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What is the role of adding NaOH to the amine tartrate salt prior to the extraction with dichloromethane?
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What is the role of adding NaOH to the amine tartrate salt prior to the extraction with dichloromethane?
Anhydrous NaOH is a salt in the sense of the definition that salts consist of anions (in this case the hydroxide ion) and cations (the sodium ion) in a regular crystalline pattern (for NaOH at room temperature, an orthorhombic structure). By convention, its basic properties are more relevant, so NaOH is usually called a base - but that doesn't mean it is not a salt.
Anhydrous NaOH is a salt in the sense of the definition that salts consist of anions (in this case the hydroxide ion) and cations (the sodium ion) in a regular crystalline pattern (for NaOH at room temperature, an orthorhombic structure). By convention, its basic properties are more relevant, so NaOH is usually called a base - but that doesn't mean it is not a salt.
Tartaric acid is reacted with an amine and leads to the formation of the amine tartrate salt and, being a salt and therefore charged, it is water soluble. When you add the NaOH, you free the amine acid salt and now it becomes the uncharged amine (aka, free basing). The uncharged or neutral amine is now soluble in dichlormethane. The tartrate is converted to sodium tartrate, still charged and very water soluble but insoluble in dichloromethane. The role, then, of the NaOH, is to neutralize the amine and render it extractable into dichloromethane.
Edit: There is some ambiguity in the question. By amine I was assuming an organic amine. In fact, tartrate salts of amines with chiral centers is typically used to resolve amines into their enantiomers. Then, after resolution (by whatever means), the amine is freed of the tartrate by the method described above. It is possible that ‘amine’ meant NH3 but then the question should have asked about the ammonium salt as pointed out by Kurt Van den Broeck. In this case, ammonia would have been liberated. Where it would go…could stay in the water or degas (bubble off), go into the dichloromethane or a little of everything, but then there would be little point to the dichloromethane extraction.
Tartaric acid is reacted with an amine and leads to the formation of the amine tartrate salt and, being a salt and therefore charged, it is water soluble. When you add the NaOH, you free the amine acid salt and now it becomes the uncharged amine (aka, free basing). The uncharged or neutral amine is now soluble in dichlormethane. The tartrate is converted to sodium tartrate, still charged and very water soluble but insoluble in dichloromethane. The role, then, of the NaOH, is to neutralize the amine and render it extractable into dichloromethane.
Edit: There is some ambiguity in the question. By amine I was assuming an organic amine. In fact, tartrate salts of amines with chiral centers is typically used to resolve amines into their enantiomers. Then, after resolution (by whatever means), the amine is freed of the tartrate by the method described above. It is possible that ‘amine’ meant NH3 but then the question should have asked about the ammonium salt as pointed out by Kurt Van den Broeck. In this case, ammonia would have been liberated. Where it would go…could stay in the water or degas (bubble off), go into the dichloromethane or a little of everything, but then there would be little point to the dichloromethane extraction.
Amine salt = ammonium salt. E.g. ammonium chloride = ammonia hydrochloride
We use "amine salt" names for compounds for which it's not convenient to name the ammonium ion. For example, we usually say "pyridinium chloride," not "pyridine hydrochloride." But we always say "cocaine hydrochloride" because there's not a convenient name for protonated cocaine.
A quaternary ammonium salt is nitrogen with four alkyl groups attached. This will necessarily have a positive charge (a formal positive charge on nitrogen), so a quaternary ammonium ion is always found as part of a quaternary ammonium salt.
Amine salt = ammonium salt. E.g. ammonium chloride = ammonia hydrochloride
We use "amine salt" names for compounds for which it's not convenient to name the ammonium ion. For example, we usually say "pyridinium chloride," not "pyridine hydrochloride." But we always say "cocaine hydrochloride" because there's not a convenient name for protonated cocaine.
A quaternary ammonium salt is nitrogen with four alkyl groups attached. This will necessarily have a positive charge (a formal positive charge on nitrogen), so a quaternary ammonium ion is always found as part of a quaternary ammonium salt.
Neither - it’s an alkali, a notoriously strong one. It will react with an acid to form a salt. It will also react with the oils on your skin and the fat in your living cells to make soap, so treat it with great respect!
Neither - it’s an alkali, a notoriously strong one. It will react with an acid to form a salt. It will also react with the oils on your skin and the fat in your living cells to make soap, so treat it with great respect!
Reaction with benzenesulfonyl chloride to form the benzenesulfonamides. Primary amine n-propylamine forms an alkali-soluble sulfonamide; secondary methylethylamine forms an alkali insoluble sulfonamide.
Reaction with benzenesulfonyl chloride to form the benzenesulfonamides. Primary amine n-propylamine forms an alkali-soluble sulfonamide; secondary methylethylamine forms an alkali insoluble sulfonamide.
Actually you cannot use primary alkyl amines to prepare useful diazonium salts because a primary alkyl diazonium salt is too unstable—they lose N2 very quickly forming unstable primary carbocations.
Primary aryl amines do form relatively stable aryl diazonium salts that are useful synthetic intermediates.
However, if your question concerns primary amines vs secondary amines, the answer is that only the primary amine can lose the TWO protons necessary to form the diazonium salt. Secondary amines in fact produce N-nitroso compounds when treated with nitrous acid (HNO2 from NaNO2 & HCl).
Actually you cannot use primary alkyl amines to prepare useful diazonium salts because a primary alkyl diazonium salt is too unstable—they lose N2 very quickly forming unstable primary carbocations.
Primary aryl amines do form relatively stable aryl diazonium salts that are useful synthetic intermediates.
However, if your question concerns primary amines vs secondary amines, the answer is that only the primary amine can lose the TWO protons necessary to form the diazonium salt. Secondary amines in fact produce N-nitroso compounds when treated with nitrous acid (HNO2 from NaNO2 & HCl).
Anhydrous NaOH is a salt in the sense of the definition that salts consist of anions (in this case the hydroxide ion) and cations (the sodium ion) in a regular crystalline pattern (for NaOH at room temperature, an orthorhombic structure).
By convention, its basic properties are more relevant, so NaOH is usually called a base - but that doesn't mean it is not a salt.
Anhydrous NaOH is a salt in the sense of the definition that salts consist of anions (in this case the hydroxide ion) and cations (the sodium ion) in a regular crystalline pattern (for NaOH at room temperature, an orthorhombic structure).
By convention, its basic properties are more relevant, so NaOH is usually called a base - but that doesn't mean it is not a salt.
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Tartaric acid is reacted with an amine and leads to the formation of the amine tartrate salt and, being a salt and therefore charged, it is water soluble. When you add the NaOH, you free the amine acid salt and now it becomes the uncharged amine (aka, free basing). The uncharged or neutral amine is now soluble in dichlormethane. The tartrate is converted to sodium tartrate, still charged and very water soluble but insoluble in dichloromethane. The role, then, of the NaOH, is to neutralize the amine and render it extractable into dichloromethane.
Edit: There is some ambiguity in the question. By amine I was assuming an organic amine. In fact, tartrate salts of amines with chiral centers is typically used to resolve amines into their enantiomers. Then, after resolution (by whatever means), the amine is freed of the tartrate by the method described above. It is possible that ‘amine’ meant NH3 but then the question should have asked about the ammonium salt as pointed out by Kurt Van den Broeck. In this case, ammonia would have been liberated. Where it would go…could stay in the water or degas (bubble off), go into the dichloromethane or a little of everything, but then there would be little point to the dichloromethane extraction.
Tartaric acid is reacted with an amine and leads to the formation of the amine tartrate salt and, being a salt and therefore charged, it is water soluble. When you add the NaOH, you free the amine acid salt and now it becomes the uncharged amine (aka, free basing). The uncharged or neutral amine is now soluble in dichlormethane. The tartrate is converted to sodium tartrate, still charged and very water soluble but insoluble in dichloromethane. The role, then, of the NaOH, is to neutralize the amine and render it extractable into dichloromethane.
Edit: There is some ambiguity in the question. By amine I was assuming an organic amine. In fact, tartrate salts of amines with chiral centers is typically used to resolve amines into their enantiomers. Then, after resolution (by whatever means), the amine is freed of the tartrate by the method described above. It is possible that ‘amine’ meant NH3 but then the question should have asked about the ammonium salt as pointed out by Kurt Van den Broeck. In this case, ammonia would have been liberated. Where it would go…could stay in the water or degas (bubble off), go into the dichloromethane or a little of everything, but then there would be little point to the dichloromethane extraction.
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The NaOH dissolves the protein so that protein can then react with the copper salts.
If you directly add copper sulphate in NaOH they will react to form copper hydroxide.
If you add too much NaOH, the copper hydroxide is the major reaction, and the purple colour is hidden.
If you use commercially available Biuret you get a deeper colour by dissolving the food in NaOH first.
Source: https://www.biuret-test.com
The NaOH dissolves the protein so that protein can then react with the copper salts.
If you directly add copper sulphate in NaOH they will react to form copper hydroxide.
If you add too much NaOH, the copper hydroxide is the major reaction, and the purple colour is hidden.
If you use commercially available Biuret you get a deeper colour by dissolving the food in NaOH first.
Source: https://www.biuret-test.com
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Amine salt = ammonium salt. E.g. ammonium chloride = ammonia hydrochloride
We use "amine salt" names for compounds for which it's not convenient to name the ammonium ion. For example, we usually say "pyridinium chloride," not "pyridine hydrochloride." But we always say "cocaine hydrochloride" because there's not a convenient name for protonated cocaine.
A quaternary ammonium salt is nitrogen with four alkyl groups attached. This will necessarily have a positive charge (a formal positive charge on nitrogen), so a quaternary ammonium ion is always found as part of a quaternary ammonium salt.
Amine salt = ammonium salt. E.g. ammonium chloride = ammonia hydrochloride
We use "amine salt" names for compounds for which it's not convenient to name the ammonium ion. For example, we usually say "pyridinium chloride," not "pyridine hydrochloride." But we always say "cocaine hydrochloride" because there's not a convenient name for protonated cocaine.
A quaternary ammonium salt is nitrogen with four alkyl groups attached. This will necessarily have a positive charge (a formal positive charge on nitrogen), so a quaternary ammonium ion is always found as part of a quaternary ammonium salt.
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Neither - it’s an alkali, a notoriously strong one. It will react with an acid to form a salt. It will also react with the oils on your skin and the fat in your living cells to make soap, so treat it with great respect!
Neither - it’s an alkali, a notoriously strong one. It will react with an acid to form a salt. It will also react with the oils on your skin and the fat in your living cells to make soap, so treat it with great respect!
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Reaction with benzenesulfonyl chloride to form the benzenesulfonamides. Primary amine n-propylamine forms an alkali-soluble sulfonamide; secondary methylethylamine forms an alkali insoluble sulfonamide.
Reaction with benzenesulfonyl chloride to form the benzenesulfonamides. Primary amine n-propylamine forms an alkali-soluble sulfonamide; secondary methylethylamine forms an alkali insoluble sulfonamide.
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Actually you cannot use primary alkyl amines to prepare useful diazonium salts because a primary alkyl diazonium salt is too unstable—they lose N2 very quickly forming unstable primary carbocations.
Primary aryl amines do form relatively stable aryl diazonium salts that are useful synthetic intermediates.
However, if your question concerns primary amines vs secondary amines, the answer is that only the primary amine can lose the TWO protons necessary to form the diazonium salt. Secondary amines in fact produce N-nitroso compounds when treated with nitrous acid (HNO2 from NaNO2 & HCl).
Actually you cannot use primary alkyl amines to prepare useful diazonium salts because a primary alkyl diazonium salt is too unstable—they lose N2 very quickly forming unstable primary carbocations.
Primary aryl amines do form relatively stable aryl diazonium salts that are useful synthetic intermediates.
However, if your question concerns primary amines vs secondary amines, the answer is that only the primary amine can lose the TWO protons necessary to form the diazonium salt. Secondary amines in fact produce N-nitroso compounds when treated with nitrous acid (HNO2 from NaNO2 & HCl).
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Well, if you’re talking about an aqueous solution, the following two options are possible;-
Note: These two options are also true for any non-aqueous solvent which does not chemically react with NaOH and in which it is sufficiently soluble.
Well, if you’re talking about an aqueous solution, the following two options are possible;-
Note: These two options are also true for any non-aqueous solvent which does not chemically react with NaOH and in which it is sufficiently soluble.
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