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Organic Chemistry : Help with Oxidation-Reduction Reactions

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Example Question #1 : Help With Oxidation Reduction Reactions

Which of the following is not capable of oxidizing a secondary alcohol to a ketone?

Possible Answers:

$\displaystyle K_2Cr_2O_7$

CrO_3 $\displaystyle CrO_3$, H_2SO_4 $\displaystyle H_2SO_4$, acetone

Lithium aluminum hydride $\displaystyle (LiAlH_4)$

Pyridinium chlorochromate (PCC)

All of these answers can oxidize secondary alcohols to ketones

Correct answer:

Lithium aluminum hydride $\displaystyle (LiAlH_4)$

Explanation:

Lithium aluminum hydride is correct because it is a reducing agent, and is therefore not capable of oxidizing secondary alcohols. Instead, LAH could be used to perform the reverse reaction, reducing a ketone to an alcohol. The other answer choices are oxidizing agents.

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Example Question #1 : Help With Oxidation Reduction Reactions

In the following equation, which element is reduced?

$MnO{_{4}}^{-} (aq) + I^{-}(aq) \rightarrow Mn^{2+} (aq) + I_{2} (s)$ $\displaystyle MnO{_{4}}^{-} (aq) + I^{-}(aq) \rightarrow Mn^{2+} (aq) + I_{2} (s)$

Possible Answers:

$\displaystyle Mn$

$\displaystyle O$

Manganese, iodine, and oxygen are all oxidized.

$\displaystyle I$

Correct answer:

$\displaystyle Mn$

Explanation:

In an oxidation-reduction (redox) reaction, reduction and oxidation both occur. Thus, not all of the elements can be oxidized, and not all of them can be reduced. In this equation, the oxidation number of oxygen is $\displaystyle -2$. Multiplying that by the number of oxygen atoms ($\displaystyle 4$), the overall charge on oxygen is $\displaystyle -8$. $MnO_{4}$ $\displaystyle MnO_{4}$ has an overall $\displaystyle -1$ charge, so the oxidation number on $\displaystyle Mn$ must combine with $\displaystyle -8$ to form $\displaystyle -1$. Thus, the oxidation number of $\displaystyle Mn$ at the beginning of the reaction is $\displaystyle +7$. The iodine at the beginning of the reaction has an oxidation number of $\displaystyle -1$, as seen by the negative superscript.

At the end of the reaction, $\displaystyle Mn$ has an oxidation number of $\displaystyle +2$, as seen by the positive superscript. Iodine has an oxidation number of $\displaystyle 0$ (there is no charge on the $I_{2}$ $\displaystyle I_{2}$).

Thus, $\displaystyle Mn$ went from having an oxidation number of $\displaystyle +7$ to one of $\displaystyle +2$. Iodine went from having an oxidation number of $\displaystyle -1$ to one of $\displaystyle 0$. Oxidation occurs when electrons are lost (the number becomes more positive), and reduction occurs when electrons are gained (the number becomes more negative). Because the oxidation number of iodine became more positive, iodine was oxidized. Because the oxidation number of manganese became more negative (less positive), manganese was reduced.

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Organic Chemistry : Help with Oxidation-Reduction Reactions

Study concepts, example questions & explanations for Organic Chemistry

All Organic Chemistry Resources

Example Questions

Example Question #1 : Help With Oxidation Reduction Reactions

Example Question #1 : Help With Oxidation Reduction Reactions

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