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shtirl [24]
4 years ago
11

Draw the primary alcohol with molecular formula C5H12O that cannot be synthesized by the hydroboration–oxidation of an alkene.

Chemistry
1 answer:
alexandr402 [8]4 years ago
3 0

Answer:

The answer is neopentyl alcohol. (CH₃)₃C-CH₂-OH

Explanation:

This alcohol cannot be synthesized by the hydroboration–oxidation of an alkene since the reaction of hydroboration–oxidation go through steps that involved the addition of BH₃ into the original structure. In the case of neopentyl alcohol, if we carry out retro synthetic by the hydroboration–oxidation, we will find the original alkene is (CH₃)₃C=CH₂ which is impossible since the center carbon atom has 5 bonds.

The condition for hydroboration–oxidation reaction is that the alkene must have the structure of R-CH=CH₂

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Assuming ideal behavior of the gas for a fixed amount when temperature is held constant, the pressure and volume are inversely proportional as given by the expression
     P1V1 = P2V2 
where the terms with subscripts of one represent the initial conditions for pressure and volume of the gas while for terms with subscripts of two represent the final  conditions. 
Rearranging the Boyle's law equation to calculate for the final volume V2:
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3 years ago
Calcium oxide (CaO) can react with carbon dioxide (CO2) in a synthesis reaction. Select the most likely products of this reactio
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Answer: CaCO3 reaction.

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2 years ago
Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+
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<u>Answer:</u> The spontaneous cell reaction having smallest E^o is I_2+Cu\rightarrow Cu^{2+}+2I^-

<u>Explanation:</u>

We are given:

E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V

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

The equation used to calculate electrode potential of the cell is:

E^o_{cell}=E^o_{oxidation}+E^o_{reduction}

The combination of the cell reactions follows:

  • <u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)

Oxidation half reaction:  Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-   E^o_{oxidation}=0.45V

Reduction half reaction:  I_2(s)+2e^-\rightarrow 2I_-(aq.)   E^o_{reduction}=0.54V

E^o_{cell}=0.45+0.54=0.99V

Thus, this cell will not give the spontaneous cell reaction with smallest E^o_{cell}

  • <u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)

Oxidation half reaction:  Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-   E^o_{oxidation}=-0.34V

Reduction half reaction: I_2(s)+2e^-\rightarrow 2I_-(aq.)   E^o_{reduction}=0.54V

E^o_{cell}=-0.34+0.54=0.20V

Thus, this cell will give the spontaneous cell reaction with smallest E^o_{cell}

  • <u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)

Oxidation half reaction:  Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-   E^o_{oxidation}=0.45V

Reduction half reaction:  Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)   E^o_{reduction}=0.34V

E^o_{cell}=0.45+0.34=0.79V

Thus, this cell will not give the spontaneous cell reaction with smallest E^o_{cell}

Hence, the spontaneous cell reaction having smallest E^o is I_2+Cu\rightarrow Cu^{2+}+2I^-

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