Alcohols + Haloalkanes Flashcards

Question

Describe the hydrolysis of haloalkanes by nucleophilic substitution Chloromethane + OH- ion

Answer 1

The bond polarity in haloalkanes determines how haloalkanes react Haloalkanes react with chemicals called nucleophiles Haloalkane chloromethane This has an electron deficient carbon atom with a partial positive charge In the hydroxide ion, the oxygen atom has a lone pair of electrons and a negative charge The hydroxide ion is an example of a nucleophile The lone pair of electrons on the hydroxide ion are attracted to the electron deficient carbon atom on the chloromethane The nucleophile now donates its lone pair of electrons to form a covalent bond to the electron deficient carbon atom I've shown the hydroxide ion forming the covalent bond from the opposite side to the halogen atom (arrow opposite side) By approaching from this side, there is less repulsion between the negative hydroxide ion and the negative halogen atom movement of electrons are represented by using a curly arrow must start at the lone pair of electrons and end at the electron deficient carbon atom A carbon atom can have a maximum of four covalent bonds So at the same time, the covalent bond between the carbon atom and the halogen atom breaks with the pair of electrons moving onto the halogen atom This is an example of heterolytic fission - both e- in covalent bond moving onto the halogen At the end of the reaction, we have produced an alcohol molecule. In this case methanol. We have also released a halide ion, in this case the chloride ion Because this has left the haloalkane, we call the halide ion a leaving group We have substituted the halogen atom in the haloalkane with a hydroxyl group This is called a hydrolysis reaction This is an example of nucleophilic substitution hydrolysis takes place via nucleophilic substitution hydroxide ion acting as a nucleophile -------------------------------- involves the breakage of the carbon-chlorine bond and the formation of a new bond between the carbon and the hydroxyl group (-OH), essentially "hydrolyzing" the chloromethane molecule by breaking it down with water, or in this case, the hydroxide ion Hydrolysis is a chemical reaction where a molecule is broken down by reaction with water (or another molecule with a nucleophile, like the hydroxide ion) In the reaction between chloromethane (CH3Cl) and hydroxide ions (OH-), the hydroxide ion attacks the carbon atom in chloromethane, displacing the chlorine atom https://cdn-ilcgdmp.nitrocdn.com/ZNZGaPUCQqgkHHGdlFKJfPiIeULNhire/assets/images/optimized/rev-8bf78c6/studymind.co.uk/wp-content/uploads/2022/04/1-47.png

Answer 2

To carry out the hydrolysis of a haloalkane, we use an aqueous solution of hydroxide ions E.g. NaOH or KOH dissolved in water However, there is a problem with this Haloalkanes are insoluble in water So we also need to add ethanol to the reaction The ethanol solubilises the haloalkane so it can react with the aqueous hydroxide ions To increase the rate, we heat the reaction under reflux cool temperature aqueous ethanolic NaOH 50:50 water-ethanol solvent

Answer 3

CH3Cl (l)+ NaOH (aq) ---> CH3OH + NaCl(aq) ethanol solvent heat under reflux ---->

Answer 4

how the rate of hydrolysis depends on the halogen atom in the haloalkane comparing rate of hydrolysis of 1-chloro/bromo/iodopentane All primary haloalkanes to measure the rate of hydrolysis, we monitor the formation of the halide ion using aqueous silver nitrate take 3 test tubes into each test tube, place 1cm3 of ethanol haloalkanes are insoluble in water so the ethanol acts as a solvent, allowing the haloalkanes to mix with aqueous solutions Next we add 0.1cm^3 of haloalkane to each test tube We now place each test tube into a water bath at 60 degrees celcius at cooler temperatures, hydrolysis is very slow In a separate test tube, we add aqueous silver nitrate and place this test tube into the same water bath now wait for ten minutes for all of the solutions to reach the same temperature now add 1 cm^3 of aqueous silver nitrate to each test tube and start timing In aqueous solutions, the solvent is water and water is a nucleophile So the water molecule reacts with the haloalkanes by nucleophilic substitution and hydrolysis takes place During hydrolysis, the halide ion will be released from the haloalkanes. The halide ion will then react with the silver ions to form an insoluble precipitate of silver halide Cl- (aq) + Ag+ (aq) --> AgCl(s) chloride ion will form a white precipitate of silver chloride bromide ion - cream precipitate iodide ion - yellow precipitate In each case, we time how long it takes for a precipitate to form We are using water to carry out hydrolysis rather than hydroxide ions That is because hydroxide ions react with silver ions to form insoluble silver hydroxide So we cannot use hydroxide ions for this experiment

Answer 5

Results 1-iodopentane forms a yellow precipitate very rapidly 1-bromopentane forms a cream precipitate this reaction is slower than 1-iodopentane 1-chloropentane forms a white precipitate 1-chloropentane reacts much more slowly than either 1-iodopentane or 1-bromopentane iodoalkane > bromoalkane > chloroalkane Iodoalkanes undergo hydrolysis most rapidly followed by bromo then chloro This can be explained by looking at the bond enthalpies involved During hydrolysis, we have to break the carbon to halogen bond The carbon to chlorine bond has a relatively high bond enthalpy This means it takes a lot of energy to break the carbon to chlorine bond Because of this, the hydrolysis of chloroalkanes is a very slow reaction The carbon to iodine bond has a relatively low bond enthalpy and does not take a lot of energy to break This explains why iodoalkanes react rapidly The carbon-to bromine bond enthalpy lies between the other two (Medium bond enthalpy) which is why bromoalkanes react more rapidly than chloroalkanes but less rapidly than iodoalkanes Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Notice that bond strength falls as you go from C-F to C-I, and notice how much stronger the carbon-fluorine bond is than the rest. The electronegativities of carbon and iodine are equal and so there will be no separation of charge on the bond.