Which C4H10O Alcohols Do Not React with K2Cr2O7?

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SUMMARY

The discussion centers on the reactivity of C4H10O alcohols with potassium dichromate (K2Cr2O7), a known oxidizing agent. It is established that tertiary alcohols do not undergo oxidation with K2Cr2O7, while secondary alcohols can be oxidized to ketones. The participants suggest that the inquiry involves identifying all possible isomers of C4H10O and determining which can be oxidized. The conclusion is that only secondary alcohols will react, while tertiary alcohols will not.

PREREQUISITES
  • Understanding of alcohol classification: primary, secondary, and tertiary alcohols
  • Knowledge of oxidation reactions and oxidizing agents
  • Familiarity with the molecular formula C4H10O and its isomers
  • Basic organic chemistry concepts, including isomer drawing
NEXT STEPS
  • Research the oxidation of secondary alcohols to ketones using K2Cr2O7
  • Study the structural isomers of C4H10O and their properties
  • Learn about the mechanisms of alcohol oxidation reactions
  • Explore other oxidizing agents and their reactivity with different alcohol types
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Chemistry students, organic chemists, and anyone interested in the reactivity of alcohols and oxidation reactions.

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Question

How many alcohols do not react with K_2Cr_2O_7? Alcohols have the molecular formula C_4 H_{10} O.

Attempt at solution

K_2Cr_2O_7 is an oxidazing agent. The tertiary alcohol cannot be oxidised with it, but you can oxidise secondary alcohols (Source: http://en.wikipedia.org/wiki/Oxidation_of_secondary_alcohols_to_ketones). I think other molecules can be oxidized. So the amount of oxidised alcohols is one.
 
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Not sure what the question really asks. Perhaps they want you to count all possible alcohols with a given formula and check, how many will be oxidised and how many will not?
 
Borek said:
Not sure what the question really asks. Perhaps they want you to count all possible alcohols with a given formula and check, how many will be oxidised and how many will not?

Thanks! I clarified the question. Hopefully, it is now easier to understand.
 
The formula indicates that there are no rings or unsaturation. Draw every isomer of this compound and use what you know.
 

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