The weekly homework found on this blog has resumed. Since we are still on the topic of hydrocarbon, let's do a review of the concepts which we have learnt from this chapter.
The following picture provides this week's question.
The following picture provides this week's question.
The explanation (the answer to the question):
During nitration the NO2+ electrophile substitutes the H on either C2 or C4, since OH is an electron donating substituent, hence it directs the electrophile to the 2nd or 4th position.
Since there are two H to replace to obtain the 2-substituted product and one to replace to obtain the 4-substituted product; by probability, we should get a 66.7% of the former and 33.3% of the latter.
However, there is steric replusion experience when the 2nd electrophile apporaches C2 to do the substitution as compared to when the electrophile approaches C4. Hence, substitution on C4's H is favourable. This accounts to why the percentage of 4-substitution to be higher than expected and for the 2-substitution to be lower than expected.
To account for why we obtain equal distribution of both products could be due to these two reasons, (1) probability and (2) steric hinderance, being significance and not one being important and the other unimportant - If the second reason is important and the first isn't, you will see more of the 4-substituted product (which we do observe for many other cases).
Comments:
Congratulations to Berlinda. I am most satisfied with her answer.
This question is a trick question. It uses the illusion that there is no outright majority and tries make you argue for the distribution of the products. In fact, it tries to try you to explain that the 2-substituted product is more stable. Which, is something that many of you fell for.
If the 2-substituted product is more stable, we would obtain more than 66.7% of the 2-substituted product (e.g. maybe 70 - 80%?), since by probability it already aids the chances of the 2-substituted product to be formed.
Hence, in most cases, application of one principle is sufficient to answer a question. However, if that principle used does not seem to account for a given observation, try to think if it is (1) a combination of different principles which play equal importance or (2) an entirely different chemistry principle is used.
During nitration the NO2+ electrophile substitutes the H on either C2 or C4, since OH is an electron donating substituent, hence it directs the electrophile to the 2nd or 4th position.
Since there are two H to replace to obtain the 2-substituted product and one to replace to obtain the 4-substituted product; by probability, we should get a 66.7% of the former and 33.3% of the latter.
However, there is steric replusion experience when the 2nd electrophile apporaches C2 to do the substitution as compared to when the electrophile approaches C4. Hence, substitution on C4's H is favourable. This accounts to why the percentage of 4-substitution to be higher than expected and for the 2-substitution to be lower than expected.
To account for why we obtain equal distribution of both products could be due to these two reasons, (1) probability and (2) steric hinderance, being significance and not one being important and the other unimportant - If the second reason is important and the first isn't, you will see more of the 4-substituted product (which we do observe for many other cases).
Comments:
Congratulations to Berlinda. I am most satisfied with her answer.
This question is a trick question. It uses the illusion that there is no outright majority and tries make you argue for the distribution of the products. In fact, it tries to try you to explain that the 2-substituted product is more stable. Which, is something that many of you fell for.
If the 2-substituted product is more stable, we would obtain more than 66.7% of the 2-substituted product (e.g. maybe 70 - 80%?), since by probability it already aids the chances of the 2-substituted product to be formed.
Hence, in most cases, application of one principle is sufficient to answer a question. However, if that principle used does not seem to account for a given observation, try to think if it is (1) a combination of different principles which play equal importance or (2) an entirely different chemistry principle is used.
22 comments:
This reaction is an electrophilic substitution reaction where NO2+ is the electrophile. As the OH group attached to the benzene ring
is electron donating due to the presence of a lone pair of electrons, the second substitutent, NO2+, is 2 and 4 directing.
Both of the products have 50% chances of being formed. For the
2-substituted product, it is formed due to the formation of a more stable intermediate where there is localisation of electrons. The positive charge in the benzene ring is closest to the electron donating OH. Thus, it makes the intermediate stable which supports the formation of the 2-substituted product.
For the 4-substituted product, it is formed due to less steric
repulsion between the electron clouds of OH and NO2+ when they are furthest away from each other.
The 3-substituted product is not formed as the intermediate is not
stable due to the first subsitutent (OH) being electron donating and not electron withdrawing.
OH is electron donating, and it directs the other substituents into position 2 and 4 because the highest electron density is at position 2 and 4, favouring reaction at position 2 and 4. Position 4 is more favoured because of the lower steric repulsion between the electron cloud of OH and NO2. Hence, the formation of the 4-substituted product is more favourable due to the lower activation energy of the slow step as a result of lower steric repulsion. Thus, one would assume the 4-substituted product to be the major product. However, because OH is electron donating, the intermediate formed during the formation of 2-substituted product is more stable than the intermediate formed during the formation of the 4-substituted product. When the intermediate formed is more stable, the reaction for 2-substituted product is favoured. Thus, these 2 factors produce the 2-substituted and 4-substituted products in a 1:1 ratio.
As –OH substituent is electron donating, it is 2, 4-directing. Therefore –NO2 is substituted at 2nd and 4th carbon. Since substitution at the 2nd carbon is also equivalent to substitution at the 6th, therefore, theoretically, the first product should be present in 66.7%. However, as the electron cloud of –OH and –NO2 will repel each other, 2-directing substitution experiences steric hindrance. Therefore, the slow step for this substitution will have a higher activation energy. Thus less 2 intermediates are formed and the percentage of 2-directed NO2 is less than 66.7%.
The original reactant, phenol, has undergone nitration to form 2 products, 2-nitrophenol and 4-nitrophenol. Although, it would seem that 4-nitrophenol would be the major product, the ratio of which the 2 products are formed is 1: 1. Although there is steric hindrance between the –OH electron cloud and the NO2 electron cloud, the 2-position is also favoured as the -2 position gives a more stable intermediate than the 4-position. This is because when the addition takes place at the 4-position, the carbon atom where the H and the NO2 are attached to, is attached to 2 other alkyl groups. This means that the carbon atom is attached to 3 R groups and 1 H group. For the 2-position addition, the carbon atom where the H and the NO2 are attached to is also attached to 2 alkyl groups as well. However, the OH group is attached to one of the alkyl groups next to the 2-position. Since the OH group is electron donating, it donates electrons to the alkyl group which further donates the electrons to the Carbon atom in the centre. Since the intermediate is a carbocation, it has a positive charge on the carbon atom in the centre. Therefore, when the electron density of that carbon atom increases, the positive charge of the carboncation becomes more negative and thus making the intermediate less reactive and thus more stable than the 4-position addition intermediate. Therefore, the 2 factors make both positions equally preferred and thus the 2 products are produced in equal quantities.
In a normal electrophilic substitution reaction,the molecule would be directed at the 2nd and 4th position as the OH- is an electron donating pair.In that case, the 4th position would be favoured more than the 2nd reaction as there would be steric hinderance when it is placed at the 2nd position, thus favouring the 4th position.However, in this case, both positions are favoured as the molecule is NO2, of an electron withdrawing group. Thus, there would not be steric hinderance when it is placed at the 2nd position as the electron cloud of NO2 would not repel the electron cloud of OH- due to both N and O being electronegative atoms,of which both would rather draw electrons towards themselves than give away electrons.Thus, due to this property, there would not be any steric hinderance taking place and this would result in an equal distribution of NO2 at the 2nd and 4th position.
In this reaction, phenol reacts with HNO3 to form 2-nitro-phenol and 4-nitro-phenol. As the -OH group in phenol is electron donating, it pushes the electrons into the ring, causing the substitution to take place at the 2- and 4- positions. This is why substitution rarely occurs at the 3- position.
Both 2-nitro-phenol and 4-nitro-phenol are evenly produced as the steric hinderance does not affect the 2- position in this case. This is due to the -OH group being electron donating while the -NO2 group being electron withdrawing, as such both are attracted to each other, thus eliminating the repulsion between their electron clouds.
Electrophilic substitution has taken place. As –OH is electron donating, the 2, 4-positions would be favoured. In this case, both are produced in equal amounts.
For the 1st product on the right, where NO2 is at the 4th position, it is 50% because at the 4th position, there is less steric hindrance between the –OH electron cloud and the NO2 electron cloud. Hence normally, it would be more favoured.
However, this is not the case as both are produced in equal amounts.
This is because for the 1st product, there is only one possible position for the NO2 to go to. However for the second product, there are 2 positions to form the same 1st product as the benzene ring is symmetrical. Hence the products will now be produced in equal amounts.
The reaction is an electrophilic substitution reaction.
Using the Kekule structure for the 2 directing intermediate, OH will stabilize the intermediate as it is electron donating. The delocalisation of charge through the movement of pi electrons when p orbitals of the remaining non substituted C atoms are joined together also ensures that the intermediate is stable. This is applicable equally to both the 2 and 4 directing substituted intermediates as similar delocalisation of electrons will simply take place at the remaining unsubstituted C atoms for the 4 directing substituted intermediate. Hence, 50-50 distribution of products can occur as both are equally stable.
The hydroxyl group in phenol is is electron-donating due to the lone pair of electrons present in oxygen. These electrons are pushed into the benzene ring and delocalised, hence pushing the electrons that were already present in the benzene ring down. Hence there are more electrons in the 2 and 6 positions due to this, and since all the electrons are pushed down, the 4 position also has a higher concentration of electrons. However, since 2,6 positions are identical, the ratio of 2-nitrophenol to 4-nitrophenol formed should be 2:1. However, from the question, we see that the ratio of 2-nitrophenol to 4-nitrophenol formed is 1:1. This is because, when the -NO2 group approaches the 2,6 positions, there is steric repulsion between itself and the -OH group already on the benzene ring. Hence, a C-NO2 bond at 4 position is more likely to occur, this bringing down the ratio from 2:1 to 1:1.
The OH on the benzene is electron donating, therefore it follows the electron donating groups with the NO2 being 2 and 4 directing. For the second position, the NO2 electrophile is able to form a stable compound despite the repulsion of electon clouds of OH and NO2. This is due to the intermediates being stable. The 4th directing is more likely to be formed as it is easily to approach position 4 due to steric repulsion. Thus the substituent NO2 would bond at position 4 and would produced in greater quantity. However, there are to positions of the 2 directing, one left and one right of the OH molecule on the benzene. Thus the probability of forming the 2 directing is higher. Thus the ratio of 2 directing compounds increase, and both 2 directing and 4 directing compounds are formed in a 1 :1 ratio.
There are 2 ways to be 2-directing because 6-directing is the same as 2-directing. Thus, the 2-directing products formed should be 66.6% and the 4-directing product formed should be 33.3%. However, compared to 2-directing product, the 4-directing product is more favourable as there is less steric hindrance. Hence, more 4-directing products are formed and results in an equal distribution of 2-directing and 4-directing products being formed.
The phenol reacts with HNO3 via electrophilic substitution. Since -OH is electron donating,positions 2 and 4 are favoured for substitution.
Of the two positions, position 4 is more favoured as it is further away from the -OH group. Hence there is less steric hindrance and substitution is more likely to take place there.
However, in a benzene ring there are 2 position 2s (on either side of the -OH group) but only one position 4. Hence although substitution at position 2 is less favoured, having 2 of the positions increases the likelihood of substitution occuring at the 2 position, therefore making the distribution of 2 and 4 directed products 50-50.
OH- is an electron donating group and is 2, 4-directing. Due to more steric hindrance between the OH- and NO2 group for the 2-substitued product, the electron clouds of OH and NO2 would repel more at the 2-position. Therefore, the 4-substituted product is more favoured. However, by drawing the Kekule structure, the 2-substituted product would result in a more stabilising effect and is thus favoured. Hence, equal amounts of 2 and 4 substituted products is formed.
In the compound, OH is electron donating, thus during nitration, NO2 is directed to positions 2 and 4.
2-substitution is not as favoured as 4-substitution due to steric hindrance between the electron clouds of OH and NO2 which causes the activation energy of 2-substitution to be higher.
However, the distribution of both products is equal as the intermediate formed in 2-substitution is relatively stable and there are two possible positions which NO2 can substitute onto the ring, hence making up for the supposedly lower formation of 2-substituted product.
-OH is electron-donating due to the lone-pair of electrons in O. Hence positions 2 and 4 are favoured. There will not be steric repulsion between the two groups when the –NO2 group is at position 2 as the O atom from the -NO2 group (which has a lone pair of electrons) will form hydrogen bonds with the H atom from the –OH group. Thus, both products exist in equal amounts.
The following occurs because as OH is electron donating, it is strongly activating and therefore it increases the reactivity of the benzene ring,hence during electrophilic substitution with NO2+,it will direct the NO2+ into the 2 and 4 positions.The 4th position is not favoured over the 2nd position because the electron donating OH creates an overall stabilizing effect on the intermediate.Therefore, as both positions are equally possible,there is 50% of the product that is formed in the 2 position and 50% of the product formed in the 4 position.
as the O-H group is electron donating,it causes the location of the second substitution to be of either positions 2 or 4.
due to steric hinderance,the 4th position is favoured for substitution to occur as as this would allow the second substituent to be further away from the O-H group,where the second substituent experiences less repulsion.
however,since 3 Lewis structures form the actual intermediate,the 2nd substituent can be located at the 2nd position as the intermediate can be stabilised by them.
Since the OH substituent is electron donating, it is position 2 and 4 directing. Therefore, the products formed are substituted at either the 2 or 4 carbons. Since there is steric repulsion when the NO2+ approaches the OH group, technically there should be less position 2 carbon substituted product. However, there are two possible positions for NO2+ to approach in order to produce a position 2 carbon substituted product. Therefore, the ratio of position 2 carbon substituted product to position 4 carbon substituted product is 1:1.
HNO3 + 2H2SO4 <-> 2HSO4- + NO2+ + H3O+
Electrophilic substitution, NO2+ is the electrophile.
Hydroxyl groups is electron-donating, so that makes ortho and para substitutions more favoured because the intermediates formed are more stable than the one formed by meta substitution (the hydroxyl group stabilizes the intermediate). -> ortho-/para- directing.
Now let’s compare the intermediates of ortho and para substitutions. To form the orthosubstituted intermediate, the nitronium ion will experience steric hindrance by the hydroxyl group on the adjacent carbon atom as their electron clouds repel each other. This makes the activation energy for orthosubstitution higher than that of para substitution, where steric hindrance is minimized. From here, it seems that there will be a smaller proportion of the orthosubstituted intermediate formed, so less o-nitrophenol should be formed.
However, there are two carbon atoms on the benzene ring that the nitronium ion can attack in order to form o-nitrophenol (C2 or C5). This doubles the chance of forming the orthosubstituted intermediate, bringing the distribution of the products o-nitrophenol and p-nitrophenol to be about equal.
Phenol is an OH bonded to a benzene ring. OH bonded to the benzene ring is under the category of being an electron donating group.
In other words, it is a strong activating group. This means that it is a 2-,4-directing group, or ortho and para-directing respectively.
It also stabilizes the positively-charged intermediate formed during the electrophilic substitution reaction phenol undergoes. This is done by donating electrons into the benzene ring, hence the term electron-donating.
It is usually assumed that the 4-directing product is the major product because there is less steric repulsion between the two groups that are further apart, than that of the 2-directing product.
However in this particular case, both products are produced in an even 50%-50% distribution. The reason for this distribution of the products, is that the attraction force between the NO2 group and the benzene ring, and this effect offsets the steric repulsion.
The arene undergoes two reactions, electrophilic substitution and free radical substitution.
An arene would undergo electrophilic substitution to produce a substitution at the -2 and -4 carbon, which are the products shown. However, the -4 would be a major product, because there is less steric repulsion at the -4 carbon substitution than at the -2 substitution. Thus the -4 carbon substituted product has a higher percentage among the products formed. However, the above products show that the percentage of -2 and -4 carbon substituted products formed are of equal ratios. Thus showing that free radical substitution took place as well.
In free radical substitution, there is a means to allievate the electron deficiency of a carbon radical. When the radical centre is next to a carbon cabon double bond, delocalisation of resonance occurs. The stabilisation of the radical intermediate can occur through delocalisation. The p-orbital of radical centre and p-orbital of the pi bond overlaps. The electrons in the 3 orbitals then can flow through the 3 overlapped orbitals. This would then be the delocalisation of the electrons. This then makes the radical carbon less electron deficient. This would make the -2carbon substitution product more stable, hen ce more of it will be produced, thus reducing the number of -4carbon substituted products formed. Therefore the percentage would equal out into 50-50%formed.
Another possibility could be that the O atom on the -2 Carbon substituted intermediate forms hydrogen bonding with the H atom on the –OH found on the -1 carbon. The hydrogen bonds formed is a very strong fixed dipole-dipole van der Waals force. Though it maybe weaker than a covalent, ionice and metallic bond, it still strengthens the stability of the -2 Carbon substituted intermediate. Therefore, since there is increased stability of the -2 Carbon substituted intermediate formed, more of its product is formed in the reaction and hence -4 Carbon substitued product is not the major product of the reaction but instead the -2 Carbon and -4 Carbon substituted products are formed in equal proportions.
Since the OH substituent is electron donating, the addition of NO2 to the phenol is 2,4-directing.
However, the oxygen atom at the OH substituent is electronegative and has a partial negative charge. This partial negative charge attracts the NO2+ ions, so the 2-directing is favoured.
At the same time, the electron clouds of the oxygen and NO2 repel each other(steric repulsion).
These 2 factors balance out and so phenol is as 2-directing as it is 4-directing.
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