Solved Chemistry 2310 Worksheet 4 5) Classify each pair of | Chegg.com - Free Printable

Let’s go through each pair one by one, analyzing their structures and stereochemistry to classify them as constitutional isomers, enantiomers, diastereomers, identical molecules, or not isomers.

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a)

Structures:

Left:
- Chiral center with Br, H, CH₃, CH₂CH₃
- Configuration: Let's assign priorities: Br (1), CH₂CH₃ (2), CH₃ (3), H (4) → clockwise → R

Right:
- Same groups: Br, H, CH₃, CH₂CH₃
- But drawn with Br on the opposite side → if we rotate it mentally, it’s the mirror image → S

✔ They are non-superimposable mirror images → ENANTIOMERS

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b)

Both are cyclopentenes with OH and CH₃ substituents. The double bond fixes the ring conformation.

Check positions:
- In both, OH is on one carbon, CH₃ on an adjacent carbon.
- The stereochemistry at the chiral centers? Actually, the double bond makes the ring flat in that region — but look closer: both have the OH and CH₃ on the same side relative to the ring plane? Actually, they are drawn identically.

Wait — let’s check if there’s any difference. Both show:

- OH group up (wedge), CH₃ group down (dash) — or vice versa? Actually, in both drawings, the OH is on a wedge and CH₃ is on a dash, and the double bond is between same carbons.

✔ Same connectivity, same stereochemistry → IDENTICAL MOLECULES

*(Note: If the ring were flipped, it might be different, but here both are drawn with same orientation.)*

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c)

Left: 2-methylpentane? Wait — let’s name them.

Left: CH₃–CH(CH₃)–CH₂–CH₂–CH₃ → 2-methylpentane

Right: CH₃–CH₂–CH(CH₃)–CH₂–CH₃ → 3-methylpentane

✔ Same molecular formula (C₆H₁₄), different connectivity → CONSTITUTIONAL ISOMERS

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d)

This is tricky. Let’s analyze the Fischer projections.

Left:
```
OHC
|
HO–H–C–OH
|
CH₂OH
```

Right:
```
HOH₂C
|
HO–H–C–CHO
|
OH
```

Actually, these are the same molecule — just rotated 180° in the plane of paper. In Fischer projections, rotating 180° gives the identical molecule.

Also, check: both have the same groups: aldehyde, CH₂OH, and two OH groups on chiral centers.

The configuration:
In left: top is CHO, bottom is CH₂OH, left OH, right OH — so it’s D-glyceraldehyde? No, this is a 3-carbon sugar.

Actually, both are glyceraldehyde derivatives, and the Fischer projection is identical when rotated.

✔ IDENTICAL MOLECULES

*(Note: Some might think they’re enantiomers, but no — rotating 180° doesn’t invert stereochemistry; it preserves it.)*

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e)

Newman projections.

Left: front carbon has H, H, CH₃; back carbon has CH₂CH₃, CH₃, H

Right: front carbon has H, H, CH₃; back carbon has CH₂CH₃, CH₃, H — but rotated?

Actually, both are staggered conformations of the same molecule: 2-methylbutane? Wait — let’s see.

Front carbon: CH₃–CH–CH₃ (with H’s), back carbon: CH₂CH₃ and H.

Actually, both represent the same compound: 2-methylbutane in staggered conformation.

But are they identical or conformers? Since they differ only by rotation around single bond, they are conformers, which are not considered stereoisomers — they are identical molecules in different conformations.

✔ IDENTICAL MOLECULES

*(Note: Conformers are not classified as stereoisomers; they interconvert rapidly.)*

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f)

Two tetrahedral carbons with Cl, OH, H, and two methyl groups.

Left: Cl and OH are *trans* to each other? Let’s assign R/S.

Actually, both molecules have two chiral centers.

Left:
- Assign priorities: Cl > OH > C(CH₃)₂ > H
- For the chiral center: Cl (1), OH (2), C(CH₃)₂ (3), H (4)
- Looking at left structure: Cl is wedge, OH is dash → if we orient H away, sequence Cl→OH→C is clockwise → R? Wait, need to draw properly.

Alternatively, compare the two:

Left: Cl and OH are on opposite sides (one wedge, one dash) → *anti*

Right: Cl and OH are both on same side? No — in right, Cl is wedge, OH is dash — same as left!

Wait — actually, both drawings are identical — same spatial arrangement.

✔ IDENTICAL MOLECULES

*(If you rotate the right one 180°, it matches the left.)*

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g)

Fischer projections again.

Left:
```
COOH
|
HO–H–C–OH
|
COOH
```

Right:
```
COOH
|
HO–H–C–OH
|
COOH
```

Wait — both are identical? But look: in left, top COOH, then H–OH, then bottom COOH — standard tartrate.

Right: same thing.

But wait — in some representations, the right one might be drawn differently? Actually, both are meso-tartaric acid? No — meso has internal symmetry.

Actually, both are identical — same Fischer projection.

But let’s check if they are enantiomers: no, because if you flip one, it’s the same.

✔ IDENTICAL MOLECULES

*(Note: If one was the mirror image, it would be enantiomer — but here both are drawn the same way.)*

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## ✔ Final Answers:

a) Enantiomers
b) Identical molecules
c) Constitutional isomers
d) Identical molecules
e) Identical molecules
f) Identical molecules
g) Identical molecules

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## 💡 Key Concepts Recap:

- Constitutional isomers: same formula, different connectivity.
- Enantiomers: non-superimposable mirror images (chiral centers inverted).
- Diastereomers: stereoisomers that are not mirror images (e.g., cis/trans, different R/S combinations).
- Identical: same connectivity AND same spatial arrangement (including conformation if rapid interconversion).
- Not isomers: different formulas.

In this set, no diastereomers or “not isomers” appear — all pairs are either constitutional isomers, enantiomers, or identical.

Let me know if you’d like 3D models or R/S assignments for any!