What Changes When a Triple Bond Undergoes Addition?
Add one mole of bromine to one mole of ethyne, and something important happens: the carbon-carbon triple bond becomes a double bond, but the two carbon atoms remain connected. Each carbon also gains a bromine atom.
Add one mole of bromine to one mole of ethyne, and something important happens: the carbon-carbon triple bond becomes a double bond, but the two carbon atoms remain connected. Each carbon also gains a bromine atom.
Add another mole of bromine, and the remaining double bond can undergo addition as well. It becomes a single bond, producing a saturated compound.
So, what actually changes when a triple bond undergoes addition? The carbon-carbon bond loses one of its two pi bonds, then potentially loses the second one in a further addition. New atoms or groups attach to the two carbons, while the original carbon skeleton stays intact.
Understanding this process also helps you distinguish addition from substitution, two reactions that can produce very different products even when the same reagent, such as bromine, is involved.
01What happens to a carbon-carbon triple bond?
Start with ethyne, the simplest alkyne:
\[ \ce{HC\bond{#}CH} \]
The two carbon atoms share three bonding pairs of electrons. One forms a sigma (\(\sigma\)) bond, and the other two form pi (\(\pi\)) bonds.
The sigma bond holds the carbon atoms together along the line joining their nuclei. The two pi bonds provide additional bonding through sideways overlap of orbitals.
When an addition reaction occurs, one pi bond is broken. The sigma bond remains, so the carbon atoms are still joined.
Think of the triple bond as three connections between the same two carbon atoms. Addition removes one of those connections and uses the available bonding capacity to attach new atoms.
This is a useful way to picture the reaction, although electrons are not physical connectors being removed. The actual change involves the rearrangement of bonding electrons.
The first addition: triple bond to double bond
Consider ethyne reacting with bromine, \(\ce{Br2}\).
One bromine atom attaches to each carbon atom. The triple bond loses one pi bond and becomes a double bond.
\[ \ce{HC\bond{#}CH + Br2 -> BrCH=CHBr} \]
The product is 1,2-dibromoethene.
Notice three things:
- The carbon-carbon bond changes from a triple bond to a double bond.
- Each carbon gains one bromine atom.
- The two-carbon skeleton remains unchanged.
The product is still unsaturated because it contains a carbon-carbon double bond. One addition has reduced the degree of unsaturation, but it has not removed all the multiple bonding.
The second addition: double bond to single bond
The remaining double bond can undergo another addition reaction.
A second mole of bromine supplies another bromine atom to each carbon:
\[ \ce{BrCH=CHBr + Br2 -> Br2CH-CHBr2} \]
The product is 1,1,2,2-tetrabromoethane.
The carbon-carbon bond is now a single bond. Both pi bonds from the original triple bond have been consumed.
This gives us an important pattern:
| Stage | Carbon-carbon bond | Bromine atoms added per original molecule | Product |
|---|---|---|---|
| Starting material | Triple | 0 | Ethyne |
| First addition | Double | 2 | 1,2-dibromoethene |
| Second addition | Single | 4 | 1,1,2,2-tetrabromoethane |
The same general pattern applies to other addition reactions involving alkynes, although the identity of the incoming atoms and the final products depend on the reagent.
02Why can a triple bond undergo two additions?
The key is understanding what the triple bond contains.
A carbon-carbon triple bond consists of:
- One sigma (\(\sigma\)) bond.
- Two pi (\(\pi\)) bonds.
The sigma bond holds the carbon atoms together along the internuclear axis. The two pi bonds provide additional bonding through sideways orbital overlap.
During addition, the pi bonding is progressively removed. The sigma bond remains intact.
This explains why an alkyne can undergo two successive additions.
Track the bonds and carbon geometry
There is another change worth recognising. As the bond order decreases, the arrangement around each carbon changes.
| Stage | Bonding around each carbon in the ethyne example | Geometry | Approximate bond angle |
|---|---|---|---|
| Starting alkyne | One C-H bond and one C≡C bond | Linear | 180° |
| After first addition | C-H, C-Br, and C=C | Trigonal planar | 120° |
| After second addition | C-H, two C-Br bonds, and C-C | Tetrahedral | 109.5° |
The carbon-carbon bond also becomes longer as its bond order decreases.
You can see how a chemical reaction changes more than just the molecular formula. It changes the bonding arrangement and three-dimensional structure.
For a more detailed explanation of these geometries, see Carbon-Carbon Bond Shapes: Single, Double and Triple Bonds.
Check carbon valency at each stage
A common mistake when drawing addition products is accidentally giving carbon five bonds.
Track one carbon atom in ethyne:
| Stage | Bonds around the carbon | Total bond order |
|---|---|---|
| Ethyne | C-H and C≡C | 4 |
| First addition | C-H, C-Br, and C=C | 4 |
| Second addition | C-H, two C-Br bonds, and C-C | 4 |
Carbon maintains four bonds throughout these reactions.
This is a reliable way to check whether your predicted product is chemically sensible.
03Addition versus substitution: the important difference
Now consider a different hydrocarbon: ethane.
Ethane has the structural formula:
\[ \ce{CH3-CH3} \]
There is no carbon-carbon double or triple bond. Every carbon-carbon bond is a single bond.
What happens if ethane reacts with bromine?
You might predict the same reaction as ethyne, with bromine attaching to the carbon atoms. But that is not what happens under ordinary alkane bromination conditions.
Ethane undergoes substitution when exposed to bromine under ultraviolet (UV) light.
\[ \ce{CH3-CH3 + Br2 ->[UV] CH3-CH2Br + HBr} \]
The product is bromoethane and hydrogen bromide.
One hydrogen atom has been replaced by a bromine atom.
Notice the difference:
- Addition uses the multiple carbon-carbon bond as the reaction site.
- Substitution replaces an existing atom or group with another atom or group.
In ethane, the carbon-carbon single bond remains unchanged. A C-H bond is replaced by a C-Br bond.
In ethyne, the carbon-carbon triple bond loses one pi bond, and new carbon-bromine bonds form.
Direct comparison
| Feature | Addition to an alkyne | Substitution of an alkane |
|---|---|---|
| Starting material | Unsaturated hydrocarbon | Saturated hydrocarbon |
| Reaction site | C≡C bond | C-H bond |
| Main bond change | C≡C becomes C=C | C-H becomes C-Br |
| Bromine product | Dibromoalkene after first addition | Bromoalkane |
| Other product | No HBr in the idealised Br2 addition equation | HBr |
| Conditions | Usually room temperature; UV not required | UV light or sufficient heat |
| Bromine observation | Orange-brown colour disappears | Colour fades under UV as bromine is consumed |
The conditions matter.
An alkane does not normally react appreciably with bromine in the dark at room temperature. UV light provides the energy needed to initiate the free-radical substitution reaction.
By contrast, an alkyne can undergo bromine addition without UV light.
Do not confuse the disappearance of bromine with proof of addition. Bromine can be consumed in both reactions. You need to consider the starting hydrocarbon and the reaction conditions.
04How can an experiment distinguish addition from substitution?
Imagine you have two colourless hydrocarbon samples. One contains an alkyne, and the other contains an alkane.
You want to determine which one reacts with bromine through addition.
The problem is that both addition and substitution can consume bromine. Simply observing a colour change is not enough.
Setting up a fair comparison
A suitable comparison uses three matched, sealed microscale reaction vessels.
- Vessel A contains ethyne and bromine water, kept in the dark.
- Vessel B contains ethane and bromine water, kept in the dark.
- Vessel C contains ethane and bromine water, exposed to UV light.
The initial bromine concentration, volumes, temperature, and hydrocarbon amounts should be controlled as closely as possible.
What should happen?
Before observing the results, make a prediction.
If bromine reacts with ethyne in the dark, what should happen to the bromine colour?
It should disappear as bromine is consumed in the addition reaction.
If ethane is kept in the dark, the bromine colour should remain because the substitution reaction has not been initiated.
If the ethane sample is exposed to UV light, substitution can occur, and bromine will gradually be consumed.
Illustrative observations
The following table shows expected qualitative observations. These are illustrative results, not measurements from an actual experiment.
| Vessel | Hydrocarbon | Conditions | Initial observation | Expected observation |
|---|---|---|---|---|
| A | Ethyne | Dark, room temperature | Orange-brown | Rapid decolourisation |
| B | Ethane | Dark, room temperature | Orange-brown | Colour remains |
| C | Ethane | UV light | Orange-brown | Colour gradually fades |
The observations support two different reaction pathways.
For ethyne:
\[ \ce{HC\bond{#}CH + Br2 -> BrCH=CHBr} \]
The bromine is consumed through addition.
For ethane:
\[ \ce{CH3-CH3 + Br2 ->[UV] CH3-CH2Br + HBr} \]
The bromine is consumed through substitution.
What was wrong with the original experimental design?
Suppose you only tested ethane under UV light and observed the bromine colour disappear.
You might incorrectly conclude that ethane contains a carbon-carbon multiple bond.
The problem is that the experiment did not distinguish the two possible reaction pathways.
Adding a dark control solves this problem. It shows whether bromine is being consumed because the hydrocarbon is unsaturated or because light has initiated substitution.
There is still one important limitation: bromine decolourisation does not distinguish an alkene from an alkyne. Both contain carbon-carbon pi bonds and can undergo addition.
The test identifies unsaturation, not the exact type of multiple bond.
Safety: Bromine solutions and hydrocarbons require appropriate laboratory controls. This comparison is for supervised laboratory investigations, not home experimentation.
05Other reagents that add across a triple bond
Bromine is a useful example, but alkynes can undergo addition with other reagents too.
The same underlying idea applies: the triple bond loses pi bonding as new atoms or groups attach to its carbon atoms.
| Reagent | What is added? | First product | Further addition |
|---|---|---|---|
| Hydrogen, \(\ce{H2}\) | H and H | Alkene | Alkane |
| Halogen, \(\ce{Br2}\) or \(\ce{Cl2}\) | X and X | Dihaloalkene | Tetrahaloalkane |
| Hydrogen halide, \(\ce{HX}\) | H and X | Haloalkene | Dihaloalkane |
| Water, \(\ce{H2O}\) | H and OH overall | Enol intermediate | Usually a carbonyl compound after rearrangement |
A few details are worth remembering.
Hydrogenation: Complete hydrogenation requires two moles of hydrogen per mole of alkyne. A metal catalyst such as nickel, platinum, or palladium is commonly used. Stopping at the alkene requires selective conditions; simply supplying one equivalent does not guarantee that the alkene can be isolated.
Halogenation: One mole of halogen produces a dihaloalkene. A second mole can convert the remaining double bond into a single bond, producing a tetrahaloalkane.
Hydrohalogenation: Addition of HX can produce a haloalkene first, followed by a dihaloalkane with further addition. For unsymmetrical alkynes, the orientation of addition can matter. Markovnikov’s rule is useful for predicting the usual orientation in simple cases.
Hydration: Adding water to an alkyne is more complicated than simply producing a stable alcohol. The initial enol generally rearranges into a carbonyl compound, usually a ketone. Ethyne is a notable case that produces ethanal.
For a broader treatment of these reagents and product-prediction methods, see Addition Reactions of Unsaturated Hydrocarbons Explained.
06Worked examples
Worked example: Predict the product when bromine adds to but-1-yne
But-1-yne reacts with one mole of bromine under ordinary addition conditions.
Write the structural equation, identify the product, and explain the change in the carbon-carbon bond.
Step 1
But-1-yne has the structure:
\[ \ce{HC\bond{#}C-CH2-CH3} \]
The triple bond is between carbon 1 and carbon 2.
Step 2
Bromine exists as \(\ce{Br2}\).
One bromine atom attaches to carbon 1, and the other attaches to carbon 2.
Step 3
The carbon-carbon triple bond becomes a double bond because one pi bond is consumed.
Step 4
\[ \ce{HC\bond{#}C-CH2-CH3 + Br2} \]
\[ \ce{-> BrCH=CBr-CH2-CH3} \]
The product is 1,2-dibromobut-1-ene.
Step 5
Carbon 1 has one hydrogen, one bromine, and a double bond to carbon 2.
Carbon 2 has one bromine, a double bond to carbon 1, and a single bond to carbon 3.
Both carbons have four bonds when bond orders are counted.
The product still contains a double bond, so it is unsaturated.
Worked example: Calculate the bromine required for complete addition
A sample contains 2.50 mmol of ethyne. It is to undergo complete bromine addition, producing 1,1,2,2-tetrabromoethane.
Bromine solution has a concentration of 0.200 mol L\(^{-1}\).
Calculate the volume of bromine solution required.
Assume complete reaction, no side reactions, and that both pi bonds undergo addition.
Step 1
Complete addition requires two moles of bromine per mole of ethyne:
\[ \ce{HC\bond{#}CH + 2Br2 -> CHBr2-CHBr2} \]
Step 2
The given amount is:
\[ n(\ce{C2H2})=2.50\text{ mmol} \]
\[ =2.50\times10^{-3}\text{ mol} \]
Step 3
The balanced equation gives a ratio of:
\[ \ce{C2H2:Br2}=1:2 \]
Therefore:
\[ n(\ce{Br2})=2(2.50\times10^{-3}) \]
\[ =5.00\times10^{-3}\text{ mol} \]
Step 4
Use the concentration relationship:
\[ c=\frac{n}{V} \]
Rearrange:
\[ V=\frac{n}{c} \]
Substitute:
\[ V=\frac{5.00\times10^{-3}}{0.200} \]
\[ =0.0250\text{ L} \]
Convert to millilitres:
\[ V=25.0\text{ mL} \]
Final answer: 25.0 mL of 0.200 mol L\(^{-1}\) bromine solution is required for complete addition.
The important point is that 12.5 mL would provide only one mole of bromine per mole of ethyne. That would correspond to the first addition, leaving a carbon-carbon double bond in the product.
The remaining 12.5 mL is needed to consume that second pi bond.
07Common mistakes to avoid
| Incorrect prediction | Why it is wrong | Correct reasoning |
|---|---|---|
| The entire triple bond breaks during the first addition. | The sigma bond remains intact. | One pi bond is consumed, leaving a double bond. |
| One mole of bromine completely saturates an alkyne. | One equivalent only converts the triple bond into a double bond. | Two equivalents are needed for complete bromine addition. |
| Bromine decolourisation proves a triple bond is present. | Alkenes also decolourise bromine, and alkanes can consume bromine under UV light. | The test indicates reactivity towards bromine, not a unique structure. |
| Addition and substitution are the same because both can involve bromine. | They change different bonds and produce different products. | Addition uses a multiple bond; substitution replaces an existing atom or group. |
| UV light is necessary for bromine addition to an alkyne. | UV light is associated with initiating alkane halogen substitution. | Alkyne bromine addition does not require UV light. |
08Questions and solutions
Question 1
Ethyne reacts with one mole of bromine under dark conditions.
a. Write the balanced structural equation.
b. Name the organic product.
c. State what happens to the carbon-carbon bond.
d. Predict the observation when bromine water is added.
Solution 1
The product is 1,2-dibromoethene, and the carbon-carbon triple bond becomes a double bond.
The reaction is:
\[ \ce{HC\bond{#}CH + Br2 -> BrCH=CHBr} \]
a. The balanced structural equation contains one mole of ethyne and one mole of bromine.
b. Each carbon receives one bromine atom, giving 1,2-dibromoethene.
c. One pi bond is consumed. The original carbon-carbon sigma bond remains, so the carbon-carbon triple bond becomes a double bond.
d. The orange-brown bromine water becomes colourless as bromine is consumed in the addition reaction.
The product remains unsaturated because it still contains a carbon-carbon double bond. One equivalent of bromine has not completely saturated the original triple bond.
Question 2
But-1-yne reacts with two moles of hydrogen bromide under ordinary ionic addition conditions.
a. Write the structural formula of the major product after the first addition.
b. Explain the orientation of the first addition using Markovnikov’s rule.
c. Predict the product after the second addition.
d. Explain why these reactions are addition rather than substitution.
Assume no peroxide-initiated alternative pathway.
Solution 2
The first product is 2-bromobut-1-ene, and the second addition produces 2,2-dibromobutane.
Part a: First addition
The starting structure is:
\[ \ce{HC\bond{#}C-CH2-CH3} \]
One equivalent of HBr adds across the triple bond.
Part b: Orientation
For ordinary Markovnikov addition, hydrogen attaches to the carbon that already has more hydrogen atoms, while bromine attaches to the more substituted carbon.
Therefore, hydrogen attaches to the terminal carbon, and bromine attaches to the internal carbon.
The product is:
\[ \ce{CH2=C(Br)-CH2-CH3} \]
This is 2-bromobut-1-ene.
Part c: Second addition
The remaining double bond undergoes another addition with HBr.
Hydrogen attaches to the terminal carbon, and bromine attaches to the carbon already carrying bromine.
The final product is:
\[ \ce{CH3-CBr2-CH2-CH3} \]
This is 2,2-dibromobutane.
Part d: Reaction classification
Both stages are addition reactions.
The H and Br atoms from HBr attach to the two carbons that formed the original multiple bond. The pi bonding is progressively consumed.
No hydrogen atom is replaced by bromine. That would be substitution.
Question 3
A student conducts an investigation using three sealed reaction vessels.
The initial bromine water concentration and volume are identical.
The following illustrative observations are recorded:
| Vessel | Hydrocarbon | Conditions | Observation |
|---|---|---|---|
| A | Ethyne | Dark | Orange-brown colour disappears rapidly |
| B | Ethane | Dark | Orange-brown colour remains |
| C | Ethane | UV light | Orange-brown colour gradually fades |
a. Identify the reaction occurring in vessel A.
b. Explain why vessel B shows no significant reaction.
c. Write the equation for the reaction in vessel C.
d. Explain why the observation in vessel C does not prove that ethane is unsaturated.
e. State one limitation of using bromine decolourisation to distinguish an alkene from an alkyne.
Solution 3
Vessel A undergoes addition, vessel B shows no significant reaction, and vessel C undergoes substitution.
Part a: Vessel A
Ethyne contains a carbon-carbon triple bond.
Bromine adds across this bond, converting it into a double bond:
\[ \ce{HC\bond{#}CH + Br2 -> BrCH=CHBr} \]
The reaction does not require UV light.
Part b: Vessel B
Ethane is saturated and contains only carbon-carbon single bonds.
In the dark at room temperature, ethane does not normally react appreciably with bromine.
Therefore, the orange-brown colour remains.
Part c: Vessel C
UV light initiates free-radical substitution.
One hydrogen atom in ethane is replaced by bromine:
\[ \ce{CH3-CH3 + Br2 ->[UV] CH3-CH2Br + HBr} \]
The products are bromoethane and hydrogen bromide.
Part d: Interpretation
The bromine colour disappears because bromine is consumed.
However, this does not mean a carbon-carbon multiple bond must have been present.
Ethane can consume bromine through substitution when UV light is supplied.
The reaction type must be identified using both the starting structure and the reaction conditions.
Part e: Limitation
Bromine decolourisation cannot distinguish an alkene from an alkyne.
Both can undergo addition with bromine, so both can remove the orange-brown colour.
Further structural evidence or quantitative reaction data would be needed to distinguish them.
Question 4
A gas sample contains only but-1-yne and butane.
The total amount of hydrocarbon in the sample is 0.0150 mol.
The sample reacts with bromine in the dark. Product analysis confirms that only the first addition to the alkyne occurs.
A total of 0.0100 mol of bromine is consumed.
Assume that:
- Butane does not react under these conditions.
- Each but-1-yne molecule consumes one mole of bromine in the first addition.
- There are no other reactions.
a. Calculate the amount, in moles, of but-1-yne in the original sample.
b. Calculate the amount of butane.
c. Calculate the additional amount of bromine required to convert all the but-1-yne into its fully brominated product.
d. Explain why exposing the original mixture to UV light would invalidate the calculation in part (a).
Solution 4
The mixture contains 0.0100 mol of but-1-yne and 0.00500 mol of butane. A further 0.0100 mol of bromine is required for complete bromine addition.
Part a: Amount of but-1-yne
Only the but-1-yne reacts with bromine under the stated conditions.
The first addition has a mole ratio of:
\[ \ce{C4H6:Br2}=1:1 \]
Therefore:
\[ n(\text{but-1-yne})=0.0100\text{ mol} \]
Part b: Amount of butane
The total amount of hydrocarbons is:
\[ n(\text{total})=0.0150\text{ mol} \]
Subtract the amount of but-1-yne:
\[ n(\text{butane})=0.0150-0.0100 \]
\[ =0.00500\text{ mol} \]
Part c: Additional bromine
The first addition changes the triple bond into a double bond.
A second mole of bromine is needed per mole of but-1-yne to consume the remaining double bond.
Therefore:
\[ n(\text{additional }\ce{Br2})=0.0100\text{ mol} \]
The total bromine required for complete addition is:
\[ 0.0100+0.0100=0.0200\text{ mol} \]
Part d: Effect of UV light
UV light could initiate substitution between bromine and butane.
This would allow the saturated hydrocarbon to consume bromine as well.
The assumption that only but-1-yne reacts would no longer be valid. Consequently, the measured bromine consumption could not be used directly to calculate the amount of alkyne.
This question shows why controlling reaction conditions is just as important as balancing the chemical equation.
Question 5
A pure hydrocarbon, X, has the molecular formula \(\ce{C4H6}\).
Two separate 0.0100 mol samples are tested.
The following results are obtained:
- Complete hydrogenation of the first sample consumes 0.0200 mol of \(\ce{H2}\) and produces butane.
- Exhaustive bromine addition to the second sample consumes 0.0200 mol of \(\ce{Br2}\) and produces an organic compound with the molecular formula \(\ce{C4H6Br4}\).
A student confidently concludes:
“X must be but-1-yne because a triple bond can undergo two additions.”
Evaluate this conclusion.
In your answer:
a. Explain how the hydrogen and bromine quantities relate to the number of pi bonds.
b. Show how but-1-yne could produce the observed results.
c. Identify a different structural isomer of \(\ce{C4H6}\) that could also produce the observed results.
d. Explain why the experimental data do not uniquely identify X.
e. Suggest an additional measurement that could distinguish the two structures.
Solution 5
The conclusion is not justified. Both but-1-yne and buta-1,3-diene are consistent with the supplied reaction quantities and molecular formula.
The important distinction is that the experiments measure the total amount of pi bonding, not how those pi bonds are arranged.
Part a: Relating reagent quantities to pi bonds
Each mole of hydrogen consumes one mole of pi bonding in a complete hydrogenation reaction.
Similarly, each mole of bromine consumes one mole of pi bonding during halogen addition.
The experiment shows that:
\[ \frac{0.0200}{0.0100} =2.00 \]
Therefore, each molecule has a total of two pi bonds available for addition.
But two pi bonds do not necessarily mean one triple bond.
They could also be two separate double bonds.
Part b: But-1-yne
But-1-yne has the structure:
\[ \ce{HC\bond{#}C-CH2-CH3} \]
Its triple bond contains one sigma bond and two pi bonds.
Complete hydrogenation consumes two moles of hydrogen:
\[ \ce{HC\bond{#}C-CH2-CH3 + 2H2} \]
\[ \ce{-> CH3-CH2-CH2-CH3} \]
The product is butane.
Complete bromine addition consumes two moles of bromine:
\[ \ce{HC\bond{#}C-CH2-CH3 + 2Br2} \]
\[ \ce{-> CHBr2-CBr2-CH2-CH3} \]
The product is 1,1,2,2-tetrabromobutane.
Both observations match the experimental data.
Part c: An alternative structure
Consider buta-1,3-diene:
\[ \ce{CH2=CH-CH=CH2} \]
It has two separate carbon-carbon double bonds.
Each double bond contains one pi bond, giving two pi bonds in total.
Complete hydrogenation also consumes two moles of hydrogen:
\[ \ce{CH2=CH-CH=CH2 + 2H2} \]
\[ \ce{-> CH3-CH2-CH2-CH3} \]
The product is again butane.
Complete bromine addition consumes two moles of bromine:
\[ \ce{CH2=CH-CH=CH2 + 2Br2} \]
\[ \ce{-> CH2Br-CHBr-CHBr-CH2Br} \]
The product is 1,2,3,4-tetrabromobutane.
Its molecular formula is also:
\[ \ce{C4H6Br4} \]
So both candidate structures match the measured reagent quantities and product formula.
Part d: Why the conclusion is not unique
The tempting approach is to see two equivalents of hydrogen and bromine and assume that the original molecule must have contained one triple bond.
The missing consideration is that a diene contains two double bonds, which also provide two pi bonds.
The experiments count how many pi bonds react. They do not establish whether those bonds were grouped into one triple bond or separated into two double bonds.
The two candidates also produce different brominated structures:
- But-1-yne produces 1,1,2,2-tetrabromobutane.
- Buta-1,3-diene produces 1,2,3,4-tetrabromobutane.
Both have the same molecular formula, but their atoms are connected differently.
Part e: Additional measurement
Infrared spectroscopy could distinguish the starting structures.
But-1-yne has a terminal alkyne group, which can produce a characteristic sharp C-H absorption associated with the \(\ce{C\bond{#}C-H}\) group, as well as an alkyne absorption.
Buta-1,3-diene instead contains carbon-carbon double bonds and does not have a terminal alkyne C-H bond.
Alternatively, NMR spectroscopy could reveal the different hydrogen and carbon environments in the two molecules.
The key conclusion is that reaction stoichiometry alone does not always determine molecular structure. You may need structural evidence as well.
08Where this understanding becomes useful
When you encounter an addition reaction involving an alkyne, start by identifying how many pi bonds are available.
Then track what the reagent contributes, how the carbon-carbon bond order changes, and whether another addition is possible.
For the distinction between addition and substitution, always check three things:
- Which bond is reacting?
- What is attached to the carbon atoms afterwards?
- What conditions are needed for the reaction?
These checks will help you predict products, explain bromine observations, and avoid confusing two reactions that can use the same reagent.
The next useful step is connecting these reaction products to structural identification. In particular, IR and NMR spectroscopy can help distinguish compounds that have the same molecular formula and even consume the same quantities of reagents.
Course context: This article is framed around the Chemistry Stage 6 (2017) course. The Chemistry 11-12 Syllabus (2025) is scheduled to begin implementation in 2028, with the first HSC examination under the new syllabus in 2029. See the NSW syllabus implementation information.