The correct increasing order of trans-effect of the following species is

C N^{-} > B r^{-} > C_{6} H_{5}^{-} > N H_{3}

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N H_{3} > C N^{-} > B r^{-} > C_{6} H_{5}^{-}

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C N^{-} > C_{6} H_{5}^{-} > B r^{-} > N H_{3}

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B r^{-} > C N^{-} > N H_{3} > C_{6} H_{5}^{-}

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Solution

The correct option is C $C N^{-} > C_{6} H_{5}^{-} > B r^{-} > N H_{3}$
Trans effect can be defined as the effect of a ligand towards substitution on other ligand which is trans to it.

(Weak) $F^{-} < O H^{-} < H_{2} O < N H_{3} < p y < C l^{-} < B r^{-} < I^{-} < S C N^{-} < N O_{2}^{-} < S C (N H_{3})_{2} < P h^{-}$

$< S O_{3}^{2 -} < P R_{3} < A s R_{3} < S R_{2} < C H_{3}^{-} < H^{-} < N O < C O < N C^{-} < C_{2} H_{4}$ (Strong)

Trans effect is nothing but those ligands which are trans directing. That means, which prefer a trans structure. Mostly in Square planar complexes.

Hence , option C is correct .

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C_{6} H_{5} - C | H C H_{3} - B r

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Predict the order of reactivity of the following compounds in SN1 and SN2 reactions

C6H5CH2Br , C6H5CH(C6H5)Br , C6H5CH(CH3)Br , C6H5C(CH3)(C6H5)

C N^{-}

H^{+}

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C H_{3} - \oplus C = O

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\oplus N O_{2}

C_{6} H_{5} - C H O + B r - C H_{2} - C O O C_{2} H_{5} (i) Z n - -------------- \to (i i) H O H / N H_{4} C l (i i i) Δ

C_{6} H_{5} - C H = C H - C O O C_{2} H_{5}

C_{6} H_{5} - C H_{3} | C | H - B r + H_{2} O ⟶ H O - C H_{3} | C | H - C_{6} H_{5} + H B r

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