Neighbouring Group Participation (NGP) is observed in nucleophilic substitution reactions, where a neighbouring group helps in the removal of the leaving group to form a reactive intermediate that leads to the formation of the product. Increase in the reaction rate and unexpected stereo chemical outcomes are associated in reactions involving NGP.
An atom having an unshared pair of electrons and also present at least beta to the leaving group can act as a neighbouring group. Also, NGP is mostly observed on solvolysis reactions where the solvent acts as the nucleophile.
A typical reaction involving NGP is shown below.
During NGP, the neighbouring group (G) attacks the electrophilic centre to eliminate the leaving group (L). This leads to the formation of a cyclic intermediate which is very reactive. This is called anchimeric assistance from the neighbouring group. The nucleophile (Nu-)then attacks this intermediate to form the product. If the attack happens of the carbon that was having the leaving group the configuration will be retained because the configuration at that carbon will be inverted twice.
Groups like halides, hydroxides, ethers, thio ethers, amino groups, carboxylates, phenyl group, pi-bonds etc. have been indentified to act as neighbouring groups in many reactions.
Some more examples of reaction involving NGP are shown below.
Na2CO3 can be manufactured by Solvay’s process, but not K2CO3. Why?
Solvay process involves the following 3 steps.
In the second step NaHCO3 formed is insoluble in water while NH4Cl is soluble. So, the two compounds can be separated and the third reaction can be performed. If the second step is attempted with KCl in place of NaCl, the products formed will be KHCO3 and NH4Cl. Both these compounds are soluble in water and hence cannot be separated easily.
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Ibuprofen is an anti-inflammatory drug used for relief from fever and body pain. Ibuprofen’s pain killing effect takes place almost immediately after a suitable dosage is administered while its anti-inflammatory effects take a little longer. Along with the pain killing properties Ibuprofen also comes with lots of side effects and should NOT be taken without the advice of a physician.
Ibuprofen can be synthesized from a sequence of reactions starting with 2-methylpropylbenzene. The steps are shown below.
Which out of (a) Clemmensen Reduction and (b) Wolff Kishner Reduction is the best suitable reduction method for the following reaction?
Wolff Kishner Reduction is the best suitable reaction method for the given reaction. Though both Clemmensen and Wolff Kishner reduction methods bring about the same outcome, Clemmensen Reduction is not used for compounds having acid sensitive groups along with the carbonyl group.
An organic compound P of molecular formula C4H10O upon dehydrogenation gives a compound Q which forms phenyl hydrazone with phenylhydrazine and both the compounds P and Q respond positive to the Iodoform Test. What is the name of the compound P?
3 Faraday of electricity are passed through molten Al2O3 aqueous solution of CuSO4 and molten NaCl in three different electrolytic cells. What will be the molar ratio of Al, Cu, Na deposited at the respective cathodes?
1 Faraday of electricity will deposit 1 equivalent of a metal. So 3 equivalents of Al, Cu & Na will be deposited which is same as 1 mole of Al, 1.5 mole of Cu and 3 mole of Na.
Elimination reaction in Organic Compounds involves the removal of certain groups from the reactant molecule usually leading to the formation of an unsaturated compound as a product.
For example the reaction between an alcohol and a mineral acid at high temperature is an elimination reaction leading to formation of an alkene along with the elimination of a molecule of water.
The elimination reaction between an alkyl halide and a base that leads to the formation of an alkene can follow three major pathways. They are named E1, E2 and E1CB. In the name E stands for elimination (as expected) and the numbers 1 and 2 stand for unimolecular and bimolecular respectively. In E1CB, CB stands for conjugate base.
As we see in the above example, atoms from adjacent carbon atoms are removed; E1 and E2 reactions are often referred to as beta-elimination.
Let us talk about the elimination reaction through E2 mechanism.
Elimination reaction through E2 mechanism is a concerted reaction i.e. the reaction happens in a single step. Both the base and the alkyl halide take part in that step. As this reaction happens in only one step that step is the rate determining step and as both the base and the alkyl halide are involved in the rate determining step, the rate of the reaction will depend on the concentrations of both the alkyl halide and the base. So, this reaction is a second order reaction.
The mechanism of E2 elimination is shown below.
The base abstracts the hydrogen in the beta position to the leaving group and starts the reaction. At the same time a pi-bond is formed between the adjacent carbon atoms and also the halide is eliminated. A bond between the base and the hydrogen starts forming; the bond between the hydrogen and carbon starts breaking; a pi-bond between the adjacent carbon atoms start forming; the bond between the carbon and halogen starts breaking. This reaction goes through a transition state in which the bonds that start breaking become weaker and weaker and eventually break completely while the bonds that start forming become stronger and stronger and eventually become completely formed bonds.
The transition state of the reaction is shown below (the dotted lines represent a bond that is either being formed or being broken).
There are three important things related to a E2 elimination reaction.
1. The beta – hydrogen and the halogen should be on the same plane.
As seen in the mechanism a pi-bond is formed between two adjacent carbon atoms during the reaction at the same time when the C – H bond and C – Br bond are breaking. A pi-bond is formed by the sideways overlap of two p-orbitals. For maximum overlap and thereby form a strong bond both the p-orbitals should be parallel to each other on the same plane. So, the beta-hydrogen and the halogen should be on the same plane. This is necessary because the sp3 orbital of the carbon bonded to the hydrogen and the sp3 orbital of the carbon bonded to the halogen become the p-orbitals that overlap to form the pi-bond.
2. The beta-hydrogen and the halogen should preferably be anti-coplanar
Not only the beta-hydrogen and the halogen should be on the same plane they should preferably be anti (opposite sides) to each other than syn (same side). If elimination happens in anti-coplanar conformation it is called anti-elimination. This is necessary to avoid steric and electronic repulsions in the transition state which increase the energy of the transition state. The rate of the reaction depends on the energy of the transition state. The higher the energy of the transition state the difficult it is for it to be formed and slower the rate of the reaction. Any factor increasing the energy of the transition state will slow the reaction down. However, in molecules where the beta-hydrogen and the halogen are syn-coplanar due to the geometry of the molecule, syn-elimination happens.
3. More substituted alkene is formed as the major elimination product
The stability of an alkene depends on the number of alkyl substitutions present on the carbon atoms forming the pi-bond or more precisely, more the number of alpha-hydrogens on an alkene more stable it will become due to hypeconjugation. During elimination reactions, the substrates may have two different sets of beta-hydrogens, one leading to the formation of less susbtituted alkene and one leading to the formation of more substituted alkene. The more substituted alkene is formed as the major product as it is more stable. The product of the E2 elimination reaction resembles the transition state from which it is formed. The transition state leading to a more substituted alkene will be less energetic than the transition state leading to a less substituted alkene.
Apart from the three things mentioned above there are other interesting features too of E2 elimination such as the stereochemical outcome, competition with E1 and SN2 reactions etc.
Which among (a) Nitrobenzene and (b) benzene, is a suitable solvent for the Friedel - Craft alkylation of para - bromo benzene?
Among the three aromatic compounds mentioned in the question, benzene is the most reactive towards an electrophile followed by bromo benzene and then by nitro benzene. If we want the electrophile to substitute on bromo benzene then a solvent which is less reactive than itself towards the reagent should be used. So, nitrobenzene is a suitable solvent in this case.
Monosodium salt of oxalic acid is titrated with NaOH solution. In a second titration it is titrated with KMnO4 solution. What will the ratio of equivalent weights of monosodiumoxalate in both the titration be?
The equivalent weight of oxalic acid in the titration with NaOH is equal to its molecular weight and in the titration with KMnO4 it is equal to half of its molecular weight. So, the ratio of equivalent weights of monosodium oxalate in both the titrations will be 2.
An alkene, “A”, with molecular formula, C6H12 on reductive ozonolysis gives two products. One of these gives a positive iodoform test but negative Tollen’s reagent test. The other product gives a positive Tollen’s reagent test but negative iodoform test. What is the name of “A”?
Yes. In aniline the lone pair of electrons on the nitrogen atom will be in resonance with the benzene ring and will be less available to be donated to a proton as compared to the lone pair of electrons on the nitrogen atom of ethyl amine.
Vinyl chloride does not undergo nucleophilic substitution reaction but allyl chloride does. Why?
In vinyl chloride due to resonance the carbon halogen bond will have a partial double bond character. Moreover, positive charge developed on the vinyl carbon will be unstable compared to positive charge on the allyl carbon.
The scent of rain is one of the things that comes to our mind when we think about a rainy day. The smell is so intense in air that you can not miss it. Geosmin is responsible for this strong scent that is present in air when rain falls after a hot or dry weather.
Geosmin is produced by several class of microbes especially the Streptomyces. Geosmin is released when these microbes die. Geosmin translates to “Smell of Earth”. It is an organic substance that also is responsible for the earthy taste of beetroot. Geosmin is a colourless liquid which boils at 270oC. Humans are extremely sensitive to the smell of Geosmin and can detect it even at extremely low concentration.
Geosmin can be synthesized in the lab from a sequence of reactions starting with a condensation reaction between 2-methyl-cyclohexanone and ethyl vinyl ketone.
Isn’t it fascinating to know how chemistry is a part of everything around us?
Use in Chemistry: Conversion of Silver salts of carboxylic acids to alkyl halides
The widely believed mechanism involves the Silver salt of the carboxylic acid reacting with bromine to form acyl hypobromite which undergoes homolytic bond cleavage to produce bromine radical and a carboxylate radical. This carboxylate radical then releases a carbon dioxide molecule to produce an alkyl radical. This alkyl radical combines with the bromine radical from the acyl hypobromite to form alkyl bromide.
A compound A crystallizes in a hexagonal close packing system. If the compound is made of X and Y in which X occupies all the lattice position of hcp layers and Y occupies two third of the octahedral voids, what will the formula of A be?
Hydroxycitric acid (HCA) is a derivative of citric acid and is found in variety of tropical plants. It is found in Garcinia Indica commonly known as kokum and is used in Indian & Thai food as a flavoring agent.
Earlier experiments with HCA on mice showed that when mice were administered with HCA along with their food, it made them eat less.
If our food contains large amounts of carbohydrates, the ones that are in excess get converted finally into fat through enzymes. Any chemical that slows down this enzyme will help in accumulation of less fat. HCA, like citrate (natural chemical in our body) slows down the enzymes that convery carbohydrates into fatty acids which in turn become fat. HCA is also found to make liver release large amounts of sugar in the blood and this high sugar level in blood may suppress hunger.
Though the action of HCA was observed in mice, the same effect is yet to be proven in humans. While some claim HCA to be effective in reducing food intake leading to weight loss some claim that HCA causes serious health problems inculding liver damage and cardio vascular disorders.
A negatively charged species can act both as a base and also as a nucleophile. When an alkyl halide reacts with a negatively charged species there are two possible outcomes; elimination and substitution. If the outcome is elimination the negatively charged species has played the role of a base and if the outcome is substitution then the negatively charged species has acted as a nucleophile.
If we describe a chemical reaction in terms of bonds it is nothing but the breaking of old bonds and formation of new bonds. In a reaction, if a species forms a bond with (or abstracts) a proton it is acting as a base and if it forms a bond with (or attacks) a carbon atom it is acting as a nucleophile.
What is the fundamental difference between a base and a nucleophile?
Two major parameters control a chemical reaction. They are Thermodynamics and Kinetics. While thermodynamics helps us predict the spontaneity of a reaction at the given conditions, kinetics helps us understand the mechanism of a reaction.
Let us consider the following equilibrium;
As all the acid base reactions reach equilibrium after sometime, the extent to which the equilibrium has moved in the forward or backward direction will tell us about the stability of the reactants and products.
There are two bases in this equilibrium; AcO- & HO-. To compare the strengths of these two bases we can look at the their stability. A strong base is the one that is more unstable and easily gains stability by abstracting a proton.
If AcO- is more unstable than HO- the equilibrium will be more shifted in forward direction and if HO- is more unstable than AcO- then the equilibrium will be more shifted in backward direction. The direction in which the equilibrium is shifted can be known from the equilibrium constant.
For the example taken, the equilibrium constant is equal to Kw/Ka which is in the order of 10-14/10-5 = 10-9 (Ka of HOAc = 1.8 * 10-5). The small value of equilibrium constant shows that the equilibrium is highly shifted in the backward direction. So, HO- should be more unstable than AcO- and hence be a stronger base.
The strength of a base will be determined from the equilibrium constant which is in turn related to the free energy change of a process.
The above discussion shows that basic strength is a Thermodynamic Factor.
Let us consider the following two reactions;
Both the reactions are performed under identical conditions. Reaction (i) is over in 7 minutes and reaction (ii) is over in 3 minutes. With this information if you have to identify the stronger nucleophile among HO- and CN- and you are most likely to identify CN- to be the stronger nucleophile almost as a reflex action.
Yes, CN- is the stronger nucleophile among the two because the reaction involving it as a nucleophile gets over faster or in other words CN- attacks the electrophilic carbon faster than HO- does.
Nucleophilic strength is not usually related to the charge density on the negatively charged species (charge density is important to predict basic strength) but on how fast can it attack the electrophilic centre.
Nucleophilic Strength is a Kinetic Factor.
Any parameter that makes the nucleophile slow in attacking the electrophilc centre will be decreasing its strength as a nucleophile. For example, the nucleophilc strength of an anion like I- is decreased by polar protic solvents.
Polar protic solvents release protons and these protons solvate the anions and make it difficult for them to move. This reduced the speed with which they attack the electrophilic centre and hence behave as weak nucleophiles.
Aspartame is an artificial sweetener. It is used as a substitute for sugar in food and some beverages. It is the methyl ester of the dipeptide of L-aspartic acid and L-phenyl alanine. Both L-aspartic acid and L-phenylalanine are natural amino acids. It is said that aspartame is almost 200 times sweeter than sugar. This property allows us to have the same amount of sweetness as sugar but will a little quantity. As small quantity of aspartame is used, its calorific contribution becomes insignificant.
Aspartame gets hydrolysed into methanol and its constituent amino acids when hydrolysed in highly acidic conditions and also at elevated temperatures. There has been several articles on research work that shows some serious side effects of aspartame due to this break down especially due to the production of methanol. Methanol can cause permanent blindness even when consumed in little quantity.
Mr. Stanley was missing some money that he had kept in this cup board. This is the third time that he was missing money. In earlier occasions he blamed himself for having lost the money. But, this time he knew for sure that something is wrong. Someone is stealing money from him. “Who could it be?”, he asked himself. He made a list of suspects. It read 4 names. Alex, Charles, Robin and Peter.
Mr. Stanley wanted to catch the culprit during the act. He thought of a wonderful plan. He fetched the washing soda from his laundry room. He then took a little of the washing soda and spread it over some of the notes he had. He ensured that the amount was just enough to catch the culprit. He then left the notes in the same place as earlier and left the room.
Mr. Stanley is using a procedure all chemists do during a titration. During a titration, an indicator is used to indicate the end of the reaction that is being performed. The indicator can be added additionally to the reaction mixture to show completion of the reaction. In some reactions one of the products can act as an indicator. Substances commonly used as an indicator exhibit different color in different medium. Phenophthalein, a common indicator, becomes pink in a basic solution and is colorless in acidic solution.
Like other acid base indicators, phenolphthalein has different colors for its acidic (unionized) and basic (ionized) forms. It is colorless in acidic form and pink colored in basic form
When Mr. Stanley returned to his office after 10 minutes, he observed that the notes have gone as he had expected. He immediately summoned his four suspects to his office. He told them about the frequent disappearance of money from his office. All the four were surprised to hear that there has been a series of theft at their workplace. Mr. Stanley said, “I laid a trap this time to catch the culprit.” There was no nervousness or change of expression on anyone’s face. They still seemed surprised.
"Show me your Palms", he said. All the four extended their hands to show their palms. He then sprayed a solution on phenolphthalein on everyone’s palms. All but Robin’s palms were colorless. The washing soda (Sodium Carbonate) had made phenolphthalein change its colour to pink.
Robin, suspended from work, after being caught Red (pink to be precise) Handed is doing social service to compensate for all the money he had stolen.
Note: Please DO NOT handle chemicals with bare hands. Always be protected from chemicals.
Amino Acids have both amino group as well as a carboxylic acid group. The pKa values of both the groups are well separated that at any given pH a molecule of amino acids carries some charge. For example at a pH as low as 1, the amino group will be protonated and the net charge of the molecule will be positive and at a pH as high as 12 the acidic hydrogen atoms will be removed and the net charge on the molecule will be negative. So, at normal biological pH, most amino acids have the amino group protonated and the carboxylic acid group deprotonated. This ionic structure is called the zwitter ion.
The various forms that an amino acid takes with increase in pH is shown below.
At a lower pH (pH = 2) the amino acid is positively charged and as we increase the pH and move to a higher pH (pH = 12) the amino acid becomes negatively charged. At a lower pH the concentration of the cation (A)is more than that of the anion (C) and as the pH increases the concentration of the anion also increases and becomes more than that of the cation. This implies that at a particular pH the concentration of both the cation and the anion should be equal. This pH at which the concentrations of both the cation and anion are equal is called as the Isoelectric Point of the amino acid.
Generally isoelectric point of an amino acid is the average of both the pKa values.
Ka1 = ([B][H+])/[A]
So, [A] = ([B][H+])/Ka1
Ka2 = ([C][H+])/[B]
So, [C] = (Ka2 [B])/[H+]
At Isoelectric Point, [A] = [C]
([B][H+])/Ka1 = (Ka2 [B])/[H+]
Rearranging the above expressions we get the following
[H+]2 = Ka1 * Ka2
or pH = (pKa1 + pKa2) / 2
There are some amino acids which have an acid or basic side chain which will lead to an additional Ka value. In such cases, the average of the pKa values on either side of the zwitter ion is the isoelectric point.
In the above case Isoelectric point is the average of pKa2 and pKa3 values.
In the above case Isoelectric point is the average of pKa1 and pKa2 values.
Among MgSO4 and BaSO4 which one is more soluble in water at room temperature? Why?
MgSO4 is more soluble than BaSO4 in water at room temperature because among the sulphates of alkaline earth metals the decrease in hydration energy down the group is more rapid than the decrease in lattice energy down the group.