In I the unshared pair of electrons on $$N$$  always available for protonation in III due to presence of electronegative $$O$$  atom the electron density on $$N$$  is decreased . In II and IV resonance suppresses the basic character.

An electrophile is defined as electron deficient species which attacks on electron rich areas. Being electron deficient, the electrophiles behaves as Lewis acids.
$${\text{e}}{\text{.g}}{\text{.}}\,\,\,\,{H^ + },C{l^ + },B{r^ + },N\mathop {{O_2}}\limits^ + ,N\mathop O\limits^ + \,\,{\text{etc}}.$$
The electrophiles can be seen in the form of neutral molecules also

Thus, we can say that electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile.

Rate of reaction will be $$R - I > R - Br > R - Cl > R - F.$$        because $${I^ - }$$ is the best, while $${F^ - }$$ is the poorest leaving groups among halide ions.

When the attack of nucleophile takes place on the opposite side of the leaving group in the substrate molecule, a transition state is obtained. It is partially bonded to both the attacking nucleophile and the leaving group and results in the formation of product. Such reactions are called as $${S_N}2$$  ( bimolecular nucleophilic substitution reaction ) because in such reactions, rate depends on the concentration of both, the substrate and the nucleophile.

When the given mixture is shaken with $$1\,M\,HCl,$$   amine get protonated and becomes cation $$\left( {RNH_2^ \oplus } \right),$$   which does not dissolve in organic solvent but usually dissolve in $${H_2}O$$  due to its charge. So, shaking with aqueous $$HCl$$  will pull amines into the aqueous phase and leave all other compounds in organic layer.

The compounds in which asymmetric carbon atom is present, are called optically active, they rotate the plane polarised light but the compounds which do not show optical activity inspite of the presence of chiral carbon atoms are called $$meso$$  - compounds. The absence of optical activity in these compounds is due to the presence of plane of symmetry in the molecule.

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NOTE : $$ - N{O_2}$$  is an electron-attracting group where as $$ - C{H_3}$$  : is an electron-releasing group. An electron - attracting substituent tends to disperse the negative charge of the phenoxide ion and thus, makes it more stable. This, in turn, increases the acid strength of phenol. The substituent in para position is more effective than in the meta position as the former involves a resonating structure bearing negative charge on the carbon attached to the electron - withdrawing substituent.
An electron - releasing substituent tends to intensify the negative charge of the phenoxide ion and thus makes it more unstable. This, in turn, decreases the acid strength of phenol. Hence, the order of acid strength is