- We're performing reductive ozonolysis, so products should be either aldehydes or ketones, but there are carboxylic acids in the products too.
- To form the first given product, there needs to be a double bond 3 carbons away from the triple bond, only option a) satisfies this but it has a double bond in between which will also ozonolyse. Can anyone tell me what I'm doing wrong? The answer given in the key is a)
We can use reverse engineering and process of elimination to arrive at the right answer.
All the options have a triple bond. So let us name the terminal alkyne carbons with number 1 and 2. With 2 being the Carbon on the rightmost end.
Let us represent the second and third products as follows : R1-COOH and R2-COOH where R1 = CHO and R2= H.
So we get R2-COOH from carbon number 2 of the terminal alkyne carbons. To get R1-COOH there needs to be a double bond at the carbon that is attached to carbon number 1.
Therefore we can rule out options (B) and (C).
Now to further reduce our options, the option D would produce a CO2 since there is a carbon which has two double bonds. Since there is no CO2 in the products we can conclude that the correct option is (A)