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integral fraction numerator s i n invisible function application 12 theta minus s i n invisible function application 9 theta over denominator 2 c o s invisible function application 7 theta minus 1 end fraction d theta equals to

  1. fraction numerator s i n invisible function application 5 theta over denominator 5 end fraction minus fraction numerator s i n invisible function application 2 theta over denominator 2 end fraction plus k    
  2. fraction numerator s i n invisible function application 2 theta over denominator 2 end fraction minus fraction numerator s i n invisible function application 5 theta over denominator 5 end fraction plus k    
  3. fraction numerator c o s invisible function application 5 theta over denominator 5 end fraction minus fraction numerator c o s invisible function application 2 theta over denominator 2 end fraction plus k    
  4. fraction numerator c o s invisible function application 2 theta over denominator 2 end fraction minus fraction numerator c o s invisible function application 5 theta over denominator 5 end fraction plus k    

The correct answer is: fraction numerator c o s invisible function application 2 theta over denominator 2 end fraction minus fraction numerator c o s invisible function application 5 theta over denominator 5 end fraction plus k


    integral fraction numerator s i n invisible function application 12 theta minus s i n invisible function application 9 theta over denominator 2 c o s invisible function application 7 theta minus 1 end fraction d theta equals integral fraction numerator left parenthesis 2 c o s invisible function application 7 theta minus 1 right parenthesis left parenthesis s i n invisible function application 5 theta minus s i n invisible function application 2 theta right parenthesis over denominator left parenthesis 2 c o s invisible function application 7 theta minus 1 right parenthesis end fraction d theta
    equals fraction numerator negative c o s invisible function application 5 theta over denominator 5 end fraction plus fraction numerator c o s invisible function application 2 theta over denominator 2 end fraction plus K

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    Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

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    Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:

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    Which of the following is/are correctly matched?

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    Energy level diagram for allylic, vinylic and alkylic, free radicals is .given below:

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    Which of the following is/are correctly matched?

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    Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

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    In the treatment of cyclohexene with NBS, which of the following products will be least stable?

    Karl Ziegler reported that alkenes react with N-bromosuccinimide (NBS) in presence of light to give products resulting from substitution of hydrogen by bromjne at the allylic position, i.e., the position next to the double bond. Let us consider the halogenation of cyclohexene.

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    The high reactivity of alkyl halides can be explained in terms of nature of C-X bond which is highly polarised covalent bond due to large difference in the electronegativities of carbon and halogen .atom. This polarity is responsible for the nucleophilic substitution reactions of alky- halides which mostly occur by S subscript N to the power of 1 end exponent end subscript and S subscript N to the power of 2 end exponent end subscript mechanisms S subscript N to the power of 1 end exponent end subscript reaction is a two step process . and in the first step, R-X ionises to give carbocation (slow process). In the second step, the nucleophile attacks the carbocation from either side to form the product (fast process). In S subscript N to the power of 1 end exponent end subscript reaction, there can be racemization and inversion. S subscript N to the power of 1 end exponent end subscript reaction is favoured by heavy (bulky) groups on the carbon atom attached to halogens. i.e., R subscript 3 end subscript C minus X greater than R subscript 2 end subscript C H minus X greater than R minus C H subscript 2 end subscript X greater than C H subscript 3 end subscript X
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    The main product formed in the following reaction is:

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    In S subscript N to the power of 2 end exponent end subscript reaction, the strong nucleophile OH- attacks from the opposite side of the chlorine atom to give an intermediate (transition state). which breaks to yield the product (alcohol) and leavj.ng open parentheses X to the power of minus end exponent close parentheses group. The alcohol has a configuration opposite to drat 'of the bromide and is said to proceed with inversion of configuration. S subscript N to the power of 2 end exponent end subscript reaction is favoured by small groups on the carbon atom attached to halogen i.e., C H subscript 3 end subscript minus X greater than R minus C H subscript 2 end subscript X greater than R subscript 2 end subscript C H X greater than R subscript 3 end subscript C minus X
    The main product formed in the following reaction is:

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    Compound  undergoes intramolecular nucleophilic substitution of the type .... giving .

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