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Draw the structure of the product formed when 2-methylbutanal is treated with cold aqueous base. Draw...

Draw the structure of the product formed when 2-me

Draw the structure of the product formed when 2-methylbutanal is treated with cold aqueous base.


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Answer #1
Concepts and reason

• A condensation reaction is a type of chemical reaction, which has a huge importance in an organic synthesis.

• The reaction undergoes the formation of a carbon-carbon double bond. An aldol condensation reaction is a type of condensation reaction, in which two aldehydes have α-hydrogen atom, which react with each other to form an aldol product.

• The first step is the aldol reaction, in which the enolate ion is formed, then it reacts with another molecule of the aldehyde.

• The dehydration of product of an aldol reaction, forms the αβ-unsaturated carbonyl molecules (aldol). This is known as an aldol condensation reaction.

• In a base catalyzed aldol reaction, the base deprotonates the acidic hydrogen on an α-carbon atom to form the enolate.

• An enolate is a good nucleophile and carbonyl C{\rm{C}} is a good electrophile. So, the enolate reacts with the carbonyl carbon of another aldehyde.

Fundamentals

The mechanism of an aldol condensation contains of two steps. The first part is an aldol reaction and the second part is a dehydration elimination reaction. First, the base abstracts the proton to form a carbanion, and then it attacks the carbonyl group of another compound to form an oxy anion. This compound undergoes a hydrolysis and forms an aldol, which is followed by dehydration that will give an aldol product.

Mechanism:-

L-
L 1
H
base
RCH
10-
H-
C-
-CH2R
RCH,
H
H2CR—C—H
enolate
OH
H—
—
—
—CH R
B hydroxyl aldehyde
dehydration
H2O +
H—C—
=&—CHYR

OO
jОН
enolate anion

enolate anion
+
H
Yiborg

OH
بلللل
HH
aldol product

Ans:

The structure of the product that is formed, when 2-methylbutsnal is treated with a cold aqueous base is given below:

CHے
ل
NaOH, H2O
4-5°C
بل

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Answer #2

The base abstracts the proton from 2-methylbutanal and forms a carbanion, which attacks the carbonyl group of 2-methylbutanal

answered by: Happy.cat
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