CH-3 - CH - O - CH-3CH-2 - CH - O overset-OH- Delta-longrightarrow- mixture of aldols-z- - total number of aldols including stereoisomers-Report your answer as -z - 5-

Question

Toppr Admin · Accepted Answer

The products are formed as shown and total stereoisomers formed -4-4-2-2-12Z-x2212-5-12-x2212-5-7

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

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

CH3−CH=O+CH3CH2−CH=OOH−,Δ⟶ mixture of aldols(z) = total number of aldols including stereoisomers.Report your answer as (z - 5).

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

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

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