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Chlorine of vinyl chloride $$\left( {C{H_2} = CHCl} \right)$$    is non-reactive ( less reactive ) towards nucleophile ( in nucleophilic substitution reaction ) because it shows the following resonating structure due to $$+M$$ - effect of $$-Cl$$ - $$atom.$$

In structure II, $$Cl$$ - $$atom$$  have positive charge and partial double bond character with $$C$$ of vinyl group, so it is more tightly attracted towards the nucleus and does not get replaced by nucleophile in $${S_N}$$ reaction.

Chiral conformation will not have plane of symmetry. Since twist boat does not have plane of symmetry it is chiral.

In molecules (A), (C) and (D), the carbon atom has a multiple bond, only (B) has $$s{p^3}$$ hybridization.

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Option (C) is most probable answer. Out of compound (c) or (d), the (d) results in formation of racemic product due to chirality. Compound (c) although not chiral but can form racemic product. After removal of $$C{l^ - }$$  gives a carbocation which in turn undergo racemisation after rearrangement.

Benzene having any activating group, i.e. $$OH,R,$$  etc, undergoes electrophilic substitution very easily as compared to benzene itself. Thus, toluene $$\left( {{C_6}{H_5}C{H_3}} \right),$$   phenol $$\left( {{C_6}{H_5}OH} \right)$$   undergo elecrophilic substitution very readily than benzene. Chlorine with $$ + E$$  and $$ + M$$ - effect deactivates the ring due to strong $$-I$$ - effect. So, it is difficult to carry out the substitution in chlorobenzene than in benzene, so the correct order is
$$\mathop {{\text{Phenol}}}\limits_{\left( {{\text{iv}}} \right)} {\text{ > }}\mathop {{\text{Toluene}}}\limits_{\left( {{\text{ii}}} \right)} {\text{ > }}\mathop {{\text{Benzene}}}\limits_{\left( {\text{i}} \right)} {\text{ > }}\mathop {{\text{Chlorobenzene}}}\limits_{\left( {{\text{iii}}} \right)} $$

Carbanions are stabilised by electron withdrawing groups. $$ - N{O_2}$$  is stronger electron withdrawing group as compared to $$ - CHO.$$   At $$ortho$$  - position, the effect is more pronounced.

Key Idea Presence of electron withdrawing substituent increases the acidity while electron releasing substituent, decreases the acidity.
Phenyl is an electron withdrawing substituent while $$ - C{H_3}$$  is an electron releasing substituent. Moreover, phenoxide ion is more resonance stabilised as compared to benzyloxide ion, thus releases proton more easily. That’s why phenol is a strong acid among the given compounds.

Due to presence of $$ > C = C < $$   in $$but$$  $$-2-ene,$$   it shows restricted rotation. Hence, give two types of arrangements around the space of $$ > C = C < $$   as $$cis$$  and $$trans$$  - forms.