Draw the structure of the product formed when 2-methylbutanal is treated with cold aqueous base.
• 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 is a good electrophile. So, the enolate reacts with the carbonyl carbon of another aldehyde.
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:-
The structure of the product that is formed, when 2-methylbutsnal is treated with a cold aqueous base is given below:
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-methylbutanal is treated with cold aqueous base. NaOH H2O 4-5 °C
Map Sapling Learning Draw the structure of the product formed when 2-methylbutanal is treated with cold aqueous base. NaOH, H,0 4-5 Previous E Check Answer Give Up & View Solution Anu iBooks Hint
Draw the structure of the product formed when the following compound is heated in aqueous base. The formula for the product is C8H12O.
Draw the structure of the product that is formed when the
compound shown below is treated with the following reagents:1) NaOCH3; 2)
CH3CH2CH2CH2Br.
Draw the structure of the product that is formed when the
compound shown below is treated with the following reagents:
1) NaOCH3; 2) CH3I; 3)
H3O+, heat.
Draw the structure of the organic product expected when BrCH2CH2CH2OH is treated with base.
Draw the structure of the product that is formed when the compound shown below is treated with the following reagents: 1) Na, NH3(liq); 2) D2, Pd, including its relative stereochemistry. Interactive 3D display mode
Draw the structure of the product that is formed when the compound shown below is treated with the following reagents: 1) LDA, THF; 2) CH3l. Interactive 3D display mode
Draw the structure of the product that is formed when the compound shown below is treated with the following reagents: 1) BH3/THF; 2) H2O2, HO- , H2O. Interactive 3D display mode
1) Which base yields the most SN2 when treated with the same
starting material? (draw this structure with stereochemistry)
2)Which base would have the least yield of SN2?
3) What conclusions are made about the amount of SN2 product
formed as a function of base strength?
4) Discuss stereospecificity of an SN2 reaction and draw the
SN2 transition state intermediate formed when benzoate is treated
with starting material.
4. When the starting material shown below is treated with a variety...