Macromolecules worksheet: Fill out & sign online | DocHub - Free Printable

To solve this question, we need to understand what structural isomers are.

---

🔹 Definition: Structural Isomers

Structural isomers (also called constitutional isomers) are molecules that have the same molecular formula but different connectivity of atoms — meaning the atoms are bonded together in different orders or arrangements.

So, for two molecules to be structural isomers:
- They must have the same number and types of atoms (same molecular formula).
- But their bonding patterns differ (i.e., different structures).

---

Now let’s analyze each pair in the options:

---

🔎 Option A: Molecules 1 and 4

- Molecule 1: This is a 6-carbon sugar with an aldehyde group at the end. It's glucose in open-chain form.
- Molecule 4: Also a 6-carbon chain with an aldehyde group at one end and hydroxyl groups along the chain. This is also glucose, just drawn differently? Wait — actually, let's compare.

But wait — both 1 and 4 look like glucose in linear form. Let's check more carefully.

Actually, Molecule 1 has the structure:
```
H-C=O
|
HO-C-H
|
H-C-OH
|
HO-C-H
|
H-C-OH
|
H-C-OH
|
H
```
That's D-glucose in open-chain form.

Molecule 4:
```
H-C=O
|
H-C-OH
|
H-C-OH
|
H-C-OH
|
H-C-OH
|
H-C-OH
|
H
```
Wait — this looks like a straight chain of 6 carbons, all with OH groups, but all the hydroxyls are on the same side?

No — actually, this is not glucose. The stereochemistry is different.

But here's the key: both 1 and 4 have C₆H₁₂O₆ — same molecular formula.

But do they have different bonding arrangements?

Yes — molecule 1 has alternating OH and H positions, while molecule 4 has all OH groups on one side — but that’s not possible for glucose.

Wait — actually, looking closely, molecule 4 appears to be an aldose sugar with all hydroxyls on the right, which might represent D-glucose in Fischer projection? Actually, no — D-glucose has specific stereochemistry.

But regardless, molecules 1 and 4 both appear to be hexoses (6-carbon sugars), and if they have the same formula but different arrangement of functional groups, they could be isomers.

But upon closer inspection, molecule 1 and molecule 4 are both glucose in open-chain form, just drawn differently? Or are they?

Wait — molecule 1 has OH on C2 pointing left, C3 pointing right, etc., while molecule 4 has all OH groups on the right side — that would make it an enantiomer or diastereomer, not necessarily a structural isomer.

But structural isomers differ in connectivity, not just spatial arrangement.

Let’s move on — maybe there’s a better pair.

---

🔎 Option B: Molecules 5 and 14

- Molecule 5: This looks like phosphatidylcholine, a phospholipid. It has:
- Glycerol backbone
- Two fatty acid chains
- A phosphate group linked to a choline group (N(CH₃)₃)

- Molecule 14: This is a polysaccharide chain — multiple glucose units linked by glycosidic bonds.

Clearly, very different formulas and functions — one is a lipid, the other a carbohydrate.

Not isomers.

✘ Not structural isomers.

---

🔎 Option C: Molecules 6 and 12

- Molecule 6: This is a sugar ring — fructose in cyclic form (5-membered ring). It has a ketone group.

- Molecule 12: This is a sugar ring — glucose in cyclic form (6-membered ring). It has an aldehyde group.

Both are C₆H₁₂O₆, same molecular formula.

But different functional groups and ring sizes:
- Glucose is a pyranose (6-membered ring)
- Fructose is a furanose (5-membered ring)

And fructose is a ketose, glucose is an aldose

So, even though they have the same molecular formula, they are constitutional isomers — because the functional group placement differs (aldehyde vs ketone) and ring structure differs.

✔ So, 6 and 12 are structural isomers.

This is a strong candidate.

---

🔎 Option D: Molecules 12 and 13

- Molecule 12: Glucose (cyclic form), C₆H₁₂O₆
- Molecule 13: A nitrogen-containing ring — this is cytosine, a nitrogenous base (C₄H₅N₃O)

Different molecular formulas → not isomers

✘ Not structural isomers.

---

🔎 Option E: Molecules 14 and 15

- Molecule 14: A long chain of glucose units — likely starch or glycogen, polysaccharide
- Molecule 15: Two glucose units linked together — maltose

So:
- Molecule 14: (C₆H₁₀O₅)ₙ → polymer
- Molecule 15: C₁₂H₂₂O₁₁

They are not the same molecular formula, so not isomers.

✘ Not structural isomers.

---

✔ Final Evaluation

Only Option C (6 and 12) involves two molecules with:
- Same molecular formula (C₆H₁₂O₆)
- Different connectivity (fructose vs glucose)
- Different functional groups (ketose vs aldose)
- Different ring structures (furanose vs pyranose)

Thus, they are structural isomers.

---

✔ Correct Answer: C. 6 and 12

---

🟩 Explanation:

Molecule 6 is fructose (a ketose), and Molecule 12 is glucose (an aldose). Both have the molecular formula C₆H₁₂O₆, but differ in the position of the carbonyl group (ketone vs aldehyde) and ring structure, making them structural isomers.

---

✔ Final Answer: C. 6 and 12