How to Predict Hydrogenation Products from Alkene Structures

Take but-1-ene, with the structural formula \(\ce{CH2=CHCH2CH3}\). It reacts with hydrogen gas in the presence of a nickel catalyst. What should the product be?

The answer is butane, \(\ce{CH3CH2CH2CH3}\). The hydrogen molecule splits, and one hydrogen atom attaches to each carbon that was part of the original double bond. The carbon skeleton stays exactly the same.

This reaction is called hydrogenation. Once you understand what happens to the two carbons in a C=C bond, predicting hydrogenation products becomes a repeatable process rather than a reaction you have to memorise for every individual alkene.

01What happens when an alkene is hydrogenated?

Start by looking at the carbon-carbon double bond.

Each carbon atom forms four covalent bonds. In an alkene, the two carbon atoms share a double bond, meaning they have two bonds between them. Each carbon has fewer remaining bonding positions available for hydrogen than it would in an alkane.

Hydrogenation fills those remaining positions with hydrogen atoms.

A carbon-carbon double bond consists of:

• One sigma (\(\sigma\)) bond.
• One pi (\(\pi\)) bond.

During hydrogenation, the pi bond is broken, while the sigma bond remains. Each carbon then forms a new bond with a hydrogen atom.

The overall change is:

\[
\ce{C=C -> C-C}
\]

The double bond becomes a single bond, and the molecule gains two hydrogen atoms.

The general reaction is:

\[
\ce{Alkene + H2 -> Alkane}
\]

For an open-chain alkene containing one double bond, the molecular formula changes from:

\[
\ce{C_nH_{2n} -> C_nH_{2n+2}}
\]

The carbon count does not change. The hydrogen count increases by two.

For example:

Notice the pattern: the carbon skeleton is preserved, but the molecule becomes saturated.

This formula relationship assumes an open-chain alkene with exactly one C=C bond. A cyclic alkene, for example, would produce a cycloalkane because its carbon ring remains intact.

If you need to refresh how carbon chains, double bonds and molecular formulae fit together, see HSC Chemistry: Alkane, Alkene and Alkyne Formulae.

02A reliable method for predicting hydrogenation products

You can predict the product of almost any straightforward alkene hydrogenation using five steps.

Step 1: Find the C=C bond.

Identify the two carbon atoms connected by the double bond.

Step 2: Keep the carbon skeleton unchanged.

Do not move carbon atoms, remove branches, or rearrange the chain.

Step 3: Change the double bond into a single bond.

Replace the C=C with C-C.

Step 4: Add one hydrogen to each of the two original alkene carbons.

Each carbon receives exactly one additional hydrogen atom for every double bond being hydrogenated.

Step 5: Check the valencies and molecular formula.

Every carbon must have four bonds, and every hydrogen must have one.

There is one particularly useful detail here: the hydrogen atoms do not necessarily attach to two carbons that originally had hydrogen atoms.

If one of the alkene carbons originally had no hydrogen atoms, it still receives one during hydrogenation.

Let’s see how that works with a branched alkene.

03Why is hydrogenation an addition reaction rather than substitution?

These two reaction types are easy to confuse because both involve changes to existing bonds.

However, they do different things.

Addition reactions

In an addition reaction, atoms from another molecule join an unsaturated molecule.

For hydrogenation:

\[
\ce{Alkene + H2 -> Alkane}
\]

Two reactants combine to form one product.

The C=C bond becomes a C-C bond, and two new C-H bonds form.

No atom is displaced from the original alkene.

Substitution reactions

In a substitution reaction, one atom or group in a molecule is replaced by another.

A common HSC example is an alkane reacting with a halogen under ultraviolet (UV) light.

For example, methane reacts with chlorine:

\[
\ce{CH4 + Cl2 ->[UV] CH3Cl + HCl}
\]

One hydrogen atom in methane is replaced by chlorine.

The products are chloromethane and hydrogen chloride.

Notice that there are two products. The removed hydrogen does not disappear; it combines with chlorine to form HCl.

The important differences are:

A useful way to remember the difference is to look at the bonds.

• Addition: something joins across a multiple bond.
• Substitution: something replaces an existing atom or group.

Do not decide whether a reaction is addition or substitution just by counting reactants. Look at what happens to the bonds and where the atoms end up.

04How do we know hydrogenation has occurred?

Hydrogenation has a useful experimental history.

The original problem was straightforward: could hydrogen be added directly to an unsaturated hydrocarbon to produce a saturated hydrocarbon?

Mixing two gases does not automatically mean they will react quickly. A way of promoting the reaction was needed.

The early experiment

An early demonstration is attributed to von Wilde in 1874.

The experiment involved ethylene (ethene) and hydrogen gas being passed over finely divided platinum, known as platinum black.

A simplified representation of the apparatus is:

The prediction was that if hydrogen added across the double bond, ethylene would become ethane.

\[
\ce{CH2=CH2 + H2 -> CH3CH3}
\]

The reported observation was the conversion of ethylene into ethane.

This supported the idea that hydrogen could be added to an unsaturated hydrocarbon in the presence of a metal catalyst.

The important evidence was the identity of the product, not a colour change. The historical account does not establish a particular colour indicator or modern sensor measurement, so those should not be invented.

Why investigate other catalysts?

The first demonstration established that platinum could promote the reaction. It did not establish whether platinum was the only metal capable of doing so.

Later investigations by Paul Sabatier and Jean-Baptiste Senderens, beginning in 1897, showed that other finely divided metals could also catalyse hydrogenation. Nickel was particularly useful.

This helped establish catalytic hydrogenation as a practical chemical process.

A useful historical account appears in Nature’s 1915 review of hydrogenation. The American Chemical Society’s history of Raney nickel also describes the development and industrial importance of nickel-based catalysts.

What does the catalyst actually do?

The metal catalyst provides a surface where hydrogen and the alkene can interact.

Both reactants can temporarily attach to the metal surface. This makes it easier for bonds to break and new C-H bonds to form.

Think of the metal surface as a temporary workbench. It brings the reactants into a position where their bonds can be rearranged more easily.

The analogy has limits: the catalyst is not simply holding molecules together. Interactions with the metal surface change the energy required for the reaction.

The catalyst is not consumed overall. It can participate in many reaction cycles.

05Conditions and observations for hydrogenation

For HSC product-prediction questions, the main conditions to remember are:

Platinum and palladium can also act as hydrogenation catalysts. Nickel is the common catalyst to use in standard HSC reaction equations.

Industrial operating temperatures and pressures depend on the particular process, so do not assume one exact value applies to every hydrogenation reaction.

What would you observe?

This is where students sometimes make an incorrect prediction.

Hydrogen gas and a typical alkene are both colourless. The nickel catalyst is usually a dark solid.

Therefore, you should not expect a distinctive colour change when an alkene undergoes hydrogenation.

Instead, possible evidence includes:

• Hydrogen gas being consumed.
• A decrease in gas pressure in a suitable closed system.
• A decrease in gas volume when measured under suitable conditions.
• Heat being released because hydrogenation is generally exothermic.

There is another useful way to investigate whether the alkene has become saturated: the bromine water test.

An alkene reacts with bromine under ordinary test conditions, causing the orange-brown colour to disappear. An alkane does not decolourise bromine water in the absence of UV light.

After complete hydrogenation, the resulting alkane should therefore no longer give the positive alkene test.

Be careful with the distinction:

• Bromine water is a test for unsaturation.
• Hydrogen gas with a metal catalyst is used for hydrogenation.

They are different reactions with different purposes.

07Questions and solutions

The following questions move from straightforward product prediction to more demanding problems involving stoichiometry, reaction types, and structural isomers.

Question 1

Propene undergoes complete hydrogenation.

(a) Write the structural formula of the organic product.

(b) Write a balanced chemical equation for the reaction.

(c) State the required catalyst and reaction conditions.

Solution 1

The product is propane, \(\ce{CH3CH2CH3}\), because one hydrogen atom adds to each carbon of the original double bond.

(a) The structural formula is:

\[
\ce{CH3CH2CH3}
\]

(b) The balanced equation is:

\[
\ce{CH3CH=CH2 + H2 -> CH3CH2CH3}
\]

There are three carbon atoms on each side and eight hydrogen atoms on each side.

(c) The reaction requires:

• Hydrogen gas.
• A nickel catalyst.
• Heat.

This is an addition reaction because hydrogen adds across the C=C bond. No carbon atoms are removed or replaced.

Question 2

A chemist reacts 5.60 g of but-1-ene with 0.150 mol of hydrogen gas in the presence of a nickel catalyst and heat.

Assume the reaction proceeds until the limiting reactant is completely consumed.

Calculate:

(a) The amount of but-1-ene initially present.

(b) The limiting reactant.

(c) The amount and mass of butane produced.

(d) The amount of hydrogen remaining after the reaction.

Use \(A_r(\ce{C})=12.0\) and \(A_r(\ce{H})=1.0\).

Solution 2

But-1-ene is the limiting reactant. It produces \(0.100\) mol, or \(5.80\) g, of butane, leaving \(0.050\) mol of hydrogen unreacted.

(a) Calculate the initial amount of but-1-ene.

The molecular formula is \(\ce{C4H8}\).

Its molar mass is:

\[
M(\ce{C4H8})=4(12.0)+8(1.0)
\]

\[
M(\ce{C4H8})=56.0\ \mathrm{g\,mol^{-1}}
\]

Using \(n=m/M\):

\[
n(\text{but-1-ene})=\frac{5.60}{56.0}
\]

\[
n(\text{but-1-ene})=0.100\ \mathrm{mol}
\]

(b) Identify the limiting reactant.

The reaction equation is:

\[
\ce{C4H8 + H2 -> C4H10}
\]

The mole ratio is \(1:1\).

We have:

• \(0.100\) mol of but-1-ene.
• \(0.150\) mol of hydrogen.

Only \(0.100\) mol of hydrogen is needed to react with all the but-1-ene.

Therefore, but-1-ene is the limiting reactant, and hydrogen is in excess.

(c) Calculate the amount and mass of butane produced.

The mole ratio between but-1-ene and butane is also \(1:1\).

\[
n(\text{butane})=0.100\ \mathrm{mol}
\]

The molar mass of butane is:

\[
M(\ce{C4H10})=4(12.0)+10(1.0)
\]

\[
M(\ce{C4H10})=58.0\ \mathrm{g\,mol^{-1}}
\]

Therefore:

\[
m=0.100\times58.0
\]

\[
m(\text{butane})=5.80\ \mathrm{g}
\]

(d) Calculate the hydrogen remaining.

Hydrogen initially present:

\[
0.150\ \mathrm{mol}
\]

Hydrogen consumed:

\[
0.100\ \mathrm{mol}
\]

Therefore:

\[
n(\ce{H2}\text{ remaining})=0.150-0.100
\]

\[
\boxed{0.050\ \mathrm{mol}}
\]

The important point is that having excess hydrogen does not change the amount of butane produced. The amount of product is controlled by the limiting reactant.

Question 3

Two separate experiments are conducted using bromine.

• Experiment A: Propene is mixed with bromine solution at room temperature, without UV light.
• Experiment B: Propane is mixed with bromine under UV light.

In both experiments, the orange-brown colour of bromine fades.

A student concludes that both reactions must be addition reactions because the bromine disappears.

Explain why this conclusion is incorrect. Include the reaction type, products, and bond changes for each experiment.

Solution 3

The conclusion is incorrect because both addition and substitution reactions can consume bromine and cause its colour to fade. The reaction type depends on the starting hydrocarbon, conditions, and bonds formed.

Experiment A: Propene and bromine

Propene is an alkene, so it contains a C=C double bond.

Bromine adds across this double bond:

\[
\ce{CH3CH=CH2 + Br2 -> CH3CHBrCH2Br}
\]

The product is 1,2-dibromopropane.

The reaction is addition because:

• The C=C bond becomes a C-C bond.
• One bromine atom attaches to each original alkene carbon.
• No hydrogen atom is displaced.
• One main organic product forms.

The orange-brown bromine colour disappears because bromine is consumed.

Experiment B: Propane and bromine

Propane is an alkane, so it contains only carbon-carbon single bonds.

Under UV light, propane undergoes substitution with bromine.

One possible reaction is:

\[
\ce{CH3CH2CH3 + Br2 ->[UV] CH3CH2CH2Br + HBr}
\]

Another possible product is 2-bromopropane:

\[
\ce{CH3CH2CH3 + Br2 ->[UV] CH3CHBrCH3 + HBr}
\]

The reaction is substitution because:

• A C-H bond is broken.
• A C-Br bond forms.
• The removed hydrogen combines with bromine to form HBr.
• Different substitution positions can produce different organic products.

Bromine is consumed in this reaction as well, so its colour fades.

Why the distinction matters

The bromine colour change alone does not prove that addition has occurred.

An alkene can decolourise bromine without UV light through addition. An alkane can consume bromine under UV light through substitution.

This is why the conditions and product structures are essential when identifying a reaction.

Question 4

An unknown pure hydrocarbon has the molecular formula \(\ce{C5H10}\).

It is known to be an open-chain alkene containing exactly one C=C bond.

When the compound undergoes complete hydrogenation, the only organic product is 2-methylbutane.

(a) Identify all distinct structural isomers of the starting alkene that could produce 2-methylbutane.

(b) Explain why hydrogenation cannot determine the original position of the double bond.

Assume that the carbon skeleton does not rearrange during hydrogenation.

Solution 4

There are three distinct starting alkene structures consistent with the product: 2-methylbut-1-ene, 3-methylbut-1-ene, and 2-methylbut-2-ene.

All three hydrogenate to 2-methylbutane.

(a) Identify the possible starting structures.

Start with the product:

\[
\ce{CH3-CH(CH3)-CH2-CH3}
\]

Hydrogenation does not rearrange its carbon skeleton under the stated assumptions.

To identify possible starting alkenes, consider which carbon-carbon single bonds could originally have been double bonds.

Possibility 1: 2-methylbut-1-ene

\[
\ce{CH2=C(CH3)-CH2-CH3}
\]

Hydrogenation gives:

\[
\ce{CH3-CH(CH3)-CH2-CH3}
\]

Possibility 2: 3-methylbut-1-ene

\[
\ce{CH2=CH-CH(CH3)-CH3}
\]

Hydrogenation also gives:

\[
\ce{CH3-CH2-CH(CH3)-CH3}
\]

This is the same product, written from the opposite end of the chain.

Possibility 3: 2-methylbut-2-ene

\[
\ce{CH3-C(CH3)=CH-CH3}
\]

Hydrogenation again gives:

\[
\ce{CH3-CH(CH3)-CH2-CH3}
\]

(b) Explain why the original double-bond position cannot be identified.

Hydrogenation removes the C=C bond and replaces it with a C-C single bond.

The hydrogen atoms occupy the remaining bonding positions, but the reaction does not preserve a record of where the original double bond was located.

All three starting compounds have:

• The same molecular formula.
• The same branched carbon skeleton.
• One carbon-carbon double bond.

Consequently, each consumes one mole of hydrogen per mole of alkene and produces the same saturated hydrocarbon.

The product tells us the carbon skeleton of the starting alkene, but not necessarily the original position of its double bond.

This is an important limitation of using hydrogenation to identify an unknown organic compound.

Question 5

A chemist has an unknown hydrocarbon mixture containing only:

• Pentane, \(\ce{C5H12}\).
• Pent-1-ene, \(\ce{C5H10}\).
• Pent-2-ene, \(\ce{C5H10}\).

The mixture may contain either or both of the two alkenes.

The following experimental information is recorded:

• Total amount of hydrocarbons: \(0.100\ \mathrm{mol}\).
• Total mass of the original mixture: \(7.08\ \mathrm{g}\).
• A separate sample of the original mixture decolourises bromine solution in the absence of UV light.
• Complete hydrogenation using excess hydrogen and a nickel catalyst consumes \(0.060\ \mathrm{mol}\) of \(\ce{H2}\).
• After hydrogenation, the only carbon-containing product is pentane.

Use \(A_r(\ce{C})=12.0\) and \(A_r(\ce{H})=1.0\).

(a) Calculate the original amount of pentane.

(b) Calculate the combined original amount of pent-1-ene and pent-2-ene.

(c) Use the mass of the original mixture to check that your calculated composition is consistent with the experimental data.

(d) Can the exact ratio of pent-1-ene to pent-2-ene be determined from these experiments? Explain why or why not.

Assume that both alkenes contain exactly one C=C bond, all hydrogenation reactions go to completion, and no other reactions occur.

Solution 5

The original mixture contains \(0.040\ \mathrm{mol}\) of pentane and \(0.060\ \mathrm{mol}\) of combined pent-1-ene and pent-2-ene.

However, the individual amounts of pent-1-ene and pent-2-ene cannot be determined from the hydrogenation results.

(a) Calculate the original amount of pentane.

The key is recognising that only the alkenes consume hydrogen during hydrogenation.

Each alkene molecule contains one C=C bond, so each mole of alkene reacts with one mole of hydrogen.

The total hydrogen consumed is:

\[
n(\ce{H2})=0.060\ \mathrm{mol}
\]

Therefore, the combined amount of the two alkenes must be:

\[
n(\text{alkenes})=0.060\ \mathrm{mol}
\]

The total amount of hydrocarbons is \(0.100\) mol.

Therefore:

\[
\begin{aligned}
n(\text{pentane})&=0.100-0.060\\
&=0.040\ \mathrm{mol}
\end{aligned}
\]

(b) Calculate the combined amount of pentenes.

From the hydrogenation stoichiometry:

\[
\ce{C5H10 + H2 -> C5H12}
\]

The mole ratio is \(1:1\).

Therefore:

\[
\boxed{n(\text{pent-1-ene + pent-2-ene})=0.060\ \mathrm{mol}}
\]

This is the combined amount. It does not tell us how much of each isomer was originally present.

(c) Check the original mass.

Calculate the molar masses.

For pentane:

\[
M(\ce{C5H12})=5(12.0)+12(1.0)
\]

\[
M(\ce{C5H12})=72.0\ \mathrm{g\,mol^{-1}}
\]

For either pentene:

\[
M(\ce{C5H10})=5(12.0)+10(1.0)
\]

\[
M(\ce{C5H10})=70.0\ \mathrm{g\,mol^{-1}}
\]

Now calculate the mass of each component.

Pentane:

\[
m=0.040\times72.0
\]

\[
m(\text{pentane})=2.88\ \mathrm{g}
\]

Combined pentenes:

\[
m=0.060\times70.0
\]

\[
m(\text{pentenes})=4.20\ \mathrm{g}
\]

Total mass:

\[
m_{\text{total}}=2.88+4.20
\]

\[
\boxed{m_{\text{total}}=7.08\ \mathrm{g}}
\]

This agrees exactly with the experimental mass.

The calculated composition is therefore consistent with the supplied data.

(d) Can the ratio of pent-1-ene to pent-2-ene be determined?

No. The exact ratio cannot be established from these experiments.

Both pent-1-ene and pent-2-ene undergo hydrogenation to form pentane:

\[
\ce{CH2=CHCH2CH2CH3 + H2 -> CH3CH2CH2CH2CH3}
\]

\[
\ce{CH3CH=CHCH2CH3 + H2 -> CH3CH2CH2CH2CH3}
\]

Each mole of either alkene consumes one mole of hydrogen.

Consequently, hydrogen consumption tells us the combined amount of alkene, but not the relative amounts of the two isomers.

There are two tempting mistakes here.

The first is assuming that all \(0.100\) mol of pentane found after hydrogenation must have come from the original alkenes.

That is incorrect. The original \(0.040\) mol of pentane was already saturated, so it remained unchanged. The \(0.060\) mol of pentenes became additional pentane.

The second mistake is assuming that because the final product is pentane, the starting alkene must have been pent-1-ene.

Pent-2-ene produces exactly the same final product.

The bromine test does not resolve this ambiguity either. It confirms that an unsaturated hydrocarbon is present, but it cannot distinguish pent-1-ene from pent-2-ene.

A different measurement that is sensitive to the original double-bond position would be needed.

This is the deeper lesson: a chemical reaction can tell you a great deal about a starting structure without necessarily identifying every detail of that structure.

08What to remember when predicting hydrogenation products

When you encounter a hydrogenation question in HSC Chemistry, focus on the original double bond rather than trying to memorise individual product names.

Your essential checks are:

1. Find the C=C bond.
2. Change it into a C-C bond.
3. Add one hydrogen atom to each carbon that formed the original double bond.
4. Preserve the carbon skeleton and any branches.
5. Check that every carbon has four bonds.
6. If calculating quantities, remember the \(1:1\) mole ratio between a monoalkene and hydrogen.

Finally, keep hydrogenation separate from substitution. Hydrogenation adds hydrogen across an alkene’s double bond using a metal catalyst. Substitution replaces an atom or group, as in the reaction of an alkane with a halogen under UV light.

Once you can predict hydrogenation products confidently, the next useful step is learning to predict the products of the other alkene addition reactions, particularly halogenation, hydrohalogenation, and hydration. The same principle of adding atoms across the original double bond provides the starting point for all three.