24]

Answer:

The correct answer is option B.

Explanation:

3NaOH+2NBr_3\rightarrow 3HOBr+3NaBr+N_2

Moles of NBr_3 = 40 mol

Moles of NaOH = 48 mol

According to reaction, 3 moles of NaOH reacts with 2 moles NBr_3

Then ,48 moles of NaOH will reacts with:

\frac{2}{3}\times 48 mol=32 mol of NBr_3

Then ,40 moles of NaBr_3 will reacts with:

\frac{3}{2}\times 40 mol=60 mol of NaOH

As we can see that 48 moles of sodium will completey react with 32 moles of nitrogen tribromide.

Moles left after reaction = 40 mol - 32 mol = 8 mol

Hence, the NBr_3 is an excessive reagent.

6 0
3 years ago
Read 2 more answers
Air is made up of different gases, such as oxygen, nitrogen, and carbon dioxide.
marysya [2.9K]

Answer:

option A

Explanation:

hope it helps

4 0
2 years ago
What are the n, l, and possible ml values for the 2p and 5f sublevels?
elixir [45]

Answer:

1. 2p sublevels, n = 2, orbital <em>p</em>, l = 1, ml = 0, ±1

2. 5f sublevels n = 5, orbital <em>f</em>, l = 3, ml = 0, ±1, ±2, ±3

Explanation:

The rules for electron quantum numbers are:

1. Shell number, 1 ≤ n

2. Subshell number, 0 ≤ l ≤ n − 1

3. Orbital energy shift, -l ≤ ml ≤ l

4. Spin, either -1/2 or +1/2

So,

1. 2p sublevels,

n = 2, orbital <em>p</em>, so l = 1, ml = 0, ±1

2. 5f sublevels

n = 5, orbital <em>f</em>, so l = 3, ml = 0, ±1, ±2, ±3

7 0
3 years ago
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