Which one of the following has the most nucle | vedclass

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(A) Nucleophiles are species that have an excess of electrons. In $Pyrrole$,the lone pair of nitrogen is involved in the delocalization of the ring to

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Pyridine

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(A) Nucleophiles are species that have an excess of electrons. In $Pyrrole$,the lone pair of nitrogen is involved in the delocalization of the ring to maintain aromaticity,so it is not available for donation. In $Aniline$,the lone pair is involved in conjugation with the $\pi$ electrons of the benzene ring,which reduces its availability. In $Acetanilide$ $(C_6H_5NHCOCH_3)$,the lone pair on nitrogen is involved in resonance with the carbonyl group $(C=O)$,making it even less nucleophilic. In $Pyridine$,the lone pair on the nitrogen atom is in an $sp^2$ hybridized orbital that is perpendicular to the $\pi$ system of the ring. Therefore,it is not involved in the aromatic sextet and is relatively free for donation. Thus,the nitrogen of $Pyridine$ is the most nucleophilic.

Which one of the following has the most nucleophilic nitrogen?

Similar Questions

Identify the products $y$ and $x$ in the following reaction sequence:
$C_6H_5CONH_2 \xrightarrow{NaOBr} y$
$C_6H_5CONH_2 \xrightarrow[(ii) H_3O^+]{(i) C_6H_5SO_2Cl / py, \Delta} x$

The final product $[B]$ is :

Consider the following compounds:
$I. CH_3-CH_2-NH_2$
$II. Ph-CH_2-NH_2$
$III. Ph-NH_2$
The correct order of their basic strength is:

Suppose the following reaction:
$CH_3-CH(CH_3)-COOH + CH_3-NH_2$ $\rightarrow 'A'$ $\xrightarrow{\Delta} 'B'$ $\xrightarrow{LiAlH_4 (excess)} 'C'$
The final product '$C$' will be:

What is the product obtained in the following reaction?

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