Chemistry-
General
Easy

Question

Which statement(s) is/are wrong?

  1. Path I is Claisen-Schmidt rearrangement reaction, whereas Path II is Hofmann bromamide rearrangement reaction

  2. Both paths proceeds  the formation of acyl nitrene as an intermediate species 
  3. In Path I and Path II the intermediate compound formed is alkyl isocynate

  4. Both the paths proceed  the formation of nitrene  as a intermediate species

The correct answer is: Both the paths proceed  the formation of nitrene  as a intermediate species

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Solution to the inequation sin to the power of 6 space x plus cos to the power of 6 space x less than 7 over 16 must be

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The oxidation number of carboxylic carbon atom in CH subscript 3 COOH is:

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The oxidation number of carboxylic carbon atom in CH subscript 3 COOH is:

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If A plus B plus C equals pi comma then tan squared invisible function application A over 2 plus tan squared invisible function application B over 2 plus tan squared invisible function application C over 2 is always

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vertical line tan space x vertical line equals tan space x plus fraction numerator 1 over denominator cos begin display style space end style x end fraction left parenthesis 0 less or equal than x less or equal than 2 pi right parenthesis has

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In this question, we have to find the number of solution. In each quadrant tan value is different. Make two different case, one is where tan is positive ,[ 0 , π/2  ] U [π , 3 π/2 ] . And second case , tan is negative at [π/2 , π] U [3 π/2 , 2 π ] .

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Let a, b, c, d element of R. Then the cubic equation of the type a x cubed plus b x squared plus c x plus d equals 0 has either one root real or all three roots are real. But in case of trigonometric equations of the type sin cubed space x plus b sin squared space x plus c sin space x plus d equals 0 can possess several solutions depending upon the domain of x. To solve an equation of the type a cos space theta plus b sin space theta equals c. The equation can be written as cos space left parenthesis theta minus alpha right parenthesis equals c divided by square root of open parentheses a squared plus b squared close parentheses end root The solution is theta equals 2 straight n pi plus alpha plus-or-minus beta where tan space alpha = b divided by a comma cos space beta equals c divided by square root of open parentheses a squared plus b squared close parentheses end root

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cos to the power of 4 invisible function application pi over 8 plus cos to the power of 4 invisible function application fraction numerator 3 pi over denominator 8 end fraction plus cos to the power of 4 invisible function application fraction numerator 5 pi over denominator 8 end fraction plus cos to the power of 4 invisible function application fraction numerator 7 pi over denominator 8 end fraction equals

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