a) The way that functional groups affect the reactivity of organic compounds is because of their differences in electronegativity. For example, if a compound is more electronegative, it means that it has a tendency to attract a bonding pair of electrons. So at OH-, the alcohol would be more polar as the oxygen attracts the boiling point, the colour, solubility, etc. this is due to bonding.
b) What happens in a nucleophilic substitution reaction is that the nucleophiles attack the carbons of a carbon-halogen bond. Once the nucleophile attacks the carbon, it takes over the carbons position, causing them to switch. This is caused by the electron pairs on the nucleophiles is attracted towards the small positive charge on the electron. For example,
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These electrons can be from amines’ lone pairs of electrons. These electrons are very electronegative nitrogen atoms. Nucleophiles attack positive parts of another molecule or atom. However, the first must be attracted to the molecule or ion. The primary reactions of amines with halogen alkanes are the heating to give mixtures of products. Products of these reactions include secondary and tertiary amines. Also, salts and quaternary ammonium salts are involved. Secondary amines occurs when there is excess ethylamine is present in the mixture. This can also be a reversible reaction. Hydrogen from the diethylammonium is removed to give free secondary amines (diethylamine). Furthermore, the diethylamine would react with the bromethane to start the secondary amine with a halogenalkane. The quaternary ammonium salt is the final stage of the reaction, and this is created by the trimethylamine reacting with the bromoethane to give tetraethylammonium bromide. Amines are able to act as bases and nucleophile because of the unshared pair of electrons. When primary, secondary and tertiary amines are substituted to the number of alkyl, they bond with the nitrogen atom. For example, in the reaction of ethylamine, a ethylammonium ion is formed due to the amine taking a hydrogen ion from a hydroxonium. The solution would then contain ethylammonium chloride or a sulphuate. A white smoke would be produced on this reaction as the amine …show more content…
This is caused by the hydrogen that is present in the aldehyde, which make the aldehyde to be easily oxidised. Carboxylic acids are formed when this reaction takes place, this would take place on acidic conditions when the aldehyde is being oxidised. Salts can be formed when conditions are alkaline. On the other hand, it can be more difficult to oxidise ketones as there are no hydrogens present. They can be resistant to oxidisation. Very strong oxidising agents would be needed to oxidise ketones. An example of an oxidising agent would be potassium manganate solution. When ketones are oxidised, the carbon-carbon bonds are broken. To reduce aldehydes and ketones, reducing agents would be used. Examples of reducing agents are, lithium tetrahydridoaluminate and sodium tetrahydridoborate. The “tetrahydido” in the names mean that there are four hydrogens around the aluminium or boron in the negative ion. The structures for both reducing agents contain negative ions and there are empty orbitals on the aluminium or boron due to the co-co ordinate covalent bonds which use lone pairs of electrons from the hydride ion. The same principles would go to the reduction of aldehydes and ketones where you get the same organic compound. It does not matter whichever reducing agent is used. In the reaction of nucleophilic addition, the aldehyde and the ketone act the same way. For example, when hydrogen cyanide reacts with aldehydes and
When 1-bromobutane is reacted with potassium t-butoxide there is only one product formed, 1-butene. This is because the halide is on a primary carbon thus producing only one product.
The product was recrystallized to purify it and the unknown filtrate and nucleophile was determined by taking the melting points and performing TLC. Nucleophilic substitution reactions have a nucleophile (electron pair donor) and an sp3 electrophile (electron pair acceptor) with an attached leaving group. This experiment was a Williamson ether synthesis usually SN2, with an alkoxide and an alkyl halide. Conditions are favored with a strong nucleophile, good leaving group, and a polar aprotic solvent.
As shown in Fig. 5, the final pH of the NaClO-NH3 solution after simultaneous removal are 5.4, 6.9, 7.2, 7.5, 8.5, 9.6, 10.7, 11.5 and 12.8 with respect to the initial pH of 5, 6, 7, 8, 9, 10, 11, 12 and 13, from which, an interesting law can be concluded as that if the initial pH is an acidic, the final pH is slightly increased; but if the initial pH is an alkaline, the final pH is declined. NaClO-NH3 is macromolecule compounds with a large inter surface area. It contains abundant functional groups such as hydroxyl (OH), carboxyl (COO), quinone, amino (–NH2), etc, which determines that NaClO-NH3 is a salt of strong base and weak acid, as well the ionization equilibrium and hydrolytic equilibrium would be complicated. When the pH of the NaClO-NH3 solution was acidic, the functional groups such as OH, COO and NH2- would react with H+ to generate the NH3 sediment, resulting in a decrease of inter surface area owing to the block and a great loss of NaClO-NH3, then the NOx removal as well as the duration time was decreased. As for the increase of the final pH in the acidic conditions, this was a result of the consumption of H+ by NaClO. The decrease of the
This experiment studied the kinetics and the effects of solvent polarity of a solvolysis reaction. This reaction is a SN1 reaction in which the solvent (water) is the nucleophile. The reaction begins with the removal of a chloride ion; this is the rate determining step (slow step). Water is then added to the carbocation, forming a protonated alcohol. Lastly, a proton is removed by the present base. Since the first step is the rate determining step, it is a first-order reaction.
So, the reaction does not simply just break the H and the R, but rather -OH and -OR. This reaction is an equilibrium reaction. Applying Le Chatelier’s principle, if alcohol is used as a solvent to carboxylic acid and has a small amount of water (product), then the reaction would favor the product. On the other hand, if the reaction is to go backwards, whereas the reaction would start from an ester going to a carboxylic acid, then the water would be used as a solvent. Common acid catalysts are sulfuric acid, tosylic acid, and Lewis acids such as scandium(III) triflate.
3. By increasing the pressure (only really significant in reactions involving gases). 4. By the use of a suitable catalyst. 5.
« Catalyst - A catalyst is a substance, which can alter the rate of a
This type of reaction encourages the formation of C-C π bond by breaking two single bonds to carbon where one of them is a hydrogen atom and the other is typically a halide atom. The type of reaction it undergoes is mainly indicated by the type of alkyl halide(electrophile) it starts off as: a primary alkyl halide versus a secondary alkyl halide. The type of elimination reaction (1 or 2) is dictated by the type of strong base it is presented with; bulky versus non-bulky. Elimination 2 takes place when a primary alkyl halide reacts with a branched base.
Hydrogen peroxide oxidizes organic and inorganic compounds by free radicals (OH•, O•-2).O•-2 is produced in large no by ionization of weakly acidic aqueous hydrogen peroxide while OH• is produced by ionization of buffered hydrogen peroxide at ph 9.5-10.8.But OH• has better bleaching effect than O•-2 [35].
The Calvin cycle uses NADPH, ATP, and carbon as the reactants. In the light-dependent reactions, the electron released from water travels through the photosystems and the electron transport chain in the thylakoid membrane, then attaching to NAD+ to be carried to the Calvin cycle. When the electrons are taken from the water, the oxygen diffuses out and only hydrogen ions are left on the inside of the thylakoid, called the
Predictions may be made about the suitability of possible catalysts by assuming that the mechanism of catalysis consists of two stages, either of which can be first:
Draw the structure of the organic product for the reaction between the following compound and the phosphorus ylide shown. Solution The phosphorus ylide will only react with the carbonyl group. Final product Comment: Think of the Wittig reaction as the reverse of oxidative cleavage of alkenes. Baeyer-Viliger Oxidation The Baeyer-Viliger oxidation is an organic reaction used to convert a ketone to an ester using a peroxyacid, with an “insertion” of one oxygen atom. • A peroxyacid can be prepared by reacting a carboxylic acid and hydrogen peroxide.
As we already know, chemical reactions can be quite striking when they produce a lot of gas and noise. We definitely saw that during some of these experiments, such as during experiment 6 when the chemicals began fizzing and changing colors. I found myself wondering what kind of reaction two chemicals would have. Most strange was how quickly some of the chemicals released heat, making them feel so warm to the touch I couldn’t hold them for long. That said, some of the reactions seemed pointless, such as in experiment 1 when we mixed two dry
For example: when looking at hydrochloric acid which is a strong acid it is combined with Hydrogen and Chlorine forming its product:
Hydrogen peroxide decomposes into water and oxygen upon heating or in the presence of numerous substances, particularly salts of such metals as iron, copper, manganese, nickel, or chromium. It combines with many compounds to form crystalline solids useful as mild oxidizing agents; the best-known of these is sodium perborate (NaBO2H2O23H2O or NaBO34H2O). With certain organic compounds, hydrogen peroxide reacts to form hydroperoxides or peroxides, several of which are used to initiate polymerization reactions. In most of its reactions, hydrogen peroxide oxidizes other substances, although it is itself oxidized by a few compounds, such as potassium perm...