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SVEN [57.7K]
3 years ago
8

Compound a is an alkene that was treated with ozone (followed by dms) to yield only 4-heptanone. Draw the major product that is

expected when compound a is treated with mcpba followed by aqueous acid (h3o+).

Chemistry
1 answer:
olga nikolaevna [1]3 years ago
5 0

Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive  cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.

Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.

The reaction is as shown in the image.

The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid H_3O^{+} is shown in the image.

m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.

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When heated, lithium reacts with nitrogen to form lithium nitride: 6Li(s) + N2(g) → 2Li3N(s) What is the theoretical yield of Li
anyanavicka [17]

Answer:

The % yield of the reaction = 27.5 %

Explanation:

Step 1: Data given

Mass of Li = 12.7 grams

Mass of N2 = 34.7 grams

Actual yield of Li3N = 5.85 grams

Molar mass of  Lithium = 6.94 g/mol

Molar mass of N2 = 28 g/mol

Molar mass of LI3N = 34.83 g/mol

Step 2: The balanced equation:

6Li(s) + N2(g) → 2Li3N(s)

Step 3: Calculate moles of Lithium

Moles Li = mass Li / Molar mass Li

Moles Li = 12.7 grams / 6.94 g/mol

Moles Li = 1.83 moles

Step 4: Calculate moles of N2

Moles N2 = 34.7 g/ 28 g/mol

Moles N2 = 1.24 moles

Step 5: Limiting reactant

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

Lithium is the limiting reactant. It will completely be consumed (1.83 moles).

N2 is in excess. There will be consumed 1.83 / 6 = 0.305 moles

There will remain 1.24 - 0.305 = 0.935 moles

Step 6: Calculate moles of Li3N

For 6 moles Li consumed, we need 1 mole of N2 to produce 2 moles of Li3N

For 1.83 moles Li, we'll have 1.83/3 = 0.61 moles of Li3N

Step 7: Calculate mass of Li3N

Mass Li3N =moles LI3N * Molar Mass LI3N

Mass Li3N = 0.610 moles * 34.83 g/mol

Mass Li3N = 21.2463 grams = Theoretical yield

Step 8: Calculate % yield

% yield = actual yield / theoretical yield

% yield = (5.85 / 21.2463)*100% = 27.5%

The % yield of the reaction = 27.5 %

8 0
3 years ago
For a particular redox reaction, NO-2 is oxidized to NO-3 and Ag+ is reduced to Ag . Complete and balance the equation for this
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The following are the steps  to complete and balnce the equation for the given reaction

<u>Explanation:</u>

We are given, NO2– is oxidized to NO3– and Ag is reduced to Ag

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Step 1) Assign the oxidation state to each element reaction

NO2– + Ag+ -----> NO3– + Ag(s)

N= +3                           N = +5                        

O = -2                            O = -2

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Step 3) Balance the O by adding 1 H2O for 1 O

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Step 6) Balance the electron in both half reaction

    NO2– + H2O -----> NO3– + 2H+ + 2e-

     2 Ag+ + 2e------> 2 Ag(s)

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