What Reagent for Step 1 and Intermediate X in But-2-ene-1,4-diol Conversion

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SUMMARY

The conversion of But-2-ene-1,4-diol to ketobutanedioic acid involves two steps, with the second step utilizing hot acidified KMnO4. The discussion identifies potential reagents for the first step, including cold acidified KMnO4, steam with concentrated H2SO4, and warm acidified K2Cr2O7. The intermediate X is proposed to be HOCH2CH(OH)CH2CH2OH, which is formed from the reaction of the diol with the selected reagent. The elimination of options based on their reactivity with alkenes and diols is crucial for determining the correct pathway.

PREREQUISITES
  • Understanding of organic reaction mechanisms, specifically oxidation reactions.
  • Familiarity with reagents such as KMnO4 and K2Cr2O7 in organic synthesis.
  • Knowledge of functional group transformations, particularly diols and carboxylic acids.
  • Basic skills in interpreting chemical reaction equations and intermediates.
NEXT STEPS
  • Research the mechanism of oxidation reactions involving cold and hot KMnO4.
  • Study the properties and applications of H2SO4 in organic synthesis.
  • Explore the role of K2Cr2O7 as an oxidizing agent in organic chemistry.
  • Learn about the synthesis and reactivity of diols and their derivatives.
USEFUL FOR

Chemistry students, organic chemists, and anyone involved in synthetic organic chemistry will benefit from this discussion, particularly those focusing on oxidation reactions and functional group transformations.

suitcasespeake
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Homework Statement



But-2-ene-1,4-diol is converted in two steps through an intermediate X into ketobutanedioic acid.
Hot acidified KMnO4 is added in step 2.

HOCH2CH=CHCH2OH ----> X -----> HO2CCOCH2CO2H

What could be the reagent for step 1 and the intermediate X?

reagent for step 1
X​
A cold acidified KMnO4
HOCH2CH2CH(OH)CH2OH​
B hot acidified KMnO4 O
HCCH(OH)CH2CHO​
C steam and concentrated
H2SO4 HOCH2CH(OH)CH2CH2OH​
D warm acidified K2Cr2O7
HO2CCH=CHCO2H​

Sorry about the unalignment.

The Attempt at a Solution


I'm aware that an alkene reacts with cold KMnO4 to give a dol and hot KMnO4 to give a dicarboxylic acid but I'm stumped since the first compound is both a diol and alkene. Any help appreciated!
 
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suitcasespeake said:

Homework Statement



But-2-ene-1,4-diol is converted in two steps through an intermediate X into ketobutanedioic acid.
Hot acidified KMnO4 is added in step 2.

HOCH2CH=CHCH2OH ----> X -----> HO2CCOCH2CO2H

What could be the reagent for step 1 and the intermediate X?

reagent for step 1
X​
A cold acidified KMnO4
HOCH2CH2CH(OH)CH2OH​
B hot acidified KMnO4 O
HCCH(OH)CH2CHO​
C steam and concentrated
H2SO4 HOCH2CH(OH)CH2CH2OH​
D warm acidified K2Cr2O7
HO2CCH=CHCO2H​

Sorry about the unalignment.

The Attempt at a Solution


I'm aware that an alkene reacts with cold KMnO4 to give a dol and hot KMnO4 to give a dicarboxylic acid but I'm stumped since the first compound is both a diol and alkene. Any help appreciated!

Hi suitcasespeake! Welcome to PF! :smile:

I suppose your C option is steam and concentrated ##H_2SO_4##, right?

It is easy to solve the problem by eliminating the options. Do you see which options are not possible?