Question

Write a balanced equation for each of the following reactions: (a) $\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3+{\mathrm{O}}_2⟶$ (b) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}+2{\mathrm{H}}_2⟶$ (c) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3+2{\mathrm{Br}}_2⟶$

Short answer

(a) $2\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3+7{\mathrm{O}}_2\to 6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$; (b) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}+2{\mathrm{H}}_2\to {\mathrm{C}}_3{\mathrm{H}}_8$; (c) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3+2{\mathrm{Br}}_2\to {\mathrm{CH}}_3{\mathrm{CHBrCBrCH}}_3$."

Step-by-step solution

Identify Reactants and Products for Reaction (a)

For the reaction (a): $\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3+{\mathrm{O}}_2⟶$, we are dealing with the combustion of an alkyne (propyne). The typical products of combustion with sufficient oxygen are carbon dioxide (${\mathrm{CO}}_2$) and water (${\mathrm{H}}_2\mathrm{O}$).

Write the Unbalanced Equation for Reaction (a)

The unbalanced chemical equation for combustion of propyne can be written as:$$\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3+{\mathrm{O}}_2\to {\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$$

Balance the Equation for Reaction (a)

For the complete combustion of one mole of $\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3$, we need:- 3 carbon atoms, so $3{\mathrm{CO}}_2$- 4 hydrogen atoms, so $2{\mathrm{H}}_2\mathrm{O}$Thus, we require $4\frac{1}{2}$ moles of ${\mathrm{O}}_2$ to supply sufficient oxygen, which we can express better as integers: $7$ moles of ${\mathrm{O}}_2$ for 2 moles of alkyne. The complete balanced equation is:$$2\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3+7{\mathrm{O}}_2\to 6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$$

Identify Reactants and Products for Reaction (b)

For the reaction (b): ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}+2{\mathrm{H}}_2⟶$, the reactant is propyne undergoing hydrogenation to form a saturated hydrocarbon, which is propane (${\mathrm{C}}_3{\mathrm{H}}_8$).

Write the Unbalanced Equation for Reaction (b)

The chemical equation for the hydrogenation of propyne can be written as:$${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}+2{\mathrm{H}}_2\to {\mathrm{C}}_3{\mathrm{H}}_8$$

Balance the Equation for Reaction (b)

The equation already balances as written:$${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}+2{\mathrm{H}}_2\to {\mathrm{C}}_3{\mathrm{H}}_8$$each side of the equation contains 3 carbon atoms and 8 hydrogen atoms.

Identify Reactants and Products for Reaction (c)

For reaction (c): ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3+2{\mathrm{Br}}_2⟶$, propene undergoes halogenation to form a dibromo compound, which is 1,2-dibromo-2-butene: ${\mathrm{CH}}_3{\mathrm{CHBrCBrCH}}_3$.

Write the Unbalanced Equation for Reaction (c)

The equation for the halogenation of 2-butyne can be expressed as:$${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3+2{\mathrm{Br}}_2\to {\mathrm{CH}}_3{\mathrm{CHBrCBrCH}}_3$$

Balance the Equation for Reaction (c)

In this step, tally the number of atoms:- Carbon atoms: 4 on each side.- Hydrogen atoms: 6 on each side.- Bromine atoms: 4 on each side.The equation is already balanced as:$${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3+2{\mathrm{Br}}_2\to {\mathrm{CH}}_3{\mathrm{CHBrCBrCH}}_3$$

Key concepts

Balancing Chemical Equations

Balancing chemical equations is an essential aspect of understanding and describing chemical reactions accurately. It ensures that the law of conservation of mass is satisfied, meaning the number of atoms for each element is the same on both the reactant and product sides of the equation. To balance an equation, identify how many atoms of each element exist on either side. Use coefficients to equalize these numbers, ensuring minimal changes to the chemical formulas.
For example, in the combustion of propyne, represented as $$2\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3+7{\mathrm{O}}_2\to 6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$$, we adjust the coefficients to balance carbon, hydrogen, and oxygen atoms, reflecting the stoichiometry of the reaction.
Learning to balance equations efficiently is a fundamental skill in studying chemical reactions.

Combustion Reactions

Combustion reactions involve burning a substance, typically hydrocarbons, in the presence of oxygen, resulting in the release of energy in the form of heat and light. Generally, the products for complete combustion include carbon dioxide (${\mathrm{CO}}_2$) and water (${\mathrm{H}}_2\mathrm{O}$).
In the case of propyne, the combustion reaction is expressed as:

• Reactant: $\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3$ (propyne)
• Products: ${\mathrm{CO}}_2$, ${\mathrm{H}}_2\mathrm{O}$

Balancing these reactions involves ensuring that all carbon, hydrogen, and oxygen atoms are accounted for, which reflects the specific stoichiometric coefficients required for complete combustion as shown in: $2\mathrm{CH}\equiv \mathrm{C}-{\mathrm{CH}}_3+7{\mathrm{O}}_2\to 6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$.
This type of reaction is crucial across various applications, including energy production.

Hydrogenation

Hydrogenation involves adding hydrogen to unsaturated hydrocarbons, turning double or triple bonds into single bonds. This reaction is notably practiced to convert alkenes and alkynes into alkanes.
The hydrogenation of propyne is represented by:$${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}+2{\mathrm{H}}_2\to {\mathrm{C}}_3{\mathrm{H}}_8$$.
In this reaction:

• Propyne (${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}$) gains hydrogen to form propane (${\mathrm{C}}_3{\mathrm{H}}_8$).
• The reaction is exothermic, meaning it releases heat.

This process is widely used in industrial applications such as refining fuels and synthesizing fats, crucial for many manufacturing processes.

Halogenation

Halogenation describes the reaction where a halogen atom is added to a molecule, often replacing hydrogen atoms. With alkynes, like 2-butyne, halogens attach directly to form vicinal dihalides.
In the described reaction, ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3+2{\mathrm{Br}}_2\to {\mathrm{CH}}_3{\mathrm{CHBrCBrCH}}_3$, 2-butyne reacts with bromine (${\mathrm{Br}}_2$), producing 1,2-dibromo-2-butene.
This process involves:

Adding bromine atoms across the triple bond of the alkyne.

Conversion results in a fully saturated compound.

Halogenation plays a vital role in organic synthesis and the chemical industry, helping to produce a variety of useful compounds from polymers to pharmaceuticals.

Alkyne Reactions

Alkynes are hydrocarbons with at least one carbon-carbon triple bond, making them unsaturated and highly reactive. The unique properties of their triple bond make alkynes suitable for a range of reactions such as halogenation and hydrogenation.
Relevant reactions include:

• Combustion: As seen with propyne, resulting in ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$.
• Hydrogenation: Adding hydrogen to form alkanes, such as propyne converting to propane.
• Halogenation: Adding halogens like bromine to form compounds with dihalogen units.

These reactions not only transform alkynes into more stable products but also showcase their significance in chemical synthesis and various industrial processes. Understanding alkyne reactivity aids in creating a wide array of organic materials.