Etiket Arşivleri: Aldehydes and Ketones
An aldehyde is an organic compound containing a terminal carbonyl group, i.e., a O=CH- group attached to hydrogen or a carbon chain. This functional group, which consists of a carbon atom which is bonded to a hydrogen atom and double-bonded to an oxygen atom (chemical formula -CHO), is called the aldehyde group. The aldehyde group is also called the formyl or methanyl group.The word aldehyde seems to have arisen from alcohol dehydrogenated. In the past, aldehydes were sometimes named after the corresponding alcohols, for example vinous aldehyde foracetaldehyde. (Vinous is from Latin vinum = wine, the traditional source of ethanol; compare vinyl.)The aldehyde group is polar. Oxygen, being more electronegative, pulls the electrons in the carbon-oxygen bond towards itself, thus creating an electron deficiency at the carbon atom.Owing to resonance stabilization of the conjugate base, an α-hydrogen in an aldehyde is more acidic than a hydrogen atom in an alkane, with a typical pKa of 17.
A ketone is either the functional group characterized by a carbonyl group (O=C) linked to two other carbon atoms or a chemical compound that contains this functional group. A ketone can be generally represented by the formula: R1(CO)R2.
A carbonyl carbon bonded to two carbon atoms distinguishes ketones from carboxylic acids, aldehydes, esters, amides, and other oxygen-containing compounds. The double-bond of the carbonyl group distinguishes ketones from alcohols and ethers. The simplest ketone is acetone (also called propanone).The carbon atom adjacent to a carbonyl group is called the α-carbon. Hydrogens attached to this carbon are called α-hydrogens. In the presence of an acid catalyst the ketone is subjected to so-called keto-enol tautomerism. The reaction with a strong base gives the corresponding enolate. A diketone is a compound containing two ketone groups.
Oxygen is far more electronegative than carbon and so has a strong tendency to pull electrons in a carbon-oxygen bond towards itself. One of the two pairs of electrons that make up a carbon-oxygen double bond is even more easily pulled towards the oxygen. That makes the carbon-oxygen double bond very highly polar.
The slightly positive carbon atom in the carbonyl group can be attacked by nucleophiles. A nucleophile is a negatively charged ion (for example, a cyanide ion, CN–), or a slightly negatively charged part of a molecule (for example, the lone pair on a nitrogen atom in ammonia, NH3).During the reaction, the carbon-oxygen double bond gets broken. The net effect of all this is that the carbonyl group undergoes addition reactions, often followed by the loss of a water molecule. This gives a reaction known as addition-elimination or condensationBoth aldehydes and ketones contain a carbonyl group. That means that their reactions are very similar in this respect.
Test tubes, steam bath, ice bath, beakers, acetone, formaldehyde, ethanol, phenylhdrazine hydrochloride solution, 2,4-dinitropenylhydrazine, semicarbazide, NaOH, water, AgNO3, NH3.
DERIVATIVES OF CARBONYL COMPOUNDS
First 10 mL of formaldehyde was dissolved in 5 mL of 95 % ethanol.The acetone solution was added to 2 mL of phenylhydrazine hydrchloride solution.Then it was heated on a steam bath for 10 min.The miixture was cooled in a ice bath.
0.5 mL of stock solution of 2,4-dinitrophenylhydrazine was taken into the test tube.The solution was heated.3 mL 95% ethanol was added to the warm solution.In a separate test tube 3-4 drops of acetone was dissolved in 3 mL ethanol.2,4-dinitrophenylhydrazine solution was added to the solution of acetone and allowed to stand at room temperature until the mixture crystallization is completely.
1 mL of stock solution of semicarbazide was taken and 5 drops of formaldehyde was added and shaked gently.The test tube was placed in boiling water for 5 min.Test tube was cooled.Then test tube was transferred to on ice bath and stretch the sides of with a glass stirring rod.It was cooled until the crystallization is complete.
3 drops of acetone was dissolved in 1 mL of water.Then 3 mL of 10 % NaOH was added into test tube and dropwise 10 % iodine in potassium iodine-water.The test tube was placed in a beaker of water at 60°C.More of iodine solution was added until the color of iodine persisted for 2 min.10 % NaOH solution dropwise was added until the color disappears.Finally 10 mL of water was added.And then the same procedure was done with formaldehyde instead of acetone.
TEST FOR DISTINGUISHING BETWEEN ALDEHYDES AND KETONES
2 mL of 5 % AgNO3 was addedto the test tube.One drop of 10 % NaOH and the solution was mixed.Ammonis dropwise was diluted to the test tube with constant shaking until brown precipitate of silver oxide just dissolved.
2 mL of Fehling A solution was taken and Fehling B solution was added slowly until the initially formed light blue precipitate of cupper hydroxide dissolves.3 drops of acetone was added to the test tube.It was boiled gently for 5-10 min.Then this procedure was repeated with formaldehyde instead of acetone.
The addition of hydrogen across a C=O double bond raises several important points. First, and perhaps foremost, it shows the connection between the chemistry of primary alcohols and aldehydes. But it also helps us understand the origin of the term aldehyde. If a reduction reaction in which H2 is added across a double bond is an example of a hydrogenation reaction, then an oxidation reaction in which an H2 molecule is removed to form a double bond might be called dehydrogenation. Thus, using the symbol [O] to represent an oxidizing agent, we see that the product of the oxidation of a primary alcohol is literally an “al-dehyd” or aldehyde. It is an alcohol that has been dehydrogenated. The product of this reaction was originally called aketone, although the name was eventually softened to azetone and finally acetone. Thus, it is not surprising that any substance that exhibited chemistry that resembled “aketone” became known as a ketone. A variety of oxidizing agents can be used to transform a secondary alcohol to a ketone. A common reagent for this reaction is some form of chromium(VI) chromium in the +6 oxidation state in acidic solution. This reagent can be prepared by adding a salt of the chromate (CrO42-) or dichromate (Cr2O72-) ions to sulfuric acid. Or it can be made by adding chromium trioxide (CrO3) to sulfuric acid. Regardless of how it is prepared, the oxidizing agent in these reactions is chromic acid, H2CrO4. Aldehydes and ketones play an important role in the chemistry of carbohydrates. The term carbohydrate literally means a “hydrate” of carbon, and was introduced to describe a family of compounds with the empirical formula CH2O. Glucose and fructose, for example, are carbohydrates with the formula C6H12O6. These sugars differ in the location of the C=O double bond on the six-carbon chain. Glucose is an aldehyde; fructose is a ketone. Its functional groups include aldehyde, ether, and alcohol. It is the primary component of the extract of the vanilla bean. Synthetic vanillin is used as a flavoring agent in foods, beverages, and pharmaceuticals.
Acetone , Benzaldahyde, Distilled water, 95 % Ethanol, Phenylhydrazine hydrochloride solution, 2,4 Dinitrophenylhydrazine, semicarbazide, 10 % NaOH, 10 % iodine, 5 % AgNO3, silver oxide, Fehling A, Fehling B and Cupper hydroxide.
Steam-bath, ice-bath, Erlenmeyer flask, thermometer and test tubes.
1. How can you identify aldehydes and ketones experimentally when an unknown mixture is given?
This mixture is separated two methods; a-Fehling’s test, b-Tollens reagent
If fehling’s test or tollens reagent are used, Aldehydes are separated from the mixture, because Aldehydes are easily oxidized to caboxiylic acid,
R-CO-H + Ag2O à R-CO-OH + 2AgO
R-CO-H + C4O à R-CO-OH + C2O
But Ketones are not oxidized. So, Aldehydes and Ketones are separated with Fehling’s Test or Tollens Reagent.
2. By use of appropriate classification tests only, distinguish between pure samples of methyketone, acetaldhyde, diethylketone and proplonaldehyde?
Methylketone, acetaldehyde, diethylketone and propiolaldahyde, often a liquid is converted to several of the derivatives. They are reacted with phenylhydrazine and phenylhydrazone is producted, which are derivatives of aldehydes and ketones.
The melting points of the derivatives are determined, and they are separated with using melting point of derivatives of aldehydes and ketones.
3. Why not use ethyl alcohol as a solvent in the iodoform test?
Because ethyl alcohol is oxidized to aldehyde which is oxidized carboxyl acid. Because of this result not using ethyl alcohol as a solvent in the iodoform test.
R-OH à R-CO H à R-COOH
The only structural difference between hydrocarbons and aldehydes and ketones is the presence in the latter of the carbonyl group, and it is this group that is responsible for the differences in properties, both physical and chemical.The differences arise because the carbonyl group is inherently polar—that is, the electrons that make up the C=O bond are drawn closer to the oxygen than to the carbon. This gives the oxygen a partial negative charge and the carbon a partial positive charge, and the carbonyl group is often represented using the Greek letter to indicate a partial charge (that is, a charge less than one)
The negative end of one polar molecule is attracted to the positive end of another polar olecule, which may be a molecule either of the same substanceOr of a different substance.In this experiment, the derivatives of aldehydes and ketones was observed.
The derivatives are cointained a corbon- nitrogen double bond. They are formed by the initial addition of an H_NH_R molecule which was Phenylhydrazine (H-NH NH Ph) , the 2, 4 dinitrophenylhydrazine ( H- NH-NH C6H3(NO2)2) and semicarbazide (H-NH-NH-CO NH2) by the elimination of water to from, phenylhydrozones, semicarbazides which are bonded aldehydes.