I read one of the answers for a similar question, According to that it is due to the hemiacetale group converting into a carbonyl group. I don't get why the same thing can't be applied to sucrose though.
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
if youre looking for a rather unsophisticated answer.....it would go as....for a pyranose ring to open up...it needs to have a hemiacetal group ...such a group is present in maltose and absent in sucrose...the same thing cant be applied to sucrose as it dosent have a hemi acetal group...please check the structure again...(needs a C-OH linked to the ether linkage that holds together both rings..i guess thats how it goes in simple terms...hopefully i answered your question
P.S:The link given in the answer above mine has a better detailed explanation
