Alkene Reaction Worksheet - Warning: TT: undefined function: 32 ... - Free Printable

Let’s go step by step through this concept map. It shows reactions of alkenes and alkynes — mostly addition, oxidation, reduction, and substitution reactions from Chapters 6–7 in organic chemistry.

We’ll start from the central alkene: 1-butene (CH₂=CH–CH₂–CH₃). Actually, looking closely, it's written as CH₂ with a CH₃ attached to the double bond carbon — so that’s propene: CH₂=CH–CH₃. Wait — no, let me check again.

Actually, the structure drawn is:

```
CH₃
|
CH₂ = C - H ? No — wait, the drawing says:
```

Looking at the diagram: The central molecule has “CH₂” on one end of the double bond and “CH₃” on the other — so it’s propene: CH₂=CH–CH₃? But then there’s an arrow going down to a box labeled “X₂”, which gives a product with two CH₃ groups — that suggests the original alkene might be 2-methylpropene? Let’s re-express.

Wait — actually, looking at the top center: the molecule is drawn as:

```
CH₃
|
CH₂ = C
```

That’s 2-methylpropene (isobutylene): (CH₃)₂C=CH₂

Yes! That makes sense because many of the products have branched chains.

So we’ll treat the central alkene as (CH₃)₂C=CH₂ — 2-methylpropene.

Now let’s fill in each blank systematically.

---

Top Left Branch: Reaction to give HO–CH(CH₃)–CH₂–X

This is anti-Markovnikov addition of HX? No — wait, the product is HO–CH(CH₃)–CH₂–X — that looks like halohydrin formation but reversed? Actually, no.

Wait — the reagent box is empty, and the product is HO–CH(CH₃)–CH₂–X.

But note: the oxygen is on the more substituted carbon? In 2-methylpropene, if you do acid-catalyzed hydration, you get tertiary alcohol. Here, the OH is on the secondary carbon? Let’s think.

Actually, this must be hydroboration-oxidation followed by something? No — hydroboration gives anti-Markovnikov alcohol without X.

Wait — perhaps this is addition of hypohalous acid (HOX)? Like Br₂/H₂O → bromohydrin.

In bromohydrin formation, for unsymmetrical alkenes, OH goes to more substituted carbon, Br to less.

For (CH₃)₂C=CH₂ + Br₂/H₂O → (CH₃)₂C(OH)–CH₂Br

But here the product is drawn as HO–CH(CH₃)–CH₂–X — which would imply OH on the CH, and X on CH₂ — same thing if we write it as HO–CH(CH₃)–CH₂–X meaning the carbon with OH has one H and one CH₃, so it’s (CH₃)(H)C–OH attached to CH₂X — yes, that matches (CH₃)₂C(OH)CH₂X? No — (CH₃)₂C(OH)CH₂X would be tertiary alcohol.

Wait — confusion in notation.

The product is drawn as:

```
HO
|
CH - CH₃
|
CH₂ - X
```

No — looking back at the image description: "HO CH₃" above "CH₂ X" — probably it's HO–CH(CH₃)–CH₂–X, meaning the carbon with OH is chiral? For 2-methylpropene, adding HO-X should give (CH₃)₂C(OH)CH₂X — tertiary alcohol.

But the drawing shows HO attached to a carbon that also has H and CH₃, and then CH₂X — so that would be from propene, not 2-methylpropene.

I think I made a mistake.

Let me reinterpret the central alkene.

Looking at the diagram again: the central alkene is drawn with “CH₂” on left, double bond, then “CH₃” on right — so it’s propene: CH₂=CH–CH₃.

Then:

- Up-left: reaction gives HO–CH(CH₃)–CH₂–X → that would be from propene + HOX → Markovnikov-like for OH? No — in halohydrin, OH goes to more substituted carbon.

Propene + Br₂/H₂O → CH₃–CH(OH)–CH₂Br

Which is exactly HO–CH(CH₃)–CH₂–Br — yes! So X is Br or Cl.

Reagent: Br₂, H₂O or Cl₂, H₂O

So first blank (top left reagent): Br₂, H₂O (or similar)

Product: already given as HO–CH(CH₃)–CH₂–X

---

Top Right: Epoxide formation

From alkene to epoxide: use mCPBA or peracid.

Product is drawn as a triangle with O and CH₃ — so for propene, epoxide is methyloxirane.

Reagent: mCPBA or RCO₃H

Blank: mCPBA

---

Right side: Acid-catalyzed hydration

Alkene + H₂SO₄, H₂O → alcohol

For propene: Markovnikov addition → CH₃–CH(OH)–CH₃ (isopropanol)

Product blank: CH₃CH(OH)CH₃

---

Below that: Hydroboration-oxidation

1) BH₃, 2) NaOH, H₂O₂ → anti-Markovnikov alcohol

Propene → CH₃CH₂CH₂OH (propanol)

Product blank: CH₃CH₂CH₂OH

---

Left side: Dihydroxylation

1) OsO₄, 2) NaHSO₃, H₂O → syn diol

Propene → CH₃CH(OH)CH₂OH

Product blank: CH₃CH(OH)CH₂OH

---

Downward from alkene: Addition of X₂

Gives vicinal dihalide

Propene + Br₂ → CH₃CHBrCH₂Br

Product blank: CH₃CHBrCH₂Br (if X=Br)

---

To the right of that: Addition of HX

Propene + HX → Markovnikov addition → CH₃CHXCH₃

Product blank: CH₃CHXCH₃

---

Further right: Alkyne section

There’s a terminal alkyne: HC≡C–CH₃ (propyne)

How do we get there? From the dihalide via double dehydrohalogenation.

From CH₃CHBrCH₂Br + 2 NaNH₂ → HC≡C–CH₃

Reagent blank between dihalide and alkyne: 2 eq NaNH₂ or strong base

Then, alkyne reactions:

First: HgSO₄, H₂SO₄, H₂O → hydration of terminal alkyne → methyl ketone

HC≡C–CH₃ + H₂O → CH₃COCH₃ (acetone)

Product blank: CH₃COCH

Second: Hydroboration-oxidation of terminal alkyne

1) R₂BH (disiamylborane or similar), 2) NaOH, H₂O₂ → aldehyde

HC≡C–CH₃ → CH₃CH₂CHO (propanal)

Product blank: CH₃CH₂CHO

Third: Addition of 1 eq X₂ to alkyne → trans-dihalide

HC≡C–CH₃ + Br₂ → CH₃CBr=CBrH (trans)

Product blank: CH₃CBr=CBrH (specify trans if needed, but often just formula)

Then further addition of X₂ → tetrahalide

CH₃CBr=CBrH + Br₂ → CH₃CBr₂CBr₂H

Product blank: CH₃CBr₂CBr₂H

---

Back to alkyne: Deprotonation

Terminal alkyne + strong base → acetylide ion

HC≡C–CH₃ + NaNH₂ → ⁻CC–CH₃

Reagent blank: NaNH₂

Then, alkylation: ⁻C≡C–CH₃ + alkyl halide → internal alkyne

Here, it reacts with (CH₃)₂CH–X? Looking at the product: it becomes CH₃–C≡C–CH(CH₃)₂

So alkyl halide is (CH₃)₂CH–X (isopropyl halide)

Reagent blank: (CH₃)₂CHX

Product: CH₃C≡CCH(CH₃)₂

Then, partial hydrogenation to cis-alkene

Use Lindlar catalyst: H₂, Pd/CaCO₃, quinoline

Product: CH₃CH=CHCH(CH₃)₂ with cis configuration

Drawing shows H and H on same side? The product drawn is:

```
H₃C H
\ /
C=C
/ \
H CH(CH₃)₂
```

Wait — no, the drawing says:

```
H₃C H
\ /
C=C
/ \
H₃C CH₃? No — it says H₃C and then below H₃C–CH–CH₃? Let's see:

Actually, the product is drawn as:

H
|
H₃C - C = C - H
|
CH(CH₃)₂

With stereochemistry implied as cis? The arrows show both H on same side? In the diagram, it's written as:

H₃C H
\ /
C=C
/ \
H CH(CH₃)₂

But that would be if the groups are arranged that way.

Actually, from CH₃C≡CCH(CH₃)₂, reducing with Lindlar gives (Z)-CH₃CH=CHCH(CH₃)₂

Which can be drawn as:

CH₃ H
\ /
C=C
/ \
H CH(CH₃)₂

Yes.

Reagent blank: H₂, Lindlar catalyst

Then, full hydrogenation to alkane

Use H₂, Pt or Pd

Product: CH₃CH₂CH₂CH(CH₃)₂

Which is 2-methylbutane? CH₃CH₂CH₂CH(CH₃)₂ is actually 2-methylbutane? Let's count:

Carbon 1: CH₃– (from ethyl part?) Better: CH₃–CH₂–CH₂–CH(CH₃)₂

That’s 1-(1-methylethyl)propane? Standard name: 2-methylbutane is (CH₃)₂CHCH₂CH₃ — different.

CH₃CH₂CH₂CH(CH₃)₂ is 2-methylbutane? No:

Structure: carbon chain: C1–C2–C3–C4, with C4 having two methyls? No.

CH₃–CH₂–CH₂–CH(CH₃)₂ means:

- Carbon 1: CH₃–

- Carbon 2: –CH₂–

- Carbon 3: –CH₂–

- Carbon 4: –CH– with two methyl groups? No, CH(CH₃)₂ means one H and two methyls, so carbon 4 is CH, attached to two CH₃ groups.

So total carbons: C1,C2,C3,C4 plus two methyls on C4 → 6 carbons.

Name: 2-methylpentane? Let's number properly.

Longest chain: from one end of propyl to one methyl: CH₃–CH₂–CH₂–CH–CH₃ with another CH₃ on the CH → so chain of 5 carbons with methyl on carbon 2? If we number from the branch:

Better: the group is –CH(CH₃)₂ attached to propyl, so it's 1-isopropylpropane? Standard name: 2-methylpentane is wrong.

Actual structure: CH₃–CH₂–CH₂–CH(CH₃)₂

Carbon atoms:

- Terminal CH₃– (C1)

- –CH₂– (C2)

- –CH₂– (C3)

- –CH– (C4) with two methyl groups: so C4 attached to H, CH₃, CH₃, and C3.

So longest chain is from C1 to one of the methyls on C4: C1–C2–C3–C4–C5 (where C5 is one methyl), so pentane with a methyl on C4? That would be 2-methylpentane if numbered differently.

Standard naming: choose longest chain. Here, chain of 4 carbons in propyl plus one carbon in isopropyl? No.

Total carbons: 6. Longest continuous chain: start from one methyl of isopropyl, through C4, C3, C2, C1: that's 5 carbons. Then the other methyl on C4 is a substituent.

So: CH₃–CH(CH₃)–CH₂–CH₂–CH₃? No, our molecule is CH₃–CH₂–CH₂–CH(CH₃)₂

Which is the same as (CH₃)₂CH–CH₂–CH₂–CH₃ — yes! That's 2-methylpentane.

(CH₃)₂CH– is isopropyl, attached to CH₂CH₂CH₃ — propyl, so isopropylpropyl, but standard name: longest chain is 5 carbons: from one end of propyl to the end of isopropyl.

Numbering: CH₃–CH₂–CH₂–CH–CH₃ with a CH₃ on the CH — so carbon 1: CH₃– (of propyl), C2: CH₂, C3: CH₂, C4: CH, C5: CH₃ (one methyl), and additional CH₃ on C4.

So chain of 5 carbons with methyl on C4 — but we number to have lowest numbers, so better to number from the other end: make the branched carbon C2.

Set C1 as the CH of isopropyl: but it's tertiary? No.

Molecule: carbon A: CH₃– (first methyl of isopropyl)

Carbon B: CH– (central of isopropyl) attached to A, C, and D

Carbon C: CH₃ (second methyl of isopropyl)

Carbon D: CH₂– (first of propyl)

Carbon E: CH₂–

Carbon F: CH₃

Longest chain: A-B-D-E-F: 5 carbons, with a methyl (C) on B.

So B is carbon 2, so 2-methylpentane.

Yes.

Product blank: CH₃CH₂CH₂CH(CH₃)₂ or (CH₃)₂CHCH₂CH₂CH₃

Reagent: H₂, Pt or H₂, Pd

---

Now, back to the left side: ozonolysis

The alkene before ozonolysis is drawn as:

```
H₃C H
\ /
C=C
/ \
H CH(CH₃)₂
```

Which is the same as the cis-alkene we just made: (Z)-CH₃CH=CHCH(CH₃)₂

Ozonolysis: 1) O₃, 2) (CH₃)₂S → cleaves double bond

Products: CH₃CHO and O=CHCH(CH₃)₂

Which is acetaldehyde and 2-methylpropanal

The diagram shows:

Left product: from O₃ then (CH₃)₂S → two carbonyls

One is drawn as H₃C–C(=O)–H (acetaldehyde)

Other is H–C(=O)–CH(CH₃)₂ (2-methylpropanal)

And they are shown connected? No, in the diagram, it's a single box with both structures? Looking back:

It says:

```
H₃C H
\ /
C=C
/ \
H CH(CH₃)₂
```

Then arrow to left with 1. O₃, 2. (CH₃)₂S → and then a box containing:

On left: H₃C–C(=O)–H

On right: H–C(=O)–CH(CH₃)₂

But in the text, it's written as:

"H₃C H
\ /
C=C
/ \
H CH(CH₃)₂"

Then after ozonolysis, it should give two separate molecules.

In the diagram, the product box seems to contain both, but typically we list them separately.

However, in the blank, it might be expecting the two products.

But looking at the layout, there is a large box on the far left that receives the output of ozonolysis.

Probably, it wants the two carbonyl compounds.

But since it's a single box, perhaps write both.

Alternatively, maybe it's showing the combined structure, but that doesn't make sense.

Another possibility: the alkene being ozonized is not the one we thought.

Look at the diagram: above the ozonolysis, there is an alkene drawn as:

```
H₃C H
\ /
C=C
/ \
H CH₃? No — it says H₃C and then below H₃C–CH–CH₃? Let's read carefully.

In the user's description: "H₃C H" on top, "H CH(CH₃)₂" on bottom? Earlier I assumed CH(CH₃)₂, but let's confirm.

Actually, in the initial problem statement, it's written as:

After the alkyne alkylation, we have CH₃–C≡C–CH(CH₃)₂, then reduced to alkene: CH₃–CH=CH–CH(CH₃)₂ with cis configuration.

Then ozonolysis gives CH₃CHO and OHC–CH(CH₃)₂

OHC–CH(CH₃)₂ is 2-methylpropanal.

So the product box should contain these two.

But in the diagram, it's depicted as a single entity? Perhaps it's a typo or simplification.

Maybe the blank is for the reactant alkene, but no, the arrow is from the alkene to the box via ozonolysis.

Another look: in the diagram, there is a box to the left that is fed by ozonolysis, and it contains a structure that looks like a dialdehyde or something, but described as:

"H₃C H
\ /
C=C
/ \
H CH(CH₃)₂" — that's the reactant.

Then after ozonolysis, the products are separate.

But in the blank, since it's a single box, perhaps we need to write both products.

However, looking at the layout, there is a large rectangle on the far left that is the final product of ozonolysis, and it might be intended to hold both carbonyls.

Perhaps it's showing the combination, but I think for accuracy, we'll state the two products.

But let's see the answer format.

Perhaps in some contexts, they draw it as a single line, but I think for this, we'll put the two structures.

But to match the diagram, maybe it's expecting the molecular formulas or names.

Another idea: perhaps the alkene is symmetric, but it's not.

Let's move on and come back.

Also, there is a reaction from the internal alkyne to the cis-alkene, which we did.

And from the cis-alkene to the alkane.

Now, there is another path: from the terminal alkyne, addition of 1 eq HX.

HC≡C–CH₃ + HX → CH₃CX=CH₂ (Markovnikov addition)

For terminal alkyne, HX adds Markovnikov, giving the vinyl halide with halogen on more substituted carbon.

HC≡C–CH₃ + HBr → CH₃CBr=CH₂

Product blank: CH₃CBr=CH₂

Then, further addition of HX → geminal dihalide

CH₃CBr=CH₂ + HBr → CH₃CBr₂CH₃

Product blank: CH₃CBr₂CH₃

Reagent for second HX: HX (same as first)

In the diagram, it's shown as "1 eq HX" to give the vinyl halide, then another arrow with "HX" to give the geminal dihalide.

Yes.

Now, back to the ozonolysis product.

In the diagram, the box on the far left is receiving the output of ozonolysis of the alkene CH₃CH=CHCH(CH₃)₂ (cis).

Products: CH₃CHO and (CH₃)₂CHCHO

So the blank should contain these two.

Since it's a single box, perhaps write "CH₃CHO and (CH₃)₂CHCHO" or draw them.

But in text, we can describe.

Perhaps the diagram intends for us to recognize that it's two separate molecules.

Another possibility: the alkene being ozonized is different.

Look at the structure drawn above the ozonolysis arrow: it's shown as:

```
H₃C H
\ /
C=C
/ \
H CH₃? No — in the user's input, it's "H₃C H" on top, and "H CH(CH₃)₂" on bottom? Let's assume it's CH₃-CH=CH-CH(CH₃)2 with cis stereochemistry.

Yes.

So products are acetaldehyde and 2-methylpropanal.

Now, for the very left blank, it's a large box, so likely expects both products.

Perhaps in some maps, they combine, but I think we'll list them.

Now, let's compile all blanks.

List of blanks in order, roughly from top to bottom, left to right.

1. Top left reagent (to give halohydrin): Br₂, H₂O (or Cl₂, H₂O)

2. Top right reagent (epoxidation): mCPBA

3. Right middle product (acid-catalyzed hydration): CH₃CH(OH)CH₃

4. Below that product (hydroboration-oxidation): CH₃CH₂CH₂OH

5. Left middle product (dihydroxylation): CH₃CH(OH)CH₂OH

6. Down from alkene, product of X₂ addition: CH₃CHXCH₂X (e.g., CH₃CHBrCH₂Br)

7. To the right, product of HX addition: CH₃CHXCH₃

8. Further right, product of 1 eq HX to alkyne: CH₃CX=CH₂ (e.g., CH₃CBr=CH₂)

9. Then product of second HX: CH₃CX₂CH₃ (e.g., CH₃CBr₂CH₃)

10. Reagent for converting dihalide to alkyne: 2 eq NaNH₂

11. Product of alkyne hydration (HgSO₄ etc.): CH₃COCH₃

12. Product of alkyne hydroboration-oxidation: CH₃CH₂CHO

13. Product of 1 eq X₂ to alkyne: CH₃CX=CXH (trans, e.g., CH₃CBr=CBrH)

14. Product of second X₂ addition: CH₃CX₂CX₂H (e.g., CH₃CBr₂CBr₂H)

15. Reagent for deprotonating alkyne: NaNH₂

16. Reagent for alkylating acetylide: (CH₃)₂CHX (isopropyl halide)

17. Product of alkylation: CH₃C≡CCH(CH₃)₂

18. Reagent for partial hydrogenation to cis-alkene: H₂, Lindlar catalyst

19. Product of full hydrogenation: CH₃CH₂CH₂CH(CH₃)₂ or (CH₃)₂CHCH₂CH₂CH₃

20. Ozonolysis product box: CH₃CHO and (CH₃)₂CHCHO

Additionally, there is a blank for the reagent to convert the geminal dihalide to alkyne, which we have as 2 eq NaNH₂.

And for the conversion of the vinyl halide to geminal dihalide, reagent is HX.

In the diagram, there is also a blank for the reagent to go from the alkane back to alkene? No, the arrows are mostly forward.

Another blank: between the cis-alkene and the alkane, we have the reagent for full hydrogenation.

We have it.

Now, for the ozonolysis, since the box is singular, perhaps they want the combined formula or something, but I think listing both is fine.

Perhaps in the context, it's understood.

Let's write the final answers in a list corresponding to the blanks.

Since the diagram has specific positions, but without coordinates, we'll describe by location.

For the purpose of this response, I'll provide the filled-in values.

Final Answer:

- Reagent for halohydrin formation (top left): Br₂, H₂O

- Reagent for epoxidation (top right): mCPBA

- Product of acid-catalyzed hydration (right middle): CH₃CH(OH)CH₃

- Product of hydroboration-oxidation (below previous): CH₃CH₂CH₂OH

- Product of dihydroxylation (left middle): CH₃CH(OH)CH₂OH

- Product of X₂ addition to alkene (down from alkene): CH₃CHXCH₂X (e.g., with X=Br)

- Product of HX addition to alkene (right of previous): CH₃CHXCH₃

- Product of 1 eq HX to terminal alkyne (far right, upper): CH₃CX=CH₂

- Product of second HX addition (below previous): CH₃CX₂CH₃

- Reagent for double dehydrohalogenation (between dihalide and alkyne): 2 eq NaNH₂

- Product of alkyne hydration (HgSO₄ etc.): CH₃COCH₃

- Product of alkyne hydroboration-oxidation: CH₃CH₂CHO

- Product of 1 eq X₂ to alkyne: CH₃CX=CXH (trans)

- Product of second X₂ addition: CH₃CX₂CX₂H

- Reagent for deprotonating terminal alkyne: NaNH₂

- Reagent for alkylating acetylide ion: (CH₃)₂CHX

- Product of alkylation: CH₃C≡CCH(CH₃)₂

- Reagent for syn hydrogenation to cis-alkene: H₂, Lindlar catalyst

- Product of full hydrogenation: CH₃CH₂CH₂CH(CH₃)₂

- Product of ozonolysis (far left box): CH₃CHO and (CH₃)₂CHCHO

Note: X represents a halogen, commonly Br or Cl. In many cases, Br is used as example.

For simplicity, we can use Br in examples.

Also, for the geminal dihalide from alkyne, it's CH₃CBr₂CH₃, which is 2,2-dibromopropane.

Similarly, others.

I think this covers all blanks.