$C H_{3} - C H = O + C H_{3} C H_{2} - C H = O O H^{-}, Δ ⟶$ mixture of aldols
(z) = total number of aldols including stereoisomers.
Report your answer as (z - 5).

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Solution

The products are formed as shown and total stereoisomers formed :
$4 + 4 + 2 + 2 = 12$
$Z - 5 = 12 - 5 = 7$

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