Detailed Reaction Procedure:
Step 1: Cinnamaldehyde + Cyclopentanone (Mix for 1 min).
Step 2: Add 10% NaOH + Water to the mixture above (Mix for 3 min).
Step 3: Add Ethanol to the mixture above (Mix for 3 min). Precipitate will form.
Step 4: Collect precipitate by suction filtration and wash it with Glacial Acetic Acid + Water (Continue suction for 5 to 10 min).
Step 5: Transfer the crude product known as BPC (or (2E,5E)-2,5-bis((E)-3-phenylallylidene)cyclopentan-1-one)) into beaker with Ethanol and heat it up until it boils (or also known as "recrystallization").
Step 6: Collect the pure BPC (crystals) by suction filtration and wash it with Ice-Cold Ethanol (Continue suction for 5 min).
Overall Reaction:
a) There are 9 geometric isomers (i.e. âcisâ and âtransâ or âE â and âZ â isomers) of BPC; however in this reaction, the product mixture is composed of one major geometric isomer of BPC. Clearly any
b) Ketones with α-hydrogens can undergo a process called, "keto-enol tautomerism". Please answer the following sub-parts regarding this process.
I) Draw the keto-enol tautomers of cyclopentanone.
II) Which of the two tautomers is more stable? Why is it more stable?
III) Please show the mechanism for the formation of keto-enol tautomers of cyclopentanone in the presence of an acid.
c) The last step of the formation of BPC involves dehydration of the aldol by elimination of a hydroxide ion; which is generally considered a poor leaving group. Please justify why the elimination of OHâ occurs in this reaction.
In this experiment, (2,SE)-2,5-bis(lE)-3-phenylallylidene)cyclopentan-1-one example of a crossed aldol condensation. (BPC) is synthesized in an Initially, ë°ë¬ë°3-phenylalylidene)cyclopentan-1-one (PC) is formed dil. NaOH(aq) cinnamaldehyde cyclopentanone However, in the presence of excess cinnamaldehyde, PC rapidly undergoes a second condensation to form BPC: dil. NaOH(aq) cinnamaldehyde PC BPC OH OH- cinnamaldehyde H. H cinnamaldehyde OH Oh The first step involves the acid-base reaction between a strong base such as hydroxide ion and a hydrogen located a carbon alpha to a carbonyl group. In the subsequent step (see Step 2), the enolate attacks the carbonyl group of aldehyde and followed by the loss of OH- as a leaving group to give an enal, which has the alkene and the ketone functional groups
