All questions of Pericyclic Reactions for Chemistry Exam

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

B and d are the same ryt!!...they just hav to be in opp stereo ...why is it d??

Explain plz

I think A is the right ans

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.

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

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.

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.

**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.

C

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.

Explanation:

• 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 [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.

C

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.

Answer B or C but not A.

Most Reactive Diene in Diels-Alder Reaction

In a Diels-Alder reaction, the most reactive diene is the one that can easily form a cyclic transition state with the dienophile. Let's analyze the given options to find out the most reactive diene.

a) (4E)-1,4-hexadiene

This diene has a trans-configuration, which means that the two double bonds are on opposite sides of the molecule. It can form a cyclic transition state with a dienophile, but the reaction may be hindered due to the trans-configuration.

b) (4Z)-1,4-hexadiene

This diene has a cis-configuration, which means that the two double bonds are on the same side of the molecule. It can easily form a cyclic transition state with a dienophile, but it may not be the most reactive due to steric hindrance.

c) (2E,4E)-2,4-hexadiene

This diene has two trans-double bonds, which means that they are on opposite sides of the molecule. It can easily form a cyclic transition state with a dienophile, and it is the most reactive diene because it has the least steric hindrance.

d) (2Z,4Z)-2,4-hexadiene

This diene has two cis-double bonds, which means that they are on the same side of the molecule. It can form a cyclic transition state with a dienophile, but it may not be the most reactive due to steric hindrance.

Therefore, the correct answer is option C, (2E,4E)-2,4-hexadiene, which is the most reactive diene in a Diels-Alder reaction.

Hayyy

Endo product always predominates that's why option B is correct one..

4n condition thermal so chane the stereo so cis gives trans opetion A is right

This is an example of cope rearrangement reaction.In cope rearrangement [3,3] sigmatropic shift occurs.

B

First apply ENE rexn then removal of AcOH

C