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Questions with answers | Pericyclic, Stereochemistry

questions with solutions for CSIR NET GATE IIT JAM and other entrance examinations as per Pericyclic reaction and stereochemistry.
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Hello everyone today we will discuss a few questions with solutions for CSIR NET GATE IIT JAM and other entrance examinations as per. we have to solve the proper reason that is helpful for an entrance examination aspirant. the correct approach to two question solutions under pericyclic reactions and stereochemistry topic. These two topics are very widely discussed in another section that is provided link below. you have to must check it and better to understanding the approach to these two questions. Click Here.

Now, let’s begin solving the question-answer solution…

In the first and second questions, we can see

Pericyclic Reaction Questions with solution

Those questions are from CSIR NET DEC 2017. Those questions are previously asked in CSIR NET let’s solve this question and clear your approach to pericyclic reaction FMO theory in HOMO and LUMO concept of Symmetric principal. In presence of heat and photochemical conditions, the different types of FMO theory are produced in the section on photochemical reaction.

Stereochemistry Questions with solution

Now in looking forward to questioning number-1 you can see the stereochemistry of the Spiro compound that is given in the question. These questions are asked in CSIR NET December 2017. First of all, we talking about the spiro compound form in presence of a ring or a double bond.

Questions solve:

Those questions are easily solved, in case you have to clear the concept of pericyclic and stereochemistry chapters. must we try to understand the correct approach for solving a problem that helps you oo to solve any types of questions in the examinations hall? Firstly read the question carefully and apply the correct approach from the theoretical portion.

Q.1. The following molecule has

  1. plane of symmetry
  2. R configuration
  3. S configuration
  4. Center of symmetry
Question number 1 in stereochemistry question answer series.

Question Number one

Answer: Option-2: R configuration


Must be reminded that the system gives the even system, which means two rings, four rings, 6 rings, 8 rings …. so on, all systems are even. Similarly, in the Allene system, we can say that the event system has Two double bonds, four double bonds, six double bonds, and eight double bonds,… And so on, all are event systems.

This is the answer of Question number 1 in stereochemistry question answer series.

The answer to Question number 2 in the stereochemistry question answer series.

Note: First of all we are not in that condition between stereochemistry E showing the possibility (1) e must be even system (2) The site of head two groups are not same, corresponds to you in the tail site two groups are not same.

  • Satisfy the two conditions then show the stereochemistry R/S Otherwise E/Z.

Note: follow those conditions when processes occur numbering and find the configuration of the stereo compound (Chiral).

  • Numbering starts from the Tail.
  • Even system (1-ring, 1-double bond) Total=2; That mains stereochemistry posable.

The above structure gives the Clockwise Rotation 1-2-3-4, and 4-below ⇒”R configuration

Q.2. According to Frontier Molecular Orbital (FMO) Theory, the Highest Occupied Molecular Orbital (HOMO) of hexatriene in the following reaction is


Question number 2 in pericyclic reaction question answer series.

Question number two.

Answer: option 4


According to Frontier Molecular Orbital (FMO) Theory in presence of photochemical conditions ( i.e. hv) 6𝝅𝒆′ 𝒔 system number of node = n-1 where n= number of carbon/ number of pi electrons.

Hince, Node of 𝝍1 = 1-1=0; similarly node of 𝝍2 = 2-1 =1, node of 𝝍3 = 3-1 =2, node of 𝝍4 = 4-1 =3, node of 𝝍5 = 5-1 =4; and node of 𝝍6 = 6-1 =1;

This is the answer of Question number 2 in pericyclic reaction question answer series.
Answer Q.2. Pericyclic reaction in presence of the photochemical condition.

Now, in presence of photochemical conditions ( i.e. hv) 6𝝅𝒆′ 𝒔 system: HOMO is 𝝍4. That corresponds to node =3, which is satisfied with the option 4 orbital diagram.

  • In presence of hv 𝝍𝟒 is the HOMO
  • In 6𝝅𝒆′ 𝒔 Now, we can check the number of nodes then we satisfied the 𝝍𝟒 HOMO.
  • 𝝍_𝟒 no. of node = 4-1 =3

More details

We hope that those questions are suggestive for any national levels of examinations. first of all, we are discussing the topic-wise in another section.

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