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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 OHCCH(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!