Mechanism of Aldol condensation, Mixed aldol , Intramolecular Aldol condensation.

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Mechanism of Aldol condensation, Mixed aldol , Intramolecular Aldol condensation.
	
Document Preview: Aldol Condensation 
Assignment 2 
Q.1 The aldehyde and the ketone below are self-condensed with aqueous NaOH so that an 
unsaturated carbonyl compound is the product in both cases. Give a structure for each product and 
explain why you think this product is formed. 
 
A.1 In both cases only enolate forms which acts as a nucleophile and the same aldehyde/Ketone are 
acting as an electrophile. The giant molecule formed first followed by elimination occurs on further 
heating. This elimination occurs via E1cB mechanism and release one molecule of water. Thus the 
a,ß-unsaturated carbonyl compounds are produced. In the following reaction 2-methylpent-2-en-1-
al and Mesityl oxide/ 4-methylpent-3-en-2-one are obtained. The consecutive elimination takes 
place due to presence of a-hydrogen atom.

Q.2 What happens when the following compound is heated with dil. NaOH? 
 
A.2 In case of unsymmetrical ketone question of regioselectivity appears. i.e., which a-carbon will 
undergo enolisation. If the 3-C carbon undergoes enolisation then after condensation nodehydration will take place due lack of another a-H atom. So, the condensation product will contain 
OH group as shown below. The carbon atom marked as black dot does not contain the a-H atom. 
 
On the other hand if 1-C take part in enolisation it is due to presence of more than one a-H atom, 
dehydration occurs. Thus we get an a, ß-unsaturated product. So, as a result we shall get a mixtre of 
two products. 
 
Q.3 Explain their formation.

A.3 It is due to presence of two carbonyl like (N=O &N-O) strong groups in nitromethane it acts as 
strong base. Instead of self-condensing, it withdraws the a-H from the given aldehyde to obtain the 
enolate. The intermediate so formed is basic enough to withdraw a proton from nitromethane to 
form the nitroalcohol.The process occurs through acylation of OH group followed by E1cB elimination. The enolate forms 
is driven by nitro group. Although pyridine it is a weak base...

David · Accepted Answer

Aldol Condensation 
Assignment 2 
Q.1 The aldehyde and the ketone below are self-condensed with aqueous NaOH so that an 
unsaturated carbonyl compound is the product in both cases. Give a structure for each product and 
explain why you think this product is formed.
A.1 In both cases only enolate forms which acts as a nucleophile and the same aldehyde/Ketone are 
acting as an electrophile. The giant molecule formed first followed by elimination occurs on further 
heating. This elimination occurs via E1cB mechanism and release one molecule of water. Thus the 
α,β-unsaturated carbonyl compounds are produced. In the following reaction 2-methylpent-2-en-1-
al and Mesityl oxide/ 4-methylpent-3-en-2-one are obtained. The consecutive elimination takes 
place due to presence of α-hydrogen atom.
Q.2 What happens when the following compound is heated with dil. NaOH?
A.2 In case of unsymmetrical ketone question of regioselectivity appears. i.e., which α-carbon will 
undergo enolisation. If the 3-C carbon undergoes enolisation then after condensation no 
dehydration will take place due lack of another α-H atom. So, the condensation product will contain 
OH group as shown below. The carbon atom marked as black dot does not contain the α-H atom.

Mechanism of Aldol condensation, Mixed aldol , Intramolecular Aldol condensation. Document Preview: Aldol Condensation Assignment 2 Q.1 The aldehyde and the ketone below are self-condensed with...

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