No. of stereoisomers in aldol condensation

[Physics lover]

Homework Statement

Problem is in attempt at a solution

Relevant Equations

Aldol condensation
Total stereoisomers=$2^n$
where n =no. of double bonds

According to me 4 products will be formed with a double bond in each of them.So total no. of stereoisomers will be=$4×2^1=8$.First product will be formed due self aldol in acetaldehyde.Second will be formed due to self aldol in Propanaldehyde.And the other two will be formed due to mixed aldol.I just want somebody to check my answer.Is it correct?

 

[mjc123]

You are right about the double-bond compounds, but the question asks about the number of aldols - implying before dehydration. How many aldols, and how many stereoisomers of each?

 

[Physics lover]

You are right about the double-bond compounds, but the question asks about the number of aldols - implying before dehydration. How many aldols, and how many stereoisomers of each?

But they have mentioned delta also.So dehydration should also be there.Otherwise why would they mention delta.

 

[mjc123]

I don't know, but they say aldols. Unless they use "aldols" to mean "products of the aldol reaction, i.e. α,β-unsaturated aldehydes", which would be sloppy.

 

[Physics lover]

I don't know, but they say aldols. Unless they use "aldols" to mean "products of the aldol reaction, i.e. α,β-unsaturated aldehydes", which would be sloppy.

So is it a question fault.Please tell me whether delta should be there or not if they means before dehydration.