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All questions of Aromatic and Heterocyclic Chemistry for Chemistry Exam
The most acidic species is:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The most acidic species is:
a)
b)
c)
d)
|
Sarthak Rajendra Bande answered |
B is anti aromatic, so maximum acidic .moreover after losing Proton it will become aromatic so highly acidic
The major product of the following is:
- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The major product of the following is:
a)
b)
c)
d)
|
Dronacharya Institute answered |
The Claisen rearrangement is a powerful carbon–carbon bond-forming chemical reaction discovered by Rainer Ludwig Claisen. The heating of an allyl vinyl ether will initiate a [3,3]-sigmatropic rearrangement to give a γ,δ-unsaturated carbonyl.
Choose the correct answer from the following four choices.Statement: Pyridine is more basic than pyrrole.Reason: The nitrogen in pyrrole carries a proton while the nitrogen in pyridine does not.
Assertion: Nitrogen in trigonal geometry are generally more basic than nitrogens in a tetrahedral geometry. a)Both Reason and Assertion are correct.b)Both Reason and Assertion are wrong.c)Reason is correct and Assertion is wrong.d)Reason is wrong and Assertion is correct.The correct answer is option 'B'. Can you explain this answer?
Assertion: Nitrogen in trigonal geometry are generally more basic than nitrogens in a tetrahedral geometry.
|
Asf Institute answered |
- Pyridine is more basic than pyrrole because lone pair of electrons on N in pyridine and pyrrole are different in nature. These form a part of aromatic sextet in pyrrole, while not in pyridine.
- Pyrrole, C4H4NH (in which N contributes a lone pair) has a pKa−3.8 but pyridine (where N is part of the ring's double bond) has a pKa 5.14. Electron pair availability indicates the strength of basicity. In this case, pyridine is the stronger base.
- In pyrrole, the lone pair electrons of the nitrogen atom gets involved with the 2 carbon-carbon double bonds in the 5-member ring to form a conjugated system of pi electrons, leading to greater stability of the molecule.
- Pyridine, on the other hand, already has a stable conjugated system of 3 double bonds in the aromatic hexagonal ring, like benzene. Hence, the lone pair electrons on the N atom in pyridine can be easily donated to a H+ ion or a Lewis acid.
Pyridine undergoes electrophilic nitration at elevated temperature to given the following as a major product:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Pyridine undergoes electrophilic nitration at elevated temperature to given the following as a major product:
a)
b)
c)
d)
|
|
Dronacharya Institute answered |
There is a relatively less partial positive charge on meta positions of Pyridine. Hence, when an electrophile attacks it attacks on meta position.
Hence D is correct.
Hence D is correct.
Order of reactivity of the following X, Y and Z in Diels-Alder reaction is:
- a)X > Z > Y
- b)Y > X > Z
- c)Y > Z > X
- d)X > Y > Z
Correct answer is option 'D'. Can you explain this answer?
Order of reactivity of the following X, Y and Z in Diels-Alder reaction is:
a)
X > Z > Y
b)
Y > X > Z
c)
Y > Z > X
d)
X > Y > Z
|
Shoaib Ali answered |
X is in cic position so react easily, Z is in trans position so first heat it and than Z react, so in between Z and Y 'Y react easily than Z, so option D correct
Can you explain the answer of this question below:The major product of the reaction is:
- A:
- B:
- C:
- D:
The answer is d.
The major product of the reaction is:
|
Parvathy Vs answered |
B and d are the same ryt!!...they just hav to be in opp stereo ...why is it d??
The correct order of the basicity of the following compound is:
- a)IV > III > I > II
- b)III > VI > I > II
- c)IV > III > II > I
- d)III > VI > II > I
Correct answer is option 'A'. Can you explain this answer?
The correct order of the basicity of the following compound is:
a)
IV > III > I > II
b)
III > VI > I > II
c)
IV > III > II > I
d)
III > VI > II > I
|
|
Subha Som answered |
In iv no double bond present so lp is fully available
iii the lp does not undergO resonance
ii is less πexess than i
iii the lp does not undergO resonance
ii is less πexess than i
The most basic compound among the following is:-- a)Acetanilide
- b)Benzylamine
- c)p-Nitro aniline
- d)Aniline
Correct answer is option 'B'. Can you explain this answer?
The most basic compound among the following is:-
a)
Acetanilide
b)
Benzylamine
c)
p-Nitro aniline
d)
Aniline
|
|
Asf Institute answered |
Basicity is inversely proportional to resonance of lone pair electrons. Benzylamine is more basic. The electron pairs do not involve in resonance in benzylamine. In other amines, there is delocalization of lone pair of electron on N atom on the ring. In acetanilide, the delocalization of lone pair of electrons on N atom is due to adjacent CO group.
Hence option (B) is the answer.
Hence option (B) is the answer.
The decreasing order of the reactivity of the following compounds towards electrophiles is:
- a)II > I > III
- b)II > III > I
- c)III > I > II
- d)I > II > III
Correct answer is option 'D'. Can you explain this answer?
The decreasing order of the reactivity of the following compounds towards electrophiles is:
a)
II > I > III
b)
II > III > I
c)
III > I > II
d)
I > II > III
|
Pooja Choudhury answered |
Pyrole is most reactive towards the electrophiles because in this case it is carrying one lone pair of electron .and it is attached with one hydrogen atom , resulting less resonance stabilize than that of pyridine one the another hand thiophene is more reactive because it's carry 2 lone pair of electrons at last pyrole is more reactive than thiophene because nitrogen is smaller in size than that of sulphur so the availability of electrons cloud is maximum than sulphur ... consequently it is highly reactive towards electrophiles
The heterocyclic diene employed in cyclo – addition reactions is:- a)Furan
- b)Pyrrole
- c)2, 5-dimethylpyrrole
- d)Thiophene
Correct answer is 'A'. Can you explain this answer?
The heterocyclic diene employed in cyclo – addition reactions is:
a)
Furan
b)
Pyrrole
c)
2, 5-dimethylpyrrole
d)
Thiophene
|
Anirban Kapoor answered |
Heterocyclic diene employed in cyclo addition reactions is Furan.
Explanation:
Cycloaddition reactions involve the formation of a cyclic compound from two or more reactants. In this reaction, a diene and a dienophile undergo a cycloaddition reaction to form a cyclic compound. The diene used in this reaction is usually a heterocyclic diene, which contains at least one heteroatom (non-carbon atom) in the ring structure.
Furan is a heterocyclic diene that contains an oxygen atom in its five-membered ring structure. Furan is an excellent diene for cycloaddition reactions because it readily undergoes a Diels-Alder reaction with a dienophile to form a cyclic compound. Furan is also a useful diene because it is readily available and can be easily functionalized to give a wide range of derivatives.
Pyrrole, 2,5-dimethylpyrrole, and thiophene are also heterocyclic dienes that are used in cycloaddition reactions, but they are less commonly used than furan. Pyrrole and 2,5-dimethylpyrrole are less reactive than furan because they contain a nitrogen atom in the ring structure, which decreases the electron density of the diene. Thiophene is less reactive than furan because it contains a sulfur atom in the ring structure, which decreases the electron density of the diene and makes it more prone to oxidation.
In summary, furan is the preferred heterocyclic diene for cycloaddition reactions because it is readily available, easily functionalized, and highly reactive.
Explanation:
Cycloaddition reactions involve the formation of a cyclic compound from two or more reactants. In this reaction, a diene and a dienophile undergo a cycloaddition reaction to form a cyclic compound. The diene used in this reaction is usually a heterocyclic diene, which contains at least one heteroatom (non-carbon atom) in the ring structure.
Furan is a heterocyclic diene that contains an oxygen atom in its five-membered ring structure. Furan is an excellent diene for cycloaddition reactions because it readily undergoes a Diels-Alder reaction with a dienophile to form a cyclic compound. Furan is also a useful diene because it is readily available and can be easily functionalized to give a wide range of derivatives.
Pyrrole, 2,5-dimethylpyrrole, and thiophene are also heterocyclic dienes that are used in cycloaddition reactions, but they are less commonly used than furan. Pyrrole and 2,5-dimethylpyrrole are less reactive than furan because they contain a nitrogen atom in the ring structure, which decreases the electron density of the diene. Thiophene is less reactive than furan because it contains a sulfur atom in the ring structure, which decreases the electron density of the diene and makes it more prone to oxidation.
In summary, furan is the preferred heterocyclic diene for cycloaddition reactions because it is readily available, easily functionalized, and highly reactive.
The compound that is NOT oxidized by KMnO4 is:- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
The compound that is NOT oxidized by KMnO4 is:
a)
b)
c)
d)
|
|
Shoaib Ali answered |
If head carbon does not have any hydrogen than kmno4 doesn't oxidise this,In option D carbon doesn't have any hydrogen
The number of nodes present in the highest occupied molecular orbital of 1, 3, 5-hexatiriene in its ground state in thermal conditions is:
- a)One
- b)Two
- c)Three
- d)Four
Correct answer is option 'B'. Can you explain this answer?
The number of nodes present in the highest occupied molecular orbital of 1, 3, 5-hexatiriene in its ground state in thermal conditions is:
a)
One
b)
Two
c)
Three
d)
Four
|
|
Athul Chaudhary answered |
Explanation:
1, 3, 5-hexatriene is a conjugated hydrocarbon molecule. The highest occupied molecular orbital (HOMO) is the molecular orbital with the highest energy level that is occupied by electrons. In the ground state, the electrons fill the molecular orbitals in order of increasing energy levels.
The HOMO of 1, 3, 5-hexatriene is a pi molecular orbital. The number of nodes present in this orbital can be determined using the following formula:
Number of nodes = (n - 1) - m
Where n is the total number of atomic orbitals that combine to form the molecular orbital and m is the number of nodal planes in the molecular orbital.
For the pi molecular orbital of 1, 3, 5-hexatriene, n = 6 (since there are six carbon atoms) and m = 2 (since there are two nodal planes). Therefore, the number of nodes in the pi molecular orbital of 1, 3, 5-hexatriene is:
(6 - 1) - 2 = 3
So the correct answer is option B, which states that there are two nodes present in the HOMO of 1, 3, 5-hexatriene.
1, 3, 5-hexatriene is a conjugated hydrocarbon molecule. The highest occupied molecular orbital (HOMO) is the molecular orbital with the highest energy level that is occupied by electrons. In the ground state, the electrons fill the molecular orbitals in order of increasing energy levels.
The HOMO of 1, 3, 5-hexatriene is a pi molecular orbital. The number of nodes present in this orbital can be determined using the following formula:
Number of nodes = (n - 1) - m
Where n is the total number of atomic orbitals that combine to form the molecular orbital and m is the number of nodal planes in the molecular orbital.
For the pi molecular orbital of 1, 3, 5-hexatriene, n = 6 (since there are six carbon atoms) and m = 2 (since there are two nodal planes). Therefore, the number of nodes in the pi molecular orbital of 1, 3, 5-hexatriene is:
(6 - 1) - 2 = 3
So the correct answer is option B, which states that there are two nodes present in the HOMO of 1, 3, 5-hexatriene.
Thiophene reacts with HCHO in presence of aqueous HCl to give:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Thiophene reacts with HCHO in presence of aqueous HCl to give:
a)
b)
c)
d)
|
|
Yash Roy answered |
The CH2O/HCl reagent is suitable for chloromethylation of various polycyclic aromatic hydrocarbons as well as heterocyclic aromatic compounds. Thiophene gives 2-(chloromethyl)thiophene in 40-41% isolated yield, and 4-methylimidazole gives the 5-chloromethyl derivative (51-68%).
With respect to electrophilic aromatic substitution, reactivity order of pyrrole, pyridine and indole is:- a)Indole > Pyrrole > Pyridine
- b)Pyrrole > Pyridine > Indole
- c)Pyrrole > Indole > Pyridine
- d)Indole > Pyridine > Pyrrole
Correct answer is option 'C'. Can you explain this answer?
With respect to electrophilic aromatic substitution, reactivity order of pyrrole, pyridine and indole is:
a)
Indole > Pyrrole > Pyridine
b)
Pyrrole > Pyridine > Indole
c)
Pyrrole > Indole > Pyridine
d)
Indole > Pyridine > Pyrrole
|
Sagarika Patel answered |
The aromatic lone pair owing to nitrogen in pyrrole makes the aromatic system more electron rich, and you can get a hint of it by looking at the electron pushing mechanism there.
The electron pushing from pyridine fails, firstly because the lone pair is fixed at an orthogonal position away from the aromatic ring. Secondly, that particular intermmediate cannot be obtained because of structural geometry problems, (=N=)+ should be linear and cannot be formed while maintaining a ring.
In which of the following reaction, which sigmatropic reaction will take place:
- a) [1, 5] CH3 shift
- b) [1, 5] H shift
- c) [1, 3] CH3 shift
- d) [1, 3] H shift
Correct answer is option 'D'. Can you explain this answer?
In which of the following reaction, which sigmatropic reaction will take place:
a)
[1, 5] CH3 shift b)
[1, 5] H shift c)
[1, 3] CH3 shift d)
[1, 3] H shift
|
Yogendra Rajpoot answered |
Option b is correct because reaction is thermally take place and 1,3 hydrogen shifting takes place photochemically only and there is two time 1,5 hydrogen shifting takes place which take place thermally
The major product of the following reaction is:
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
The major product of the following reaction is:
a)
b)
c)
d)
|
Uma Bharti answered |
Here 3 pi bond invlove...means 4n +2 system in con..in substrate both double bond is trans overall cis .con chnge in stereo so pdt gives trans stereo
The major product of the reaction is:
- a)
- b)
- c)
- d)No product formation
Correct answer is option 'B'. Can you explain this answer?
The major product of the reaction is:
a)
b)
c)
d)
No product formation
|
|
Madhumita Chakraborty answered |
Structure 1 reacts with structure 2 then we get structure B as major cause here Hydrogen placed on the below of atom hence hydrogen bonding is possible between oxygen and hydrogen O - H which is not possible in other cases thus why we get B as major product.
The thermal ring opening reactions of cyclo butane are:- a)Conrotatory
- b)Disrotatory
- c)Conrotatory or disrotatory depending on the temperature at which reaction is carried
- d)Cannot be predicted
Correct answer is option 'A'. Can you explain this answer?
The thermal ring opening reactions of cyclo butane are:
a)
Conrotatory
b)
Disrotatory
c)
Conrotatory or disrotatory depending on the temperature at which reaction is carried
d)
Cannot be predicted
|
Anushka Chavan answered |
Thermal Ring Opening Reactions of Cyclo Butane
Introduction:
Cyclo butane is a cyclic alkane with four carbon atoms in a ring. It is known for its high ring strain due to its puckered structure. When cyclo butane undergoes a thermal ring opening reaction, the mechanism can either be conrotatory or disrotatory.
Conrotatory Mechanism:
In the conrotatory mechanism, the breaking of the sigma bonds in the ring occurs in a synchronous manner. This means that the two carbon-carbon sigma bonds on either side of the ring break at the same time, and the two resulting electrons move in the same direction. This leads to the formation of a planar intermediate with a pi bond. The pi bond then rotates, and the two carbon atoms on either side of the ring come closer together to form a new sigma bond. This results in the formation of an open-chain alkene.
Disrotatory Mechanism:
In the disrotatory mechanism, the breaking of the sigma bonds in the ring occurs in an asynchronous manner. This means that the two carbon-carbon sigma bonds on either side of the ring break at different times, and the two resulting electrons move in opposite directions. This leads to the formation of a twisted intermediate with two pi bonds. The pi bonds then rotate, and the two carbon atoms on either side of the ring move further apart to form two new sigma bonds. This results in the formation of an open-chain diene.
Temperature Dependence:
The mechanism of the thermal ring opening reaction of cyclo butane depends on the temperature at which the reaction is carried out. At high temperatures, the reaction is conrotatory, while at low temperatures, the reaction is disrotatory. This is due to the fact that the activation energy for the conrotatory mechanism is lower at high temperatures, while the activation energy for the disrotatory mechanism is lower at low temperatures.
Conclusion:
In conclusion, the thermal ring opening reaction of cyclo butane can occur via either a conrotatory or disrotatory mechanism, depending on the temperature at which the reaction is carried out. At high temperatures, the reaction is conrotatory, while at low temperatures, the reaction is disrotatory.
Introduction:
Cyclo butane is a cyclic alkane with four carbon atoms in a ring. It is known for its high ring strain due to its puckered structure. When cyclo butane undergoes a thermal ring opening reaction, the mechanism can either be conrotatory or disrotatory.
Conrotatory Mechanism:
In the conrotatory mechanism, the breaking of the sigma bonds in the ring occurs in a synchronous manner. This means that the two carbon-carbon sigma bonds on either side of the ring break at the same time, and the two resulting electrons move in the same direction. This leads to the formation of a planar intermediate with a pi bond. The pi bond then rotates, and the two carbon atoms on either side of the ring come closer together to form a new sigma bond. This results in the formation of an open-chain alkene.
Disrotatory Mechanism:
In the disrotatory mechanism, the breaking of the sigma bonds in the ring occurs in an asynchronous manner. This means that the two carbon-carbon sigma bonds on either side of the ring break at different times, and the two resulting electrons move in opposite directions. This leads to the formation of a twisted intermediate with two pi bonds. The pi bonds then rotate, and the two carbon atoms on either side of the ring move further apart to form two new sigma bonds. This results in the formation of an open-chain diene.
Temperature Dependence:
The mechanism of the thermal ring opening reaction of cyclo butane depends on the temperature at which the reaction is carried out. At high temperatures, the reaction is conrotatory, while at low temperatures, the reaction is disrotatory. This is due to the fact that the activation energy for the conrotatory mechanism is lower at high temperatures, while the activation energy for the disrotatory mechanism is lower at low temperatures.
Conclusion:
In conclusion, the thermal ring opening reaction of cyclo butane can occur via either a conrotatory or disrotatory mechanism, depending on the temperature at which the reaction is carried out. At high temperatures, the reaction is conrotatory, while at low temperatures, the reaction is disrotatory.
In the [4 + 2] cycloaddition of 1, 3-butadiene and ethylene:- a)Overlap of the HOMO of butadiene with the LUMO of ethylene.
- b)Overlap of the HOMO of ethylene with the LUMO of butadiene.
- c)Both (a) and (b)
- d)None of these.
Correct answer is option 'C'. Can you explain this answer?
In the [4 + 2] cycloaddition of 1, 3-butadiene and ethylene:
a)
Overlap of the HOMO of butadiene with the LUMO of ethylene.
b)
Overlap of the HOMO of ethylene with the LUMO of butadiene.
c)
Both (a) and (b)
d)
None of these.
|
|
Bijoy Patel answered |
**Explanation:**
The [4 + 2] cycloaddition reaction between 1,3-butadiene and ethylene involves the overlap of the highest occupied molecular orbital (HOMO) of one reactant with the lowest unoccupied molecular orbital (LUMO) of the other reactant. Let's analyze the overlap of the HOMO and LUMO orbitals in this reaction:
**a) Overlap of the HOMO of butadiene with the LUMO of ethylene:**
- The HOMO of butadiene is a π molecular orbital, formed by the overlap of the p orbitals of the carbon atoms in the conjugated system. It consists of two nodal planes, with electron density both above and below the π system.
- The LUMO of ethylene is also a π molecular orbital, formed by the overlap of the p orbitals of the carbon atoms in the double bond. It consists of a single nodal plane in the middle of the π system, with electron density on either side of the plane.
- The overlap of the HOMO of butadiene with the LUMO of ethylene occurs when the π system of butadiene approaches the double bond of ethylene.
- The electron density in the HOMO of butadiene interacts with the electron-deficient region of the LUMO of ethylene, resulting in the formation of a new σ bond between the two reactants.
- This overlap leads to the formation of a cyclohexene derivative as the product of the [4 + 2] cycloaddition reaction.
**b) Overlap of the HOMO of ethylene with the LUMO of butadiene:**
- The HOMO of ethylene is also a π molecular orbital, formed by the overlap of the p orbitals of the carbon atoms in the double bond. It consists of a single nodal plane in the middle of the π system, with electron density on either side of the plane.
- The LUMO of butadiene is a π* molecular orbital, formed by the overlap of the p orbitals of the carbon atoms in the conjugated system. It also consists of two nodal planes, with electron density both above and below the π system.
- The overlap of the HOMO of ethylene with the LUMO of butadiene occurs when the double bond of ethylene approaches the π system of butadiene.
- However, this overlap does not lead to the formation of a new bond or the [4 + 2] cycloaddition product. Instead, it leads to a non-productive interaction between the two reactants.
**c) Both (a) and (b):**
- As discussed above, the overlap of the HOMO of butadiene with the LUMO of ethylene results in the formation of a new σ bond and the [4 + 2] cycloaddition product.
- On the other hand, the overlap of the HOMO of ethylene with the LUMO of butadiene does not lead to the formation of a new bond or the desired cycloaddition product.
- Therefore, the correct answer is option 'c' - both (a) and (b) - as the productive overlap occurs between the HOMO of butadiene and the LUMO of ethylene.
The [4 + 2] cycloaddition reaction between 1,3-butadiene and ethylene involves the overlap of the highest occupied molecular orbital (HOMO) of one reactant with the lowest unoccupied molecular orbital (LUMO) of the other reactant. Let's analyze the overlap of the HOMO and LUMO orbitals in this reaction:
**a) Overlap of the HOMO of butadiene with the LUMO of ethylene:**
- The HOMO of butadiene is a π molecular orbital, formed by the overlap of the p orbitals of the carbon atoms in the conjugated system. It consists of two nodal planes, with electron density both above and below the π system.
- The LUMO of ethylene is also a π molecular orbital, formed by the overlap of the p orbitals of the carbon atoms in the double bond. It consists of a single nodal plane in the middle of the π system, with electron density on either side of the plane.
- The overlap of the HOMO of butadiene with the LUMO of ethylene occurs when the π system of butadiene approaches the double bond of ethylene.
- The electron density in the HOMO of butadiene interacts with the electron-deficient region of the LUMO of ethylene, resulting in the formation of a new σ bond between the two reactants.
- This overlap leads to the formation of a cyclohexene derivative as the product of the [4 + 2] cycloaddition reaction.
**b) Overlap of the HOMO of ethylene with the LUMO of butadiene:**
- The HOMO of ethylene is also a π molecular orbital, formed by the overlap of the p orbitals of the carbon atoms in the double bond. It consists of a single nodal plane in the middle of the π system, with electron density on either side of the plane.
- The LUMO of butadiene is a π* molecular orbital, formed by the overlap of the p orbitals of the carbon atoms in the conjugated system. It also consists of two nodal planes, with electron density both above and below the π system.
- The overlap of the HOMO of ethylene with the LUMO of butadiene occurs when the double bond of ethylene approaches the π system of butadiene.
- However, this overlap does not lead to the formation of a new bond or the [4 + 2] cycloaddition product. Instead, it leads to a non-productive interaction between the two reactants.
**c) Both (a) and (b):**
- As discussed above, the overlap of the HOMO of butadiene with the LUMO of ethylene results in the formation of a new σ bond and the [4 + 2] cycloaddition product.
- On the other hand, the overlap of the HOMO of ethylene with the LUMO of butadiene does not lead to the formation of a new bond or the desired cycloaddition product.
- Therefore, the correct answer is option 'c' - both (a) and (b) - as the productive overlap occurs between the HOMO of butadiene and the LUMO of ethylene.
|
Gaurav Singh answered |
Benzyne mechanism followed by diels alder the ...after treating the aduct with acid it will give the napthalene
Identify the major Product P in the following two–step reaction:
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Identify the major Product P in the following two–step reaction:
a)
b)
c)
d)
|
|
Ajit Yadav answered |
Firstly pyrrole undergoes electrophilic substitution reaction preferentially at c 2 then resulting give meta directing and then nitration give product b
Diels-Alder reaction normally yields endo-adduct as a major product. This is due to:- a) Higher stability of the product.
- b) Faster rate of formation of the endo product.
- c) Steric hindrance.
- d) Secondary orbital interactions between a diene and a dienophile.
Correct answer is option 'D'. Can you explain this answer?
Diels-Alder reaction normally yields endo-adduct as a major product. This is due to:
a)
Higher stability of the product.
b)
Faster rate of formation of the endo product.
c)
Steric hindrance.
d)
Secondary orbital interactions between a diene and a dienophile.
|
|
Saranya Mehta answered |
Introduction:
The Diels-Alder reaction is a powerful synthetic tool used to construct cyclic compounds. It involves the reaction between a conjugated diene and a dienophile to form a cyclic product known as the cycloadduct. In most cases, the major product obtained in the Diels-Alder reaction is the endo-adduct. This preference for the endo product can be explained by several factors.
Explanation:
1. Secondary orbital interactions:
The endo-adduct is favored due to the presence of secondary orbital interactions between the diene and dienophile. These interactions occur between the non-bonding electron pair on the diene and the empty orbital on the dienophile. This stabilizes the transition state leading to the formation of the endo product. The interaction is more favorable in the endo position compared to the exo position, leading to the preference for the endo-adduct.
2. Steric hindrance:
Steric hindrance also plays a role in determining the regioselectivity of the Diels-Alder reaction. The endo-product is often favored due to the lower steric hindrance in the transition state leading to its formation. The exo-product is hindered by the presence of bulky substituents that can lead to steric clashes, making the formation of the endo-product more favorable.
3. Stability of the product:
The endo-adduct is generally more stable than the exo-adduct due to the spatial arrangement of substituents. The endo-product often has substituents arranged in a more favorable conformation, leading to lower steric strain and increased stability. This stability contributes to the preference for the endo product in the Diels-Alder reaction.
4. Faster rate of formation:
In some cases, the endo-product may also be favored due to the faster rate of formation. The transition state leading to the endo-product may have a lower activation energy compared to the transition state leading to the exo-product. This kinetic preference can be attributed to a combination of factors, including orbital interactions and steric effects.
Conclusion:
In conclusion, the preference for the endo-adduct in the Diels-Alder reaction can be attributed to secondary orbital interactions, steric hindrance, stability of the product, and sometimes a faster rate of formation. These factors work together to determine the regioselectivity of the reaction, leading to the predominant formation of the endo product.
The Diels-Alder reaction is a powerful synthetic tool used to construct cyclic compounds. It involves the reaction between a conjugated diene and a dienophile to form a cyclic product known as the cycloadduct. In most cases, the major product obtained in the Diels-Alder reaction is the endo-adduct. This preference for the endo product can be explained by several factors.
Explanation:
1. Secondary orbital interactions:
The endo-adduct is favored due to the presence of secondary orbital interactions between the diene and dienophile. These interactions occur between the non-bonding electron pair on the diene and the empty orbital on the dienophile. This stabilizes the transition state leading to the formation of the endo product. The interaction is more favorable in the endo position compared to the exo position, leading to the preference for the endo-adduct.
2. Steric hindrance:
Steric hindrance also plays a role in determining the regioselectivity of the Diels-Alder reaction. The endo-product is often favored due to the lower steric hindrance in the transition state leading to its formation. The exo-product is hindered by the presence of bulky substituents that can lead to steric clashes, making the formation of the endo-product more favorable.
3. Stability of the product:
The endo-adduct is generally more stable than the exo-adduct due to the spatial arrangement of substituents. The endo-product often has substituents arranged in a more favorable conformation, leading to lower steric strain and increased stability. This stability contributes to the preference for the endo product in the Diels-Alder reaction.
4. Faster rate of formation:
In some cases, the endo-product may also be favored due to the faster rate of formation. The transition state leading to the endo-product may have a lower activation energy compared to the transition state leading to the exo-product. This kinetic preference can be attributed to a combination of factors, including orbital interactions and steric effects.
Conclusion:
In conclusion, the preference for the endo-adduct in the Diels-Alder reaction can be attributed to secondary orbital interactions, steric hindrance, stability of the product, and sometimes a faster rate of formation. These factors work together to determine the regioselectivity of the reaction, leading to the predominant formation of the endo product.
The major product of the following is:
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The major product of the following is:
a)
b)
c)
d)
|
Vedika Singh answered |
Six membered transition state of chair form will give option A.
Consider the following statements:(I) Clasisen rearrangement is a [3, 3] sigmatropic rearrangement.
(II) Cope rearrangement proceeds via is chair like transition state.
(III) In the photochemical [2 + 2] cycloaddition of ethylene overlapping of HOMO of one molecule with LUMO of the other molecular takes place.Select the correct statements.- a)I and III
- b)II and III
- c)Only II
- d)I and II
Correct answer is option 'D'. Can you explain this answer?
Consider the following statements:
(I) Clasisen rearrangement is a [3, 3] sigmatropic rearrangement.
(II) Cope rearrangement proceeds via is chair like transition state.
(III) In the photochemical [2 + 2] cycloaddition of ethylene overlapping of HOMO of one molecule with LUMO of the other molecular takes place.
(II) Cope rearrangement proceeds via is chair like transition state.
(III) In the photochemical [2 + 2] cycloaddition of ethylene overlapping of HOMO of one molecule with LUMO of the other molecular takes place.
Select the correct statements.
a)
I and III
b)
II and III
c)
Only II
d)
I and II
|
|
Bijoy Patel answered |
Explanation:
Therefore, the correct option is 'D' - I and II.
- Claisen rearrangement: It is a [3, 3] sigmatropic rearrangement, which involves the migration of an allyl group from one oxygen atom to another oxygen atom, with the formation of a new carbon-carbon bond. Hence, statement I is correct.
- Cope rearrangement: It is a [3, 3] sigmatropic rearrangement, which involves the migration of a hydrocarbon group from one carbon atom to an adjacent carbon atom, with the formation of a new carbon-carbon bond. This reaction proceeds via a chair-like transition state. Hence, statement II is correct.
- Photochemical [2+2] cycloaddition: This reaction involves the formation of a cyclobutane ring from two molecules of ethylene under the influence of light. In this reaction, the highest occupied molecular orbital (HOMO) of one molecule overlaps with the lowest unoccupied molecular orbital (LUMO) of the other molecule. Hence, statement III is also correct.
Therefore, the correct option is 'D' - I and II.
Claisen rearrangement is an example of:- a)[2, 3] sigmatropic rearrangement.
- b)[2, 4] sigmatropic rearrangement.
- c)[1, 5] sigmatropic rearrangement.
- d)[3, 3] sigmatropic rearrangement.
Correct answer is option 'D'. Can you explain this answer?
Claisen rearrangement is an example of:
a)
[2, 3] sigmatropic rearrangement.
b)
[2, 4] sigmatropic rearrangement.
c)
[1, 5] sigmatropic rearrangement.
d)
[3, 3] sigmatropic rearrangement.
|
|
Ameya Reddy answered |
Claisen rearrangement is an example of [3,3] sigmatropic rearrangement.
Explanation:
- Rearrangement reactions involve the movement of atoms or groups within a molecule, resulting in a new structure.
- Sigmatropic rearrangement is a type of rearrangement reaction that involves the simultaneous breaking and forming of sigma bonds.
- In Claisen rearrangement, a vinyl ether undergoes rearrangement to form an allyl ether, with migration of a carbon-carbon double bond from one position to another within the molecule.
- The reaction involves the simultaneous breaking and forming of three sigma bonds, hence it is a [3,3] sigmatropic rearrangement.
- Other examples of sigmatropic rearrangements include [2,3], [3,3], [1,3], [1,5], [2,5], etc. depending on the number of atoms involved in the rearrangement and the direction of migration.
In summary, Claisen rearrangement is an example of [3,3] sigmatropic rearrangement, which involves the simultaneous breaking and forming of three sigma bonds.
Explanation:
- Rearrangement reactions involve the movement of atoms or groups within a molecule, resulting in a new structure.
- Sigmatropic rearrangement is a type of rearrangement reaction that involves the simultaneous breaking and forming of sigma bonds.
- In Claisen rearrangement, a vinyl ether undergoes rearrangement to form an allyl ether, with migration of a carbon-carbon double bond from one position to another within the molecule.
- The reaction involves the simultaneous breaking and forming of three sigma bonds, hence it is a [3,3] sigmatropic rearrangement.
- Other examples of sigmatropic rearrangements include [2,3], [3,3], [1,3], [1,5], [2,5], etc. depending on the number of atoms involved in the rearrangement and the direction of migration.
In summary, Claisen rearrangement is an example of [3,3] sigmatropic rearrangement, which involves the simultaneous breaking and forming of three sigma bonds.
Hydrogen bonding is maximum in:
- a)Ethanol
- b)Ethyl chloride
- c)Diethyl ether
- d)Triethylamine
Correct answer is option 'A'. Can you explain this answer?
Hydrogen bonding is maximum in:
a)
Ethanol
b)
Ethyl chloride
c)
Diethyl ether
d)
Triethylamine
|
|
Shail Ghoshal answered |
Hydrogen bonding is maximum in ethanol.
Explanation:
What is hydrogen bonding?
Hydrogen bonding is a special type of intermolecular force that occurs between molecules containing hydrogen atoms bonded to highly electronegative elements such as oxygen, nitrogen, or fluorine. It is a strong dipole-dipole interaction that results in the formation of a partially positive hydrogen atom and a partially negative atom.
Factors affecting hydrogen bonding:
There are several factors that influence the strength of hydrogen bonding, including:
1. Electronegativity difference: The greater the electronegativity difference between hydrogen and the atom it is bonded to, the stronger the hydrogen bonding.
2. Size of the atom: Smaller atoms with higher electronegativity can form stronger hydrogen bonds.
3. Number of hydrogen bonding sites: The more hydrogen atoms bonded to electronegative atoms, the more hydrogen bonding sites are available.
4. Molecular shape: The presence of hydrogen bonding sites in a linear or planar arrangement increases the strength of hydrogen bonding.
Comparison of the given compounds:
a) Ethanol: Ethanol (CH3CH2OH) contains an -OH group, which can form hydrogen bonds. The electronegative oxygen atom attracts the hydrogen atom, resulting in the formation of strong hydrogen bonds between ethanol molecules.
b) Ethyl chloride: Ethyl chloride (CH3CH2Cl) does not contain any hydrogen bonding sites. The chlorine atom is not electronegative enough to form strong hydrogen bonds.
c) Diethyl ether: Diethyl ether (CH3CH2OCH2CH3) contains an oxygen atom, but it does not have any hydrogen bonding sites. The oxygen atom is not bonded to a hydrogen atom.
d) Triethylamine: Triethylamine (N(C2H5)3) does not contain any hydrogen bonding sites. Although it contains a nitrogen atom, it is not bonded to a hydrogen atom.
Conclusion:
Among the given compounds, ethanol has the maximum hydrogen bonding because it contains an -OH group, which can form strong hydrogen bonds. The other compounds do not have hydrogen bonding sites or have less electronegative atoms, resulting in weaker or no hydrogen bonding.
Explanation:
What is hydrogen bonding?
Hydrogen bonding is a special type of intermolecular force that occurs between molecules containing hydrogen atoms bonded to highly electronegative elements such as oxygen, nitrogen, or fluorine. It is a strong dipole-dipole interaction that results in the formation of a partially positive hydrogen atom and a partially negative atom.
Factors affecting hydrogen bonding:
There are several factors that influence the strength of hydrogen bonding, including:
1. Electronegativity difference: The greater the electronegativity difference between hydrogen and the atom it is bonded to, the stronger the hydrogen bonding.
2. Size of the atom: Smaller atoms with higher electronegativity can form stronger hydrogen bonds.
3. Number of hydrogen bonding sites: The more hydrogen atoms bonded to electronegative atoms, the more hydrogen bonding sites are available.
4. Molecular shape: The presence of hydrogen bonding sites in a linear or planar arrangement increases the strength of hydrogen bonding.
Comparison of the given compounds:
a) Ethanol: Ethanol (CH3CH2OH) contains an -OH group, which can form hydrogen bonds. The electronegative oxygen atom attracts the hydrogen atom, resulting in the formation of strong hydrogen bonds between ethanol molecules.
b) Ethyl chloride: Ethyl chloride (CH3CH2Cl) does not contain any hydrogen bonding sites. The chlorine atom is not electronegative enough to form strong hydrogen bonds.
c) Diethyl ether: Diethyl ether (CH3CH2OCH2CH3) contains an oxygen atom, but it does not have any hydrogen bonding sites. The oxygen atom is not bonded to a hydrogen atom.
d) Triethylamine: Triethylamine (N(C2H5)3) does not contain any hydrogen bonding sites. Although it contains a nitrogen atom, it is not bonded to a hydrogen atom.
Conclusion:
Among the given compounds, ethanol has the maximum hydrogen bonding because it contains an -OH group, which can form strong hydrogen bonds. The other compounds do not have hydrogen bonding sites or have less electronegative atoms, resulting in weaker or no hydrogen bonding.
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