Alkane Naming Worksheet for students to practice converting condensed structural formulas into IUPAC names.

Let’s go step by step to name each alkane using IUPAC rules.

We’ll look for:
- The longest continuous carbon chain (that’s the parent name).
- Number the chain so substituents get the lowest possible numbers.
- Name and number any branches (methyl, ethyl, etc.).
- Put it all together in alphabetical order of substituents, with numbers separated by commas and words by hyphens.

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1. CH₃CH₂CH₂CH₂CH₃

This is a straight chain of 5 carbons → pentane.
No branches.

→ pentane

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2. CH₃CH₂CH₂CHCH₃ with a CH₃ below the fourth carbon

Write it out:
Carbon 1: CH₃–
Carbon 2: –CH₂–
Carbon 3: –CH₂–
Carbon 4: –CH(CH₃)–
Carbon 5: –CH₃

Longest chain = 5 carbons → pentane
Branch on carbon #2? Wait — let’s number from the other end to give lower number.

If we number right to left:

Carbon 1: CH₃– (rightmost)
Carbon 2: –CH(CH₃)– ← branch here
Carbon 3: –CH₂–
Carbon 4: –CH₂–
Carbon 5: –CH₃

So methyl group is on carbon #2.

→ 2-methylpentane

Wait — actually, let me double-check structure:

Original: CH₃CH₂CH₂CHCH₃
|
CH₃

That means the fourth carbon has a methyl branch. So if we number left to right:

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

Branch on C4. But if we number right to left:

C1: CH₃– (was C5)
C2: –CH(CH₃)– (was C4) ← now branch on C2
C3: –CH₂–
C4: –CH₂–
C5: –CH₃

Yes! Better to have branch on C2 than C4.

→ 2-methylpentane

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3. CH₃CH₂CH₂CHCH₂CH₃ with CH₂CH₃ below the fourth carbon

Structure:

Main chain: CH₃–CH₂–CH₂–CH–CH₂–CH₃
|
CH₂CH₃

So the branch is an ethyl group attached to carbon #4 (if numbered left to right).

But let’s find the longest chain. Is there a longer chain going through the branch?

Try: start at left CH₃, go to CH₂, CH₂, then down to CH₂CH₃? That would be:

CH₃–CH₂–CH₂–CH–CH₂–CH₃
|
CH₂–CH₃

If you go from left end, through the branch:
Left CH₃ (C1) – CH₂ (C2) – CH₂ (C3) – CH (C4) – CH₂ (of branch, C5) – CH₃ (C6) → that’s 6 carbons.

Then what about the original right part? From C4 to CH₂–CH₃ is only 2 more, but we already used C4 as branching point.

Actually, the longest chain is 6 carbons: either going straight across or including the ethyl branch.

Wait — let’s redraw mentally:

The main horizontal chain is 6 carbons: positions 1 to 6.

At position 4, there’s an ethyl group (–CH₂CH₃).

But if we include that ethyl into the main chain, can we make 7?

Start at left: C1–C2–C3–C4–(down to C5–C6 of ethyl) → that’s 6 atoms.

From C4 to right: C4–C5–C6 → also 3 atoms.

So total longest chain is still 6? No — wait:

Alternative path: Start at bottom of ethyl: CH₃–CH₂– (branch) – attach to C4 – then go left to C3–C2–C1 → that’s 6 carbons.

Or go right from C4 to C5–C6 → that’s 3 more, so total from bottom ethyl to right end: CH₃–CH₂–C4–C5–C6 → 5 carbons.

Not better.

Actually, the longest chain is 6 carbons horizontally. Branch is ethyl on carbon #3 if we renumber.

Let’s number to give lowest number to substituent.

Option 1: Left to right:

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

Substituent on C4.

Option 2: Right to left:

C1: CH₃– (right end)
C2: –CH₂–
C3: –CH(CH₂CH₃)– ← same as above, now on C3
C4: –CH₂–
C5: –CH₂–
C6: –CH₃

Better! Substituent on C3.

So parent chain: hexane
Substituent: ethyl on carbon 3

→ 3-ethylhexane

Check: yes, 6-carbon chain, ethyl on #3.

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4. CH₃CH₂CHCH₂CHCH₃ with CH₃ under third carbon and CH₃ under fifth carbon

Structure:

CH₃–CH₂–CH–CH₂–CH–CH₃
| |
CH₃ CH₃

Numbering: try left to right:

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

Methyl groups on C3 and C5.

Now try right to left:

C1: CH₃– (right)
C2: –CH(CH₃)– ← methyl on C2
C3: –CH₂–
C4: –CH(CH₃)– ← methyl on C4
C5: –CH₂–
C6: –CH₃

Now substituents on C2 and C4 → lower numbers than 3 and 5.

So preferred numbering: right to left → methyls on 2 and 4.

Parent: hexane
Two methyl groups → dimethyl

Alphabetical: “dimethyl” comes before... well, only one type.

Numbers: 2,4-dimethylhexane

→ 2,4-dimethylhexane

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5. CH₃CCH₂CH₃ with two CH₃ groups on the second carbon?

Structure:

CH₃
|
CH₃–C–CH₂–CH₃
|
CH₃

So central carbon (let’s call it C2) has three methyls? Wait:

Written as:

CH₃
|
CH₃CCH₂CH₃
|
CH₃

So the second carbon is bonded to:
- One CH₃ (left)
- One CH₃ (top)
- One CH₃ (bottom)
- And CH₂CH₃ (right)

Wait — that carbon has four bonds: to three CH₃ and one CH₂CH₃? That’s impossible — carbon can’t have five bonds.

Ah, typo in my reading.

Look again:

It says:

CH₃
|
CH₃–C–CH₂–CH₃
|
CH₃

So the central carbon is C, bonded to:

- Top: CH₃
- Left: CH₃
- Bottom: CH₃
- Right: CH₂CH₃

That’s four bonds — okay.

So this carbon is tertiary butyl-like, but attached to ethyl.

Longest chain: start from right: CH₃–CH₂–C(CH₃)₃

So chain: CH₃–CH₂–C– then three methyls.

Longest continuous chain: from right CH₃ through CH₂ to C, then to one of the methyls → that’s 3 carbons? No.

Actually: pick the longest path.

Path 1: CH₃–CH₂–C–CH₃ (one of the methyls) → 4 carbons.

Path 2: same for others.

Is there a 5-carbon chain? No, because the central carbon is branched.

So longest chain is 4 carbons: e.g., CH₃–CH₂–C(CH₃)–CH₃ but wait, the central carbon has two extra methyls.

Standard way: this is 2,2-dimethylbutane.

Let’s number properly.

Choose longest chain: must include the ethyl group and one methyl.

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

But C3 has two additional methyl groups.

So parent chain: butane (4 carbons)
Two methyl groups on C2? Wait, in this numbering, C3 is the branched carbon.

Better to number so branched carbon gets low number.

Set:

C1: one of the methyl groups on central carbon
C2: central carbon
C3: CH₂
C4: CH₃

Then C2 has two other methyl groups? Messy.

Standard rule: choose longest chain that includes most substituents.

Here, longest chain is 4 carbons: for example, take CH₃–CH₂–C(CH₃)₂–CH₃? No, the central carbon is only connected to one CH₂CH₃ and three CH₃.

Actually, the molecule is:

Central carbon: bonded to three CH₃ and one CH₂CH₃.

So the longest chain is from one CH₃ through central C to CH₂ to CH₃ → that’s 4 carbons.

And the central carbon has two additional methyl groups.

In standard naming, we call this 2,2-dimethylbutane.

How?

Chain: C1–C2–C3–C4 where:

C1: CH₃– (end of ethyl)
C2: –CH₂–
C3: –C– (central)
C4: –CH₃ (one methyl)

But then C3 has two more methyl groups → so they are substituents on C3.

To minimize numbers, we should number so that the branched carbon is C2.

Set:

C1: CH₃– (one of the methyl groups on central)
C2: central carbon
C3: –CH₂–
C4: –CH₃

Then C2 has two other methyl groups → so substituents on C2: two methyls.

Parent chain: butane (C1-C2-C3-C4)
Substituents: two methyl groups on C2 → 2,2-dimethylbutane.

Yes.

Confirm: molecular formula C6H14, which matches 2,2-dimethylbutane.

→ 2,2-dimethylbutane

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6. CH₃CH₂CHCH₂CHCH₃ with CH₃CH₂ under third carbon and CH₃ under fifth carbon

Structure:

CH₃–CH₂–CH–CH₂–CH–CH₃
| |
CH₂CH₃ CH₃

First, find longest chain.

Current horizontal: 6 carbons.

But at C3, there’s an ethyl group (CH₂CH₃), so maybe longer chain going down.

Try: start at bottom of ethyl: CH₃–CH₂– (branch) – attach to C3 – then go left to C2–C1 → that’s 5 carbons.

Or go right from C3 to C4–C5–C6 → 4 carbons.

Total: from bottom ethyl to right end: CH₃–CH₂–C3–C4–C5–C6 → 6 carbons.

Same as horizontal.

But if we go from left end through branch: C1–C2–C3–CH₂–CH₃ (branch) → 5 carbons.

Still 6 is max.

Now, within 6-carbon chains, which has more substituents or lower numbers?

Option: use the chain that includes the ethyl branch as part of main chain.

Define new chain: start at left CH₃ (C1) – CH₂ (C2) – CH (C3) – then instead of going right, go down to CH₂ (C4) – CH₃ (C5). That’s only 5.

Not better.

Perhaps: start at the ethyl branch's end: CH₃–CH₂– (call this C1–C2) – attach to current C3 – then go to current C4–C5–C6.

So chain: C1 (ethyl CH₃) – C2 (ethyl CH₂) – C3 (current C3) – C4 (current C4) – C5 (current C5) – C6 (current C6) → 6 carbons.

Now, what substituents? At C3 (which is now C3 in new chain), there is a methyl group? Original had at C5 a methyl, but in this chain, current C5 is C5, which has a methyl.

Also, at C3 (new chain), what else? In original, C3 was attached to C2 (left), C4 (right), and ethyl (down). In new chain, C3 is attached to C2 (up, ethyl part), C4 (right), and originally to C2-left, which is now a substituent.

This is getting messy.

Better to stick with horizontal chain and see.

Horizontal chain: 6 carbons.

Substituents: at C3: ethyl group
At C5: methyl group

Numbering left to right: substituents on C3 and C5.

Numbering right to left: C1: right CH₃
C2: CH (with methyl) → so methyl on C2
C3: CH₂
C4: CH (with ethyl) → ethyl on C4
C5: CH₂
C6: CH₃

So substituents on C2 (methyl) and C4 (ethyl)

Compare: left-to-right: 3-ethyl,5-methyl
Right-to-left: 2-methyl,4-ethyl

Which set of numbers is lower? Compare first differing number: 2 vs 3 → 2 is lower, so right-to-left is better.

Now, alphabetize substituents: ethyl before methyl? E before M, so ethyl first.

But in naming, we list substituents in alphabetical order, ignoring prefixes like di-, tri-.

So "ethyl" and "methyl" → ethyl comes first.

Numbers: 4-ethyl and 2-methyl

So write as: 4-ethyl-2-methylhexane

But check if chain can be longer.

Earlier I thought 6 is max, but let's count atoms.

Total carbons: horizontal 6 + ethyl has 2, but ethyl is branch, so total carbons = 6 (chain) + 2 (ethyl) + 1 (methyl) - overlap? No, the ethyl and methyl are additional.

Molecular formula: C9H20 probably.

Longest chain: suppose we go from the ethyl branch's end through to the methyl branch's end.

Start at ethyl's CH₃ – CH₂ – C3 – C4 – C5 – and C5 has a methyl, so to that methyl.

Chain: CH₃ (ethyl end) – CH₂ – C3 – C4 – C5 – CH₃ (methyl on C5) → that's 6 carbons.

Same as before.

Or from ethyl end to left end: CH₃–CH₂–C3–C2–C1 → 5 carbons.

So 6 is indeed longest.

Thus, with numbering giving 2-methyl and 4-ethyl.

Alphabetical: ethyl before methyl, so 4-ethyl-2-methylhexane.

But is there a chain of 7? Let's try:

Suppose: start at left C1 – C2 – C3 – then down ethyl's CH₂ – CH₃ → that's 5.

From there, no connection to right.

Another path: from right methyl on C5 – C5 – C4 – C3 – down ethyl's CH₂ – CH₃ → that's 6 carbons: C(methyl)-C5-C4-C3-C(ethyl CH2)-C(ethyl CH3)

Yes! 6 carbons.

Same length.

In this chain, what are substituents?

Define chain: C1: CH₃ (methyl on original C5)
C2: C5 (original)
C3: C4
C4: C3
C5: CH₂ (of ethyl)
C6: CH₃ (end of ethyl)

Now, at C2 (original C5), what is attached? Originally, C5 was attached to C4, C6 (horizontal), and a methyl. In this chain, C2 is attached to C1 (the methyl we started with), C3 (C4), and originally to C6, which is now a substituent.

Similarly, at C4 (original C3), attached to C3 (C4), C5 (ethyl CH2), and originally to C2, which is a substituent.

So substituents: at C2: a methyl group? No, C1 is already in chain.

At C2: the group that was C6-horizontal is now a substituent: which is CH₃ (since C6 is CH₃).

Original C6 is CH₃, attached to C5.

In this new chain, C2 is original C5, which was attached to: C4 (now C3), C6 (CH₃), and the methyl we made C1.

So the remaining attachment is to C6, which is a methyl group.

Similarly, at C4 (original C3), attached to: C3 (C4), C5 (ethyl CH2), and originally to C2, which is CH₂CH₃ (since C2 is CH₂, C1 is CH₃).

So substituent at C4: ethyl group.

Also, at C3 (original C4), what is attached? Originally, C4 was between C3 and C5, and had two H's, so no substituent.

In this chain, C3 is original C4, attached to C2 and C4, and two H's, so no issue.

So substituents: at C2: methyl group (original C6)
At C4: ethyl group (original C2-C1)

Numbers: C2 and C4.

Same as before: 2-methyl and 4-ethyl.

And the chain is still hexane.

So regardless, we have 4-ethyl-2-methylhexane.

But let's confirm the numbering direction.

In this last chain, if we number from the other end: C1: ethyl CH₃, C2: ethyl CH₂, C3: C3, C4: C4, C5: C5, C6: methyl on C5.

Then substituents: at C3: ethyl? No.

At C3 (original C3): attached to C2, C4, and original C2, which is ethyl group? Original C2 is CH₂CH₃, so yes, ethyl substituent on C3.

At C5 (original C5): attached to C4, C6, and original C6, which is methyl, so methyl on C5.

So substituents on C3 and C5.

Compare to previous numbering where we had on C2 and C4.

2 and 4 vs 3 and 5 → 2,4 is lower, so better to have substituents on 2 and 4.

Thus, 4-ethyl-2-methylhexane.

But in alphabetical order, ethyl before methyl, so 4-ethyl-2-methylhexane.

However, I recall that when choosing numbering, we compare the sets digit by digit: 2,4 vs 3,5 — 2<3, so 2,4 wins.

Yes.

→ 4-ethyl-2-methylhexane

Wait, but let's make sure about the chain selection.

I think I made a mistake.

In the structure:

CH₃–CH₂–CH–CH₂–CH–CH₃
| |
CH₂CH₃ CH

The carbon at position 3 (from left) has an ethyl group, and position 5 has a methyl.

The longest chain is actually 7 carbons if we go through the ethyl group.

Let's try: start at the end of the ethyl group: CH₃–CH₂– (this is the branch) – attach to C3 – then go to C4 – C5 – and C5 has a methyl, so to that methyl.

Chain: CH₃ (ethyl end) – CH₂ – C3 – C4 – C5 – CH₃ (methyl on C5) → that's 6 carbons.

From C3, we can also go left to C2 – C1, so another path: CH₃ (ethyl) – CH₂ – C3 – C2 – C1 → 5 carbons.

Still not 7.

Unless... is there a way to connect?

No, because C3 is only connected to C2, C4, and the ethyl.

So maximum chain is 6 carbons.

But let's count the atoms in a potential 7-carbon chain.

Suppose: C1: CH₃- (left end)
C2: -CH2-
C3: -CH- (with ethyl)
Then from C3, instead of to C4, go to the ethyl's CH2, then to its CH3 — that's C4 and C5.

Then from C3, we can't go further.

From the ethyl's CH3, no other connections.

So no 7-carbon chain.

Thus, 6 is correct.

But I recall that in such cases, sometimes the chain including the branch is longer.

Let's list all carbons:

Label:

- A: CH3- (left end)
- B: -CH2-
- C: -CH- (branch point)
- D: -CH2-
- E: -CH- (another branch)
- F: -CH3 (right end)
- G: -CH2- (of ethyl on C)
- H: -CH3 (end of ethyl)
- I: -CH3 (methyl on E)

So carbons: A,B,C,D,E,F,G,H,I — 9 carbons.

Longest chain: for example, A-B-C-G-H : 5 carbons
A-B-C-D-E-F : 6 carbons
A-B-C-D-E-I : 6 carbons (since E to I is direct)
H-G-C-D-E-F : 6 carbons
H-G-C-D-E-I : 6 carbons
B-C-G-H : 4
etc.

What about H-G-C-B-A : 5
Or I-E-D-C-G-H : 6 carbons: I-E-D-C-G-H

Yes, 6.

Is there a 7? Try I-E-D-C-B-A : 6 carbons.

Or H-G-C-D-E-I : 6.

All paths are 6 or less.

So longest chain is 6 carbons.

With substituents, we have to choose the chain that gives the most substituents or lowest numbers, but in this case, all 6-carbon chains will have two substituents.

For example, if we take chain A-B-C-D-E-F, then substituents are: on C: ethyl (G-H), on E: methyl (I) — so two substituents.

If we take chain H-G-C-D-E-I, then substituents are: on C: the group B-A, which is ethyl (since B is CH2, A is CH3), and on E: the group F, which is methyl.

Same thing.

So in both cases, we have an ethyl and a methyl on the chain.

Now, for chain A-B-C-D-E-F, numbering A to F: substituents on C3 (ethyl) and C5 (methyl)

Numbering F to A: C1=F, C2=E (with methyl), C3=D, C4=C (with ethyl), C5=B, C6=A — so substituents on C2 (methyl) and C4 (ethyl)

As before.

For chain H-G-C-D-E-I, numbering H to I: C1=H, C2=G, C3=C, C4=D, C5=E, C6=I

Substituents: on C3: the group B-A, which is ethyl (B is CH2, A is CH3)
On C5: the group F, which is methyl

So on C3 and C5.

If number I to H: C1=I, C2=E, C3=D, C4=C, C5=G, C6=H

Substituents: on C2: F (methyl)
On C4: B-A (ethyl)

So on C2 and C4 — same as before.

So consistently, we can achieve substituents on C2 and C4.

Thus, the name is 4-ethyl-2-methylhexane.

But let's write it as 2-methyl-4-ethylhexane? No, alphabetical order: ethyl before methyl, so 4-ethyl-2-methylhexane.

In IUPAC, we ignore di-, tri-, but for alphabetical, we use the full name of the substituent, so "ethyl" before "methyl".

Yes.

→ 4-ethyl-2-methylhexane

I think this is correct.

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7. CH₃CH₂CHCH₂CHCH₃ with CH₃CH₂ under third carbon and CH₃ under fifth carbon, and also CH₃ under fifth carbon? Wait, let's read.

Structure:

CH₃–CH₂–CH–CH₂–CH–CH₃
| |
CH₂CH₃ CH₃
|
CH₃

Oh! On the fifth carbon, there are two groups: one CH₃ and another CH₃? Or is it written as:

Looking back at user input:

7. CH₃CH₂CHCH₂CHCH₃
| |
CH₃CH₂ CH₃
|
CH₃

So on the fifth carbon (from left), there is a CH₃ and also another CH₃ below it? That would mean the fifth carbon has two methyl groups.

Let's parse:

The condensed formula is CH₃CH₂CHCH₂CHCH₃, with branches.

Under the third CH (which is the third carbon), there is CH₃CH₂, so ethyl group.

Under the fifth CH (fifth carbon), there is CH₃, and then below that CH₃, there is another CH₃? That doesn't make sense.

Probably, it's meant that on the fifth carbon, there is a branch that is CH(CH₃)₂ or something.

Look at the text: "CH₃CH₂CHCH₂CHCH₃" with "CH₃CH₂" under the third carbon, and "CH₃" under the fifth carbon, and then "CH₃" under that "CH₃"? That might be a formatting issue.

In the image description, it's likely that on the fifth carbon, there is a single branch that is a group, but written as:

Perhaps it's CH₃CH₂CHCH₂CHCH₃ with under third: CH₃CH₂, and under fifth: CH(CH₃)₂ or something.

Let's assume from common problems.

Typically, for problem 7, it might be:

CH₃–CH₂–CH–CH₂–CH–CH₃
| |
CH₂CH₃ CH(CH₃)₂

Because if it's "CH₃" under fifth and then "CH₃" under that, it might mean the fifth carbon has a branch that is -CH- with two methyls, i.e., isopropyl group.

In the text: "CH₃CH₂CHCH₂CHCH₃" and then below the fifth CH, it says "CH₃" and below that "CH₃", but probably it's aligned under the fifth carbon, meaning the fifth carbon has a substituent that is -CH(CH₃)₂, but written poorly.

Perhaps it's:

The fifth carbon has two methyl groups attached, so it's -C(CH₃)2- but then it would be quaternary, but in the formula it's "CHCH₃", which suggests it's CH with one H, so if it has two methyls, it should be C with no H.

I think there's a misinterpretation.

Let me look at the original user input for #7:

"7. CH₃CH₂CHCH₂CHCH₃
| |
CH₃CH₂ CH₃
|
CH₃"

So visually, the "CH₃" under the fifth carbon, and then another "CH₃" under that "CH₃", which likely means that the substituent on the fifth carbon is a group that has a carbon with two methyls, i.e., the fifth carbon is attached to a CH group that has two methyls, so -CH(CH₃)2.

In other words, the branch on C5 is isopropyl group.

Because if it were two separate methyls, it would be written as two branches from C5, but here it's stacked, suggesting a single branch that is branched.

So, structure:

CH₃–CH₂–CH–CH₂–CH–CH₃
| |
CH₂CH₃ CH(CH₃)₂

So on C3: ethyl group
On C5: isopropyl group (1-methylethyl)

Now, find longest chain.

Total carbons: chain 6 + ethyl 2 + isopropyl 3 = 11, but shared.

Longest chain: could be through the isopropyl.

For example, start at left A-B-C-D-E-F, but E has isopropyl.

Isopropyl is -CH(CH₃)2, so from E, to G (the CH), then to H and I (methyls).

So chain: say, from H-G-E-D-C-B-A : 7 carbons: H (methyl) - G (CH) - E - D - C - B - A

Yes! 7 carbons.

Similarly, from I-G-E-D-C-B-A : also 7.

Or from the ethyl side: from ethyl's end G'-C'-C-D-E-G-H : let's see.

Ethyl on C3: say J-K-C, where K is CH2, J is CH3.

Then C to D to E to G to H : J-K-C-D-E-G-H : 7 carbons.

Same length.

So longest chain is 7 carbons.

Now, choose the chain that has the most substituents or lowest numbers.

Take chain: A-B-C-D-E-G-H, where G is the CH of isopropyl, H is one methyl of isopropyl.

So C1: A (CH3-)
C2: B (-CH2-)
C3: C (-CH-)
C4: D (-CH2-)
C5: E (-CH-)
C6: G (-CH-)
C7: H (CH3)

Now, what substituents?

At C3: originally, C3 was attached to B, D, and ethyl group (J-K). In this chain, C3 is attached to C2, C4, and the ethyl group, so ethyl substituent on C3.

At C5: E is attached to C4, C6, and originally to F (the right CH3). In this chain, C5 is attached to C4 and C6, and the group F is a substituent: which is CH3, so methyl on C5.

At C6: G is attached to C5, C7, and the other methyl of isopropyl, say I. So substituent on C6: methyl group (I).

So substituents: on C3: ethyl
On C5: methyl
On C6: methyl

Numbers: 3,5,6

If we number the other way: C1: H, C2: G, C3: E, C4: D, C5: C, C6: B, C7: A

Then substituents: on C3: the group F (methyl)
On C5: the ethyl group (J-K)
On C2: the other methyl of isopropyl (I) — since G is C2, attached to C1, C3, and I.

So on C2: methyl (I)
On C3: methyl (F)
On C5: ethyl (J-K)

Numbers: 2,3,5

Compare to previous 3,5,6 — 2,3,5 is lower because 2<3.

So better numbering: C1=H, C2=G, C3=E, C4=D, C5=C, C6=B, C7=A

Substituents:
- On C2: methyl (I)
- On C3: methyl (F)
- On C5: ethyl (J-K)

So two methyl groups and one ethyl.

Names: ethyl, methyl, methyl

Alphabetical: ethyl before methyl, and for multiple methyl, use di-.

So substituents: 5-ethyl, 2,3-dimethyl

Parent: heptane

So 5-ethyl-2,3-dimethylheptane

Check if this is correct.

We can verify with another chain.

Take the chain through the ethyl: say J-K-C-D-E-G-H

C1: J (CH3 of ethyl)
C2: K (CH2 of ethyl)
C3: C
C4: D
C5: E
C6: G
C7: H

Substituents: on C3: the group B-A, which is ethyl (B is CH2, A is CH3)
On C5: the group F, methyl
On C6: the other methyl of isopropyl, I

So on C3: ethyl
On C5: methyl
On C6: methyl

Numbers: 3,5,6

If number H to J: C1=H, C2=G, C3=E, C4=D, C5=C, C6=K, C7=J

Substituents: on C3: F (methyl)
On C5: B-A (ethyl)
On C2: I (methyl) — since G is C2, attached to C1, C3, and I.

So on C2: methyl (I)
On C3: methyl (F)
On C5: ethyl (B-A)

Same as before: 2,3-dimethyl and 5-ethyl.

So yes.

Thus, 5-ethyl-2,3-dimethylheptane

---

8. CH₃CCH₂CH₂CHCH₃ with CH₃ under second carbon, CH₃ under second carbon, and under fifth carbon: CH with two CH₃ groups

Structure:

CH₃
|
CH₃–C–CH₂–CH₂–CH–CH₃
| |
CH₃ CH
/ \
CH₃ CH₃

So on C2: two methyl groups (so C2 is quaternary)
On C5: a group that is CH with two methyls, so -CH(CH₃)2, isopropyl.

Find longest chain.

Possible chain: from left, but C2 has three methyls? Let's see.

Carbons:

- C1: CH3- (left)
- C2: C (central)
- Attached to C2: C1, two CH3 (say C3 and C4), and C5 (CH2)
- C5: CH2
- C6: CH2
- C7: CH (with isopropyl)
- C8: CH3 (right end)
- On C7: attached to C6, C8, and CH (say C9), which has two CH3 (C10 and C11)

So longest chain: for example, from C3 (methyl on C2) - C2 - C5 - C6 - C7 - C9 - C10 : 7 carbons.

C3-C2-C5-C6-C7-C9-C10

Yes.

Similarly, other combinations.

So 7-carbon chain.

Choose one: say C10-C9-C7-C6-C5-C2-C3

C1: C10 (CH3)
C2: C9 (CH)
C3: C7 (CH)
C4: C6 (CH2)
C5: C5 (CH2)
C6: C2 (C)
C7: C3 (CH3)

Now, substituents:

At C2: C9 is attached to C1, C3, and C11 (the other methyl of isopropyl), so on C2: methyl group (C11)

At C3: C7 is attached to C2, C4, and C8 (the right CH3), so on C3: methyl group (C8)

At C6: C2 is attached to C5, C7, C1, and C4. In this chain, C6 is attached to C5 and C7, and the groups C1 and C4 are substituents.

C1 is CH3, C4 is CH3, so two methyl groups on C6.

So substituents: on C2: methyl (C11)
On C3: methyl (C8)
On C6: two methyl groups (C1 and C4)

Numbers: 2,3,6,6

If number the other way: C1: C3, C2: C2, C3: C5, C4: C6, C5: C7, C6: C9, C7: C10

Then substituents: on C2: C1 and C4 (two methyls)
On C5: C8 (methyl)
On C6: C11 (methyl) — since C9 is C6, attached to C5, C7, and C11.

So on C2: two methyls
On C5: methyl
On C6: methyl

Numbers: 2,2,5,6

Compare to previous 2,3,6,6 — 2,2,5,6 is lower because at second digit, 2<3.

So better numbering: C1=C3, C2=C2, C3=C5, C4=C6, C5=C7, C6=C9, C7=C10

Substituents:
- On C2: two methyl groups (C1 and C4) — but C1 is already C1 in chain? No.

In this numbering:

C1: C3 (CH3)
C2: C2 (quaternary carbon)
C3: C5 (CH2)
C4: C6 (CH2)
C5: C7 (CH)
C6: C9 (CH)
C7: C10 (CH3)

Now, attachments:

At C2: bonded to C1, C3, and two other groups: C1_left and C4_methyl. Since C1 is already in chain (C1 is C3), the other attachments are: the left CH3 (call it C1a) and the bottom CH3 (C4).

So two methyl groups on C2.

At C5: C7 is bonded to C4, C6, and C8 (the right CH3). So on C5: methyl group (C8)

At C6: C9 is bonded to C5, C7, and C11 (the other methyl of isopropyl). So on C6: methyl group (C11)

So substituents: on C2: two methyls → 2,2-dimethyl
On C5: methyl → 5-methyl
On C6: methyl → 6-methyl

So altogether: 2,2,5,6-tetramethyl? But we have four methyl groups, but in terms of naming, we list them.

The groups are all methyl, so tetramethyl.

Numbers: 2,2,5,6

Parent: heptane

So 2,2,5,6-tetramethylheptane

But is that correct? We have an isopropyl group, but in this chain, we've incorporated it.

Yes.

To confirm, the carbon C9 is included in the chain as C6, and its attachments are handled.

So final name: 2,2,5,6-tetramethylheptane

But let's make sure about the numbering.

In this numbering, substituents on 2,2,5,6.

If we number from the other end: C1=C10, C2=C9, C3=C7, C4=C6, C5=C5, C6=C2, C7=C3

Then substituents: on C2: C11 (methyl)
On C3: C8 (methyl)
On C6: C1a and C4 (two methyls) — since C2 is C6, bonded to C5, C7, C1a, C4.

So on C2: methyl (C11)
On C3: methyl (C8)
On C6: two methyls (C1a, C4)

Numbers: 2,3,6,6

Compare to 2,2,5,6 — at second digit, 2<3, so 2,2,5,6 is better.

Yes.

So 2,2,5,6-tetramethylheptane

But in the structure, on C5 of the original, we have the isopropyl, which is accounted for.

Okay.

Now, summarize all answers.

Final Answer:

1. pentane
2. 2-methylpentane
3. 3-ethylhexane
4. 2,4-dimethylhexane
5. 2,2-dimethylbutane
6. 4-ethyl-2-methylhex