Naming Alkanes - Worksheet #1: Identify and name the given branched alkane structures.

Let’s go through each alkane one by one. We’ll name them using IUPAC rules:

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).
4. Put it all together: [substituent position]-[substituent name][parent chain name]

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Problem 1:

Structure:
H₃C–CH–CH₃
  │
  CH₃

Longest chain = 3 carbons → propane
Branch: methyl group on carbon #2
→ 2-methylpropane

*(Note: This is also called “isobutane”, but we use systematic naming here.)*

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Problem 2:

Structure:
H₃C–CH–CH₃
  │
 CH₂–CH₃

Longest chain = 4 carbons → butane
Branch: methyl on carbon #2
Numbering from left: branch at C2
Numbering from right: branch at C3 → choose lower number → C2
→ 2-methylbutane

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Problem 3:

Structure:
H₃C–CH₂–CH₂–CH–CH₂–CH₂–CH₃
     │
     CH₂–CH₃

Longest chain = 7 carbons → heptane
Branch: ethyl group attached to carbon #4
(If you number from either end, the branch is at C4)
→ 4-ethylheptane

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Problem 4:

Structure:
H₃C–CH₂–CH₂–CH–CH–CH₂–CH₃
    │  │
     CH₃ CH₂–CH₃

Longest chain = 7 carbons → heptane
Two branches:
- Methyl on carbon #4
- Ethyl on carbon #5? Wait — let’s renumber!

Actually, if we number from the RIGHT:

Right to left:
C1: CH₃ (end of ethyl branch?) No — better to find longest straight chain.

Wait — look again:

The main chain can be drawn as:

Start from leftmost CH₃–CH₂–CH₂– then the CH with CH₃, then CH with CH₂CH₃, then CH₂–CH₃.

That’s 7 carbons in a row? Let’s count:

Carbon 1: left CH₃
C2: CH₂
C3: CH₂
C4: CH (with CH₃ branch)
C5: CH (with CH₂CH₃ branch)
C6: CH₂
C7: CH₃ → yes, 7-carbon chain.

Now assign numbers to give lowest positions to substituents.

If we number left to right:
Methyl on C4, ethyl on C5 → positions 4 and 5

If we number right to left:
Then ethyl would be on C3, methyl on C4 → positions 3 and 4 → better!

So renumber:

From right:
C1: CH₃ (right end)
C2: CH₂
C3: CH (has ethyl group)
C4: CH (has methyl group)
C5: CH₂
C6: CH₂
C7: CH₃

Substituents: ethyl on C3, methyl on C4

Alphabetical order: ethyl before methyl

→ 3-ethyl-4-methylheptane

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Problem 5:

Structure:
H₃C–CH₂–CH–CH₂–CH–CH₂–CH₃
   │    │
    CH₃ CH₂–CH₂–CH₃

Longest chain? Let’s see.

Main horizontal chain: 7 carbons? But there’s a propyl group on the right side.

Wait — actually, if we go down the propyl branch, we might get a longer chain.

Try this path:

Start from left CH₃–CH₂–CH(CH₃)–CH₂–CH(propyl)–CH₂–CH₃

But the propyl group is –CH₂–CH₂–CH₃, so if we include that, the chain becomes:

From leftmost CH₃ → CH₂ → CH → CH₂ → CH → then instead of going to CH₂–CH₃, go down the propyl: CH₂–CH₂–CH₃

That gives:
C1: left CH₃
C2: CH₂
C3: CH (methyl branch)
C4: CH₂
C5: CH
C6: CH₂ (from propyl)
C7: CH₂
C8: CH₃ → 8 carbons!

Is that valid? Yes — because the branch is part of the longest continuous chain.

So parent chain = octane

Now, where are the branches?

In this 8-carbon chain:

We have a methyl group on what was originally C3 — now in new numbering?

Let’s define the chain properly:

Chain:
C1 – C2 – C3 – C4 – C5 – C6 – C7 – C8
Where:
C1 = leftmost CH₃
C2 = CH₂
C3 = CH (originally had a methyl branch)
C4 = CH₂
C5 = CH (originally had the propyl branch — but now it's part of main chain)
Wait — no! If we made the propyl part of the main chain, then the original "propyl" is now C6-C7-C8.

But then what’s attached to C5? The original right-side CH₂–CH₃ becomes a branch!

Original structure:

Left: H₃C–CH₂–CH–CH₂–CH–CH₂–CH₃
    │    │
      CH₃ CH₂–CH₂–CH₃

If we take the chain going through the propyl:

Start at top-left CH₃–CH₂–CH–CH₂–CH– then instead of going to CH₂–CH₃, go down to CH₂–CH₂–CH₃

So atoms:

C1: CH₃ (left)
C2: CH₂
C3: CH (has methyl group)
C4: CH₂
C5: CH (has ethyl group? Because the original right end is CH₂–CH₃, which is now a 2-carbon branch)

Yes!

So longest chain = 8 carbons → octane

Branches:
- Methyl on C3
- Ethyl on C5

Numbering: if we go left to right: methyl on 3, ethyl on 5
If we go right to left: ethyl on 4, methyl on 6 → worse

So left to right is better.

Alphabetical: ethyl before methyl? No — e comes before m, so ethyl first.

But positions: 3-methyl and 5-ethyl → when writing, we list substituents alphabetically regardless of position.

Ethyl starts with 'e', methyl with 'm' → ethyl first.

So: 5-ethyl-3-methyloctane

Wait — check numbering again.

Is there a way to get lower numbers?

What if we start from the other end of the 8-carbon chain?

Define chain from the bottom-right:

C1: CH₃ (end of former propyl)
C2: CH₂
C3: CH₂
C4: CH (was the branching point)
C5: CH₂
C6: CH (has methyl)
C7: CH₂
C8: CH₃

Then branches:
At C4: ethyl group (the original right-end CH₂–CH₃)
At C6: methyl group

Positions: 4 and 6 → higher than 3 and 5 → so previous numbering is better.

Thus: 5-ethyl-3-methyloctane

But wait — standard rule: lowest set of locants. Compare 3,5 vs 4,6 → 3,5 is lower.

Also, when listing, alphabetical order: ethyl before methyl.

So final: 5-ethyl-3-methyloctane

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Problem 6:

Structure:
Top: H₃C–CH₂–CH₂–CH₂–CH₂
Bottom: H₃C–CH₂–CH₂–CH₂–C–CH₂–CH₃
        │
        CH₃

This looks like two chains connected at a central carbon.

Central carbon has:
- Up: CH₂–CH₂–CH₂–CH₂–CH₃ (pentyl)
- Down: CH₂–CH₂–CH₂–CH₃ (butyl)? Wait no:

Bottom chain: H₃C–CH₂–CH₂–CH₂–C–CH₂–CH₃
So from left: CH₃–CH₂–CH₂–CH₂–C– then to CH₂–CH₃ and down to CH₃

Actually, the central carbon is bonded to:

- A pentyl group above: –CH₂–CH₂–CH₂–CH₂–CH₃
- A butyl group to the left: –CH₂–CH₂–CH₂–CH₃? Wait no:

Let’s write connections clearly.

The central carbon (let’s call it C*) is bonded to:

1. –CH₂–CH₂–CH₂–CH₂–CH₃ (above) → pentyl
2. –CH₂–CH₂–CH₂–CH₃ (left) → butyl? From the bottom line: H₃C–CH₂–CH₂–CH₂–C* → so that’s a butyl group attached to C*
3. –CH₂–CH₃ (right) → ethyl
4. –CH₃ (down) → methyl

So C* has four alkyl groups: pentyl, butyl, ethyl, methyl.

To name: find the longest chain that includes C*.

Possible chains:

Option 1: Go up pentyl + C* + right ethyl → 5 + 1 + 2 = 8 carbons?
Pentyl is 5, C* is 1, ethyl is 2 → total 8.

Chain: CH₃–CH₂–CH₂–CH₂–CH₂–C*–CH₂–CH₃

That’s 8 carbons: C1 to C8, with C* being C6? Let’s number:

Set C1 = top end: CH₃– (of pentyl)
C2: CH₂
C3: CH₂
C4: CH₂
C5: CH₂
C6: C*
C7: CH₂
C8: CH₃

At C6, we have two additional groups:
- Left: –CH₂–CH₂–CH₂–CH₃ (butyl)
- Down: –CH₃ (methyl)

So parent chain = octane

Substituents on C6: butyl and methyl

But butyl is larger — however, we must use the longest chain as parent, which we did.

Now, are there longer chains? What if we go left butyl + C* + up pentyl? Same length: 4 + 1 + 5 = 10? No — butyl is 4 carbons, but when attached, the chain is from end of butyl to end of pentyl via C*: that’s 4 (butyl) + 1 (C*) + 5 (pentyl) = 10 carbons!

I think I missed that.

Let’s define:

Start from left end of bottom chain: H₃C–CH₂–CH₂–CH₂–C*–CH₂–CH₂–CH₂–CH₂–CH₃ (up)

That’s:
C1: CH₃ (left)
C2: CH₂
C3: CH₂
C4: CH₂
C5: C*
C6: CH₂ (up)
C7: CH₂
C8: CH₂
C9: CH₂
C10: CH₃

Yes! 10-carbon chain → decane

Now, what’s attached to C5?
- Right: –CH₂–CH₃ (ethyl)
- Down: –CH₃ (methyl)

So substituents: ethyl and methyl on C5

Numbering: if we go left to right, substituents on C5
If we go right to left: C1 would be top CH₃, then C10 is left CH₃, so C5 becomes C6? Let’s see:

Reverse:
C1: top CH₃
C2: CH₂
C3: CH₂
C4: CH₂
C5: CH₂
C6: C*
C7: CH₂
C8: CH₂
C9: CH₂
C10: CH₃ (left)

Then substituents on C6: ethyl and methyl → positions 6,6

Compare to original numbering: positions 5,5 → 5<6, so better to number from left.

Thus: 5-ethyl-5-methyldecane

Alphabetical: ethyl before methyl.

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Problem 7:

Structure:
  CH₂–CH₂–CH₃
  │
H₂C–CH–CH₂–CH–CH₃
│   │
CH₃  CH₂–CH₂–CH₃

Let’s parse this.

It seems like:

There’s a central chain: let’s identify atoms.

Write it as:

Carbon A: H₂C– (with CH₃ below) → so it’s CH– with H₂C meaning? Probably typo or shorthand.

Looking closely:

It says:
  CH₂–CH₂–CH₃
  │
H₂C–CH–CH₂–CH–CH₃
│    │
CH₃ CH₂–CH₂–CH₃

I think "H₂C" means a methylene group, but in context, it’s likely that the first group is:

A carbon with three bonds shown:
- Bond up to CH₂–CH₂–CH₃ (propyl)
- Bond right to CH– (which is next)
- Bond down to CH₃

Similarly, the last CH has bond down to CH₂–CH₂–CH₃ (another propyl)

So let’s map:

Call the carbons:

C1: the leftmost carbon — it’s written as H₂C–, but with a CH₃ below and connected to CH above? Confusing.

Perhaps it’s:

The backbone is: C – C – C – C – C

With branches.

Assume the main horizontal chain is 5 carbons:

Position 1: carbon with a methyl group (below) and connected to position 2
Position 2: carbon with a propyl group (above: CH₂–CH₂–CH₃) and connected to 1 and 3
Position 3: CH₂
Position 4: carbon with a propyl group (below: CH₂–CH₂–CH₃) and connected to 3 and 5
Position 5: CH₃

But position 1 is written as H₂C– with CH₃ below — so if it’s H₂C, that suggests two hydrogens, but it’s also bonded to CH₃ and to next carbon — so it should be CH–, not H₂C.

Probably a notation issue. In many worksheets, "H₂C" is used for terminal methylene, but here it’s branched.

I think it’s meant to be:

The first carbon is a CH group (since it has three substituents: H, CH₃, and next carbon), but written as H₂C by mistake? Or perhaps it’s indicating the group.

To avoid confusion, let’s count all atoms.

List all carbon atoms:

- Left branch: CH₃– (attached to first carbon)
- First carbon: let’s call it C1 — bonded to: CH₃ (left-down), H (implied), C2 (right), and also to a propyl group above? The diagram shows:

Above C1: CH₂–CH₂–CH₃
Below C1: CH₃
Right: to C2

So C1 is bonded to four things:
1. CH₃ (below)
2. CH₂– (of propyl above)
3. C2 (right)
4. H? But carbon must have four bonds — if it’s shown with three explicit bonds, the fourth is H.

In organic structures, if not specified, hydrogens are implied.

So C1 is a carbon with:
- One H (not shown)
- One CH₃ (below)
- One CH₂CH₂CH₃ (above)
- One bond to C2

Similarly, C2 is CH (since it’s between C1 and C3, and no other branches shown? Wait, the diagram shows only the connections given.

Look back:

The structure is:

  CH₂–CH₂–CH₃ ← this is attached to the second carbon? Or first?

The way it's written:

Line 1:   CH₂–CH₂–CH₃
Line 2:   │
Line 3: H₂C–CH–CH₂–CH–CH₃
Line 4: │   │
Line 5: CH₃  CH₂–CH₂–CH₃

So, the vertical line from CH₂–CH₂–CH₃ goes down to the "CH" in line 3, which is the second character.

Similarly, the vertical line from CH₃ (line 5) goes up to "H₂C" in line 3.

And another vertical line from CH₂–CH₂–CH₃ (line 5) goes up to the fourth character in line 3, which is "CH".

So, the backbone in line 3 is: H₂C – CH – CH₂ – CH – CH₃

With:

- Above the second atom (CH): CH₂–CH₂–CH₃
- Below the first atom (H₂C): CH₃
- Below the fourth atom (CH): CH₂–CH₂–CH₃

Now, "H₂C" typically means -CH₂-, but here it's at the end, and has a substituent below, so it must be that the first carbon is actually a CH group, not CH₂.

Probably, "H₂C" is a misnomer; it should be understood as a carbon with two H's only if no other substituents, but here it has a CH₃ below, so it's a methine carbon.

To clarify, let's assign:

Denote the five-carbon chain as C1–C2–C3–C4–C5

Where:

C1: the leftmost — written as H₂C, but with a CH₃ attached below → so C1 is carbon with bonds to:
- Two H's? No, if it's H₂C, it implies two hydrogens, but then attaching CH₃ would make it five bonds — impossible.

I think there's a formatting error. In standard notation, when they write:

  R
  |
H₂C–CH–...

It means that the H₂C is the first carbon, which is CH₂ group, but then it has a substituent R attached, which would require it to be CH, not CH₂.

This is confusing. Perhaps "H₂C" here is meant to be the group, but in context, it's likely that the first carbon is a chiral center or something.

Another interpretation: perhaps "H₂C" is a typo and it's meant to be "CH₃–" or something.

Let me try to redraw mentally.

From the diagram:

- There is a carbon (call it C_a) that has:
- A methyl group (CH₃) attached (below)
- A propyl group (CH₂CH₂CH₃) attached (above)
- And bonded to another carbon (C_b)

Then C_b is bonded to C_a, to C_c, and has a hydrogen (implied)

Then C_c is CH₂, bonded to C_b and C_d

C_d is bonded to C_c, to C_e, and to a propyl group (CH₂CH₂CH₃) below

C_e is CH₃

So the main chain could be C_a – C_b – C_c – C_d – C_e, which is 5 carbons.

But C_a has two alkyl groups: methyl and propyl, so it's a tertiary carbon.

Similarly, C_d has a propyl group.

To find the longest chain, consider including the propyl groups.

For example, from the top propyl through C_a to C_e:
Propyl is 3 carbons, C_a, C_b, C_c, C_d, C_e — that's 3+1+1+1+1+1=8 carbons? Let's count:

Start from end of top propyl: CH₃–CH₂–CH₂– (attached to C_a)
Then C_a – C_b – C_c – C_d – C_e (CH₃)

So atoms:
C1: CH₃ (top propyl end)
C2: CH₂
C3: CH₂
C4: C_a
C5: C_b
C6: C_c
C7: C_d
C8: C_e (CH₃)

8-carbon chain.

At C4 (C_a), there is a methyl group attached (the one below)
At C7 (C_d), there is a propyl group attached (the one below: CH₂–CH₂–CH₃)

So substituents:
- Methyl on C4
- Propyl on C7

Numbering: if we go this way, positions 4 and 7

If we go the other direction: start from C_e as C1, then C_d=C2, etc., up to top propyl.

C1: C_e (CH₃)
C2: C_d
C3: C_c
C4: C_b
C5: C_a
C6: CH₂ (of top propyl)
C7: CH₂
C8: CH₃

Then substituents:
At C2 (C_d): propyl group
At C5 (C_a): methyl group

Positions: 2 and 5

Compare to previous 4 and 7 — 2 and 5 is lower.

Also, 2<4, so better.

So parent chain = octane

Substituents: methyl on C5, propyl on C2

Alphabetical: methyl before propyl (m before p)

So: 2-propyl-5-methyloctane

But is "propyl" correct? The group is n-propyl, yes.

However, we need to ensure that the chain is indeed the longest.

Is there a 9-carbon chain? For example, if we go from the bottom propyl on C_d through to the top propyl on C_a.

From bottom propyl end: CH₃–CH₂–CH₂– (attached to C_d)
Then C_d – C_c – C_b – C_a – then to top propyl: CH₂–CH₂–CH₃

So:
C1: CH₃ (bottom propyl end)
C2: CH₂
C3: CH₂
C4: C_d
C5: C_c
C6: C_b
C7: C_a
C8: CH₂ (top propyl)
C9: CH₂
C10: CH₃

10 carbons! Oh! I missed that.

Yes! So longest chain is 10 carbons.

Chain: from end of bottom propyl to end of top propyl via C_d–C_c–C_b–C_a

Atoms:
C1: CH₃ (bottom)
C2: CH₂
C3: CH₂
C4: C_d
C5: C_c
C6: C_b
C7: C_a
C8: CH₂ (top)
C9: CH₂
C10: CH₃

Now, what's attached to this chain?

At C4 (C_d): originally, it was bonded to C_e (CH₃), which is now a methyl group branch.

At C7 (C_a): bonded to a methyl group (the one that was below it).

In the original structure, C_a had a methyl below, and C_d had the bottom propyl, but now the bottom propyl is part of the main chain, so the branch at C_d is the original C_e, which is CH₃ — so methyl group.

Similarly, at C_a, the branch is the methyl group that was attached.

So substituents:
- Methyl on C4
- Methyl on C7

Both are methyl groups.

Numbering: if we go from bottom to top as above, positions 4 and 7

If we go from top to bottom:
C1: top CH₃
C2: CH₂
C3: CH₂
C4: C_a
C5: C_b
C6: C_c
C7: C_d
C8: CH₂
C9: CH₂
C10: CH₃ (bottom)

Then substituents:
At C4 (C_a): methyl
At C7 (C_d): methyl

Same positions: 4 and 7

But we can choose the direction that gives lower numbers — same either way.

Since both substituents are methyl, we combine them.

Locants: 4 and 7

So: 4,7-dimethyldecane

Is that correct? Let's verify the chain.

Main chain: 10 carbons in a row: from one end of a propyl, through the central part, to the other end of the other propyl.

Specifically:
- Start at the end of the propyl that was attached to C_d (bottom): call it P1-P2-P3-C_d
- Then C_d to C_c to C_b to C_a
- Then C_a to Q1-Q2-Q3 (top propyl)

So sequence: P1-P2-P3-C_d-C_c-C_b-C_a-Q1-Q2-Q3

That's 10 carbons.

Attached to C_d: besides P3 and C_c, it was also attached to C_e (CH₃) — so yes, a methyl group on C_d, which is C4 in this numbering.

Attached to C_a: besides C_b and Q1, it was attached to a methyl group — so methyl on C_a, which is C7.

Yes.

And no other branches.

So the name is 4,7-dimethyldecane

We don't need to specify "n-" for propyl since it's incorporated.

Final answer for problem 7: 4,7-dimethyldecane

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Now, compiling all answers:

1. 2-methylpropane
2. 2-methylbutane
3. 4-ethylheptane
4. 3-ethyl-4-methylheptane
5. 5-ethyl-3-methyloctane
6. 5-ethyl-5-methyldecane
7. 4,7-dimethyldecane

Double-check problem 5: earlier I said 5-ethyl-3-methyloctane, but let me confirm the chain.

In problem 5, the structure is:

H₃C–CH₂–CH–CH₂–CH–CH₂–CH₃
   │    │
    CH₃  CH₂–CH₂–CH₃

Longest chain: if we go from left CH₃–CH₂–CH(CH₃)–CH₂–CH(CH₂CH₂CH₃)–CH₂–CH₃, but the propyl can be extended.

As I did, taking the propyl as part of main chain: so chain is 8 carbons: from left CH₃ to the end of propyl.

Specifically: C1: left CH₃
C2: CH₂
C3: CH (with methyl)
C4: CH₂
C5: CH (with ethyl? Because the original right end is CH₂–CH₃, which is now a branch)

When we make the propyl part of main chain, the original right-end CH₂–CH₃ becomes an ethyl group attached to C5.

Yes.

And positions: methyl on C3, ethyl on C5.

Alphabetical: ethyl before methyl, so 5-ethyl-3-methyloctane.

Some might argue about the numbering, but 3 and 5 is correct.

Problem 6: 5-ethyl-5-methyldecane — yes.

Problem 7: 4,7-dimethyldecane — yes.

All seem correct.

Final Answer:
1. 2-methylpropane
2. 2-methylbutane
3. 4-ethylheptane
4. 3-ethyl-4-methylheptane
5. 5-ethyl-3-methyloctane
6. 5-ethyl-5-methyldecane
7. 4,7-dimethyldecane