Okay, so I need to figure out the order of increasing weight fraction of the para-isomer yield for these substances undergoing electrophilic substitution with bromine. Let's start by recalling some chemistry concepts here. Electrophilic substitution reactions on aromatic rings, like benzene derivatives, are influenced by the directing effects of substituents. The substituents can activate or deactivate the ring towards electrophilic substitution. 

First, I remember that substituents that are electron-donating groups activate the ring and direct incoming electrophiles to the positions ortho and para relative to themselves. Electron-withdrawing groups deactivate the ring but can still direct the incoming electrophile to certain positions depending on the nature of the group. For example, nitro groups are strong deactivating groups and meta directors. 

The question mentions that only one monobromo derivative is formed, which suggests that each substituent directs the incoming bromine to a single position. For substituents that are meta directors, like nitro groups, the reaction might form only the meta isomer, but in this case, since it's specified that only the para isomer is considered, maybe the substituents here are ortho/para directors. Wait, but some substituents are meta directors. Hmm, perhaps I need to check each substituent's directing effect.

Looking at each substance:

1) C6H5-CH3: Methyl group attached to benzene. Methyl is an electron-donating group via the inductive effect (since it's an alkyl group). So it's an ortho/para director. 

2) C6H5-COOC2H5: This is a benzene ring with a benzoate ester group. The ester group is an electron-withdrawing group (due to the electron-withdrawing effect of the carbonyl), but wait, ester substituents are ortho/para directors. Wait, no—wait, esters are deactivating (electron-withdrawing) groups but they are ortho/para directors. Because the carbonyl is electron-withdrawing through resonance, but the adjacent alkyl group (from the ester's two carbons) is electron-donating. So overall, the ester group is a meta director? Wait, no. Wait, benzoyl groups (like C6H5-CO-) are meta directors. Wait, but benzoate is different. Let me think. The benzoate ester group (COO-) is an electron-withdrawing group because the oxygen is more electronegative. So the substituent is deactivating. But since the substituent is attached to the benzene ring via an oxygen, which is part of a carbonyl group, the overall effect is that it is a meta director. Because the electron-withdrawing effect dominates. Wait, but I'm confused. Wait, substituents can be classified as activators or deactivators. Electron-donating groups (like -OCH3, -CH3, -NH2) are activators (ortho/para directors). Electron-withdrawing groups (like -NO2, -COOH, -SO3H) are deactivators. But there are exceptions. For example, the ester group (-COOR) is a deactivating (electron-withdrawing) group, but it's an ortho/para director. Wait, no, wait. Wait, the ester group is ortho/para directing but deactivating. Let me check: the ester group is -OOCR. The carbonyl oxygen withdraws electrons through resonance, but the adjacent alkyl group donates electrons through induction. So overall, the ester group is ortho/para directing but deactivating. So when there's an ester group, the ring is less reactive than benzene, but the substituent directs incoming electrophiles to ortho and para positions. So in this case, substance 2 would form ortho and para products. But the question says that only one monobromo derivative is formed. Hmm, wait. If the substituent is an ortho/para director, but the ring is activated, then bromine would substitute in ortho and para. However, the question states that only one monobromo derivative is formed. That suggests that the substituent is a meta director, so that the bromine would only substitute in the meta position. Wait, but if the substituent is meta directing, then the bromine would go to meta, but if the substituent is ortho/para directing, then it would substitute in ortho and para. But the problem states that only one monobromo derivative is formed. So maybe the substituent is such that the ring is deactivating but meta directing? Wait, but that would mean that the substituent is a strong deactivator and a meta director. For example, nitro groups. Wait, but the first substituent is a methyl group, which is activating. So perhaps the question is considering that in some cases, even though a substituent is ortho/para directing, due to the ring's activation, the bromine might only substitute in the para position. Wait, but that doesn't make sense. Wait, maybe the question assumes that only the para isomer is formed, so all substituents here are meta directors? But then none of the substituents here would be meta directors. Let me check each substituent again.

Substance 1: CH3. Ortho/para director, activating. So bromine would substitute in ortho and para. But the question states that only one monobromo is formed. Hmm. Maybe the question is considering that only the para isomer is formed in some cases. But why would that be? Wait, perhaps the substituents are such that the para position is more substituted than the ortho. Wait, but in electrophilic substitution, the para product is generally the major product when there's no steric hindrance. Wait, but in this case, maybe the substituents are such that the para isomer is less favored? Or perhaps the question is considering that when a substituent is such that the para position is less activated than the ortho. Wait, maybe some substituents deactivate the para position more than the ortho. Hmm. Alternatively, perhaps the question is considering that some substituents direct the incoming electrophile to only the para position. But that would require the substituent to be a meta director. Wait, but all the substituents here: 1 is ortho/para, 2 is ortho/para, 3 is Cl (which is ortho/para), 4 is NO2 (meta), 5 is C2H5 (ortho/para), 6 is COOH (which is deactivating, ortho/para). Wait, except 4 (NO2) is a meta director. So only 4 is a meta director here. So for substituent 4, the bromine would substitute in the meta position. But the question states that only the para isomer is formed. So perhaps the question is considering that each substituent directs the bromine to only the para position. But that's not possible unless the substituent is a meta director. Wait, maybe the substituent is such that the ring is deactivated but the substituent is meta directing. But all substituents except 4 are ortho/para directors. So this is confusing. Wait, perhaps the question is considering that each substituent directs the bromine to a single position, which is the para position. But that's only possible if the substituent is a meta director. So maybe the question is considering that each substituent is a meta director, but that's not the case. Hmm. Alternatively, perhaps the substituent directs the bromine to the para position because that's the only position available due to steric hindrance. For example, in some cases, the ortho position is hindered, so only the para product is formed. For example, in benzene rings with bulky substituents in the ortho positions. But none of these substituents here are bulky. Substituent 1 is CH3, which is small. Substituent 2 is a benzoate ester, which is also small. Substituent 3 is Cl, which is small. Substituent 5 is ethyl, small. Substituent 6 is COOH, which is also small. Substituent 4 is NO2, which is a small group. So maybe the question is not considering steric effects here. So perhaps the key is that for substituents that are meta directors, the bromine would only substitute in the meta position, so the para isomer is not formed. But the question states that only the para isomer is formed. Therefore, perhaps the substituents here are all ortho/para directors, but the para isomer is the major product. But the question is asking about the weight fraction of the para-isomer. So if the substituent is an ortho/para director, then the para isomer would be the major product. However, the question is asking for the order of increasing weight fraction of the para-isomer. So substances with substituents that are ortho/para directors would have higher para yields, while those with substituents that are meta directors (like NO2) would have lower para yields (since they only form the meta isomer, so the para isomer is not formed). Wait, but NO2 is a meta director. So for substance 4, which is NO2, the bromine would substitute in the meta position, so the para isomer would not form. Therefore, the weight fraction of the para-isomer for substance 4 would be zero. Similarly, if any substituent is a meta director, then the para isomer would have zero yield. But looking at the substituents:

1) CH3: ortho/para director. So para isomer is formed. The yield would depend on substituent strength. 

2) COO2C6H5: ester group. Ortho/para director. So para isomer formed.

3) Cl: ortho/para director. Para formed.

4) NO2: meta director. Para not formed.

5) C2H5: ortho/para. Para formed.

6) COOH: ortho/para. Para formed.

So only substance 4 (NO2) does not form the para isomer at all. Therefore, the order of increasing weight fraction of the para-isomer would start with substance 4 (0% para), followed by substances where the para yield is low. Wait, but the question is about the order of increasing weight fraction. So substances with higher para yields come later.

Wait, but the question is about the order of increasing weight fraction of the para-isomer. So we need to arrange the substances from the one with the least para yield to the one with the highest. 

But for each substituent:

Substance 1: CH3 (activating, ortho/para director). The para position is more activated than the ortho. So the para isomer is formed in a higher yield. Wait, but the yield depends on the activation strength. Stronger activating groups would lead to higher para yields. For example, -CH3 is a stronger activating group than -COOH (which is deactivating). Wait, no. Wait, CH3 is an activating group (donating electrons), whereas COOH is deactivating (electron-withdrawing). So the activating strength: -CH3 > -C2H5 (ethyl) > -COOH (since COOH is deactivating). Wait, but ethyl is an activating group as well. Wait, but the substituent -C2H5 (ethyl) is an alkyl group, which is activating. So substituent 5 is ethyl, which is activating. So substances 1 (CH3), 5 (C2H5), 2 (COO2C6H5), 3 (Cl), 6 (COOH), and 4 (NO2). Wait, but substituent 2 is an ester group, which is deactivating but ortho/para directing. So substituent 2 is deactivating but directs to para. So substituent 2's para yield would be lower than substituent 1 (CH3), which is activating. Similarly, substituent 3 (Cl) is an ortho/para director but deactivating (since Cl- is a weaker deactivating group than NO2, but still deactivating). Substituent 4 (NO2) is a meta director, so para isomer not formed. Substituent 5 (C2H5) is activating. Substituent 6 (COOH) is deactivating. 

Wait, but substituents can be ordered based on their activating/deactivating strength. Activating groups: -CH3 (strong), -C2H5 (moderate), -Cl (weaker, but still activating?), -COOH (deactivating), -COO2C6H5 (deactivating), -NO2 (deactivating). Wait, but Cl is actually a deactivating group (due to the electronegativity of Cl), but it's an ortho/para director. So substituents:

1) CH3 (strong activating, ortho/para)

5) C2H5 (activating, ortho/para)

2) COO2C6H5 (deactivating, ortho/para)

3) Cl (deactivating, ortho/para)

6) COOH (deactivating, ortho/para)

4) NO2 (deactivating, meta)

So the order of activating strength from strongest to weakest:

1 (CH3) > 5 (C2H5) > 2 (COO2C6H5) > 3 (Cl) > 6 (COOH) > 4 (NO2)

But wait, Cl is deactivating but still an ortho/para director. So its activating strength is low compared to the activating groups (1 and 5). But among the deactivating groups, the order from strongest to weakest deactivating would be NO2 (strongest), then perhaps COOH, then COO2C6H5, then Cl? Wait, Cl is a weaker deactivator than NO2 and COOH. Because the electronegativity of Cl is less than O. Wait, but Cl is attached directly to the benzene ring. The deactivating effect of Cl is due to the inductive effect (Cl is more electronegative than H, so it withdraws electrons through induction), but the ortho/para directing effect is due to resonance (Cl has lone pairs that can donate electrons through resonance into the ring). So Cl is an ortho/para director but deactivating. So in terms of activating strength, Cl is less activating than the activating groups (1 and 5) but more activating than the deactivating groups (since it's deactivating but still directs to ortho/para). 

So the activating strength order is:

1 (CH3) > 5 (C2H5) > 2 (COO2C6H5) > 3 (Cl) > 6 (COOH) > 4 (NO2)

Because NO2 is the strongest deactivating. COOH is weaker than COO2C6H5? Wait, no. The substituents:

- COO2C6H5 is an ester group. The ester group is ortho/para directing but deactivating. 

- COOH is a carboxylic acid group. The COOH group is deactivating (because of the -I effect of the carbonyl) but still ortho/para directing (due to resonance). 

Now, comparing the deactivating strength between COO2C6H5 and COOH. The ester group (COO2C6H5) has a methoxy group attached to the oxygen. The methoxy group is an electron-donating group (due to resonance), so the ester group is less deactivating than a pure carboxylic acid group. Wait, no. Wait, the ester group is OOCR. The carbonyl oxygen withdraws electrons, but the adjacent alkyl group donates electrons. So the overall effect is that the ester group is deactivating (since the carbonyl is more dominant), but it's still ortho/para directing. The COOH group has a -COOH substituent. The COOH group is a stronger deactivating group than the ester group. Because the COOH group's carbonyl is more electronegative, so the inductive effect is stronger. Therefore, the ester group (COO2C6H5) is less deactivating than the carboxylic acid group (COOH). Wait, but wait: the ester group has two oxygen atoms. Let's think. The ester group is O-C-O-R. The carbonyl group (C=O) withdraws electrons, but the adjacent O-R group (where R is alkyl) donates electrons through resonance. So the overall effect is that the ester group is deactivating (due to the carbonyl's inductive withdrawal) but still ortho/para directing. The COOH group is more deactivating because the -COOH group is a stronger electron-withdrawing group. So the order of deactivating strength (from strongest to weakest) would be:

NO2 > COOH > COO2C6H5 > Cl > ... Wait, no. Wait, Cl is a weaker deactivator than COO2C6H5. Wait, no. Wait, Cl is an ortho/para director but deactivating. COO2C6H5 is deactivating but ortho/para. COOH is deactivating and ortho/para. So in terms of deactivating strength, NO2 is the strongest, followed by COOH, then COO2C6H5, then Cl, then perhaps the ethyl group is activating, so not in the deactivating list. Wait, but the question is about the para-isomer yield. 

So the para-isomer yield for each substituent depends on the substituent's activating strength. The more activating the group, the higher the para yield. For deactivating groups, the lower the para yield (or zero if meta-directing). 

So let's list the substituents and their para yields:

1) CH3: strong activating. High para yield.

2) COO2C6H5: deactivating but ortho/para. Lower than activating but higher than moderately deactivating.

3) Cl: deactivating, ortho/para. Lower than COO2C6H5.

4) NO2: meta director. Para yield is zero.

5) C2H5: activating. Higher than COO2C6H5 and Cl.

6) COOH: deactivating. Lower than COO2C6H5. Wait, but COOH is a stronger deactivator than COO2C6H5. So the order from highest para yield to lowest:

1 (CH3) > 5 (C2H5) > 2 (COO2C6H5) > 3 (Cl) > 6 (COOH) > 4 (NO2)

But the question asks for increasing order of the weight fraction of the para-isomer. So the order would be from lowest to highest:

4 (NO2) < 6 (COOH) < 3 (Cl) < 2 (COO2C6H5) < 5 (C2H5) < 1 (CH3)

Wait, but wait. Let's clarify. The substituent with NO2 (4) has zero para yield. The substituent with COOH (6) is deactivating but still directs to para. So the para yield for COOH would be lower than activating groups but higher than more deactivating groups. 

But let's think about the activation strength. The activating groups (1 and 5) have higher para yields than the deactivating substituents (2,3,6), except that substituent 4 (NO2) is meta, so zero. 

So the order from least to most para yield:

4 (NO2, 0) < 6 (COOH, some low yield) < 3 (Cl, lower than COOH) < 2 (COO2C6H5, lower than Cl) < 5 (C2H5, higher than 2) < 1 (CH3, highest)

Wait, but substituent 5 (C2H5) is an activating group, so its para yield should be higher than substituents 2,3,6. Substituent 2 (COO2C6H5) is deactivating but directs to para. So substituent 2's para yield is lower than substituent 5 but higher than substituents 3 and 6. Substituent 3 (Cl) is deactivating but directs to para, so its para yield is lower than substituent 2. Substituent 6 (COOH) is deactivating but directs to para, so its para yield is lower than substituent 3. 

So the order from lowest to highest para yield:

4 (NO2) < 6 (COOH) < 3 (Cl) < 2 (COO2C6H5) < 3? Wait, substituent 3 is Cl, substituent 2 is COO2C6H5. Wait, substituent 3 (Cl) is deactivating, substituent 2 (COO2C6H5) is also deactivating. Which one has a higher para yield? 

The substituent's para yield depends on how strong the deactivating effect is. Stronger deactivating groups would have lower para yields. So between substituent 2 (COO2C6H5) and substituent 3 (Cl), which has a higher para yield? Cl is a weaker deactivator than COO2C6H5. Because the ester group (COO2C6H5) is more deactivating than Cl. Therefore, substituent 2 has a lower para yield than substituent 3. Wait, no. Wait, if a substituent is more deactivating, it would deactivating the ring more, making the para substitution less favored. So substituent 2 (COO2C6H5) is more deactivating than substituent 3 (Cl). Therefore, substituent 2 would have a lower para yield than substituent 3. Wait, but substituent 3 is Cl (which is a weaker deactivator than COO2C6H5). So substituent 3 is less deactivating than substituent 2. Therefore, substituent 3's para yield is higher than substituent 2's. 

So substituent 6 (COOH) is the least deactivating among the deactivating groups. Wait, no. Let's compare the deactivating strength:

Strongest deactivating: NO2 (4)

Next: COOH (6)

Then: COO2C6H5 (2)

Then: Cl (3)

Then: C2H5 (5) is activating, so higher than all deactivating

CH3 (1) is activating, highest. 

Wait, substituent 5 is ethyl, which is activating. So substituent 5 has higher para yield than the deactivating substituents. 

Therefore, the order of para yields from lowest to highest is:

4 (NO2, 0) < 6 (COOH, low) < 2 (COO2C6H5, lower than 3) < 3 (Cl, higher than 2) < 5 (C2H5, higher than 3) < 1 (CH3, highest)

Wait, but substituent 2 is COO2C6H5, which is less deactivating than Cl (substituent 3). So substituent 2 has a higher para yield than substituent 3. Wait, but substituent 3 is Cl, which is a weaker deactivator than COO2C6H5. So substituent 3 (Cl) would have a higher para yield than substituent 2 (COO2C6H5). Therefore, substituent 3's para yield is higher than substituent 2's. 

Therefore, the order from lowest to highest para yield is:

4 (NO2) < 6 (COOH) < 2 (COO2C6H5) < 3 (Cl) < 5 (C2H5) < 1 (CH3)

Wait, but substituent 6 (COOH) is less deactivating than substituent 2 (COO2C6H5), so substituent 6 has a higher para yield than substituent 2? Wait, substituent 6 (COOH) is less deactivating than substituent 2 (COO2C6H5). Wait, no. The order of deactivating strength (from strongest to weakest):

NO2 (4) > COOH (6) > COO2C6H5 (2) > Cl (3) > ... 

So substituent 2 (COO2C6H5) is more deactivating than substituent 3 (Cl), which is more deactivating than substituent 6 (COOH). Therefore, substituent 6's para yield is higher than substituent 2's. Therefore, substituent 6's para yield > substituent 2's para yield > substituent 3's para yield. 

Wait, but substituent 6 (COOH) is less deactivating than substituent 2 (COO2C6H5). So substituent 6 would have a higher para yield than substituent 2. Therefore, substituent 6's para yield is higher than 2's. 

So the order from lowest to highest would be:

4 (NO2) < 6 (COOH) < 2 (COO2C6H5) < 3 (Cl) < 5 (C2H5) < 1 (CH3)

Wait, but substituent 3 (Cl) has a higher para yield than substituent 2 (COO2C6H5), which is correct. Then substituent 6 (COOH) is less deactivating than substituent 2, so substituent 6 has a higher para yield than substituent 2. Wait, this is getting confusing. Let's make a table:

Substituent | Para Yield (relative)
--- | ---
4 (NO2) | 0% (meta director)
6 (COOH) | low (deactivating)
2 (COO2C6H5) | lower than 6 (since more deactivating)
3 (Cl) | higher than 2 (since less deactivating)
5 (C2H5) | high (activating)
1 (CH3) | highest (activating)

So the order from least to most para yield:

4 (0) < 6 (low) < 2 (lower than 6) < 3 (higher than 2) < 5 (high) < 1 (highest)

Therefore, the order is 4 < 6 < 2 < 3 < 5 < 1. 

But looking at the answer choices, the options are:

A. 4<6<2<1<5<3

B. 3<5<1<6<2<4

C. 4<2<6<3<1<5

D. 6<2<4<5<1<3

Wait, according to our analysis, the order should be 4 < 6 < 2 < 3 < 5 < 1. But none of the options match this. Let me check again.

Wait, perhaps I made a mistake in the order. Let's re-examine:

Para yield order from lowest to highest:

4 (NO2) has 0% yield.

Then, substituent 6 (COOH) has some low yield.

Substituent 2 (COO2C6H5) has higher yield than substituent 6.

Substituent 3 (Cl) has higher yield than substituent 2.

Substituent 5 (C2H5) has higher yield than substituent 3.

Substituent 1 (CH3) has the highest yield.

So the order is 4 < 6 < 2 < 3 < 5 < 1.

Looking at the answer choices:

Option C: 4<2<6<3<1<5 → No, because after 4 comes 2, but according to our analysis, 6 comes before 2.

Option A: 4<6<2<1<5<3 → No, because 1 should be last.

Option D: 6<2<4<5<1<3 → No.

Option B: 3<5<1<6<2<4 → No.

Wait, none of the options match. So perhaps my analysis is wrong. Let me think again.

Wait, perhaps the question is about the weight fraction of the para-isomer. The para-isomer's yield depends on the substituent's ability to direct to the para position. For substituents that are deactivating, the para yield is lower than the activating ones. But for the deactivating substituents, the para yield depends on the substituent's activating strength. Wait, no. The substituent's activating strength affects the overall reactivity of the ring towards electrophilic substitution. So a stronger activating group would lead to higher reactivity, thus higher para yield (assuming para is more favored under certain conditions). But in this case, all substituents except NO2 would form para isomers, but their yields may vary based on substituent strength.

Alternatively, perhaps the question is considering the position of the substituent. For example, substituents with more activating groups would have higher para yields. For substituents that are deactivating, the para yield is lower. But among the deactivating groups, which one is less deactivating (i.e., more activating)? 

Wait, substituent 3 is Cl, which is less deactivating than substituent 2 (COO2C6H5). So substituent 3 has a higher para yield than substituent 2. Substituent 6 (COOH) is less deactivating than substituent 2, so substituent 6 has a higher para yield than substituent 2. Therefore, substituent 6's yield is higher than substituent 2's. 

So the order from lowest to highest para yield is:

4 (NO2) < 2 (COO2C6H5) < 6 (COOH) < 3 (Cl) < 5 (C2H5) < 1 (CH3)

Wait, that's different from before. Wait, substituent 2 is COO2C6H5, which is more deactivating than substituent 6 (COOH). So substituent 6's para yield is higher than substituent 2's. Therefore, the order is 4 < 2 < 6 < 3 < 5 < 1. 

Looking at the options again:

Option C: 4<2<6<3<1<5 → Yes, this matches. 

Let me check:

4 (NO2) < 2 (COO2C6H5) < 6 (COOH) < 3 (Cl) < 1 (CH3) < 5 (C2H5). Wait, but substituent 5 (C2H5) is ethyl, which is activating. So substituent 5's para yield should be higher than substituent 1 (CH3), which is more activating. Wait, but that's not correct. 

Wait, substituent 5 is C2H5 (ethyl), which is an activating group. CH3 (substituent 1) is also activating, but ethyl is a stronger activating group than methyl. Because ethyl has more alkyl groups, which donate electrons through induction. So substituent 5 (C2H5) is more activating than substituent 1 (CH3). Therefore, substituent 5's para yield is higher than substituent 1's. 

Therefore, the order should be:

4 < 2 < 6 < 3 < 5 < 1.

But substituent 5 is ethyl, which is more activating than methyl. So substituent 5's yield is higher than substituent 1's. Therefore, the correct order is 4 < 2 < 6 < 3 < 5 < 1. 

Looking at the answer options, option C is 4<2<6<3<1<5. But according to our analysis, substituent 5 (C2H5) should come before substituent 1 (CH3). Therefore, the correct order would be 4 < 2 < 6 < 3 < 5 < 1. 

But none of the answer options match this. Wait, let's check the options again:

Option C: 4<2<6<3<1<5 → This would mean substituent 5 (C2H5) comes before substituent 1 (CH3), which is correct. But according to our analysis, substituent 5 is more activating than substituent 1, so substituent 5's yield should be higher than substituent 1's. Therefore, the order should be 4 < 2 < 6 < 3 < 5 < 1. But this order isn't present in the options. 

Wait, perhaps I made a mistake in the order. Let's re-examine:

Substituent 5 (C2H5) is ethyl, which is a stronger activating group than methyl (CH3). Therefore, substituent 5's para yield is higher than substituent 1's. So the order after substituent 3 (Cl) should be substituent 5 (C2H5) then substituent 1 (CH3). Therefore, the correct order is 4 < 2 < 6 < 3 < 5 < 1. 

But looking at the answer options:

Option C: 4<2<6<3<1<5 → No, because after 3 comes 1, which should be after 5. 

Option D: 6<2<4<5<1<3 → No. 

Option A: 4<6<2<1<5<3 → No. 

Option B: 3<5<1<6<2<4 → No. 

Hmm. None of the options match. Therefore, perhaps my analysis is incorrect. Alternatively, perhaps the question is considering the substituent's effect on the ring's reactivity, such that more reactive rings have higher yields. For example, substituents that are better at activating the ring (like ethyl) would lead to higher para yields. 

Alternatively, perhaps the para yield is determined by the substituent's ability to stabilize the transition state. For example, substituents that are more electron-withdrawing might stabilize the negative charge on the ring, leading to higher para yields. Wait, but that's opposite. Electron-withdrawing groups would deactivate the ring, making the electrophilic substitution less favored. 

Alternatively, maybe the para yield is higher for substituents that are ortho/para directors because they allow for more substitution in the para position. But for deactivating groups, the para position might be less favored. 

Alternatively, perhaps the question is considering that the more substituted the benzene ring, the higher the para yield. Wait, but the substituent's position (ortho, meta, para) is fixed. 

Alternatively, perhaps the question is considering that substituents that are stronger deactivators lead to lower para yields. 

Wait, perhaps the para yield is higher for substituents that are less deactivating. So substituents that are more activating (like ethyl and methyl) would have higher para yields. For deactivating substituents, the less deactivating they are (i.e., the more activating they are), the higher their para yield. 

So substituent 5 (C2H5) is ethyl, which is activating, so higher para yield than substituent 1 (CH3). Substituent 2 (COO2C6H5) is deactivating but ortho/para directing. So its para yield is lower than substituents 5 and 1, but higher than substituents 6,3,4. 

Substituent 6 (COOH) is less deactivating than substituent 2, so higher para yield than substituent 2. Substituent 3 (Cl) is less deactivating than substituent 2, so higher para yield than substituent 2. 

Therefore, the order from lowest to highest para yield:

4 (NO2) < 2 (COO2C6H5) < 6 (COOH) < 3 (Cl) < 5 (C2H5) < 1 (CH3)

But this order isn't present in the options. The closest option is option C: 4<2<6<3<1<5. But according to this, substituent 5 comes before substituent 1, which is correct. However, substituent 5 (C2H5) is more activating than substituent 1 (CH3). Therefore, substituent 5 should have a higher para yield than substituent 1. Therefore, the correct order should be 4 < 2 < 6 < 3 < 5 < 1. 

But none of the options match this. Therefore, perhaps there is a mistake in the options. Alternatively, perhaps the question is considering the substituent's effect on the ring's activation in a different way. 

Alternatively, perhaps the para isomer is the major product for substituents that are ortho/para directors, and the question is about the order of the para isomer's yield. For substituents that are meta directors, the para isomer is not formed, so their yield is zero. For the others, the order is based on their activating strength.

Wait, maybe the para isomer's yield is determined by the substituent's ability to stabilize the transition state. For example, substituents that are better at stabilizing the negative charge in the transition state would lead to higher para yields. 

But I'm not sure. Alternatively, perhaps the question is considering that the para isomer's yield is higher for substituents that are more activating, and lower for deactivating substituents. So the order from lowest to highest para yield is:

4 (NO2) < 6 (COOH) < 2 (COO2C6H5) < 3 (Cl) < 5 (C2H5) < 1 (CH3)

But according to our analysis, substituent 2 (COO2C6H5) is more deactivating than substituent 6 (COOH), so substituent 6's para yield is higher than substituent 2's. Therefore, substituent 2 comes after substituent 6. 

Therefore, the correct order is 4 < 6 < 2 < 3 < 5 < 1. 

Looking again at the options:

Option A: 4<6<2<1<5<3 → No. 

Option B: 3<5<1<6<2<4 → No. 

Option C: 4<2<6<3<1<5 → No. 

Option D: 6<2<4<5<1<3 → No. 

None of the options match. Therefore, perhaps the correct answer is not among the options, but since we have to choose, perhaps there's a mistake in the analysis. 

Alternatively, perhaps the para isomer's yield is higher for substituents that are better at activating the ring. Substituent 5 (C2H5) is ethyl, which is more activating than substituent 1 (CH3). Therefore, substituent 5 should have a higher para yield than substituent 1. 

But substituent 5 is in ethyl, which is a stronger activating group than methyl. Therefore, substituent 5's para yield is higher than substituent 1's. Therefore, the order after substituent 3 (Cl) should be substituent 5 (C2H5) then substituent 1 (CH3). Therefore, the order is 4 < 2 < 6 < 3 < 5 < 1. 

But again, this order isn't in the options. The closest is option C: 4<2<6<3<1<5. But substituent 5 is after substituent 1 in option C, which is incorrect. 

Alternatively, perhaps the question is considering the number of substituents that can form the para isomer. Substituent 4 (NO2) cannot form the para isomer, so its yield is zero. The rest form the para isomer. The order of the remaining substituents is based on their activating strength. 

So substituent 1 (CH3) is the strongest activating, followed by substituent 5 (C2H5), then substituent 2 (COO2C6H5), substituent 3 (Cl), substituent 6 (COOH), and substituent 4 (NO2) cannot. Wait, no. Wait, substituent 6 (COOH) is deactivating but can form the para isomer. 

I think I need to conclude that the correct order is 4 < 2 < 6 < 3 < 5 < 1, but since it's not an option, perhaps the intended answer is option C: 4<2<6<3<1<5, but that would have substituent 5 before substituent 1, which is incorrect. 

Alternatively, perhaps the question is considering that substituent 5 (C2H5) is more activating than substituent 1 (CH3), but substituent 5's para yield is higher than substituent 1's. Therefore, substituent 5 comes before substituent 1 in the order. 

But looking at the options, none match. Perhaps the intended answer is option C, but according to the analysis, substituent 5 should come after substituent 1. 

Alternatively, maybe the question is considering that substituent 2 (COO2C6H5) has a higher para yield than substituent 6 (COOH) because the ester group is more deactivating. But I'm not sure. 

Alternatively, perhaps the para yield is higher for substituents that are ortho/para directors. Substituent 5 (C2H5) is activating and ortho/para. Substituent 1 (CH3) is activating and ortho/para. Substituent 2 (COO2C6H5) is deactivating but ortho/para. Substituent 3 (Cl) is deactivating but ortho/para. Substituent 6 (COOH) is deactivating but ortho/para. Substituent 4 (NO2) is meta. 

Therefore, the para isomer is formed for substituents 1,2,3,5,6. Substituent 4 doesn't form. 

The order of para yield among these would be based on their activating strength. 

Among these, substituent 1 (CH3) is the strongest activating, followed by substituent 5 (C2H5), then substituent 2 (COO2C6H5), substituent 3 (Cl), substituent 6 (COOH). 

So the order from lowest to highest para yield: 4 < 2 < 6 < 3 < 5 < 1. 

But again, this order isn't present in the options. 

Perhaps the intended answer is option C: 4<2<6<3<1<5. But according to the analysis, substituent 5 should come before substituent 1. 

Alternatively, the question might be considering that the para isomer is the major product for substituents that are ortho/para directors. For substituents that are meta directors, the para isomer is not formed. So substituent 4 (NO2) can't form the para isomer. 

The para isomer is formed for substituents 1,2,3,5,6. The order of their para yields would be determined by their activating strength. 

Therefore, substituent 1 (CH3) is the strongest activating, leading to the highest para yield. Substituent 5 (C2H5) is next, then substituent 2 (COO2C6H5), then substituent 3 (Cl), then substituent 6 (COOH). 

So the order from lowest to highest para yield is:

4 (NO2) < 2 (COO2C6H5) < 6 (COOH) < 3 (Cl) < 5 (C2H5) < 1 (CH3)

But since this order isn't present in the options, perhaps there's a mistake. 

Alternatively, perhaps the para isomer's yield is higher for substituents that are deactivating but ortho/para directing. For example, substituent 2 (COO2C6H5) and substituent 3 (Cl) are deactivating but ortho/para directors, so their para yields are lower than substituent 1 (CH3) and 5 (C2H5). 

But substituent 6 (COOH) is less deactivating than substituent 2, so substituent 6's para yield is higher than substituent 2's. Therefore, substituent 2's para yield is lower than substituent 6's. 

This is getting too complicated. Given the time I've spent, I think the intended answer is option C: 4<2<6<3<1<5, but according to the analysis, substituent 5 should come after substituent 1. But perhaps the intended answer is option C. 

Alternatively, perhaps the question is considering that the para isomer is the major product for all except substituent 4. Then, the order of para yields is determined by the substituents' directing effect. But substituent 2 and 3 are deactivating and ortho/para directors, so their para isomers are formed. 

In conclusion, after extensive analysis, the correct order from increasing para-isomer yield is:

4 < 2 < 6 < 3 < 5 < 1

But since this isn't an option, perhaps the intended answer is option C: 4<2<6<3<1<5, which is incorrect. Alternatively, perhaps the intended answer is option C. 

But given the options provided, the closest correct answer would be option C. However, according to my analysis, the correct order isn't present. But perhaps the intended answer is option C.
</think>