7.3 Alkanes, Alkenes and Alkynes online exercise for - Free Printable

Let's go through each question in the "7.3 Alkanes, alkenes and alkynes quiz" step by step and provide the correct answers with explanations.

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1. Identify alkane, alkene or alkyne

We are given three structures:

#### Structure 1:
```
H H H
| | |
H–C–C=C–H
| | |
H H H
```

- This has a double bond between two carbon atoms.
- So it is an alkene.

✔ Answer: Alkene

---

#### Structure 2:
```
H H H H H H
| | | | | |
H–C–C–C–C–C–C–H
| | | | | | |
H H H H H H H
```

- All bonds are single bonds; saturated hydrocarbon.
- So it is an alkane.

✔ Answer: Alkane

---

#### Structure 3:
```
H–C≡C–H
```

- Triple bond between two carbons.
- So it is an alkyne.

✔ Answer: Alkyne

---

2. Name the hydrocarbons in part 1

Now we name the molecules from Question 1.

#### First molecule (alkene):
- Longest chain: 3 carbons → propane
- Double bond starts at carbon 1 → propene
- But since the double bond is between C1 and C2, it's propene
- However, standard naming: we number to give the double bond lowest numbers → it’s just propene
- But note: the structure shows the double bond between the second and third carbon? Wait — let's recheck.

Wait! The structure is:
```
H H H
| | |
H–C–C=C–H
| | |
H H H
```
So carbon 1: CH₃–
Carbon 2: –CH₂–
Carbon 3: =CH₂

Actually, the double bond is between C2 and C3, so it's propene (no need for number because only one possibility).

But wait — this is actually propene (C₃H₆), but the double bond is between C2 and C3 → same as propene.

✔ Answer: Propene

But technically, since the double bond is between C2 and C3, and there's no branching, it's simply propene.

However, if we number the chain from left to right: C1–C2=C3 → double bond starts at C2 → prop-1-ene? No!

Wait: In propene, the double bond is always between C1 and C2 — but here it's drawn as C1–C2=C3, which is still propene, but the numbering should be such that the double bond gets the lowest number.

If you number from right to left: C1=C2–C3 → then double bond is at C1 → better.

So correct name: prop-1-ene?

But propene is acceptable and commonly used.

Actually, the IUPAC name for this is propene, and it's implied the double bond is at position 1.

But in this case, the structure is CH₃–CH=CH₂, which is prop-1-ene.

Yes, that’s correct.

✔ Answer: Prop-1-ene

---

#### Second molecule (alkane):
Six carbon chain, all single bonds → hexane

✔ Answer: Hexane

---

#### Third molecule (alkyne):
H–C≡C–H → Two carbons, triple bond → ethyne

✔ Answer: Ethyne

---

3. How many carbons in the parent chain of 2,3-dimethylcyclopentene?

- "Cyclopentene" means a 5-membered ring with a double bond.
- So the parent chain (ring) has 5 carbons.

The methyl groups are substituents.

✔ Answer: 5

---

4. How many carbons are there in the branch of 5-ethyl-2-hexyne?

- The branch is ethyl group.
- Ethyl = –CH₂–CH₃ → 2 carbons

✔ Answer: 2

---

5. How many carbons are there in total in 2-ethyl-3-methylbutene?

Let’s break it down.

First, the base chain is butene → 4 carbons.

But: 2-ethyl-3-methylbutene

Wait — butene is 4 carbons. But adding ethyl and methyl branches.

But “butene” suggests the longest chain is 4 carbons.

But 2-ethyl on a butene chain would make a longer chain.

Let’s think carefully.

The name 2-ethyl-3-methylbutene might be incorrect according to IUPAC rules because the longest chain may be longer than 4.

But let’s assume the name is as given.

- Butene: 4 carbons
- 2-ethyl: adds 2 carbons (ethyl group)
- 3-methyl: adds 1 carbon

Total carbons: 4 + 2 + 1 = 7

But wait — is that correct?

Actually, the ethyl group at carbon 2 of butene would extend the chain. For example:

If you have:

```
CH₃
|
CH₃–C–CH=CH₂
|
CH₂–CH₃
```

Then the longest chain is not 4 — it's 5 or more.

But regardless, total number of carbons:

- Butene: 4
- Ethyl: 2
- Methyl: 1

Total: 7 carbons

✔ Answer: 7

---

6. Name the following branched alkanes

#### First structure:
```
H H H H H
| | | | |
H–C–C–C=C–C–C–H
| | | | | | |
H H H H H H H
```

This is a 6-carbon chain with a double bond between C3 and C4.

So it's a hexene.

Numbering: double bond at C3–C4 → lowest number is 3 → hex-3-ene

But we can number from the other end → double bond at C3–C4 → same.

So name: hex-3-ene

But wait — is there any branching?

No, all carbons are in chain.

So yes: hex-3-ene

✔ Answer: hex-3-ene

---

#### Second structure:
```
H CH₃
| |
H₃C–C=C–C–C–H
| |
H H
```

Let’s write it clearly:

- Carbon 1: CH₃–
- Carbon 2: =C– (attached to CH₃ and H)
- Carbon 3: –C– (attached to CH₃ and H)
- Carbon 4: –C– (attached to CH₃ and H)

Wait — let's assign:

Actually:

```
H CH₃
| |
H₃C–C=C–C–C–H
| |
H H
```

Wait — that’s not balanced.

Better:

It looks like:

- Left: CH₃–C(H)=C(CH₃)–CH(CH₃)–H

Wait — the middle carbon is double bonded to both sides?

No — it's:

Carbon 1: CH₃–
Carbon 2: =C– (with H)
Carbon 3: =C– (with CH₃)
Carbon 4: –C– (with CH₃)
Carbon 5: –CH₃?

Wait — the structure is:

```
H CH₃
| |
H₃C–C=C–C–C–H
| |
H H
```

Wait — this seems to have 5 carbons.

Let’s count:

- Left CH₃– (C1)
- Then C2: carbon with H and double bond
- Then C3: carbon with CH₃ and double bond
- Then C4: carbon with CH₃ and H
- Then C5: CH₃

But C4 has two bonds shown: one to C3, one to CH₃, and one to H? And another bond?

Wait — the structure is:

```
H CH₃
| |
H₃C–C=C–C–C–H
| |
H H
```

Wait — that’s confusing.

Looking again:

The central part is:

- Carbon A: CH₃– (left)
- Carbon B: –C(H)= (double bond)
- Carbon C: =C(CH₃)–
- Carbon D: –C(CH₃)–
- Carbon E: –CH₃

But carbon D has three bonds shown: to C, to CH₃, and to CH₃? No.

Wait — the diagram says:

```
H CH₃
| |
H₃C–C=C–C–C–H
| |
H H
```

So:

- Carbon 1: CH₃–
- Carbon 2: –C(H)= (bonded to C1, H, and double bond to C3)
- Carbon 3: =C(CH₃)– (bonded to C2 via double bond, CH₃, and single bond to C4)
- Carbon 4: –C– (bonded to C3, CH₃, and CH₃?) No — it's written as –C–C–H

Wait — the last part is: –C–C–H with CH₃ above and H below?

Wait — the structure is:

```
H CH₃
| |
H₃C–C=C–C–C–H
| |
H H
```

So:

- C1: CH₃–
- C2: –C(H)=
- C3: =C(CH₃)–
- C4: –C– (with CH₃ attached)
- C5: –C–H (with H below)

Wait — C4 has: bonded to C3, CH₃, and C5 → so three bonds.

C5: bonded to C4 and H → but needs another bond.

Wait — probably it's:

C4 is –CH(CH₃)–CH₃? But it's written as –C–C–H with H below.

Ah — likely: C4 is a carbon with a methyl group and bonded to C5, which is CH₃.

So full chain: C1–C2=C3–C4–C5

With:

- C1: CH₃–
- C2: –CH= (has H)
- C3: =C–CH₃ (has methyl)
- C4: –CH–CH₃ (has methyl group)
- C5: –CH₃

Wait — but C4 has two methyl groups? No — only one methyl shown.

Looking again:

The diagram shows:

```
H CH₃
| |
H₃C–C=C–C–C–H
| |
H H
```

And the bottom of the last carbon has H.

So:

- C1: CH₃–
- C2: –CH= (H)
- C3: =C–CH₃ (methyl)
- C4: –CH– (with CH₃ attached)
- C5: –CH₃

So C4 has: bonded to C3, CH₃, and C5 → so it's a CH group.

So the chain is: CH₃–CH= C(CH₃)–CH(CH₃)–CH₃

So:

- Parent chain: 5 carbons → pentene
- Double bond between C2 and C3
- Substituents:
- C3 has a methyl → 3-methyl
- C4 has a methyl → 4-methyl

So name: 3,4-dimethylpent-2-ene

But check numbering: double bond at C2–C3 → if we number from right to left:

C1' = CH₃–
C2' = –CH(CH₃)–
C3' = –C(CH₃)=
C4' = –CH–
C5' = –CH₃

Double bond between C3' and C4'? That would be higher number.

So better to keep original: double bond at C2–C3 → lower number.

Substituents at C3 and C4 → both methyls.

So 3,4-dimethylpent-2-ene

But is that correct?

Wait — C3 already has a methyl → so it's a tertiary carbon.

Yes.

✔ Answer: 3,4-dimethylpent-2-ene

---

#### Third structure:

```
CH₃–CH=C–CH₃
|
CH₃
```

So:

- Chain: CH₃–CH=C–CH₃
- With a methyl group on the third carbon.

So:

- Four carbons in chain
- Double bond between C2 and C3
- Methyl on C3

So: 3-methylbut-2-ene

But let’s number properly.

Chain: C1–C2=C3–C4

C3 has a methyl group → so it’s a branched chain.

But the longest chain is still 4 carbons.

Double bond at C2–C3.

Methyl on C3 → so 3-methylbut-2-ene

But wait — if we number from right to left:

C1’ = CH₃–
C2’ = –C(CH₃)=
C3’ = –CH–
C4’ = –CH₃

Double bond between C2’ and C3’ → same as before.

But now methyl is on C2’ → so 2-methylbut-2-ene

Which is better — because the double bond gets lower number? No — double bond is at C2–C3 in both cases.

But in first numbering: double bond at C2–C3, methyl at C3 → 3-methylbut-2-ene

In second: double bond at C2–C3, methyl at C2 → 2-methylbut-2-ene

Which is better? We want the double bond to get the lowest number, but both are same.

But the substituent should get lowest number if possible.

So if we number from right: methyl is on C2 → better.

So name: 2-methylbut-2-ene

Yes — that’s the correct IUPAC name.

✔ Answer: 2-methylbut-2-ene

---

7. What is the correct name for this hydrocarbon?

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

Wait — the structure is:

```
CH₃–C≡C–C≡C–C–CH–CH₃
| |
CH₂ CH₃
|
CH₃
```

So:

- Chain: CH₃–C≡C–C≡C–C–CH–CH₃
- On the sixth carbon: CH₂–CH₃ (ethyl group) and CH₃ (methyl group)? No — it's:

Wait — the carbon after the second triple bond has:

- Bond to previous carbon
- Bond to CH₂–CH₃ (ethyl)
- Bond to CH₃ (methyl)
- Bond to CH–CH₃ (which has CH₃)

Wait — let’s list:

- C1: CH₃–
- C2: ≡C–
- C3: –C≡
- C4: ≡C–
- C5: –C≡
- C6: –C– (with CH₂–CH₃ and CH₃)
- C7: –CH– (with CH₃)
- C8: –CH₃

Wait — no — C6 is bonded to:

- C5
- CH₂–CH₃ (ethyl)
- CH₃ (methyl)
- C7

So C6 is quaternary.

C7: bonded to C6, CH₃, and CH₃? No — it says CH–CH₃, and CH₃ above.

So C7: –CH– with one methyl group, and bonded to C8: CH₃

So C7 has: C6, CH₃, CH₃ → so it's –CH(CH₃)–CH₃

But the structure is:

```
CH₃–C≡C–C≡C–C–CH–CH₃
| |
CH₂ CH₃
|
CH₃
```

So:

- C1: CH₃–
- C2: ≡C–
- C3: –C≡
- C4: ≡C–
- C5: –C≡
- C6: –C– (with ethyl and methyl)
- C7: –CH– (with methyl)
- C8: –CH₃

So chain is 8 carbons.

Double bonds? No — triple bonds.

Triple bonds at C2–C3 and C4–C5 → so diynes.

So it's an octadiyne.

Longest chain: 8 carbons.

Triple bonds at positions 2 and 4.

Now, substituents:

- At C6: ethyl and methyl → but ethyl is CH₂–CH₃, methyl is CH₃

But C6 has two groups: ethyl and methyl → so it's 6-ethyl-6-methyl?

But wait — ethyl and methyl on same carbon → so two substituents.

But look: C6 has:

- One methyl group
- One ethyl group

So substituents: methyl and ethyl at C6.

But also, C7 has a methyl group.

Wait — C7 is –CH– with one methyl → so it has a methyl group.

But C7 is also bonded to C8 (CH₃), so it's –CH(CH₃)–CH₃ → so it has a methyl group.

So:

- C6: methyl and ethyl
- C7: methyl

But C7 is CH(CH₃)–CH₃ → so it's a methyl group attached to C7.

But the chain is:

C1–C2≡C3–C4≡C5–C6–C7–C8

So C6 is carbon with two alkyl groups: methyl and ethyl.

C7 has one methyl group.

But wait — C7 is –CH–CH₃ with CH₃ above → so it's –CH(CH₃)–CH₃ → so C7 has a methyl substituent.

So total:

- C6: methyl and ethyl → two substituents
- C7: methyl → one substituent

But wait — C6 has two groups: methyl and ethyl → so we name them.

But the structure shows:

At C6: CH₂–CH₃ and CH₃

At C7: CH₃

But C7 is –CH– with CH₃ above → so it's a methyl group on C7.

But C7 is also bonded to C8 (CH₃) → so C7 is CH(CH₃)–CH₃ → so it's a methyl group.

But the carbon at C7 is CH(CH₃)–CH₃ → so it's a methyl substituent.

But wait — the carbon is C7, bonded to C6, CH₃, and CH₃ → so it's –CH(CH₃)–CH₃ → so it's a methyl group on C7.

But actually, C7 has two methyl groups? No — it's bonded to:

- C6
- CH₃ (one methyl)
- CH₃ (the terminal CH₃)

So it's –CH(CH₃)–CH₃ → so it's a methyl group.

But the group is ethyl? No — it's two methyls.

Wait — C7 is –CH– with one methyl group and bonded to CH₃ → so it's –CH(CH₃)–CH₃ → so it's a methyl group.

But the carbon is CH(CH₃)–CH₃ → so it's a methyl substituent.

But actually, it's not a substituent — it's part of the chain.

Wait — the chain is C1–C2–C3–C4–C5–C6–C7–C8

C7 is CH(CH₃)–CH₃ → so it's a carbon with a methyl group.

So the chain is 8 carbons long.

Substituents:

- At C6: methyl and ethyl
- At C7: methyl

But C7 has a methyl group → so it's 7-methyl

But C6 has two groups: methyl and ethyl → so 6-methyl and 6-ethyl

So name: 6-ethyl-6,7-dimethyl-2,4-octadiyne

But wait — the options are:

a. 6-ethyl-6,7-methyl-2,4-octadiene
b. 6,7-dimethyl-6-ethyl-2,4-octene
c. 6-ethyl-6,7-dimethyl-2,4-octadiyne
d. 6-ethyl-6,7-dimethyl-2,4-octadiene

Our compound has two triple bonds → so diyne, not diene.

So eliminate a, b, d.

Only c has diyne.

Also, substituents: 6-ethyl, 6-methyl, 7-methyl → so 6-ethyl-6,7-dimethyl

Yes.

So c. 6-ethyl-6,7-dimethyl-2,4-octadiyne

✔ Answer: c

---

8. Name the cyclohydrocarbons

#### First structure:
A square with a double bond → cyclobutene

- Four-membered ring with one double bond → cyclobutene

✔ Answer: Cyclobutene

---

#### Second structure:
A six-membered ring with one double bond and one methyl group.

- Cyclohexene with a methyl group → methylcyclohexene

But where is the methyl?

The methyl is on a carbon adjacent to the double bond.

To give the double bond lowest number, we number so that the double bond is between C1 and C2.

Then methyl is on C3 → so 3-methylcyclohexene

But if methyl is on C1, it would be 1-methylcyclohexene, but that would be same as methylcyclohexene.

But the double bond must be between C1 and C2.

So if methyl is on C1, it’s 1-methylcyclohexene, but that’s not correct because the carbon with methyl is sp³, but in cyclohexene, C1 and C2 are sp².

So the methyl cannot be on a double bond carbon.

So if methyl is on a carbon adjacent to double bond, it’s 3-methylcyclohexene

But wait — the structure shows a methyl on a carbon that is not part of the double bond.

So it’s 3-methylcyclohexene

But is it 3 or 4? Since the ring is symmetric, 3-methyl is same as 4-methyl.

But we use the lowest number → so 3-methylcyclohexene

✔ Answer: 3-methylcyclohexene

---

✔ Final Answers:

1. Alkene, Alkane, Alkyne
2. Prop-1-ene, Hexane, Ethyne
3. 5
4. 2
5. 7
6. Hex-3-ene, 3,4-dimethylpent-2-ene, 2-methylbut-2-ene
7. c. 6-ethyl-6,7-dimethyl-2,4-octadiyne
8. Cyclobutene, 3-methylcyclohexene

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