[swmmr1928]

Experimental Details:
-When treated with base [hydroxide ion] and after appropriate workup, Compound F was converted into Compound G, which is the one and only product of this reaction.
-Compound F, when subjected to an elemental analysis experiment, gave the following results: C - 71.10% and H - 5.25%.
- Compound G, when subjected to an elemental analysis experiment, gave the following results: C –66.65% and H –5.60%

Questions:
-Based upon the experimental observations, determine the structure of Compound F.
- If you cannot unambiguously determine the structure of Compound F, explain why.
-Based upon the experimental observations, determine the structure of Compound G.
- If you cannot unambiguously determine the structure of Compound G, explain why.
-In what [mechanistic] way is the conversion of Compound F to Compound G similar
to the Cannizzaro reaction?

========================================================
My attempt:
Compound F:
C 71 H 5.25 O=100-71-5.25=23.5

0.71 g C/12.011 g C=0.0591 mol C
0.0525 g H/1.007 g H=0.0521 mol H
0.235 g O/15.99 g O=0.01469mol O

Compound F
C16H14O4

=======================
Compound G
C=66.65 H=5.6 O=100-66.65-5.6=27.75

66.65 g C/12.011 g C=5.549 mol C
5.6 g H/1.007 g H=5.561 mol H
27.75 g O/15.99 g O=1.735mol O

Compound G
C16H16O5

net addition of H2O?

Perhaps this is the carboxylic acid product of the Cannizzaro Reaction.
In order for only the carboxylic acid product to form, it is a crossed Cannizzaro Reaction.
If the reaction occurs at an aldehyde of the starting chemical, how can the other oxygens be arranged so they are not affected? All ethers,esters,etc?

Thankyou

[Extracted]

Experimental Details:
-When treated with base [hydroxide ion] and after appropriate workup, Compound F was converted into Compound G, which is the one and only product of this reaction.
-Compound F, when subjected to an elemental analysis experiment, gave the following results: C - 71.10% and H - 5.25%.
- Compound G, when subjected to an elemental analysis experiment, gave the following results: C –66.65% and H –5.60%

Questions:
-Based upon the experimental observations, determine the structure of Compound F.
- If you cannot unambiguously determine the structure of Compound F, explain why.
-Based upon the experimental observations, determine the structure of Compound G.
- If you cannot unambiguously determine the structure of Compound G, explain why.
-In what [mechanistic] way is the conversion of Compound F to Compound G similar
to the Cannizzaro reaction?

========================================================
My attempt:
Compound F:
C 71 H 5.25 O=100-71-5.25=23.5

0.71 g C/12.011 g C=0.0591 mol C
0.0525 g H/1.007 g H=0.0521 mol H
0.235 g O/15.99 g O=0.01469mol O

Compound F
C16H14O4

=======================
Compound G
C=66.65 H=5.6 O=100-66.65-5.6=27.75

66.65 g C/12.011 g C=5.549 mol C
5.6 g H/1.007 g H=5.561 mol H
27.75 g O/15.99 g O=1.735mol O

Compound G
C16H16O5

net addition of H2O?

Perhaps this is the carboxylic acid product of the Cannizzaro Reaction.
In order for only the carboxylic acid product to form, it is a crossed Cannizzaro Reaction.

If that's true this is a tricky question.  The question doesn't mention the sacrifical aldehyde.  It does mention that G is the only product formed.

If the reaction occurs at an aldehyde of the starting chemical, how can the other oxygens be arranged so they are not affected? All ethers, esters,etc?

Would an ester survive hydroxide?

[swmmr1928]

Quote
If the reaction occurs at an aldehyde of the starting chemical, how can the other oxygens be arranged so they are not affected? All ethers, esters,etc?


Would an ester survive hydroxide?

Ah, an ester will hydrolyze to Carboxylic Acid.  So, some of the oxygens can be in Carboxylic Acids and then are deprotonated in base, and reprotonated in the workup. 

But the question still remains, without a sacrificial aldehyde the crossed Cannizzaro Reaction cannot be performed.
What adds net H2O?

[orgopete]

I'd say it would be difficult for anyone to come up with THE structure for C16H14O4 which is converted to C16H16O5. Some context is needed or missing. Was it contained in Compounds A-E? Is a crossed Cannizzaro reaction a required element in this reaction? Was that part of the classroom discussion surrounding this problem set? Oh, we missed this class.

[Extracted]

But the question still remains, without a sacrificial aldehyde the crossed Cannizzaro Reaction cannot be performed.
What adds net H2O?

Well if you stick with the ester idea… hydrolysis adds H2O.  If only one product forms what type of ester are we dealing with?