Naming Alkynes by IUPAC Nomenclature Rules with Practice Problems - Free Printable

Let’s go through each structure one by one and name them using IUPAC rules for organic compounds. We’ll focus on finding the longest carbon chain that includes the multiple bond (double or triple), number it to give the multiple bond the lowest possible number, and then add substituents in alphabetical order.

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a)
Structure: A 5-carbon chain with a triple bond between C2 and C3, two methyl groups on C4, and a chlorine on C1? Wait — let’s redraw mentally:

Actually, looking at the drawing:
It’s CH₃–C≡C–C(CH₃)₂–CH₂Cl

So the longest chain including the triple bond is 5 carbons:
Start from the Cl end to give the triple bond lower number? Let’s try numbering both ways.

Option 1: Number from left (triple bond starts at C2):
C1–C2≡C3–C4–C5 → but C4 has two methyls and C5 is CH₂Cl? That doesn’t match.

Wait — better way: The group attached to the triple bond is –C(CH₃)₂CH₂Cl. So the main chain should include the triple bond and the CH₂Cl.

Longest continuous chain with triple bond: Start from CH₃– (left) through triple bond to the quaternary carbon, then to CH₂Cl → that’s 5 atoms:
C1 (CH₃) – C2 ≡ C3 – C4 (with two CH₃) – C5 (CH₂Cl)

But C4 has two methyl groups → so it’s a 5-carbon chain with triple bond between C2-C3, and two methyls on C4, and Cl on C5.

Numbering: If we number from right (Cl end):
C1 = CH₂Cl, C2 = C(CH₃)₂, C3 ≡ C4 – C5 (CH₃) → triple bond between C3-C4 → position 3.

If we number from left: triple bond between C2-C3 → position 2 → better!

So preferred numbering: left to right → triple bond at 2.

Substituents: two methyl groups on C4 → “4,4-dimethyl”

And chloro on C5 → “5-chloro”

Alphabetical: chloro before dimethyl.

So name: 5-chloro-4,4-dimethylpent-2-yne

Check: Pent = 5 carbons, -2-yne = triple bond at 2, 4,4-dimethyl, 5-chloro → yes.

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b)
Structure: Looks like a chain with triple bond in middle, ethyl group on one side, bromine on other.

Draw: Left side: CH₃CH₂–CH(CH₃)–C≡C–CH₂–CH(CH₃)–CH(Br)–CH₃

Wait — let’s count longest chain including triple bond.

From left: start from ethyl branch? Better to find longest continuous chain with triple bond.

The triple bond is central. Go left: from triple bond, one side goes to CH(CH₃)CH₂CH₃ → that’s 3 carbons plus the branch.

Other side: CH₂–CH(CH₃)–CH(Br)–CH₃ → 4 carbons.

So total chain: if we go from far left ethyl through the branched carbon, triple bond, then to the bromo end.

Define:
Left part: CH₃–CH₂–CH(CH₃)– → this is a 3-carbon chain with methyl on C2? Actually, the carbon attached to triple bond is chiral? Not important.

Best: Longest chain including triple bond:
Start from the ethyl group’s end: C1–C2–C3(where methyl is)–C4≡C5–C6–C7(where methyl is)–C8–C9(where Br is)–C10? Too long.

Wait — look again:
The structure is:
On left: a carbon with H, CH₃, and CH₂CH₃ → so it’s a –CH(CH₃)CH₂CH₃ group attached to triple bond.

On right: –CH₂–CH(CH₃)–CH(Br)–CH₃

So the triple bond connects two groups. The longest chain must include the triple bond and as many carbons as possible.

Try: Start from the ethyl group on left: CH₃–CH₂–CH– (this CH is attached to triple bond and also to a methyl) → so that carbon is tertiary.

Then triple bond, then –CH₂–CH– (this CH has a methyl) –CH(Br)–CH₃

So chain: from left ethyl’s terminal carbon:
C1 (CH₃) – C2 (CH₂) – C3 (CH) – C4 ≡ C5 – C6 (CH₂) – C7 (CH) – C8 (CH) – C9 (CH₃)

C3 has a methyl substituent.
C7 has a methyl substituent.
C8 has Br.

So 9-carbon chain? But C8 is CH(Br)CH₃, so C9 is methyl.

Chain: C1 to C9: nonane derivative.

Triple bond between C4-C5 → so yne at 4.

Substituents:
- Methyl on C3
- Methyl on C7
- Bromo on C8

Numbering: Should we number from left or right to give triple bond lower number?

Current: triple bond at 4.

If number from right:
C1 = CH₃ (end of bromo side)
C2 = CH(Br)
C3 = CH(CH₃)
C4 = CH₂
C5 ≡ C6
C7 = CH(CH₃)
C8 = CH₂
C9 = CH₃? Wait no — left side was CH₃CH₂CH– so if we number from right, the left end becomes C9 which is part of ethyl.

Actually, the left group is –CH(CH₃)CH₂CH₃, so when attached, the carbon directly attached to triple bond is the CH, which has H, CH₃, and CH₂CH₃.

So in the chain, if we make the longest chain go through the ethyl group:
Start from the ethyl’s CH₃: C1–C2–C3(attached to triple bond and methyl)–C4≡C5–C6–C7(attached to methyl)–C8(attached to Br)–C9

Yes, 9 carbons.

Triple bond between C4-C5 → position 4.

Substituents:
- Methyl on C3
- Methyl on C7
- Bromo on C8

Now, if we number from the other end:
C1 = C9 above (the CH₃ of ethyl? No — better to define.

Set C1 as the CH₃ of the ethyl group on left:
C1–C2–C3–C4≡C5–C6–C7–C8–C9
Where:
C3 has a methyl group → so 3-methyl
C7 has a methyl group → 7-methyl
C8 has Br → 8-bromo

Triple bond at 4.

If we number from right:
C1 = the CH₃ that is attached to C8 (which has Br)
C2 = C8 (has Br)
C3 = C7 (has methyl)
C4 = C6
C5 ≡ C6? Wait confusion.

Better: Right end is CH₃–CH(Br)–CH(CH₃)–CH₂–C≡C–CH(CH₃)–CH₂–CH₃

So if number from right:
C1 = CH₃ (of the bromo end)
C2 = CH(Br)
C3 = CH(CH₃)
C4 = CH₂
C5 ≡ C6
C7 = CH(CH₃)
C8 = CH₂
C9 = CH₃

Triple bond between C5-C6 → position 5.

Previously from left it was position 4 → 4 < 5, so better to number from left.

So triple bond at 4.

Substituents:
- Methyl on C3
- Methyl on C7
- Bromo on C8

Alphabetical: bromo, methyl, methyl → so 8-bromo-3,7-dimethylnon-4-yne

Check: Non = 9 carbons, -4-yne, 3,7-dimethyl, 8-bromo → yes.

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c)
Structure: Triple bond in middle, left side has a branch: CH₃CH₂–CH– with another CH(CH₃)₂? Wait.

Drawing: Left: a carbon with H, CH₂CH₃, and CH(CH₃)₂? Or what?

Actually: It looks like:
Left group: CH₃–CH₂–CH– (this CH is attached to triple bond and also to CH(CH₃)₂) → so it's –CH[CH(CH₃)₂]CH₂CH₃

Right group: –CH(CH₃)CH₂CH₂CH₃? Wait, right side is CH(CH₃)–CH₂–CH₂–CH₃? No.

Looking: Right side is –CH(CH₃)–CH₂–CH₂–CH₃? But in diagram it's drawn as a chain: after triple bond, carbon with methyl, then propyl?

Standard interpretation:
The molecule is: (CH₃CH₂)( (CH₃)₂CH )CH–C≡C–CH(CH₃)CH₂CH₂CH₃

So longest chain including triple bond:
Go from left ethyl through the branched carbon, triple bond, then to the pentyl-like chain.

Define chain:
Start from left: the ethyl group’s end: C1–C2–C3(attached to triple bond and isopropyl)–C4≡C5–C6(attached to methyl)–C7–C8–C9

C3 has a substituent: CH(CH₃)₂ → which is 1-methylethyl or isopropyl.

C6 has a methyl group.

Chain length: C1 to C9 = 9 carbons.

Triple bond between C4-C5 → position 4.

Substituents:
- On C3: isopropyl group → but isopropyl is a substituent, so we name it as (1-methylethyl) or commonly isopropyl, but IUPAC prefers systematic.

Isopropyl is acceptable, but let's see if we can include it in chain? Probably not, since it's branched.

Alternatively, the group on C3 is –CH(CH₃)₂, so it's a 1-methylethyl group.

Similarly, on C6: methyl group.

Also, the left part: C3 is attached to C2 (ethyl) and to the isopropyl.

In the chain C1-C2-C3-C4≡C5-C6-C7-C8-C9, C3 has a substituent which is –CH(CH₃)₂, so that's a 1-methylethyl group.

C6 has a methyl group.

So substituents:
- 3-(1-methylethyl)
- 6-methyl

But 1-methylethyl is isopropyl, and we can use "isopropyl" in common naming, but for strict IUPAC, it's (1-methylethyl).

However, often in such problems, they accept isopropyl.

But let's check numbering.

Triple bond at 4.

If we number from right:
C1 = C9 (end of propyl)
C2 = C8
C3 = C7
C4 = C6 (has methyl)
C5 ≡ C6? Wait.

Right end: CH₃–CH₂–CH₂–CH(CH₃)–C≡C–... so if number from right:
C1 = CH₃ (of propyl)
C2 = CH₂
C3 = CH₂
C4 = CH(CH₃)
C5 ≡ C6
C7 = CH[CH(CH₃)₂]CH₂CH₃? Messy.

The carbon attached to triple bond on right is CH(CH₃)CH₂CH₂CH₃, so it's a 1-methylbutyl group? But for chain, better to have straight chain.

Longest chain is still 9 carbons as before.

With numbering from left: triple bond at 4.

Substituents: at C3: isopropyl, at C6: methyl.

Alphabetical: isopropyl before methyl? I for isopropyl, M for methyl → I before M.

But isopropyl is "i", methyl is "m", so isopropyl first.

Name: 3-isopropyl-6-methylnon-4-yne

But is isopropyl allowed? In many textbooks, yes. To be precise, 3-(1-methylethyl)-6-methylnon-4-yne.

But perhaps for this level, isopropyl is fine.

Note: The group on C3 is actually –CH(CH₃)₂, which is isopropyl, and it's attached to C3, so yes.

Another way: the carbon C3 has three groups: H (assumed), the chain to C2, the chain to C4 (triple bond), and the isopropyl? No, carbon can have only four bonds.

Mistake: in the structure, the carbon attached to the triple bond on left is a chiral center? Let's think.

In diagram c): it's drawn as a carbon with three lines: one to ethyl, one to isopropyl (which is CH with two methyls), and one to triple bond. So that carbon has no H; it's tertiary carbon.

So the group is –C(ethyl)(isopropyl)–, so when we make the chain, if we go through ethyl, then that carbon is C3, and it has a substituent isopropyl.

Yes.

So name: 3-isopropyl-6-methylnon-4-yne

But let's confirm the chain length.

Left: ethyl is C2H5-, so two carbons, attached to a carbon that is also attached to isopropyl (three carbons) and to triple bond.

So the carbon attached to triple bond is quaternary? No, it has three carbons attached: ethyl, isopropyl, and the triple bond carbon. So it's a tertiary carbon, but in terms of chain, when we select the longest chain, we can choose to go through the ethyl or through the isopropyl.

Isopropyl has three carbons, ethyl has two, so better to go through isopropyl to make longer chain.

Ah! Important point.

The group on left is: the carbon attached to triple bond is bonded to:
- H? Or not? In skeletal structure, if it's a junction, it might not have H.

In standard skeletal drawing, a carbon with three lines shown has one H implied if it's sp3, but here it's attached to three groups: ethyl, isopropyl, and the triple bond carbon. So it has no hydrogen; it's a tertiary carbon.

The isopropyl group is –CH(CH₃)₂, so it has a carbon with one H and two methyls.

So the atom attached to the triple bond is a carbon that is bonded to:
- The triple bond carbon
- A CH₂CH₃ group (ethyl)
- A CH(CH₃)₂ group (isopropyl)

So this carbon has three carbon atoms attached, so it's tertiary.

For the longest chain, we can choose to go from the isopropyl's end through this carbon to the triple bond and then to the right.

Isopropyl: the CH group is attached to two methyls and to our central carbon.

So if we start from one methyl of isopropyl: C1 (CH₃) – C2 (CH) – C3 (central) – C4 ≡ C5 – C6 – C7 – C8 – C9

C2 has a methyl group (the other methyl of isopropyl).

C6 has a methyl group (from the right side).

Right side: after triple bond, it's –CH(CH₃)CH₂CH₂CH₃, so C6 is CH, with a methyl, then C7=CH2, C8=CH2, C9=CH3.

So chain: C1 to C9: 9 carbons.

Triple bond between C4-C5 → position 4.

Substituents:
- On C2: methyl group (since C2 is the CH of isopropyl, and it has an additional methyl)
- On C6: methyl group

So 2-methyl and 6-methyl.

Also, on C3, there is an ethyl group attached? C3 is the central carbon, which is bonded to C2, C4, and also to ethyl group.

So ethyl group is a substituent on C3.

So substituents:
- Ethyl on C3
- Methyl on C2
- Methyl on C6

Now, numbering: triple bond at 4.

If we number from the other end: from right, C1=C9, C2=C8, C3=C7, C4=C6 (has methyl), C5≡C6? Same issue.

C1 = CH3 (right end)
C2 = CH2
C3 = CH2
C4 = CH(CH3)
C5 ≡ C6
C7 = C(central)
C8 = CH (of isopropyl)
C9 = CH3 (one methyl of isopropyl)

Then C7 has ethyl group, C8 has methyl group (the other methyl of isopropyl), C4 has methyl group.

Triple bond between C5-C6 → position 5.

Previously from left it was 4, so better to number from left.

So with left numbering: C1 to C9 as above.

Substituents:
- Methyl on C2
- Ethyl on C3
- Methyl on C6

Alphabetical: ethyl, methyl, methyl → so 3-ethyl-2,6-dimethylnon-4-yne

Yes, this is better because we included the isopropyl's carbon in the main chain.

So correct name: 3-ethyl-2,6-dimethylnon-4-yne

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d)
Structure: Double bond at one end, triple bond at other, with three methylenes in between.

So: CH2=CH–CH2–CH2–CH2–C≡CH

Longest chain: 7 carbons, with double bond at 1-2, triple bond at 6-7.

Numbering: should give lowest numbers to multiple bonds. If number from left: double bond at 1, triple bond at 6.

If number from right: triple bond at 1, double bond at 6.

Same set of numbers: 1 and 6.

Rule: when both ends have multiple bonds, give lowest number to the one that comes first alphabetically? No, the rule is to give the lowest number to the multiple bond regardless of type, but if tie, then double bond gets preference? Actually, IUPAC says: number so that the multiple bonds have the lowest numbers, and if there is a choice, give the lowest number to the double bond.

Recall: for enynes, the chain is numbered to give the lowest numbers to the multiple bonds combined, and if there is a tie, the double bond gets the lower number.

Here, if number from left: double bond at 1, triple bond at 6 → positions 1,6

If number from right: triple bond at 1, double bond at 6 → positions 1,6 same.

But since double bond should get lower number if tie, we number so that double bond is at 1.

So name: hept-1-en-6-yne

Substituents: none.

So hept-1-en-6-yne

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e)
Structure: Has a double bond with Br, and a triple bond.

Drawing: Left: Br attached to a carbon that is part of =CH2? Or what.

It's: Br–C(=CH2)–CH(ethyl)–CH(methyl)–CH(methyl)–C≡CH

More precisely:
The left part is a carbon with Br, double bond to CH2, and single bond to next carbon.

So it's (Br)(H2C=)C– then –CH(C2H5)–CH(CH3)–CH(CH3)–C≡CH

So longest chain: include both double and triple bonds.

Start from the =CH2 end: C1 (=CH2) – C2 (with Br) – C3 (CH) – C4 (CH) – C5 (CH) – C6 ≡ C7

C2 has Br
C3 has ethyl group
C4 has methyl group
C5 has methyl group

Chain: 7 carbons.

Double bond between C1-C2 → position 1
Triple bond between C6-C7 → position 6

Numbering: if we number from the other end, triple bond at 1, double bond at 6.

Same as before, but now we have substituents.

Rule: give lowest numbers to multiple bonds. Positions 1 and 6 either way.

But since double bond should get lower number if tie, we number from the double bond end.

So C1=C2 (double bond), C3, C4, C5, C6≡C7

Substituents:
- Br on C2
- Ethyl on C3
- Methyl on C4
- Methyl on C5

Alphabetical: bromo, ethyl, methyl, methyl

So name: 2-bromo-3-ethyl-4,5-dimethylhept-1-en-6-yne

Check: hept = 7 carbons, 1-en, 6-yne, substituents at 2,3,4,5.

Yes.

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f)
Structure: Br attached to a carbon that is part of double bond, then chain to triple bond.

Drawing: Br–C(CH3)=CH–CH(CH3)–CH(CH3)–C≡CH

So: the double bond is between C1 and C2, with Br and CH3 on C1, H on C2? Standard.

C1 is the carbon with Br and CH3, double bond to C2, which is CH, then C3 is CH(CH3), C4 is CH(CH3), C5≡C6

Longest chain: 6 carbons.

Double bond between C1-C2
Triple bond between C5-C6

Numbering: if from left, double bond at 1, triple bond at 5.

If from right, triple bond at 1, double bond at 5.

Positions 1 and 5 vs 1 and 5 — same.

But double bond should get lower number, so number from left.

Substituents:
- On C1: methyl group (since C1 has Br, CH3, and double bond to C2 — so the methyl is a substituent)
- On C3: methyl group
- On C4: methyl group

C1 is part of double bond, and has a methyl group attached, so it's 1-bromo-1-methyl? But in naming, the carbon of the double bond is included.

The chain is C1=C2–C3–C4–C5≡C6

C1 has substituents: Br and CH3
C3 has CH3
C4 has CH3

So substituents:
- Bromo on C1
- Methyl on C1
- Methyl on C3
- Methyl on C4

But C1 has two substituents: bromo and methyl.

In naming, we list them.

Alphabetical: bromo, methyl, methyl, methyl

So 1-bromo-1,3,4-trimethylhex-1-en-5-yne? Hex because 6 carbons.

Chain: C1 to C6 = hex.

Double bond at 1, triple bond at 5.

Substituents on C1: bromo and methyl → so 1-bromo-1-methyl

On C3: methyl → 3-methyl

On C4: methyl → 4-methyl

So altogether: 1-bromo-1,3,4-trimethylhex-1-en-5-yne

But is the methyl on C1 considered a substituent? Yes.

Sometimes written as such.

Note: the carbon C1 is sp2, with atoms: double bond to C2, single bond to Br, single bond to CH3 — so yes.

So name: 1-bromo-1,3,4-trimethylhex-1-en-5-yne

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Now, compiling all answers:

a) 5-chloro-4,4-dimethylpent-2-yne
b) 8-bromo-3,7-dimethylnon-4-yne
c) 3-ethyl-2,6-dimethylnon-4-yne
d) hept-1-en-6-yne
e) 2-bromo-3-ethyl-4,5-dimethylhept-1-en-6-yne
f) 1-bromo-1,3,4-trimethylhex-1-en-5-yne

I think that's correct.

Final Answer:
a) 5-chloro-4,4-dimethylpent-2-yne
b) 8-bromo-3,7-dimethylnon-4-yne
c) 3-ethyl-2,6-dimethylnon-4-yne
d) hept-1-en-6-yne
e) 2-bromo-3-ethyl-4,5-dimethylhept-1-en-6-yne
f) 1-bromo-1,3,4-trimethylhex-1-en-5-yne