Which of the following is cabanion expected to be most stable?

p - {N O}_{2} C_{6} H_{4} - C H_{2}

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o - {N O}_{2} C_{6} H_{4} - C H_{2}

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o - C H O C_{6} H_{4} - C H_{2}

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p - C H O C_{6} H_{4} - C H_{2}

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Solution

The correct option is B $o - {N O}_{2} C_{6} H_{4} - C H_{2}$
$o {- N O_{2} C_{6} H_{4} C H_{2}}^{-}$ (Due to more electron withdrawing ability of $- N O_{2}$ than $- C H O$ ).

Stability of carbanion is directly proportional to electron withdrawing ability.

$- N O_{2}$ shows $- R$ effect which is stronger at ortho position. It will stabilise carbanion more.

Option B is correct.

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Similar questions

I C_{6} H_{5} N H_{2}

II p - N O_{2} C_{6} H_{4} N H_{2}

III m - N O_{2} C_{6} H_{4} N H_{2}

IV m - C H_{3} O C_{6} H_{4} N H_{2}

o - B r - C_{6} H_{4} - {N H}_{2}

N O_{2} - C H_{2} - C H_{2} - O^{⊖}

C H_{3} - C H_{2} - C H_{2} - O^{⊖}

C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ (- C H_{2} - C H_{2} -)_{n}

C a C_{2}

C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ {(- {C H}_{2} - {C H}_{2} -)}_{n}

64.1 k g

C a C_{2}

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