I'm researching the chemistry of soap making and I would like to know if I have the reactions correct.
Step 1.
Adding NaOH (sodium hydroxide) to H2O (water).
The reaction: Breaking the bond between Sodium (Na) and hydroxide (OH) with water.
The sodium molecule (Na) is positively charged and the hydroxide molecule (OH) is negatively charged.
Because they have opposite charges they have a ionic bond (like two magnets stuck to one another).
The H2O molecule will break the ionic bond between Na+OH- and release heat.
The Na+ and the H in the OH- will attract to the negatively charged oxygen molecules in the H2O.
The O- in the OH- will attract to the positively charged hydrogen molecules in the H2O.
(Na+OH-) + (H2O) = Na+(aq) and (OH-(aq)
(aq)= aqueous= dissolved in water
In silly terms: the lye is broken apart and arrested by the water.
Step 2.
Adding the Na+(aq) OH-(aq) (lye water solution) to triglycerides (fat/oil)
The reaction: Saponification (Base promoted ester hydrolysis)
Stage 1:
The OH- will attack the carbon with the double bond (nucleophilic attack).
In the attack the OH- will attach it's self to the carbon (nucleophilic addition).
The addition of OH- will kick the negative charge on the OH though the carbon to the double bonded oxygen. Making that oxygen O-.
Because of the movement of the negatively charged ion and the addition of OH, the double bond between the oxygen and the carbon will break making a single bond.
Making a tetrahedral intermediate.
Stage 2:
The newly made O- will move the negatively charged ion through the carbon to the oxygen that is connected to the triglyceride backbone (RO group).
Because of the movement of the negatively charged ion the double bond reforms and the triglyceride backbone (RO group) gets cut away from the rest of the molecule.
The triglyceride backbone (RO group) now has a negatively charged ion on the oxygen in the RO.
Making carboxylic acid (fatty acid) and the RO- (which is strong base)
Stage 3:
O- in the RO- group will kick the negatively charged ion to the oxygen on the original OH in the carboxylic acid (fatty acid).
Making a acid based proton transfer.
The oxygen in the OH will exchange its' hydrogen for the negatively charged ion.
Making a carboxylic anion and glycerol.
Stage 4.
The lonely sodium (Na+) hasn't reacted though all these reactions so far.
But now the sodium (Na+) will form an ionic bond with the carboxylic anion to make a molecule that has a sodium head and fatty acid tail (soap).
In simple terms: The hydroxide (OH) in the NaOH (sodium hydroxide) will attack the fat/oil and after a bunch of back and forth with some negative charges you eventually cut the fat/oil into glycerin and soap.
Did I get it right?
Here is where I got most of the info from:
https://www.khanacademy.org/test-pr...saponification-base-promoted-ester-hydrolysis
Step 1.
Adding NaOH (sodium hydroxide) to H2O (water).
The reaction: Breaking the bond between Sodium (Na) and hydroxide (OH) with water.
The sodium molecule (Na) is positively charged and the hydroxide molecule (OH) is negatively charged.
Because they have opposite charges they have a ionic bond (like two magnets stuck to one another).
The H2O molecule will break the ionic bond between Na+OH- and release heat.
The Na+ and the H in the OH- will attract to the negatively charged oxygen molecules in the H2O.
The O- in the OH- will attract to the positively charged hydrogen molecules in the H2O.
(Na+OH-) + (H2O) = Na+(aq) and (OH-(aq)
(aq)= aqueous= dissolved in water
In silly terms: the lye is broken apart and arrested by the water.
![Step 1.jpg Step 1.jpg](https://cdn.imagearchive.com/soapmakingforum/data/attach/50/50885-Step-1.jpg)
Step 2.
Adding the Na+(aq) OH-(aq) (lye water solution) to triglycerides (fat/oil)
The reaction: Saponification (Base promoted ester hydrolysis)
Stage 1:
The OH- will attack the carbon with the double bond (nucleophilic attack).
In the attack the OH- will attach it's self to the carbon (nucleophilic addition).
The addition of OH- will kick the negative charge on the OH though the carbon to the double bonded oxygen. Making that oxygen O-.
Because of the movement of the negatively charged ion and the addition of OH, the double bond between the oxygen and the carbon will break making a single bond.
Making a tetrahedral intermediate.
![Step 2 stage 1.jpg Step 2 stage 1.jpg](https://cdn.imagearchive.com/soapmakingforum/data/attach/50/50886-Step-2-stage-1.jpg)
Stage 2:
The newly made O- will move the negatively charged ion through the carbon to the oxygen that is connected to the triglyceride backbone (RO group).
Because of the movement of the negatively charged ion the double bond reforms and the triglyceride backbone (RO group) gets cut away from the rest of the molecule.
The triglyceride backbone (RO group) now has a negatively charged ion on the oxygen in the RO.
Making carboxylic acid (fatty acid) and the RO- (which is strong base)
![Step 2 stage 2.jpg Step 2 stage 2.jpg](https://cdn.imagearchive.com/soapmakingforum/data/attach/50/50889-Step-2-stage-2.jpg)
Stage 3:
O- in the RO- group will kick the negatively charged ion to the oxygen on the original OH in the carboxylic acid (fatty acid).
Making a acid based proton transfer.
The oxygen in the OH will exchange its' hydrogen for the negatively charged ion.
Making a carboxylic anion and glycerol.
![Step 2 stage 3.jpg Step 2 stage 3.jpg](https://cdn.imagearchive.com/soapmakingforum/data/attach/50/50890-Step-2-stage-3.jpg)
Stage 4.
The lonely sodium (Na+) hasn't reacted though all these reactions so far.
But now the sodium (Na+) will form an ionic bond with the carboxylic anion to make a molecule that has a sodium head and fatty acid tail (soap).
![Step 2 stage 4.jpg Step 2 stage 4.jpg](https://cdn.imagearchive.com/soapmakingforum/data/attach/50/50891-Step-2-stage-4.jpg)
In simple terms: The hydroxide (OH) in the NaOH (sodium hydroxide) will attack the fat/oil and after a bunch of back and forth with some negative charges you eventually cut the fat/oil into glycerin and soap.
Did I get it right?
Here is where I got most of the info from:
https://www.khanacademy.org/test-pr...saponification-base-promoted-ester-hydrolysis