Hydrolysis of halogenalkanes essay

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Hydrolysis of halogenalkanes essay

A primary Hydrolysis of halogenalkanes essay has two hydrogen atoms and just one alkyl group attached to the carbon atom, a secondary haloalkane has only got one hydrogen atom and two alkyl groups attached to the carbon atom and finally a tertiary haloalkane has no hydrogen atoms attached to the carbon atom but three alkyl groups.

Stability of the carbocation increases with the number of alkyl groups attached to the positively charged carbon atom.

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Tertiary carbocations are more stable and readily formed than secondary carbocations and so if the carbocation is more readily formed, a halide ion is more likely to leave.

SN1 reactions and elimination reactions normally do not occur if a primary carbocation is formed. To control this factor, I will have to keep the carbocation in my experiment constant.

I will do this by using three primary haloalkanes known to be 1- Chlorobutane, 1-Bromobutane and 1-Bodobutane. This will make my experiment more reliable by testing each haloalkane under the same hydrolysis reaction and to be more specific, a SN2 mechanism will take place for each of the three reactions.

This is because when alkyl groups are bonded to the carbon ion, they stabilise the positive charge into their sigma bonds. They delocalise a positive charge over a wider area. When the numbers of alkyl groups on the carbon ion are increased, this will lead to an increase in stability.

Hydrolysis of halogenalkanes essay

Iodine comes out to be the best leaving group out of the three halogens. This is because it is at the bottom of the group and so has more electron shells causing the outer shell electrons to be further away from the halogen. The rate of hydrolysis does increase as you go down group 7 in the periodic table because of the fact that bond enthalpy decreases due to the atomic radius increasing, producing a weaker bond.

Using this information I can say that iodine will be able to stabilise the carbon ion better than bromine and chlorine. In a general fact, the weaker the C-X bond then the better the leaving group because the halogens are more electronegative than carbon making the halide ion leave the alkyl group.

For instance the iodine ion will be more stable than the fluorine ion. Prediction I think bond enthalpy has the greatest influence on the rate of hydrolysis because it relates to bond polarisation and electronegativity. The C-X bond is polarised due to a difference in electronegativity where the halogen X atom pulls the electrons towards it self.

This causes the carbon atom to gain a partial positive charge making in electron deficient. Because of the bond being polar, the haloalkane now has a permanent dipole therefore causing the carbon atom to be attacked by a nucleophile.

As you go down the halogen group, electronegativity decreases. The halogens at the top of the group have a higher electronegativity and so have a higher bond polarity and the halogens at the bottom of the group have lower electronegativity and so have a lower bond polarity.

C-F will more polar than the rest and therefore have the slowest rate of hydrolysis. The more polar the bond, the harder it is to break it because the bond is becoming ionic.

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This is because a higher dissociation enthalpy is required to break the bond. If more energy is needed to break the C-X bond, than the hydrolysis reaction will occur slower meaning that it will take longer for reactants to turn into products. This therefore means that it will take longer periods of time to break.

My prediction can be backed up by what I mentioned before about the shielding effect and nucleur charge. As you go down group 7, the electronegativity decreases and shielding increases due to the number of energy levels also increasing. The C-Cl bond will need a higher amount of energy because the chlorine atom has less energy levels causing a greater nuclear attraction.

However the C-I bond will need a lower amount of energy because the iodine atom has more energy levels and so has a greater shielding effect causing a weaker nucleur attraction.

The ability for halogens to depart electrons in a covalent bond decreases down the group. The C-Cl bond holds on to the electrons in the covalent bond so strongly so that it takes a lot of energy to break the bond, and the opposite occurs in a C-I bond.

In my experiment, I will predict that 1-iodobutane will take the shortest time to form a silver precipitate and 1-chlorobutane will take the longest time to produce a silver precipitate.

Therefore 1-iododbutane will undergo hydrolysis much faster than 1-chlorobutane because it acts as a better leaving group. It is a better leaving group because an iodine atom has more shielding, weaker electronegativity and a lower bond enthalpy.

Iodine is less electronegative and therefore has a smaller bond polarity with carbon compared to chlorine. The temperature needs to be high because it will increase the rate of reaction due to more particles moving around with greater kinetic energy.

Pulling up or pushing down helps getting accurate as possible while measuring. Use propanane to clean off all distilled water remaining on equipment so no other side reactions take place. Make sure a mm thermometer is in the water bath to check the temperature 4] Collect two cm3 beakers and add LR ethanol and 0.

Collect 3 test tubes and put them in a test tube rack.General formula- Rhal, (r is alkyl group, and Hal is a halogen atom) In this experiment I will investigate the effects of the nature of the halogen atom (Hal) has on the reactivity of the halogenoalkane.

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H1 Chem urbanagricultureinitiative.com para más tarde. guardar. Relacionado. retardants D Plastics 23 A compound X. 10 22 Halogenalkanes have a wide variety of commercial uses. The organic product is then refluxed with LiAlH4 in dry ether.

Upon prolonged hydrolysis of B. showing clearly all the deductions that you make from the information that have. Comparing the rate of the hydrolysis of different hal- groups of halogenoalkanes Outline of the method: 1cm3 ethanol, 2 drops of a halogenoakane ( GAMSAT is one of the toughest and feared medical university admission exams.

With about questions and 2 essay writing activities spread over more than hours, this exam tests an aspirant for knowledge, aptitude, application as well as physical stamina. This page looks at how silver nitrate solution can be used as part of a test for halogenoalkanes (haloalkanes or alkyl halides), and also as a means of measuring their relative reactivities.

Hydrolysis of Halogenalkanes | Essay Example