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taurus [48]
4 years ago
13

Question 044 Which of the following sequences converts 2-methylpropene and sodium acetylide into 3-methylbutanal? 1) HBr; 2) NaC

CH; 3) O3; 4) DMS 1) NaCCH; 2) H2/Ni2B; 3) O3; 4) DMS 1) HBr, ROOR; 2) NaCCH; 3) H2/Ni2B 4) O3; 5) DMS 1) HBr, ROOR; 2) NaCCH; 3) O3; 4) H2O 1) HBr; 2) NaCCH; 3) O3; 4) H2O
Chemistry
2 answers:
Minchanka [31]4 years ago
8 0

Answer:

1) HBr; 2) NaCCH; 3) O3; 4) H2O

Explanation:

The first step is formation of alkyl halide followed by reaction with sodium acetylide, to form 3-methylbutene, which is then followed by oxidation reaction with O3& H2O to 3-methylbutanal

Kruka [31]4 years ago
6 0

Answer:

The sequence of reagents' incorporation for the conversion of 2-methylpropene into 3-methylbutanal is <u>option (C.):</u>

1. HBr,ROOR ; 2.  NaCCH,H_{2}/Ni_{2}B ; 3. O_{3}, DMS

Explanation:

The summary of this reaction's mechanism is as follows:

  • <em>Radical hydro-halogenation</em> of 2-methylpropene through a technique known as <em>"peroxide effect"</em> in which the HBr addition causes free radical formation of the alkene
  • <em>Nucleophilic SN2 substitution </em>reaction (acetylide)
  • <em>Hydrogenation</em>
  • <em>Ozonolysis of the alkene</em>: after the addition reaction, the intermediate ozonide is converted to aldehyde using dimethyl sulfide (DMS)
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SSSSS [86.1K]

<u>Answer:</u> The molecules of oxygen gas that will be reduced to water are 42 molecules

<u>Explanation:</u>

We are given:

E^o_{(NO_2^-/NH_4)}=-0.41V\\E^o_{(O_2/H_2O)}=0.82V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, oxygen will undergo reduction reaction will get reduced.

NH_4 will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

The half reactions follows:

<u>Oxidation half reaction:</u>  NH_4\rightarrow NO_2^-+6e^-     ( × 4)

<u>Reduction half reaction:</u>  O_2+4e^-\rightarrow 2H_2O   ( × 6)

<u>Overall reaction:</u> 4NH_4+6O_2\rightarrow 4NO_2^-+12H_2O

We are given:

Molecules of NH_4 = 28

By Stoichiometry of the reaction:

4 molecules of NH_4 reacts with 6 molecules of oxygen gas

So, 28 molecules of NH_4 will react with = \frac{6}{4}\times 28=42 molecules of oxygen gas

Hence, the molecules of oxygen gas that will be reduced to water are 42 molecules

4 0
3 years ago
What is the Na+ concentration in each of the following solutions:
iren [92.7K]

Sodium Sulfate = Na2(SO4) meaning there are two ions of Na+ in one mole of Sodium Sulfate the M stands for Molarity, defined as Molarity = (moles of solute)/(Liters of solution), So if the Na2SO4 solution is 3.65M that means one Liter of has 3.65 moles of Na2SO4, the stoichiometry of Na2SO4 shows that there would be two Na+ ions in solution for every one Na2SO4.

Therefore if 3.65 moles of Na2SO4 was to dissolve, it would produce 7.3 moles of Na+, and since this is still a theoretical solution, we can assume 1 L of solution.

Finally we find [Na+] = 2*3.65 = 7.3M

Use the same logic for parts b and c




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3 years ago
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Answer:

3.45 × 10^5

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I hope this helps

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3 years ago
What type of radioactivity occurs in nuclear reactor if a neutron and uranium are the typical reactants
prisoha [69]

Answer:

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Explanation:

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