Last week's homework has been checked. I have posted all your answers, my suggestions and comments in that post. Please take some time to read through your classmates' input as well. Please click here if you want to read them.
This week's homework:
With reference to the last content post which I have posted, I discussed about the different classification of chemical reactions.
The above reaction scheme (reaction map) shows 4 different reaction which cholesterol is undergoing. This week's assignment requires you to:(A) State the name of the reactions (e.g. electrophilic substitution reaction).
(B) Give reason(s) to support your answer.
Answers
Notes: Reaction D actually shows the O on -OH being the nucleophile which attacks the PCl5. A Cl- is released which then attacks the carbon of the C-O bond, thus eventually replaces the -OH with a -Cl.Comments
I think many of you have made use of the video and conceptualise the patterns to classify the chemical reactions. Some of you have also added more details in your patterns which maybe good.
However, do take note that having too much details might make it more difficult to classify the reactions. You will be able to observe that my reason for the classification was simple and succinct .
Some of the misconceptions/errors noted:
(1) Not stating how the reaction is taking place. e.g. Free radical, electrophilic, nucleophilic. Please avoid terms like halogenation or hydrohalogenation as it does not tell you the "how" bit. This is important because halogenation can occur in three manners; electrophilic substitution, electrophilic addition and free radical substitution.
(2) Thinking that PCl5 is a nucleophile. There is no available lone pair on P to do the replacement. From the video, you will notice that when NH3 is the nucleophile, it attacks the polar C-X bond and the product has the N attached to the C.
(3) A few, but notably, actually recognised the wrong chemical bonds broken. This is grave. Because the strategy which I will like all of you to know the organic reactions is by knowing the bonds that are broken in the chemical reaction. The characteristics of the bonds will help you decide how the reaction is taking place.
(4) As this assignment actually requires little details, as you can observe from the above answers, I cannot help to stress that by writing too much you will be subjected to possible content errors. - If there are content flaws, it will be worse.
26 comments:
A: Electrophilic Addition Reaction
HBr is an electrophile as it is a polar molecule that has a slight positive charge on H of HBr. Therefore, it is attracted to the electron rich C=C bond. The reaction follows Markovnikov’s Rule that the hydrogen atom of HBr will form a covalent bond with the carbon atom having the greater number of hydrogen atoms.
B: Substitution by Halogen (Cl)
In the presence of sunlight, Cl will replace H through substitution. Hydrogen atoms are replaced one by one and the amount of substitution depends on the amount of halogen.
C: Dehydration of alcohol (Elimination Reaction)
By heating the molecule with the dehydrating agent, concentrated sulfuric acid, at 170 0C, OH and H that are found on adjacent carbons atoms can be dehydrated. This is an elimination reaction, whereby H2O is eliminated.
D: Electrophilic Substitution Reaction
When PCl5 move towards the molecule, the electrons in pcl5 will repel to 1 side, making Cl electrophilic. As Cl is more electronegative than O, it has a greater ability to attract bonding electrons to it. Therefore, it is attracted to H and substitutes O atom.
A: Electrophilic Addition reaction
Reason: Two molecules react to form one larger molecule and it involves the breaking of pi bond in the C=C bond.
B: Free Radical Substitution reaction
Reason: Two products are formed and the non-polar C-H bond is broken to form a C-Cl bond.
C: Dehydration Elimination reaction
Reason: H2SO4 is a dehydrating agent removing OH from the carbon atoms. Moreover, the molecule undergoes the reaction to form two molecules in which the degree of saturation is decreased after reaction.
D: Nucleophilic Substitution reaction
Reason: A C-O polar bond is broken and the Cl in PCl4 has lone pair of electrons. The Cl atom replaces the OH molecule to form two products.
For part A, it is an electrophilic addition reaction. This is because the cholesterol molecule reacts with the HBr molecule and the double bond in the cholesterol molecule is broken and the H and Br atom is bonded to the cholesterol molecule. Since there were at first 2 reactants but only 1 product, the reaction is an addition one. Since, Br2 is an electrophile; the reaction is an electrophilic addition one.
For part B, it is a free radical substitution reaction. This is because the chlorine molecule reacted with the cholesterol molecule is substituted with one H atom in the cholesterol molecule. The products should be the cholesterol molecule and a HCl molecule as 1 Cl radical would be left to react with a H atom, forming HCl. This takes place in the presence of UV light, therefore it can be concluded that the reaction is a substitution one. Since the reaction involves the Cl-Cl bond to be broken in the presence of light to form Cl radicals to react with the cholesterol molecule, the reaction is thus a free radical substitution reaction.
For part C, the reaction is a dehydration of alcohol reaction. Since, the alcohol functional group of cholesterol, the –OH group, has been reacted with H2SO4, it is eliminated and a double bond is formed in its place. This reaction is actually an elimination reaction in nature as the H2O is eliminated from the cholesterol molecule.
For part D, the reaction is an electrophilic substitutive chlorination reaction. One of the Cl atoms in the PCl5 molecule is substituted with the -OH functional group in cholesterol. This reaction occurs due to the electrophilic nature of the PCl5 molecule which thus causes the Cl to substitute with the –OH group.
A) Electrophilic addition rxn
-Pi-bond of C=C is broken and H and Br atoms are added.
B) Free radical substitution rxn
-non-polar C-H bond is broken and H atom is replaced by Cl.
C) Elimination rxn
-OH group and 1 H atom is removed and a double bond is formed.
D) Nucleophilic substitution rxn
-Polar covalent bond(C-O) is broken. PCl5 is nucleophile.
A is an electrophilic addition reaction. This is because Br was added to the molecule and as such there was a breaking of a pi bond in C=C double bond.
B is a free radical substitution reaction. This is because there was the breaking of the C-H non polar bond so that Cl substituted H.
C is an elimination. This is because OH was eliminated and a double bond was formed.
D is a nucleophilic substitution reaction. This is because OH was substituted by Cl and as such the polar covalent bond of C-O was broken.
The reaction in A is eletrophilic addition. From the product, it can be seen that a double bond was broken and each of the molecules of HBr was added to the electon rich cholesterol.
The reaction B is a free radical substitution. During the reaction, UV light was needed for the reaction to proceed. The photon from the UV light helps break up the chlorine molecule into chlorine atoms. The chlorine atoms then reacts with the cholesterol, replacing the H atom.
The reaction C is a dehydration of the cholesterol. The concentrated H2SO4 acts as a dehydrating agent, where it reacts with the OH and H in the cholesterol resulting in another double bond in the product. Thus this reaction removes H2O from the cholesterol.
The reaction D is a nucleophilic substitution. Cl is a nucleophile due to the presence of a lone pair of electrons for election donation. Thus the Cl from PCl5 is reacted with the cholesterol, substituting the OH in the cholesterol.
(A) An electrophilic addition reaction is involved.
H-Br is polar and the H side of the molecule has a more net positive charge. During the reaction, heterolytic fission occurs on H-Br, leaving Br with the pair of bond electrons and H with an empty orbital. H becomes an electrophile because of the empty orbital. The double bond in the cholesterol is electron-rich. The 2 electrons in the pi bond is used to form a sigma bond with hydrogen. Following that, bromine forms a sigma bond with the adjacent 2o carbocation to form the new molecule.
(B)This is a free radical substitution reaction. The reaction is initiated when Cl2 forms 2Cl free radicals in the presence of UV light. The propagation step involes a free radical reacting with a molecule that is not a free radical. The termination step involes 2 free radical coming together to form 1 molecule.
(C)This is an elimination reaction. Water is eliminated.
(D)Nucleophilic substitution reaction is involved. The Cl atoms in PCl5 are nucleophilic because they have lone pairs of electrons. The carbon bonded to the OH in the cholesterol molecule is short of an electron because oxygen is very electronegative and pulls the bond pair of electrons near to itself. A Cl atom from PCl5 thus undergoes nucleophilic substitution with the cholesterol molecule.
Reaction A undergoes electrophile addition reaction as the pi bond of the C=C is broken leaving only the sigma bond. Br also has empty orbitals making it electrophilic and. H and Br become bonded to the C-C single bond.
Reaction B undergoes free radical substitution reaction as the hydrogen from a non polar C-H bond is replaced by a chlorine atom.
Reaction C undergoes elimination as a double bond is being formed from a single bond.
Reaction D undergoes nucleophilic substitution reaction as the polar bond between C and OH is broken and chlorine of PCl5, which has a lone pair of electrons and is nucleophilic, replaced the OH.
Reaction A is an electrophilic addition reaction. It is an addition reaction because the C=C double bond is broken and HBr attaches itself to the molecule. Moreover, as the C=C double bond is rich in electron density, the hydrogen from the HBr being slightly positive charged and electron deficient will attack the electron rich region, therefore it is an electrophilic addition reaction.
Reaction B is a nucleophilic substitution reaction. This is because the C-H bond is broken and therefore the Carbon atom is slightly positively charged; Moreover, Chlorine under light will break up into negatively charged 2Cl- ions which are rich in electron density and therefore the chlorine ions will attack the positively charged carbon atom hence attaching itself to the molecule.
Reaction C is an elimination reaction. Concentrated sulphuric acid under 170 degrees acts as a dehydrating agent and therefore will cause the H and OH on adjacent carbon atoms, which would form water, will be eliminated.
Reaction D is a neucleophilic substitution reaction. This is because the C-OH bond is broken and therefore the Carbon atoms is slightly positively charged and therefore as the Cl on PCl5 has a lone pair of electrons, it is rich in electron density and hence the chlorine will attach itself to the cholesterol molecule and kick the OH out.
(A)
Addition reaction
Double bonds are present . The pi bond between the C=C bond is broken but the sigma bond remains. Therefore the H and Br atoms form bonds with the singly-bonded C-C bond. It is also an addition reaction because it two molecules react to form one larger molecule. The degree of saturation also increased after the reaction.
It is not considered an electrohpilic addition reaction because it requires the breaking of a pi bond of the polar double bond, but since only Carbon atoms are double bonded together, they are of equal electronegativity, thus it is not a polar bond.
(B)
Free Radical Substitition reaction
The presence of C-H bond present. The H atom in the C-H bond is replaced by the Cl atom from Cl2 to form a C-Cl bond. The reaction happens in the presence of UV light, which is a factor for substitution reaction.
It is not an electrophilic substitution reaction because it requires the presence of a benzene ring, which isnt present in the compound of involved in the reaction.
(C)
Elimination reaction
A small molecule, -OH atom, is removed from the larger molecule. A double bond forms between two Carbon atoms since the –OH atom was removed. The degree of saturation also decreases after reaction.
(D)
Nucleophilic Substitution reaction
The bond between Carbon and OH is polar, due to the difference in their electronegativities. When PCl2 is added t the organic molecule, the pi bond of the C-OH polar bond is broken and a C-Cl bond is formed. PCl4 is then the nucleophile.
A is hydrohalogenation, via electrophilic addition. This is because the hydrogen bromide possesses electrophilic properties, as it has less electrons than the high electron density area of the carbon-carbon double bond. The hydrogen and bromide ions will then form covalent bonds with one carbon each.
B is a free radical substitution reaction. This is because there is there is UV light present for initiation of the reaction to happen, which is when chlorine is homolytically broken. The chlorine free radical then reacts with the cholesterol to give a radical of cholesterol. The radical of cholesterol then reacts with another chlorine free radical.
C is dehydration. Cholesterol, as the name has stated, is an alcohol. The conditions for dehydration are concentrated sulphuric acid and 170 degrees Celsius, which are present. Furthermore, end product displays a new carbon-carbon double bond, in place of the original hydroxyl substituent, and an adjacent hydrogen atom has been removed to form that double bond.
D is electrophilic addition. This is because the phosphorus pentachloride, which has a very large relative molecular mass, has a very large and polarisable electron cloud. It is drawn to the 2 lone pairs, where the phosphorus pentachloride is now delta positive, and the hydroxyl substituent is delta negative. The electrophilic phosphorus chloride then replaces the hydroxyl group with a chloride ion.
A: addition of hydrogen halide (hydohalogenation)
the addition of HBr causes the double bond to be broken, forming a single bond which allows H and Br to bond to the original molecule,forming a single molecule.
B: Free radical substitution (chlorination)
Cl-Cl is broken through homolytic fission, through the use of light energy. the 2Cl radicals are very reactive as the unpaired electron joins up with another elctron to form an electron pair readily. in this case, 1Cl substitutes H in the molecule, while the other Cl pairs up with the displaced H,forming HCl.
C: Condensation reaction
concentrated H2SO4 is used as a dehydrating agent, causing the formation or H2O.
D: Substitution reaction
Cl from PCl5 displaces OH from the original molecule.
A: Electrophilic addition reaction
HBr has a slight positive charge, hence it is electrophile. The benzene ring is electron rich, hence the reaction is electrophilic. It is an addition reaction because the 2 molecules reacted to form a larger molecule with the breaking of the pi bond of the benzene ring.
B: Electrophilic substitution reaction
Cl2 is an electrophile and benzene ring is electron rich, hence the reaction is electrophilic. It is a substitution reaction because H is replaced by Cl.
C: Elimination reaction
It is an elimination reaction because a molecule of H2O is formed and a pi bond is formed on the benzene ring.
D: Free radical substitution reaction
Cl- is a nucleophile and the benzene ring is electron rich, and O is replaced by Cl, thus the reaction is a free radical substitution reaction.
(A) Electrophilic Addition Reaction.
The cholesterol molecule reacts with HBr. There is an addition across the C-C double bond of the cholesterol molecule and its pi bond is broken while the sigma bond remains. H and Br are now bonded to the molecule. H is delta positive and Br is delta negative. There are 2 reactants but 1 product. Thus the reaction is an Electrophilic Addition reaction.
(B) Free Radical Substitution Reaction.
The cholesterol molecule reacts with Cl2 in the presence of sunlight. The non-polar C-H bond of the cholesterol molecule is broken. In the presence of sunlight, Cl-Cl bond is broken to form 2 Cl radicals. One of the Cl radicals reacts with the cholesterol molecule to form a C-Cl bond while the other reacts with H to form H-Cl. Hence the reaction is a Free Radical Substitution Reaction.
(C) Dehydration.
The cholesterol molecule undergoes heating at 170 degrees Celsius with a dehydrating agent (conc. H2SO4). The cholesterol molecule contains an alcohol functional group. H and OH on adjacent carbon atoms are dehydrated and a C-C double bond is formed. This is also an elimination reaction whereby H20 is eliminated. However it removes H2O and hence is a Dehydration reaction.
(D) Electrophilic Substitution Reaction.
The cholesterol molecule reacts with PCl5. As PCl5 is electrophilic, and the cholesterol molecule is electron rich, one of the Cl atoms in PCl5 substitutes OH in the molecule. Therefore it is an Electrophilic Substitution reaction.
(A) Electrophilic addition- two molecules react to form on larger molecule, the reacion involved one molecule adding across an unsaturated functional group and the degree of saturation is increased after reaction.
(B)Free Radical substitution- one atom or a group of atoms replaces another different atom. Free radical CL is involved
(C)Elimination reaction-The degree of unsaturation of the molecule increases and a small molecule is removed.
(D)? Subsition reaction- one atom or a group of atoms replaces another different atom....
Reaction A : Electrophilic Addition.
Reason: The only pi bond in the C-C double bond in the cholesterol molecule is broken when reacted with the HBr molecule. This is because the cholesterol molecule is electron-rich, causing the electrophilic HBr to be attracted to it.
Reaction B : Free Radical Substitution Reaction.
Reason: The UV light had caused the Cl-Cl bond in Cl2 to be broken and in turn, one of the Cl atoms reacted with cholesterol to break a C-H non-polar bond, causing the Hydrogen ion to be displaced and the Cl atom to replace the H ion, forming a polar C-Cl bond.
Reaction C : Elimination Reaction, specifically Dehydration.
Reason: There has been a removal of H+ and OH- ions to form a C-C double bond in cholesterol. Also, there has been an elimination and removal of a water molecule from cholesterol.
Reaction D : Electrophilic Substitution Reaction.
Reason: The PCl5 molecule is polar, with most of the electrons being closer to the P atom. Hence, the Cl atoms have gained a slightly positive charge, causing PCl5 to be an electrophile. As the cholesterol molecule is electron-rich, the PCl molecule, specifically the Cl atoms in it, are attracted to the cholesterol molecule. At the same time, the OH- particle in cholesterol has a negative charge and is attracted to the PCl molecule when it is near cholesterol. Hence, the OH- and Cl particles are displaced from cholesterol and PCL5 respectively, causing a Cl atom to substitute the OH- ion in the cholesterol molecule.
A - Electrophilic addition reaction. The pi bond of the non-polar C=C double was broken to form one larger molecule.
B - Free-redical substitution reaction. The non-polar C-H bond was broken to form a bond with a chlorine atom.
C - Condensation reaction. O-H was removed to form a double bond while producing H20. A bigger organic molecule was also produced.
D - Nuleophilic substitution reaction. Polar covalent bond was broken and Phosphorous pentachloride (PCl5) was a nucleophile as it had more electronegative (chlorine) atoms.
A)It is an electrophilic addition reaction as the addition of HBr to the compound produces a halogenoalkane.
B)It is a free radical substitution as the reaction occurs in the presence of uv light
and this reaction requires light energy to occur.
C)It is a elimination reaction(dehydration of alcohol).This is because H2O is eliminated during the reaction.
D)It is a electrophilic substitution reaction as OH- is removed and Cl- has replaced it at the end of the reaction.
opps...accidentally chose my identity as anonymous...the previous anonymous comment was mine...
Reaction (A)
(A)is an electrophilic addition reaction as there is first a carbon-carbon double bond present, which is where the addition reaction takes place. The pi bond breaks but the sigma bond remains. H-Br is also the electrophile in this case.
Reaction (B)
(B)is a free radical substitution reaction, as the bond broken is the non-polar C-H bond, whereby H is replaced by Cl.
Reaction (C)
(C)is a elimination reaction, as it results in the formation of a double-bond.
Reaction (D)
(D)is a substitution reaction as the result is the functional group OH being replaced by the atom Cl.
A: Electrophilic Addition Reaction.
It is an addition reaction as a Br atom is added across the C=C double bond.
The alcohol is an electrophile as it is electron deficient after losing 1 electron to a Br.
B: Free Radical Substitution Reaction
Cl atom is a radical as it has 1 unpaired electron.
It is a substitution reaction as a H atom is replaced by a Cl atom.
C : Elimination Reaction (Dehydration of alcohol)
OH is removed from the reactant and water is formed as a product.
D: Substitution Reaction.
OH is replaced by a Cl atom.
Reaction A:
This is an Electrophilic Addition Reaction.
The pi bond of the compound is broken in the addition reaction. The hydrogen atom of HBr forms a covalent bond with a carbon atom and Br is bonded to the carbon atom lacking in electron.
Reaction B:
This is a Free Radical Substitution Reaction.
Chlorine atom is a radical as it has an unpaired electron in its outer 3p orbital. Hydrogen is substituted by Chlorine in the reaction. Furthermore, free radical reactions take place in the presence of sunlight, hence the UV condition in the reaction.
Reaction C:
This is a dehydration reaction.
In this reaction, conc sulphuric acid acts as the dehydrating agent, whereby H and OH are removed in the dehydration process to form the resulting compound.
Reaction D:
This is a nucleophilic substitution reaction.
Cl replaces OH in the substitution reaction and a polar covalent bond is broken in the reaction. Also, the reactant Cl has a lone pair of electrons, making it a nucleophile, hence this is a nucleophilic substitution reaction.
(A) Electrophilic addition reaction
The two molecules, cholesterol and HBr, react to form one larger molecule. Breaking of pi bond of C=C.
(B) Free radical substitution reaction
The Cl atom replaces the H atom in cholesterol. Breaking of a C-H bond (non-polar).
(C) Elimination reaction
The OH molecule is eliminated from cholesterol. A double bond is formed.
(D) Nucleophilic substitution reaction
The Cl atom replaces the OH molecule in cholesterol. A polar covalent bond of C-O is broken. PCl5 is the nucleophile due to the lone pair of electrons in Cl.
The reaction at A is a Hydrohalogenation rection via Electrophilic Addition mechanism. This is concluded as the polar hydrogen bromide reacts with the molecule to form a similar product but with the double bonds broken and hydrogen and bromine added into the molecule, hence saturating the molecule.
Free-radical substitution reaction is taking place in reaction B as UV light, a requirement for such a reaction, is used and a radical,chlorine, is involved in the reaction in which it displaces a hydrogen atom in the molecule.
Reaction C is the reaction of Dehydration of alcohol as firstly, concentrated H2SO4 at 170 degrees celsius is a requirement for such a reaction and secondly, water is being removed from the molecule after reaction B has occurred.
Reaction D is a nucleophilic substitution reaction as PCL5 is a nucleophile due to the presence of an extra lone pair of electrons. Hence, it displaces the hydrogen atom in the molecule.
(A)Electrophilic addition reaction
It is an addition reaction because HBr is added across a C=C bond, causing the degree of saturation to increase. HBr has a slight positive charge, hence the reaction is electrophilic.
(B)Free radical substitution reaction
One of the Cl atoms replaced an H atom.
(C)Condensation reaction
Water is formed as a product.
(D)Free radical substitution reaction
The Cl atom replaced OH.
A: Electrophilic Addition.
The H atom in HBr is the electrophile and reacts with the electron-rich region which is the unsaturation(double bond) on the cholesterol molecule. This occurs in the second leftmost ring in the diagram.
B: Free Radical Substitution
One of the H atoms is substituted with a Cl atom in presence of UV light. This occurs on the rightmost ring in the diagram.
C: Dehydration/Elimination
H and OH on adjacent carbon atoms are removed by heating to 170°C with conc. H2SO4. This occurs on the leftmost ring in the diagram.
D: Electrophilic substitution
PCl5 is an electrophile and it substitutes a Cl for the OH group. This occurs at the leftmost ring in the diagram.
Note: I'm sorry I'm late.
Post a Comment