Question

What is the correct order of $\dfrac{o}{p}$ ratio when ${E^ + }$ attacks the following system?
(A) $PhF\;$
(B) $PhCl\;$
(C) $PhBr\;$
(D) $PhI\;$
1. $A < B < C < D$
2. $A = B = C = D$
3. $D < C < B < A$
4. $D < B < A < C$

Answer 1

Hint : Here, in this question $\dfrac{o}{p}$ ratio is written which means $\dfrac{ortho}{para}$ ratio and ${E^ + }$ indicates an electrophile. It is the ratio of ortho as well as para products formed when the substitution reaction occurs due to the attack of an electrophile.

Complete Step By Step Answer:
When an electrophilic aromatic substitution occurs in a mono- substituted benzene ring. If a group is already present on the benzene ring then that group will decide the position of upcoming electrophile i.e. called as directive influence and also determine the rate of reaction i.s. called as activate- deactivated influence.
These groups can be ortho- para activating or deactivating groups. There are some groups which are ortho- para activating or we can say, they give directive influence towards ortho-para. Some examples of these activating groups such as $- OH$ , $- OR$ , $- N{R_2}$ , $- NHR$ etc.
Some are ortho-para deactivating groups or we can say, they do not give directive influence towards ortho and para positions. Examples of this group are $- N{O_2}$ , $- CN$ etc.
Also, some groups are meta directing groups. For example- $- I$ groups, $- H$ groups, $- M$ groups.
So, $\dfrac{ortho}{para}$ ratio is the ratio of the ortho and para products formed when electrophilic aromatic substitution occurs on a benzene ring. This ratio depends on factors such as size of upcoming electrophile, size of already present group on benzene ring and chelation.
In this question, the four options are given, all have a phenyl ring but the halogens are different. We know that halogens show effect $- I$ effect and formation of ortho and para product depend on the $- I$ effect.
If the halogen has higher $- I$ effect, the percentage of para product will be higher i.e. in case of fluorine $- F$ .
If the halogen has lower $- I$ effect, the percentage of ortho product will be higher i.e. in case of iodine $- I$
So, in the case of iodine, $\dfrac{o}{p}$ ratio will be higher.
Therefore, the $\dfrac{o}{p}$ ratio follows the order $A < B < C < D$
Hence, the correct answer is option (1).

Note :
Always remember that the $\dfrac{o}{p}$ ratio is directly proportional to chelation but inversely proportional to the size of the already group on benzene and size of upcoming electrophile. When the size of the group is already present and the size of electrophile is larger, then the $\dfrac{o}{p}$ decreases.