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

  • Thread starter Thread starter suitcasespeake
  • Start date Start date
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
Sign up Log in
1 reply · 4K views
suitcasespeake
Messages
1
Reaction score
0

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!
 
Physics news on Phys.org

Discussion

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?