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Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Oxidation of alcohols

Here's some evidence that supports this pathway. One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

Aqueous methods like the Jones oxidation [n.b.: the Jones oxidation is $\ce{CrO3}/\text{aq. }\ce{H2SO4}$] are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

The actual mechanism for the oxidation step is as follows (Clayden, p. 195):

Mechanism

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


So, is the mechanism presented by my teacher valid, or its just a way to facilitate comprehension of students (because it is pretty intuitive)?

  • It's not valid.
  • Honestly, I don't see how the wrong mechanism is any more intuitive than the correct mechanism. The $\ce{[O]}$ symbol does not necessarily mean it is adding oxygen to the product; it just refers to a generic oxidation, which could either be adding oxygen or removing hydrogen. In this case, the mechanism involves removal of $\ce{H2}$ from the alcohol to form the aldehyde, not addition of $\ce{O}$ to form the hydrate.

I find it sad that we teach blatantly wrong things for no good reason. Some things, e.g. the Bohr model or d-orbital participation in "hypervalent" molecules, have their place in introductory chemistry - but there's no justification for this.

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Oxidation of alcohols

Here's some evidence that supports this pathway. One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

Aqueous methods like the Jones oxidation [n.b.: the Jones oxidation is $\ce{CrO3}/\text{aq. }\ce{H2SO4}$] are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

The actual mechanism for the oxidation step is as follows (Clayden, p. 195):

Mechanism

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


So, is the mechanism presented by my teacher valid, or its just a way to facilitate comprehension of students (because it is pretty intuitive)?

  • It's not valid.
  • Honestly, I don't see how the wrong mechanism is any more intuitive than the correct mechanism. The $\ce{[O]}$ symbol does not necessarily mean it is adding oxygen to the product; it just refers to a generic oxidation, which could either be adding oxygen or removing hydrogen. In this case, the mechanism involves removal of $\ce{H2}$ from the alcohol to form the aldehyde, not addition of $\ce{O}$ to form the hydrate.

I find it sad that we teach blatantly wrong things for no good reason. Some things, e.g. the Bohr model or d-orbital participation in "hypervalent" molecules, have their place in introductory chemistry - but there's no justification for this.

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Oxidation of alcohols

Here's some evidence that supports this pathway. One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

Aqueous methods like the Jones oxidation [n.b.: the Jones oxidation is $\ce{CrO3}/\text{aq. }\ce{H2SO4}$] are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

The actual mechanism for the oxidation step is as follows (Clayden, p. 195):

Mechanism

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.

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source | link

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Oxidation of alcohols

Here's some evidence that supports this pathway. One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

Aqueous methods like the Jones oxidation [n.b.: the Jones oxidation is $\ce{CrO3}/\text{aq. }\ce{H2SO4}$] are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

Here's theThe actual mechanism for the oxidation step is as follows (Clayden, p. 195):

Mechanism

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


So, is the mechanism presented by my teacher valid, or its just a way to facilitate comprehension of students (because it is pretty intuitive)?

  • It's not valid.
  • Honestly, I don't see how the wrong mechanism is any more intuitive than the correct mechanism. The $\ce{[O]}$ symbol does not necessarily mean it is adding oxygen to the product; it just refers to a generic oxidation, which could either be adding oxygen or removing hydrogen. In this case, the mechanism involves removal of $\ce{H2}$ from the alcohol to form the aldehyde, not addition of $\ce{O}$ to form the hydrate.

I find it sad that we teach blatantly wrong things for no good reason. Some things, e.g. the Bohr model or the octet ruled-orbital participation in "hypervalent" molecules, have their place in introductory chemistry - but there's no justification for this.

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Oxidation of alcohols

One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

Aqueous methods like the Jones oxidation are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

Here's the actual mechanism for the oxidation step (Clayden, p. 195):

Mechanism

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


So, is the mechanism presented by my teacher valid, or its just a way to facilitate comprehension of students (because it is pretty intuitive)?

  • It's not valid.
  • Honestly, I don't see how the wrong mechanism is any more intuitive than the correct mechanism. The $\ce{[O]}$ symbol does not necessarily mean it is adding oxygen to the product; it just refers to a generic oxidation, which could either be adding oxygen or removing hydrogen. In this case, the mechanism involves removal of $\ce{H2}$ from the alcohol to form the aldehyde, not addition of $\ce{O}$ to form the hydrate.

I find it sad that we teach blatantly wrong things for no good reason. Some things, e.g. the Bohr model or the octet rule, have their place in introductory chemistry - but there's no justification for this.

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Oxidation of alcohols

Here's some evidence that supports this pathway. One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

Aqueous methods like the Jones oxidation [n.b.: the Jones oxidation is $\ce{CrO3}/\text{aq. }\ce{H2SO4}$] are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

The actual mechanism for the oxidation step is as follows (Clayden, p. 195):

Mechanism

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


So, is the mechanism presented by my teacher valid, or its just a way to facilitate comprehension of students (because it is pretty intuitive)?

  • It's not valid.
  • Honestly, I don't see how the wrong mechanism is any more intuitive than the correct mechanism. The $\ce{[O]}$ symbol does not necessarily mean it is adding oxygen to the product; it just refers to a generic oxidation, which could either be adding oxygen or removing hydrogen. In this case, the mechanism involves removal of $\ce{H2}$ from the alcohol to form the aldehyde, not addition of $\ce{O}$ to form the hydrate.

I find it sad that we teach blatantly wrong things for no good reason. Some things, e.g. the Bohr model or d-orbital participation in "hypervalent" molecules, have their place in introductory chemistry - but there's no justification for this.

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source | link

Clayden et al., Organic Chemistry (2ed), p. 545 gives the reaction pathway as:

Oxidation of alcohols

One of the most well-known methods to selectively oxidise primary alcohols to aldehydes, without further oxidation to the carboxylic acid, is by using pyridinium chlorochromate in dichloromethane as solvent. This presumably works because water is excluded, which prevents the hydrate from being formed. Clayden writes:

Aqueous methods like the Jones oxidation are no good for this, since the aldehyde that forms is further oxidized to acid via its hydrate. The oxidizing agent treats the hydrate as an alcohol, and oxidizes it to the acid. The key thing is to avoid water, so PCC in dichloromethane works quite well. The related reagent PDC (pyridinium dichromate) is particularly suitable for oxidation to aldehydes.

If the reaction pathway was as your teacher taught you, then there would be no point in excluding water, since in that pathway water is not needed for the over-oxidation to the carboxylic acid.

Here's the actual mechanism for the oxidation step (Clayden, p. 195):

Mechanism

If water is present, then the aldehyde product simply forms the hydrate and the mechanism for oxidation to the carboxylic acid is exactly the same, except that one of the hydrogens is replaced with an $\ce{-OH}$. Note that you need an $\ce{-OH}$ group on the starting material to form a chromate ester - that means that aldehydes will not undergo oxidation, but their hydrates (which are geminal diols) will.


So, is the mechanism presented by my teacher valid, or its just a way to facilitate comprehension of students (because it is pretty intuitive)?

  • It's not valid.
  • Honestly, I don't see how the wrong mechanism is any more intuitive than the correct mechanism. The $\ce{[O]}$ symbol does not necessarily mean it is adding oxygen to the product; it just refers to a generic oxidation, which could either be adding oxygen or removing hydrogen. In this case, the mechanism involves removal of $\ce{H2}$ from the alcohol to form the aldehyde, not addition of $\ce{O}$ to form the hydrate.

I find it sad that we teach blatantly wrong things for no good reason. Some things, e.g. the Bohr model or the octet rule, have their place in introductory chemistry - but there's no justification for this.