Worksheet for naming branched alkanes in organic chemistry.

Let’s go through each structure one by one and name them using IUPAC rules for branched alkanes.

General Steps:
1. Find the longest continuous carbon chain (parent chain).
2. Number the chain so that substituents get the lowest possible numbers.
3. Name and number the branches (alkyl groups like methyl, ethyl, etc.).
4. List substituents in alphabetical order (ignore prefixes like di-, tri- when alphabetizing).
5. Combine everything: [substituents] + [parent chain name].

---

Problem 1:
Structure:
```
CH₃
|
H₃C–CH–CH₃
```
Longest chain = 3 carbons → propane
One methyl group on carbon #2 → 2-methylpropane

But wait — this is actually a common molecule called “isobutane”, but IUPAC name is 2-methylpropane.

✔ Final: 2-methylpropane

---

Problem 2:
```
CH₂–CH₃
|
H₃C–CH–CH₃
```
Longest chain? Let’s count:
From left to right: C–C–C = 3, but if we go down the branch: C–C–C–C = 4! So parent chain is butane.

Numbering: Start from end closest to branch. If we number left to right:
Carbon 1: CH₃–
Carbon 2: –CH– with ethyl group? Wait — let’s redraw mentally:

Actually, the structure is:
```
CH₂–CH₃
|
CH₃–CH–CH₃
```
That’s a central carbon attached to: H (implied), CH₃ (left), CH₃ (right), and CH₂CH₃ (down). So longest chain is 4 carbons: start at bottom ethyl → up to center → to either side methyl → total 4.

So: chain is butane. The branch is a methyl group on carbon #2.

Wait — better way: pick longest straight chain. Here, you can have:
- Left CH₃ – center CH – right CH₃ → 3 carbons
- Or bottom CH₂–CH₃ – center CH – left or right CH₃ → still 4 carbons? Actually, no — it's only 3 in a row unless you bend.

Actually, correct longest chain: start at top-left CH₃, go to center CH, then down to CH₂–CH₃ → that’s 4 carbons: C1–C2–C3–C4.

So:
C1: CH₃– (from left)
C2: –CH– (center)
C3: –CH₂–
C4: –CH₃

And there’s a methyl group attached to C2 (the other CH₃ on the right).

So: parent = butane
Substituent: methyl on carbon #2 → 2-methylbutane

✔ Final: 2-methylbutane

---

Problem 3:
```
CH₂–CH₃
|
H₃C–CH₂–CH₂–CH–CH₂–CH₂–CH₃
```
Longest chain: horizontal has 7 carbons. Branch is ethyl on carbon #4.

Numbering: left to right gives substituent at #4; right to left also gives #4 → same.

Parent: heptane
Substituent: ethyl at #4 → 4-ethylheptane

✔ Final: 4-ethylheptane

---

Problem 4:
```
CH₂–CH₃
|
H₃C–CH₂–CH₂–CH–CH–CH₂–CH₃
|
CH₃
```
Find longest chain. Horizontal: 7 carbons? Let’s see:

Left to right: C1–C2–C3–C4–C5–C6–C7 → yes, 7 carbons.

At C4: has a methyl group (CH₃)
At C5: has an ethyl group (CH₂–CH₃)

Wait — actually, looking again:

The structure is:
```
CH₂–CH₃ ← this is on C5?
|
...–CH–CH–...
| |
CH₃ ?
```

Actually, written as:
```
CH₂–CH₃
|
H₃C–CH₂–CH₂–CH–CH–CH₂–CH₃
|
CH₃
```

So the main chain is 7 carbons: positions 1 to 7.

On carbon #4: a methyl group (CH₃)
On carbon #5: an ethyl group (CH₂CH₃)

But we must number to give lowest numbers to substituents.

If we number left to right: substituents at 4 and 5
If we number right to left: then what was C5 becomes C3, C4 becomes C4 → substituents at 3 and 4 → better!

So renumber:

Right to left:
C1: rightmost CH₃
C2: CH₂
C3: CH (with ethyl group)
C4: CH (with methyl group)
C5: CH₂
C6: CH₂
C7: CH₃

So substituents: ethyl at C3, methyl at C4.

Alphabetical: ethyl before methyl → 3-ethyl-4-methylheptane

✔ Final: 3-ethyl-4-methylheptane

---

Problem 5:
```
CH₂–CH₂–CH₃
|
H₃C–CH₂–CH–CH₂–CH–CH₂–CH₂–CH₃
|
CH₃
```
Longest chain: let’s trace.

Horizontal: from left CH₃ to right CH₃ → 8 carbons? Let’s count:

C1: left CH₃
C2: CH₂
C3: CH (with CH₃)
C4: CH₂
C5: CH (with propyl group: CH₂–CH₂–CH₃)
C6: CH₂
C7: CH₂
C8: CH₃ → yes, 8 carbons → octane

Substituents:
- Methyl on C3
- Propyl on C5

Numbering: left to right → 3 and 5
Right to left → would be 4 and 6 → worse → so keep left to right.

Alphabetical: methyl before propyl → 3-methyl-5-propyloctane

Wait — propyl is "propyl", methyl is "methyl" → m before p → yes.

But note: sometimes people write in numerical order regardless, but IUPAC says alphabetical ignoring prefixes.

So: 3-methyl-5-propyloctane

✔ Final: 3-methyl-5-propyloctane

---

Problem 6:
```
CH₃ CH₃
| |
H₃C–CH₂–C–CH₂–CH–CH₂–CH₃
|
CH₃
```
Longest chain: let’s find.

Start from left: C1–C2–C3–C4–C5–C6–C7 → 7 carbons? But look:

C3 has two methyls and connects to C4.

Actually, chain could be longer? No, max straight is 7.

Structure:
C1: CH₃–
C2: –CH₂–
C3: –C– (with two CH₃ groups)
C4: –CH₂–
C5: –CH– (with one CH₃)
C6: –CH₂–
C7: –CH₃

So parent: heptane

Substituents:
- Two methyls on C3 → dimethyl
- One methyl on C5

So: 3,3-dimethyl and 5-methyl

Combine: 3,3,5-trimethylheptane? Wait — three methyl groups total.

List: 3,3-dimethyl and 5-methyl → so 3,3,5-trimethylheptane

Check numbering: if we number right to left:

C1: right CH₃
C2: CH₂
C3: CH (with methyl) → now at C3
C4: CH₂
C5: C (with two methyls) → at C5
C6: CH₂
C7: CH₃

Then substituents at 3 and 5,5 → same set: 3,5,5 vs original 3,3,5 → which is lower? Compare first different number: 3,3,5 vs 3,5,5 → 3,3,5 is lower because second number 3 < 5.

So original numbering is better.

✔ Final: 3,3,5-trimethylheptane

---

Problem 7:
```
CH₃
|
H₃C–CH₂–C–CH₃
|
CH₃
```
Central carbon with three methyls and one ethyl? Wait:

Structure:
```
CH₃
|
CH₃–CH₂–C–CH₃
|
CH₃
```

So central C bonded to: CH₂CH₃, CH₃, CH₃, CH₃ → four groups.

Longest chain: from ethyl through central to any methyl → 3 carbons? But wait:

Chain: CH₃–CH₂–C–CH₃ → that’s 4 carbons if you include one methyl.

Actually: pick longest continuous chain. You can have:

Option 1: CH₃–CH₂–C(CH₃)–CH₃ → but that’s not straight.

Better: the longest straight chain is 3 carbons? No — let’s think.

Actually, the central carbon is tertiary butyl-like.

Standard way: the longest chain is 3 carbons? But we can make a 4-carbon chain.

Try: start at left CH₃ (of ethyl) → CH₂ → central C → one of the methyls → that’s 4 carbons.

So parent: butane

Now, the central carbon is C2 or C3?

Set chain: C1–C2–C3–C4

Where:
C1: CH₃– (from ethyl part)
C2: –CH₂–
C3: –C– (central)
C4: –CH₃ (one of the methyls)

But C3 also has two more methyl groups → so two methyl substituents on C3.

So: 3,3-dimethylbutane? But wait — that would imply chain of 4, with two methyls on C3.

But standard name for this is actually 2,2-dimethylbutane? Let me check.

I think I messed up numbering.

Better approach: the molecule is (CH₃)₃C–CH₂CH₃? No, here it’s CH₃CH₂–C(CH₃)₃? Wait no:

Given: H₃C–CH₂–C(CH₃)₂–CH₃? That’s not right.

Looking back:

It’s written as:
```
CH₃
|
H₃C–CH₂–C–CH₃
|
CH₃
```

So the central carbon is bonded to:
- CH₂CH₃ (ethyl)
- CH₃ (top)
- CH₃ (bottom)
- CH₃ (right)

So it’s C with four alkyl groups: one ethyl and three methyls.

Longest chain: take ethyl + central + one methyl → 3 carbons? But we can do better.

Actually, the longest continuous chain is 3 carbons only if you don’t bend, but in organic naming, we allow choosing the longest path even if bent? No — it must be continuous without branching.

Standard rule: find the longest straight chain of atoms connected in a row.

Here, you can have: CH₃–CH₂–C–CH₃ → that’s 4 atoms in a row: C–C–C–C.

Yes! So chain of 4 carbons.

Call it: C1–C2–C3–C4

Set C1 = left CH₃ (of ethyl)
C2 = CH₂
C3 = central C
C4 = right CH₃

Then C3 has two additional methyl groups → so two methyl substituents on C3.

Thus: 3,3-dimethylbutane

But wait — that’s incorrect because in butane, C3 is equivalent to C2 if numbered from other end.

If we number from right: C1 = right CH₃, C2 = central C, C3 = CH₂, C4 = CH₃

Then substituents: two methyls on C2 → 2,2-dimethylbutane

Which is correct? And which numbering gives lower numbers?

2,2 vs 3,3 → 2,2 is lower → so 2,2-dimethylbutane

Yes, that’s the standard name for this compound.

✔ Final: 2,2-dimethylbutane

---

Problem 8:
```
CH₃
|
CH₃–C–CH₃
|
CH₃
```
Central carbon with four methyl groups → tetramethylmethane? But methane has only one carbon.

Longest chain: any two methyls make ethane? But we need longest continuous chain.

Actually, the longest chain is 2 carbons? No — you can’t have a chain longer than 1 carbon because all are attached to central.

Wait — this is neopentane.

Standard: the parent chain is propane? How?

Actually, you can consider: pick any three carbons in a row? But they’re not in a row.

Correct way: the longest continuous chain is 3 carbons? For example: CH₃–C–CH₃, but the middle C is quaternary.

In IUPAC, for (CH₃)₄C, the longest chain is 3 carbons: for instance, take one methyl as C1, central as C2, another methyl as C3 → but then C2 has two more methyls.

So: parent = propane
Substituents: two methyls on C2 → 2,2-dimethylpropane

Yes, that’s standard.

✔ Final: 2,2-dimethylpropane

---

Problem 9:
```
CH₂–CH₃
|
H₃C–CH₂–C–CH₂–CH₃
|
CH₂–CH₃
```
Central carbon with: ethyl (top), ethyl (bottom), ethyl (right), and CH₂CH₃ left? Wait:

Structure:
Left: H₃C–CH₂–
Attached to central C
Central C also attached to: CH₂–CH₃ (top), CH₂–CH₃ (bottom), CH₂–CH₃ (right)

So central C bonded to four ethyl groups? No:

Written as:
```
CH₂–CH₃
|
H₃C–CH₂–C–CH₂–CH₃
|
CH₂–CH₃
```

So groups on central C:
- Left: CH₂CH₃ (since H₃C–CH₂– is ethyl)
- Top: CH₂CH₃
- Bottom: CH₂CH₃
- Right: CH₂CH₃

So it’s C(CH₂CH₃)₄ — tetraethylmethane? But again, longest chain.

Longest continuous chain: take one ethyl group through central to another ethyl → e.g., CH₃–CH₂–C–CH₂–CH₃ → that’s 5 carbons: C1–C2–C3–C4–C5

Where C3 is central.

Then C3 has two more ethyl groups → each is a substituent.

So parent: pentane
Substituents: two ethyl groups on C3 → 3,3-diethylpentane

Check numbering: symmetric, so 3,3 is fine.

✔ Final: 3,3-diethylpentane

---

Problem 10:
```
CH₂
|
H₃C–CH₂–CH₂–CH₂–CH₂
|
H₃C–CH₂–CH₂–CH₂–C–CH₂–CH₃
|
CH₃
```
This looks messy. Let me interpret.

It seems like:
There is a central carbon that has:
- Up: CH₂ (which might be part of a chain?)
- Down: CH₃
- Left: CH₂–CH₂–CH₂–CH₂–CH₃? Wait no.

Looking closely:

The structure is drawn as:

Top line: H₃C–CH₂–CH₂–CH₂–CH₂– (and then a vertical bond down to a CH₂? But it says "CH₂" above, then below that the main part.

Actually, reading the diagram:

It appears to be:

A carbon atom (let's call it C*) that is bonded to:
- A CH₃ group (below)
- A CH₂–CH₃ group (right)
- A CH₂–CH₂–CH₂–CH₂–CH₃ group (left? but written as H₃C–CH₂–CH₂–CH₂–CH₂– attached to C*)
- And above, a CH₂ group, which is probably meant to be part of another chain? But it's written as "CH₂" alone.

Wait, perhaps it's:

The central carbon is bonded to four groups:
1. CH₃ (methyl)
2. CH₂CH₃ (ethyl)
3. CH₂CH₂CH₂CH₂CH₃ (pentyl)
4. CH₂– something? But it says "CH₂" and then nothing — likely a typo or misdrawn.

Looking back at user input:

"H₃C–CH₂–CH₂–CH₂–CH₂
|
H₃C–CH₂–CH₂–CH₂–C–CH₂–CH₃
|
CH₃"

And above the central C, there is "CH₂" with a bond down to the central C.

So central C is bonded to:
- Above: CH₂– ? But it doesn't connect further — probably it's a methylene group, but that would mean it's -CH₂-, implying it's part of a chain.

Perhaps it's intended to be:

The group above is -CH₂-CH₃ or something? But it's written as just "CH₂".

Another interpretation: maybe the "CH₂" above is meant to be a ethyl group or something.

Let me assume that the central carbon has four substituents:
- From top: CH₂–R, but R is not specified — in the text, it's "CH₂" alone, which is incomplete.

Perhaps it's a mistake, and it's supposed to be a ethyl or methyl.

Looking at the drawing style, in problem 10, it's likely that the central carbon is bonded to:
- A pentyl group: H₃C–CH₂–CH₂–CH₂–CH₂– (so 5 carbons)
- An ethyl group: –CH₂–CH₃
- A methyl group: –CH₃
- And a butyl group? The left part is H₃C–CH₂–CH₂–CH₂– which is butyl, but it's attached how?

The way it's written:

First line: H₃C–CH₂–CH₂–CH₂–CH₂– [this is pentyl, 5 carbons]
Then a vertical bond down to a carbon (call it C1)
Then from C1, it goes to: H₃C–CH₂–CH₂–CH₂–C–CH₂–CH₃, with C having a CH₃ below.

This is confusing.

Perhaps it's:

The central carbon is the one with the CH₃ below and CH₂CH₃ right, and it is also bonded to a CH₂– (above) which is part of a chain going left: but the left chain is written as H₃C–CH₂–CH₂–CH₂–CH₂– attached to that CH₂? That would make it a hexyl group or something.

Let's try to parse the connectivity.

Assume the central atom is a carbon (C*) that has four bonds:
1. Bond to CH₃ (below)
2. Bond to CH₂–CH₃ (right)
3. Bond to CH₂– (above) — and this CH₂ is further bonded to CH₂–CH₂–CH₂–CH₃? But in the text, it's "H₃C–CH₂–CH₂–CH₂–CH₂–" which is five carbons, so if attached to the CH₂ above, then the group is -CH₂–CH₂–CH₂–CH₂–CH₃, which is pentyl, but that would be six carbons including the CH₂.

If the group above is -CH₂–R where R is CH₂CH₂CH₂CH₃, then it's -CH₂–CH₂–CH₂–CH₂–CH₃, which is pentyl group (5 carbons in chain, but the first carbon is the CH₂ attached to C*).

Standard: the group -CH₂–CH₂–CH₂–CH₂–CH₃ is pentyl.

Similarly, the left part: "H₃C–CH₂–CH₂–CH₂–" is butyl, but it's attached to C* directly? In the diagram, it's written as:

"H₃C–CH₂–CH₂–CH₂–CH₂–" on top line, then vertical bond down to C*, and then "H₃C–CH₂–CH₂–CH₂–C–CH₂–CH₃" on bottom, with C* being the C in "C–CH₂–CH₃" and also bonded to CH₃ below.

So C* is bonded to:
- The CH₂ from the top chain: which is part of H₃C–CH₂–CH₂–CH₂–CH₂– , so that's a pentyl group (since H₃C–CH₂–CH₂–CH₂–CH₂– is C5H11-, pentyl)
- The left part: H₃C–CH₂–CH₂–CH₂– is butyl group (C4H9-)
- Right: CH₂–CH₃ (ethyl)
- Below: CH₃ (methyl)

But that's four groups: pentyl, butyl, ethyl, methyl.

So the central carbon has four different alkyl groups.

To name, find the longest chain among these.

The longest group is pentyl (5 carbons), but we can make a longer chain by going through the central carbon.

For example, take the pentyl group: CH₃–CH₂–CH₂–CH₂–CH₂– attached to C*, then from C* go to butyl group: –CH₂–CH₂–CH₂–CH₃, so the chain would be CH₃–CH₂–CH₂–CH₂–CH₂–C*–CH₂–CH₂–CH₂–CH₃, which is 10 carbons? Let's count:

From end of pentyl: C1 (CH₃)-C2-C3-C4-C5 (CH₂)-C6 (C*)-C7 (CH₂)-C8-C9-C10 (CH₃) — yes, 10 carbons.

And C6 (central) also has two other groups: ethyl and methyl.

So parent chain: decane

Substituents on C6: ethyl and methyl

Numbering: the chain is symmetric? From left to right or right to left, C6 is the same position.

If we number from left: C1 to C10, C6 is carbon 6.

Substituents: ethyl and methyl on C6.

So: 6-ethyl-6-methyldecane

Alphabetical: ethyl before methyl → 6-ethyl-6-methyldecane

Is that correct? Let's confirm the chain length.

Pentyl group: 5 carbons, but when attached, the carbon attached is included in the chain.

In the chain I described: starting from the methyl end of the "pentyl" part:
- C1: CH₃- (end of what was called pentyl)
- C2: CH₂
- C3: CH₂
- C4: CH₂
- C5: CH₂- (attached to C*)
- C6: C* (central)
- C7: CH₂- (first of butyl group)
- C8: CH₂
- C9: CH₂
- C10: CH₃

Yes, 10 carbons.

And on C6, besides being part of the chain, it has two additional groups: ethyl (CH₂CH₃) and methyl (CH₃).

So yes, 6-ethyl-6-methyldecane

✔ Final: 6-ethyl-6-methyldecane

---

Now, compiling all answers:

1. 2-methylpropane
2. 2-methylbutane
3. 4-ethylheptane
4. 3-ethyl-4-methylheptane
5. 3-methyl-5-propyloctane
6. 3,3,5-trimethylheptane
7. 2,2-dimethylbutane
8. 2,2-dimethylpropane
9. 3,3-diethylpentane
10. 6-ethyl-6-methyldecane

Final Answer:
1. 2-methylpropane
2. 2-methylbutane
3. 4-ethylheptane
4. 3-ethyl-4-methylheptane
5. 3-methyl-5-propyloctane
6. 3,3,5-trimethylheptane
7. 2,2-dimethylbutane
8. 2,2-dimethylpropane
9. 3,3-diethylpentane
10. 6-ethyl-6-methyldecane