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Other Synthesis Methods

R 1- C ≡ N + ...

Question

$R_{1} - C \equiv N + R_{2} M g B r (i) D r y e t h e r - ------ \to (i i) H_{3} O^{+} 1 - p h e n y l - 1 - p r o p a n o n e$
$R_{1}$ and $R_{2}$ groups are respectively:

A

R_{1} = C_{2} H_{5}; R_{2} = C_{6} H_{5}

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B

R_{1} = C_{6} H_{5}; R_{2} = C H_{3}

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C

R_{1} = C_{6} H_{5}; R_{2} = C_{6} H_{5}

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D

R_{1} = C H_{3}; R_{2} = C_{6} H_{5}

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Solution

The correct option is A $R_{1} = C_{2} H_{5}; R_{2} = C_{6} H_{5}$
$R_{1} = C_{2} H_{5}; R_{2} = C_{6} H_{5}$

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Similar questions

R_{1} - C \equiv N + R_{2} M g B r (i) D r y e t h e r - ------ \to (i i) H_{3} O^{+} 1 - p h e n y l - 1 - p r o p a n o n e

R_{1}

R_{2}

groups are respectively:

Q. For the following reactions, the correct reaction rates

r_{1}, r_{2}, r_{1}^{'}

r_{2}^{'}

Let $c_{1} & c_{2}$ be the centers and $r_{1} & r_{2}$ be the radius of two circles. Then

Cases Conditions

p. 1. $| r_{1} - r_{2} | < c_{1} c_{2} < r_{1} + r_{2}$

q. 2. $| r_{1} - r_{2} | = c_{1} c_{2}$

r. 3. $c_{1} c_{2} < | r_{1} + r_{2} |$

s.

4. $r_{1} + r_{2} < c_{1} r_{2}$

t. 5. $r_{1} + r_{2} = c_{1} r_{2}$

Q. Two resistances

R_{1}

R_{2}

provide series to parallel equivalents as

\frac{n}{1}

. Then the correct relationship is

R_{1}

R_{2}

be two relations defined on a non empty set A. Which of the following statements is false? Give reasons in support of your answer.
(a) If

R_{1}

R_{2}

are reflexive, then so is

R_{1} \cap R_{2}

R_{1}

R_{2}

are symmetric, then so is

R_{1} \cup R_{2}

R_{1}

R_{2}

are transitive, then so is

R_{1} \cup R_{2}

R_{1}

R_{2}

are transitive, then so is

R_{1} \cap R_{2}

R_{1}

R_{2}

are symmetric, then so is

R_{1} \cap R_{2}

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