Naming branched alkanes worksheet with structural formulas.

Let's solve each of these branched alkane naming problems step by step using IUPAC nomenclature rules.

---

Rules for Naming Branched Alkanes:

1. Find the longest continuous carbon chain (parent chain).
2. Number the chain to give substituents the lowest possible numbers.
3. Identify and name the alkyl groups (branches).
4. List substituents in alphabetical order.
5. Use prefixes like di-, tri-, etc., for identical groups.
6. Write the name with substituents in order, followed by the parent chain name.

---

Problem 1:

```
H₃C
|
H₃C—CH—CH₃
```

- Longest chain: 3 carbons → propane
- A methyl group is attached to carbon #2
- Name: 2-methylpropane

✔ Answer: 2-methylpropane

---

Problem 2:

```
H₃C
|
H₃C—CH—CH₃
|
CH₂—CH₃
```

- Let’s count the longest chain:
- From left: H₃C–CH–CH₃ → only 3
- But if we go down the branch: H₃C–CH–CH₂–CH₃ → that’s 4 carbons!
- So the longest chain is 4 carbons: butane

Now re-draw:

```
CH₃
|
CH₃–CH–CH₂–CH₃
```

- The longest chain is 4 carbons → butane
- Methyl group on carbon #2
- Numbering from left or right? Both ends are equivalent → methyl on C2
- Name: 2-methylbutane

✔ Answer: 2-methylbutane

---

Problem 3:

```
H₃C—CH₂—CH₂—CH—CH₂—CH₂—CH₃
|
CH₂—CH₃
```

- Longest chain: 7 carbons → heptane
- Branch: ethyl group (–CH₂–CH₃) on carbon #4
- Numbering: start from left or right?
- Left: ethyl at C4
- Right: ethyl at C4 → same
- So it's 4-ethylheptane

✔ Answer: 4-ethylheptane

---

Problem 4:

```
CH₂—CH₃
|
H₃C—CH₂—CH₂—CH—CH—CH₂—CH₃
| |
CH₃ CH₃
```

- Longest chain: let's see
- Main chain: H₃C–CH₂–CH₂–CH–CH–CH₂–CH₃ → 7 carbons
- But look at the branches: one has a –CH₂–CH₃ and one has –CH₃
- Can we find a longer chain?

Try going through the ethyl group:
- Start from left: H₃C–CH₂–CH₂–CH–CH₂–CH₃ → then down to CH₃ → no
- Or: H₃C–CH₂–CH₂–CH–CH–CH₂–CH₃ → still 7

Wait — what if we go from the ethyl branch?
- CH₃–CH₂–CH–CH–CH₂–CH₃ → that's not longer

So longest chain is 7 carbons → heptane

Now number the chain so substituents get lowest numbers.

Current numbering:
- Carbon 1: leftmost CH₃
- C2: CH₂
- C3: CH₂
- C4: CH (with CH₃)
- C5: CH (with CH₂–CH₃ and CH₃)
- C6: CH₂
- C7: CH₃

Substituents:
- At C4: methyl group
- At C5: methyl and ethyl

But wait — two substituents on C5? No — actually:
- C4 has a methyl
- C5 has a methyl and an ethyl? No — C5 is bonded to:
- C4
- C6
- CH₃
- CH₂–CH₃ → yes! So C5 has two alkyl groups: methyl and ethyl

But we can’t have two different groups on one carbon unless we name them properly.

Let’s try numbering from the other end:

Reverse:
- C1: rightmost CH₃
- C2: CH₂
- C3: CH (with CH₃)
- C4: CH (with CH₂–CH₃ and CH₃)
- C5: CH₂
- C6: CH₂
- C7: CH₃

Now:
- Substituents:
- C3: methyl
- C4: methyl and ethyl

So positions: methyl at C3 and C4, ethyl at C4

Compare both directions:
- Original: methyl at C4, methyl and ethyl at C5 → substituents at C4, C5
- Reverse: methyl at C3, methyl and ethyl at C4 → substituents at C3, C4

Better! Lower numbers: C3 and C4 vs C4 and C5

So use reverse numbering.

Now list substituents:
- At C3: methyl
- At C4: methyl and ethyl

Alphabetical order: ethyl before methyl

So: 3-methyl-4-ethylheptane? Wait — but ethyl and methyl both on C4?

No — the carbon at position 4 has two groups: methyl and ethyl → so it’s a carbon with two substituents.

But we need to assign names correctly.

Actually, the structure is:

```
CH₃ CH₂CH₃
| |
CH₃CH₂CH₂–CH–CH–CH₂CH₃
```

Wait — no, original:

```
H₃C–CH₂–CH₂–CH–CH–CH₂–CH₃
| |
CH₃ CH₂–CH₃
```

Wait — the second CH is bonded to:
- CH₃ (methyl)
- CH₂–CH₃ (ethyl)

So it's:

```
H₃C–CH₂–CH₂–CH–CH–CH₂–CH₃
| |
CH₃ CH₂–CH₃
```

So carbon 4: CH with CH₃
Carbon 5: CH with CH₂–CH₃ and CH₃? No — wait, it's:

Wait — the diagram shows:

```
CH₂—CH₃
|
H₃C—CH₂—CH₂—CH—CH—CH₂—CH₃
| |
CH₃ CH₃
```

Ah! So:
- Carbon 4: bonded to CH₃ (one methyl)
- Carbon 5: bonded to CH₃ and CH₂–CH₃ → so it has a methyl and ethyl group

So carbon 5 has two alkyl groups: methyl and ethyl

So the chain is 7 carbons.

Numbering from left:
- C1: left CH₃
- C2: CH₂
- C3: CH₂
- C4: CH(CH₃)
- C5: CH(CH₃)(CH₂CH₃)
- C6: CH₂
- C7: CH₃

Substituents:
- C4: methyl
- C5: methyl and ethyl

Now number from right:
- C1: right CH₃
- C2: CH₂
- C3: CH(CH₃)(CH₂CH₃) → now this is C3
- C4: CH(CH₃)
- C5: CH₂
- C6: CH₂
- C7: CH₃

So:
- C3: methyl and ethyl
- C4: methyl

So substituents at C3 and C4

Compare:
- Left: C4 and C5 → higher numbers
- Right: C3 and C4 → lower numbers → better

So number from right.

Now:
- At C3: methyl and ethyl → so we name as ethyl and methyl groups
- At C4: methyl

So total: two methyls and one ethyl

But one methyl is at C3, one at C4, and ethyl at C3

So:
- Ethyl and methyl at C3
- Methyl at C4

Now list alphabetically:
- Ethyl comes before methyl

So: 3-ethyl-3-methyl-4-methylheptane → but that's redundant

We can combine: 3-ethyl-4,3-dimethylheptane → no, must be alphabetical

Better: 3-ethyl-3,4-dimethylheptane

But wait — two methyls: one at C3, one at C4 → so 3,4-dimethyl

And ethyl at C3

So: 3-ethyl-3,4-dimethylheptane

But IUPAC: when multiple substituents, list in alphabetical order, and use di-, tri-

So: 3-ethyl-3,4-dimethylheptane

But is that correct?

Wait — the carbon at C3 has two substituents: ethyl and methyl → so it's a tertiary carbon

But naming: we list all substituents.

So:
- Ethyl group at C3
- Methyl at C3
- Methyl at C4

So: 3-ethyl-3-methyl-4-methylheptane → but we combine identical groups

So: 3-ethyl-3,4-dimethylheptane

Yes.

But is there a longer chain?

Alternative: could we make a chain including the ethyl group?

For example:
- From the ethyl group: CH₃–CH₂–CH–CH–CH₂–CH₃
- That’s only 6 carbons

Or: from left CH₃–CH₂–CH₂–CH–CH–CH₂–CH₃ → 7

No longer chain.

So 7-carbon chain is longest.

Final name: 3-ethyl-3,4-dimethylheptane

Wait — but check if numbering can be better.

Is there a way to get lower numbers?

If we number from left:
- C1: left CH₃
- C2: CH₂
- C3: CH₂
- C4: CH(CH₃)
- C5: CH(CH₃)(CH₂CH₃)
- C6: CH₂
- C7: CH₃

Then:
- Methyl at C4
- Methyl and ethyl at C5

So: 4-methyl-5-ethyl-5-methylheptane → becomes 4,5-dimethyl-5-ethylheptane

Now compare:
- Option 1: 3-ethyl-3,4-dimethylheptane → substituents at 3 and 4
- Option 2: 4,5-dimethyl-5-ethylheptane → at 4 and 5

Lower numbers: 3 < 4, so first option is better.

So: 3-ethyl-3,4-dimethylheptane

But wait — is this correct?

Wait — the carbon at C3 has two substituents: methyl and ethyl → so it's 3-ethyl-3-methyl, and another methyl at C4 → so 3-ethyl-3,4-dimethylheptane

Yes.

But let’s confirm the structure:

- Chain: C1–C2–C3–C4–C5–C6–C7
- C3 has a methyl group
- C5 has a methyl and ethyl group? No — in our reversed numbering:

Wait — confusion arises.

Let me re-label with correct numbering.

Let’s draw the molecule clearly:

```
CH₃ CH₃
| |
H₃C–CH₂–CH₂–CH–CH–CH₂–CH₃
| |
CH₂–CH₃
```

Wait — no, the diagram says:

```
CH₂—CH₃
|
H₃C—CH₂—CH₂—CH—CH—CH₂—CH₃
| |
CH₃ CH₃
```

So:
- The central carbon (say C4) has:
- CH₃ (methyl)
- CH₂–CH₃ (ethyl)
- and connected to C3 and C5

Wait — no — it's two separate carbons:

- First CH (after CH₂–CH₂) has a CH₃ → so it's CH(CH₃)
- Next CH has two groups: CH₃ and CH₂–CH₃

So the chain is:

C1: CH₃–
C2: –CH₂–
C3: –CH₂–
C4: –CH(CH₃)–
C5: –CH(CH₃)(CH₂CH₃)–
C6: –CH₂–
C7: –CH₃

So C4 has a methyl group
C5 has a methyl and an ethyl group

So substituents:
- Methyl at C4
- Methyl at C5
- Ethyl at C5

Now, number from left:
- C1 to C7: substituents at C4, C5

Number from right:
- C1' = C7 (CH₃)
- C2' = C6 (CH₂)
- C3' = C5 (CH(CH₃)(CH₂CH₃))
- C4' = C4 (CH(CH₃))
- C5' = C3 (CH₂)
- C6' = C2 (CH₂)
- C7' = C1 (CH₃)

So:
- C3': methyl and ethyl
- C4': methyl

So substituents at C3' and C4'

Numbers: 3 and 4 → same as before

But now:
- C3': ethyl and methyl
- C4': methyl

So: 3-ethyl-3-methyl-4-methylheptane → 3-ethyl-3,4-dimethylheptane

Yes.

But is there a longer chain?

Try to go through the ethyl group:

From C4: CH(CH₃)–CH(CH₃)(CH₂CH₃)–CH₂–CH₃

Can we make a chain from the ethyl group?

Start from the end of ethyl: CH₃–CH₂–CH(CH₃)–CH(CH₃)–CH₂–CH₂–CH₃

That’s:
- CH₃–CH₂–CH–CH–CH₂–CH₂–CH₃
| |
CH₃ CH₃

So that’s 7 carbons again → same length

No gain.

So heptane is correct.

Final name: 3-ethyl-3,4-dimethylheptane

But wait — in IUPAC, we list substituents in alphabetical order.

"ethyl" before "methyl"

So: 3-ethyl-3,4-dimethylheptane is correct.

✔ Answer: 3-ethyl-3,4-dimethylheptane

---

Problem 5:

```
H₃C—CH₂—CH—CH₂—CH—CH₂—CH₃
| |
CH₃ CH₂—CH₂—CH₃
```

- Longest chain: 7 carbons → heptane
- Substituents:
- Methyl at C3
- Propyl at C5

Now number from left:
- C1: CH₃
- C2: CH₂
- C3: CH(CH₃)
- C4: CH₂
- C5: CH(CH₂CH₂CH₃)
- C6: CH₂
- C7: CH₃

Substituents: methyl at C3, propyl at C5

Now number from right:
- C1: CH₃ (right)
- C2: CH₂
- C3: CH(CH₂CH₂CH₃)
- C4: CH₂
- C5: CH(CH₃)
- C6: CH₂
- C7: CH₃

So: propyl at C3, methyl at C5

Now compare:
- Left: methyl at C3, propyl at C5 → positions 3 and 5
- Right: propyl at C3, methyl at C5 → positions 3 and 5

Same.

But which gives lower numbers for first substituent?

Both have first substituent at C3 → same

But now, propyl comes before methyl alphabetically

So we want propyl to come first in name

In left numbering: methyl at C3, propyl at C5 → name: 3-methyl-5-propylheptane

In right numbering: propyl at C3, methyl at C5 → 3-propyl-5-methylheptane

Now, since propyl comes before methyl, we prefer 3-propyl-5-methylheptane

But wait — is that correct?

Yes, because we list substituents in alphabetical order.

So: 3-propyl-5-methylheptane

But is there a longer chain?

Try going through the propyl group:

From propyl: CH₃–CH₂–CH₂–CH–CH₂–CH₂–CH₃ → that’s 7 carbons again

No longer chain.

So yes.

But wait — the propyl group is n-propyl, but is it attached to C5?

Yes.

But can we call it pentyl? No — it’s three carbons.

So propyl is fine.

But note: propyl is not preferred; n-propyl is acceptable.

But in IUPAC, we just say propyl.

So name: 3-propyl-5-methylheptane

But wait — is this correct?

Wait — the chain is:

C1–C2–C3–C4–C5–C6–C7

With methyl on C3, propyl on C5

But is there a better numbering?

We already tried both directions — same locants.

But let’s see: can we number so that both substituents get lower numbers?

Currently: 3 and 5 → sum = 8

Is there a way to get 2 and 4? No.

So 3-propyl-5-methylheptane is correct.

But wait — is the propyl group really on C5?

Yes.

But let’s double-check: the structure is:

```
H₃C–CH₂–CH–CH₂–CH–CH₂–CH₃
| |
CH₃ CH₂–CH₂–CH₃
```

So:
- C3: CH(CH₃)
- C5: CH(CH₂CH₂CH₃)

Yes.

So name: 3-methyl-5-propylheptane or 3-propyl-5-methylheptane

Since propyl comes before methyl, we use 3-propyl-5-methylheptane

But wait — in IUPAC, we don't prioritize based on location, but we must list in alphabetical order

So even if propyl is at C3 and methyl at C5, we write: 3-propyl-5-methylheptane

Yes.

✔ Answer: 3-propyl-5-methylheptane

---

Problem 6:

```
CH₂
|
H₃C—CH₂—CH₂—CH₂—C—CH₂—CH₃
| | |
CH₂—CH₂—CH₂—CH₃ CH₃
```

This is complex.

Let’s analyze:

Central carbon: bonded to:
- CH₂–CH₂–CH₂–CH₃ (butyl group)
- CH₃ (methyl)
- CH₂–CH₃ (ethyl?)
- CH₂–CH₂–CH₂–CH₃ (butyl again?)

Wait — look:

The carbon is bonded to:
- CH₂–CH₂–CH₂–CH₃ (on left side)
- CH₃ (bottom)
- CH₂–CH₃ (on right)
- CH₂–CH₂–CH₂–CH₃ (on top)? Wait — no

Wait — the diagram says:

```
CH₂
|
H₃C—CH₂—CH₂—CH₂—C—CH₂—CH₃
| | |
CH₂—CH₂—CH₂—CH₃ CH₃
```

So:
- Central carbon (let’s call it C1) is bonded to:
- CH₂–CH₂–CH₂–CH₃ → butyl
- CH₃ → methyl
- CH₂–CH₃ → ethyl
- CH₂–CH₂–CH₂–CH₃ → butyl

Wait — two butyl groups?

But the top is CH₂–CH₂–CH₂–CH₃ → butyl

Left: CH₂–CH₂–CH₂–CH₃ → butyl

Right: CH₂–CH₃ → ethyl

Bottom: CH₃ → methyl

So the central carbon has four groups:
- butyl
- butyl
- ethyl
- methyl

But the chain is symmetric?

Wait — the left butyl is: CH₃–CH₂–CH₂–CH₂– (so 4 carbons)
Top butyl: CH₃–CH₂–CH₂–CH₂– (4 carbons)
Ethyl: CH₃–CH₂– (2 carbons)
Methyl: CH₃– (1 carbon)

Now, find the longest chain.

Try going from left butyl to top butyl:
- CH₃–CH₂–CH₂–CH₂–C–CH₂–CH₃ → but wait, the central carbon is bonded to both

So: CH₃–CH₂–CH₂–CH₂–C–CH₂–CH₂–CH₂–CH₃

That’s:
- C1: CH₃
- C2: CH₂
- C3: CH₂
- C4: CH₂
- C5: C (central)
- C6: CH₂
- C7: CH₂
- C8: CH₂
- C9: CH₃

So 9 carbons → nonane

Yes!

So the longest chain is 9 carbons → nonane

Now, the central carbon is part of this chain.

But what about the other groups?

The central carbon also has:
- CH₃ (methyl)
- CH₂–CH₃ (ethyl)

But in this chain, we have:
- Left: CH₃–CH₂–CH₂–CH₂–C–CH₂–CH₂–CH₂–CH₃
- But the central carbon has two additional bonds: to CH₃ and CH₂–CH₃

So the central carbon has three attachments: to left chain, right chain, methyl, and ethyl → no, carbon can have only 4 bonds.

Wait — the central carbon is bonded to:
1. CH₂–CH₂–CH₂–CH₃ (left butyl)
2. CH₂–CH₂–CH₂–CH₃ (top butyl)
3. CH₂–CH₃ (right ethyl)
4. CH₃ (bottom methyl)

So it's a quaternary carbon with four alkyl groups.

But we can make a chain of 9 carbons by connecting the two butyl groups through the central carbon.

So chain: CH₃–CH₂–CH₂–CH₂–C–CH₂–CH₂–CH₂–CH₃

That’s 9 carbons.

Now, the central carbon has two extra groups: methyl and ethyl.

But in this chain, the central carbon is C5.

So at C5, we have:
- Two chains: one to left (butyl), one to right (butyl)
- But we are using both as part of the main chain

So the substituents are:
- Methyl at C5
- Ethyl at C5

So the compound is: 5-ethyl-5-methylnonane

But wait — is that correct?

Yes.

But we can number from either end — same due to symmetry.

So: 5-ethyl-5-methylnonane

But is there a longer chain? No.

Now, alphabetical order: ethyl before methyl

So: 5-ethyl-5-methylnonane

✔ Answer: 5-ethyl-5-methylnonane

---

✔ Final Answers:

| Problem | Name |
|--------|------|
| 1 | 2-methylpropane |
| 2 | 2-methylbutane |
| 3 | 4-ethylheptane |
| 4 | 3-ethyl-3,4-dimethylheptane |
| 5 | 3-propyl-5-methylheptane |
| 6 | 5-ethyl-5-methylnonane |

---

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