In which of the following pairs is $A$ more stable than $B$?
$A$ $B$

  • A
    Option A
  • B
    Option B
  • C
    Option C
  • D
    $Ph_{3}\dot{C}, (CH_{3})_{3}\dot{C}$

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Arrange the following anions in decreasing order of their stability:
$CH_3^{\Theta}, NH_2^{\Theta}, OH^{\Theta}, Cl^{\Theta}$

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Rank the transition states that occur during the following reaction steps in order of increasing stability (least $\to$ most stable):
$1.$ $CH_3-O^+H_2 \to CH_3^+ + H_2O$
$2.$ $(CH_3)_3C-O^+H_2 \to (CH_3)_3C^+ + H_2O$
$3.$ $(CH_3)_2CH-O^+H_2 \to (CH_3)_2CH^+ + H_2O$

Which free radical from the following is least stable?

What is the correct order of stability for the given carbocations?
$(i) \ CH_3^+ \quad (ii) \ CH_3CH_2^+ \quad (iii) \ CH_3OCH_2^+$

The decreasing order of stability of the following carbocations is:
$A$. $m-CH_3O-C_6H_4-CH_2^+$
$B$. $p-CH_3O-C_6H_4-CH_2^+$
$C$. $C_6H_5-CH_2^+$
$D$. $p-NO_2-C_6H_4-CH_2^+$

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