In experiment 2, we carried out experiment to observe the reaction of halogenoalkanes with aqueous alkali and water which contains dissolved silver nitrate. Halogenoalkanes are alkanes which have one or more hydrogen atoms replaced by halogen atoms such as fluorine(F), chlorine(Cl), bromine(Br) and iodine(I) which are the elements in group VII in periodic table. Halogenoalkanes have the general formula, RX, whereby R is an alkyl or substituted alkyl group and X is any of the halogen atom. Besides, halogenoalkanes can also be classified into three categories according to what is attached to the functional group such as primary, secondary and tertiary halogenoalkane.
In experiment 2(a), we use 2-chloro-2-methylpropane as a halogenoalkane
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to react with aqueous alkali which is potassium hydroxide. 2-chloro-2-methylpropane is a tertiary halogenoalkane because it contains three alkyl groups(methyl-CH3) and one halogen atom(Cl) attached to the functional group. So, 2-chloro-2-methylpropane undergoes elimination reaction in the presence of potassium hydroxide because tertiary halogenoalkane will tend to get mainly the elimination reaction than substitution reaction. In this elimination reaction, 2-chloro-2-methylpropane reacts with potassium hydroxide to produce a colourless soluble solution, which is potassium chloride(KCl),water and a colourless gas, which is an alkene,2-methylpropene. A hydrogen atom has been removed from one of the end of a carbon atom together with chlorine atom from 2-chloro-2-methylpropane. Therefore, when the hydrogen and chlorine atoms are being removed are on adjacent carbon atoms, a new double bond is formed between those carbon atoms and an alkene is formed. The hydrogen atom removed reacts with hydroxide ion to form water molecule whereas chlorine reacts with potassium to form potassium chloride. In order to test the gas produced, the alkene gas,2-methylpropene is tested with glowing wooden splinter for its flammability. As a result, glowing wooden splinter rekindles with green flame. This proves that the gas is alkene gas because it supports combustion reaction. Moreover, the gas produced is also tested with bromine water. As a result, yellowish bromine water decolourised. The double bond in 2-methylpropene breaks down and form two new single bonds with bromine atoms in order to get a new product, 1,2-dibromo-2-methylpropane which is an alkane. This reaction is an addition reaction involving halogens. In experiment 2(b), we compare the rate of reaction between haloganoalkanes.1-chlorobutane, 1-bromobutane and 1-iodobutane are reacted with silver nitrate in test tubes and placed in hot water bath.
The water in the mixture acts as a nucleophile and the hydrolysis reaction occurs. During the hydrolysis reaction, halogenoalkanes are broken down in the presence of water to form alcohol, hydrogen ion and also halide …show more content…
ion. 1-bromobutane: 〖CH〗_3 〖CH〗_2 Br+H_2 O→〖CH〗_3 〖CH〗_2 OH+H^++〖Br〗^- 1-chlorobutane: 〖CH〗_3 〖CH〗_2 Cl+H_2 O→〖CH〗_3 〖CH〗_2 OH+H^++〖Cl〗^- 1-iodobutane:〖CH〗_3 〖CH〗_2 I+H_2 O→〖CH〗_3 〖CH〗_2 OH+H^++I^- Various precipitates may be formed from the reaction between the silver and halide ions: 〖Ag〗^+ (aq)+〖Cl〗^- (aq)→AgCl(s) 〖Ag〗^+ (aq)+〖Br〗^- (aq)→AgBr(s) 〖Ag〗^+ (aq)+I^- (aq)→AgI(s) As a result, silver chloride(white precipitate),silver bromide(pale cream precipitate) and silver iodide(pale yellow precipitate) are formed respectively.
Pale yellow precipitate appears the fastest, following with pale cream precipitate and white precipitate takes the longest time to appear. This shows that 1-iodobutane is most reactive whereas 1-chlorobutane is least reactive. The order of reactivity reflects the strengths of the carbon-halogen bonds. The carbon-iodine bond is the weakest and the carbon-chlorine the strongest of the three bonds. In order for a halide ion to be produced, the carbon-halogen bond has to be broken. The weaker the bond, the easier that is. Carbon atom is slightly positive when it attached to the halogen. It is slightly positive because most of the halogens are more electronegative than carbon, and so pull electrons away from the carbon. Chlorine is most electronegative compare to bromine and iodine, hence it drags most of the electrons away from carbon to form a strongest bonding. The bond between iodine and carbon is the weakest as iodine is least electronegative. So, the bond energy of carbon-chlorine is the highest whereas the bond energy of carbon-iodine is the
lowest.
The goal of this experiment is to determine which products are formed from elimination reactions that occur in the dehydration of an alcohol under acidic and basic conditions. The process utilized is the acid-catalyzed dehydration of a secondary and primary alcohol, 1-butanol and 2-butanol, and the base-induced dehydrobromination of a secondary and primary bromide, 1-bromobutane and 2-bromobutane. The different products formed form each of these reactions will be analyzed using gas chromatography, which helps understand stereochemistry and regioselectivity of each product formed.
The goal of this two week lab was to examine the stereochemistry of the oxidation-reduction interconversion of 4-tert-butylcyclohexanol and 4-tert-butylcyclohexanone. The purpose of first week was to explore the oxidation of an alcohol to a ketone and see how the reduction of the ketone will affect the stereoselectivity. The purpose of first week is to oxidize the alcohol, 4-tert-butylcyclohexanol, to ketone just so that it can be reduced back into the alcohol to see how OH will react. The purpose of second week was to reduce 4-tert-butylcyclohexanol from first week and determine the effect of the product's diastereoselectivity by performing reduction procedures using sodium borohydride The chemicals for this lab are sodium hypochlorite, 4-tert-butylcyclohexanone
Alcohol, which is the nucleophile, attacks the acid, H2SO4, which is the catalyst, forming oxonium. However, the oxonium leaves due to the positive charge on oxygen, which makes it unstable. A stable secondary carbocation is formed. The electrons from the conjugate base attack the proton, henceforth, forming an alkene. Through this attack, the regeneration of the catalyst is formed with the product, 4-methylcyclohexene, before it oxidizes with KMnO4. In simpler terms, protonation of oxygen and the elimination of H+ with formation of alkene occurs.
The goal of this lab is to exemplify a standard method for making alkyne groups in two main steps: adding bromine to alkene groups, and followed by heating the product with a strong base to eliminate H and Br from C. Then, in order to purify the product obtained, recrystallization method is used with ethanol and water. Lastly, the melting point and IR spectrum are used to determine the purity of diphenylacetylene.
Hydration of alkenes is characterized by the addition of water and an acid-catalyst to a carbon-carbon bond leading to an alcohol. Dehydration is exactly the opposite in which dehydration of an alcohol requires water to be removed from the reactant. Equilibrium is established between the two processes when the rate of the forward reaction equals the rate of the reverse reaction. The alkene that is used in this experiment is norbornene. Through hydration of norbornene, an alcohol group should be present on the final product yielded what is known as exo-norborneol. Percent yield is a numerical indication of how much of the reactant was actually reacted to yield product. The equation for percent yield is shown below:
Benzyl bromide, an unknown nucleophile and sodium hydroxide was synthesized to form a benzyl ether product. This product was purified and analyzed to find the unknown in the compound. A condenser and heat reflux was used to prevent reagents from escaping. Then the solid product was vacuum filtered.
The weight of the final product was 0.979 grams. A nucleophile is an atom or molecule that wants to donate a pair of electrons. An electrophile is an atom or molecule that wants to accept a pair of electrons. In this reaction, the carboxylic acid (m-Toluic acid), is converted into an acyl chlorosulfite intermediate. The chlorosulfite intermediate reacts with a HCL. This yields an acid chloride (m-Toluyl chloride). Then diethylamine reacts with the acid chloride and this yields N,N-Diethyl-m-Toluamide.
Halogenoalkanes react with an aqueous solution of hydroxide ions according to the equation: RX + OH- Ù ROH + X- (X = Cl, Br or I) The following experiments examine the effect on the rate of hydrolysis when (a) the halogen is changed and (b) primary, secondary and tertiary halogenoalkanes and a halogenoarene are used. Ethanol is used as a common solvent for the halogenoalkane and for the silver nitrate solution. Water (from the silver nitrate solution) is used as hydrolysis agent in place of hydroxide ions, which tend to react too quickly for comparisons to be made. Environmental care:
The hydrolysis of salts can be determined on the basis of the strength of the acid or base which forms it. If the salt is formed from a strong acid and a strong base, such as NaCl, the salt will form a neutral solution, since the anions of the acid and the cations of the base will not react with the water. A salt from a weak base and a strong acid, with NH4Cl as an example, will form an acidic solution. This is due to the cations from the base that increase the hydrogen ion concentration, by donating protons, which is known as a Bronsted acid. When concerning a salt formed by a weak acid and a strong base, such as Na C2H3O2, a basic solution will form. The anions of a weak acid in water will generate hydroxide ions, since the molecule will accept protons. It is termed as a Bronsted base. Though no examples were present, the salt that forms from a weak acid and a weak base can be determined by comparing the Ka (cation) and the Kb (anion) values. Most metallic ions, those found in groups 1A and 2A on the periodic table, such as Ca2+, a strong base, will have no reaction with water. However, all other metallic ions will undergo hydrolysis to form an acidic solution, such as KAl (SO4)2. As the Al is the molecule that was hydrolyzed, the spectator ions would not be present in the hydrolysis reaction, as is shown in the net ionic equations
Louis Jacque Thenard discovered hydrogen peroxide in 1818. Hydrogen peroxide is a common compound that people use in their everyday lives. It is a strong oxidizing agent and a weak acid. Hydroxide peroxide does not come in 100% concentrations, however, it mostly comes in concentrations of three percent, ten percent and thirty-five percent. The concentrations vary anywhere from 3% - 90%. The chemical formula is similar to the formula of water however, it has an extra oxygen atom. Its formula is H2O2. It is also known as HO-OH, dihydrogen peroxide, Hioxy and Proxy. This compound is a molecular compound as it contains two elements that are non-metals. Hydrogen peroxide has covalent bonds because unlike ionic compounds that transfer electrons, they share their electrons in order to receive a full outer shell of eight electrons.
Chloromethane, more commonly referred to as methyl chloride, is an organic compound categorized as a haloalkane functional group. Functional groups are atoms that control how the molecule will react with other molecues. Haloalkane functional groups consist of a carbon atom with four single bonds and one of the bonds in occupied by an element in the Halogen family; in this case chlorine. The structural formula is CH3Cl. This formula effects some of the physical properties which include the boiling point to be -22.22°C and the melting point to be -97.7°C. Other physical properties consist of a faint sweet smell, colorless gas, and it is soluble in water.
This reaction uses steam and a catalyst; another way is by heating ethanol with a dilute sulphuric acid catalyst. A low temperature favours a high yield.
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:
The concentration of the unknown HCl solution was found to be .102 M, and the concentration of the unknown acetic acid was found to be .053 M. As shown in Graph 2, the equivalence point volume can be shown by the maximum slope, and this volume was found to be 25.49 mL; the equivalence point volume of acetic acid was found to be 13.89 mL. The equivalence point volume was found by using the first derivative plot and looking at the highest slope. Phenolphthalein would be a good indicator for giving a reliable determination of the equivalence point. In the forward reaction of the HCl + NaOH and acetic acid + NaOH, the phenolphthalein clearly changed color from fairly clear to a pinkish color once the solution started becoming more basic. If an unknown weak ammonia solution was titrated with the strong acid HCl, then the phenolphthalein would not be an ideal indicator to use because the color change for the reaction between the weak ammonia (weak base) and HCl (strong acid) should occur at the likely equivalence point between 4 and 6; phenolphthalein changes color at the pH range of 8-10, so it would not be ideal use phenolphthalein because it does not have a color change at a pH range near the equivalence point pH of the weak base (ammonia) and strong acid (HCl).
The reaction which occurs is a neutralization reaction because the H+ and OH- ions react to form water.