Key Idea Electron withdrawing substituent deactivates the benzene nucleus towards electrophilic substitution while electron releasing substituent activates the ring towards electrophilic substitution.
Among the given, $$-OH$$  has the higher electron donating tendency and thus, activates the ring more towards electrophilic substitution.
Hence,
Show $$ + M$$ - effect, due to this benzene ring becomes activate.
It is more reactive towards electrophilic reagent.

$$Na + C + N + S \to NaSCN$$

Nitro group is electron withdrawing group, so it deactivates the ring towards electrophilic substitution.

No explanation is given for this question. Let's discuss the answer together.

3-methyl hexane contains chiral carbon atom. Here it exhibits chirality.

2,3-dimethylpentane

shows the property of optical isomerism due to presence of an asymmetric $$\mathop C\limits^* - atom.$$

The possible isomers of $${C_7}{H_8}O$$  are

When treated with standard solution of alkali nitrogen is liberated as $$N{H_3}$$  which is absorbed in $${H_2}S{O_4}$$  to get percentage of nitrogen.

Due to $$+I$$  as well as $$+R$$ - effect of $$ - OC{H_3}$$  group, it activates the benzene ring. While $$ - N{O_2}$$  deactivates the benzene ring due to its $$-I$$ - effect and also decrease the reaction rate, as well as $$-R$$ - effect.
So, order of $${S_E}$$  reaction is

$$\eqalign{ & C - H:0.109\,nm \cr & C = C:\,0.134\,nm \cr & C - O:\,0.143\,nm \cr & C - C:\,0.154\,nm \cr & \therefore \,\,{\text{Bond length order is}} \cr & C - H < C = C < C - O < C - C \cr} $$