Cyclic anhydride recuction mechanism

Can someone please help me/give me some clues with this mechanism. Ive tried it a few times and cant do it



Thanks

Can someone please help me/give me some clues with this mechanism. Ive tried it a few times and cant do it



Thanks

Okay so I think the best way to rationalise this, as I can't see any other way of it happening is to use LiAlH4 to reduce 4 to a chain with an aldehyde and a carboxylic acid.

5 is of course an ester. One way of making esters is of course by using a alcohol and a carboxylic acid. So I further reduced to the the aldehyde to a primary alcohol. Then using acid converted the -OH into a water leaving group and then pushed the electrons from carboxylate anion to release the leaving group.

I'm not sure if that's right as I can't see any reason why LiAlH4 would selectively reduce the aldehyde only. But that's my attempt and maybe you'll crack it even further?

Okay so I think the best way to rationalise this, as I can't see any other way of it happening is to use LiAlH4 to reduce 4 to a chain with an aldehyde and a carboxylic acid.

5 is of course an ester. One way of making esters is of course by using a alcohol and a carboxylic acid. So I further reduced to the the aldehyde to a primary alcohol. Then using acid converted the -OH into a water leaving group and then pushed the electrons from carboxylate anion to release the leaving group.

I'm not sure if that's right as I can't see any reason why LiAlH4 would selectively reduce the aldehyde only. But that's my attempt and maybe you'll crack it even further?

OK thanks! So does this look right?

Ring closing is wrong. Alcohol will attack the carbonyl (Which will be activated by acid) rather than a non nucleophilic carboxylate anion kicking out the poor leaving group OH-

But then you get this don't you?