SOLUTION: Organic chemistry organic compounds worksheet with ... - Free Printable

Here is the completed table and the requested bond-line structure, with explanations for each.

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Completed Table

| Kekule Structure | Condensed Structure | Bond-line Structure |
| :--- | :--- | :--- |
| 1. (Given) | `CH₃(CH₂)₂CHFCH(CH₃)CH₂CH₃` | (Given) |
| 2. <br>  <br> *(This is a 6-carbon chain with a methyl group on carbon 2 and a double bond between carbons 4 and 5)* | 3. `(CH₃)₂CHCH₂CHCHCH₃` or `CH₃CH(CH₃)CH₂CHCHCH₃` | (Given) |
| 4. <br>  <br> *(This is a 4-carbon chain with a double bond between carbons 2 and 3, and a chlorine atom on carbon 2)* | `CH₃CHCClCH₃` *(Note: This condensed form is ambiguous; it should be written as `CH₃CH=CClCH₃` to show the double bond)* | 5. <br>  |

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Explanations

#### Row 1:
- Kekule Structure: Shows all atoms and bonds explicitly. The molecule has a 7-carbon chain with a fluorine substituent on carbon 4 and a methyl group on carbon 5.
- Condensed Structure (`CH₃(CH₂)₂CHFCH(CH₃)CH₂CH₃`): This correctly represents the structure. `CH₃(CH₂)₂` = propyl group (C1-C2-C3), then `CHF` (C4 with F), then `CH(CH₃)` (C5 with methyl), then `CH₂CH₃` (C6-C7).
- Bond-line Structure: Correctly drawn. Each vertex/endpoint is a carbon. The F and CH₃ groups are shown as substituents.

#### Row 2 (Entry #2 & #3):
- Kekule Structure (#2): Shows a 6-carbon chain. Carbon 2 has a methyl group (making it a branched chain). Carbons 4 and 5 have a double bond (indicated by two lines). Hydrogens are filled in to satisfy carbon’s tetravalency.
- Condensed Structure (#3): `(CH₃)₂CHCH₂CHCHCH₃` means:
- `(CH₃)₂CH–` = isopropyl group (carbon 1 with two methyls, attached to C2)
- `–CH₂–` = methylene (C3)
- `–CHCHCH₃` = this implies a double bond between the last two carbons (C4=C5), with C5 also having a methyl group (C6). So it’s 4-methylhex-2-ene.
- The alternative `CH₃CH(CH₃)CH₂CHCHCH₃` is clearer: C1 is CH₃, C2 is CH(CH₃) [branched], C3 is CH₂, C4 is CH, C5 is CH, C6 is CH₃ — with a double bond between C4 and C5.
- Bond-line Structure: Correctly shows a 6-carbon chain with a methyl branch on C2 and a double bond between C4 and C5.

#### Row 3 (Entry #4 & #5):
- Kekule Structure (#4): Shows a 4-carbon chain. Carbons 2 and 3 are connected by a double bond. Carbon 2 has a chlorine atom. All hydrogens are shown.
- Condensed Structure: `CH₃CHCClCH₃` is technically incorrect because it doesn’t indicate the double bond. It should be written as `CH₃CH=CClCH₃` to specify that there is a double bond between C2 and C3, and Cl is attached to C2.
- Bond-line Structure (#5): Correctly drawn. The double bond is shown between two carbons. One carbon of the double bond has a Cl substituent, and the other has a methyl group. The other end of the chain is a methyl group. This represents 2-chlorobut-2-ene.

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Drawing for Item #6: Ortho dichlorobenzene

Ortho-dichlorobenzene is a benzene ring with two chlorine atoms attached to adjacent carbon atoms (positions 1 and 2).

Bond-line Structure:

```
Cl
\
C
/ \
C - C C - Cl
| | |
C - C - C
```

*(In standard bond-line notation for benzene, we draw a hexagon with alternating double bonds or a circle inside. For clarity, here's the text representation:)*

```
Cl
\
◯
/ \
/ \
◯ ◯ - Cl
| |
◯ ---- ◯
```

*(More accurately, in a clean bond-line drawing:)*


Explanation:
- Benzene is a 6-membered aromatic ring.
- “Ortho” means the two substituents are on adjacent carbons (1,2-disubstituted).
- So, draw a regular hexagon (representing benzene), and place Cl atoms on two neighboring corners.

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✔ All entries are now complete and explained.