If 1/3 Of An Unknown Alkene Reacts With Bromine To Form 64.8 G Of A Dibromoalkane, And The Remainder Reacts With Hydrogen Chloride To Form 55.5 G Of A Monochloroalkane, Determine The Alkene. A) Pentene B) Propene
In the captivating realm of organic chemistry, alkenes stand out as vibrant molecules characterized by their carbon-carbon double bonds, the very sites of their reactivity. These unsaturated hydrocarbons eagerly engage in addition reactions, welcoming atoms or groups of atoms to their molecular embrace. Among the most intriguing of these reactions are those involving halogens like bromine and hydrogen halides like hydrochloric acid. When an alkene dances with bromine, it gives birth to a dihaloalkane, a molecule adorned with two halogen atoms. Conversely, the reaction with hydrochloric acid leads to a monohaloalkane, a compound graced with a single halogen atom. These reactions aren't just chemical transformations; they're stories whispered in the language of stoichiometry, each gram and mole a character in the narrative.
The challenge before us is to decipher this story, to unveil the identity of an unknown alkene through the clues left behind in its reactions with bromine and hydrochloric acid. A third of this enigmatic alkene embarks on a journey with bromine, yielding 64.8 grams of a dibromoalkane. The remaining two-thirds, not to be left out, react with hydrochloric acid, producing 55.5 grams of a monochloroalkane. Our quest is to determine, from these quantitative whispers, the true nature of the alkene. Is it pentene, with its five-carbon backbone, or propene, the simplest alkene with just three carbons? The answer lies hidden within the stoichiometric relationships, waiting to be revealed through careful calculation and insightful interpretation. We must delve into the heart of the reaction, tracing the atoms and moles, to unveil the identity of our mystery alkene.
Let's begin our investigation by focusing on the reaction between the unknown alkene and bromine. This reaction, a classic example of electrophilic addition, transforms the alkene into a dibromoalkane. The key to unlocking the alkene's identity lies in the stoichiometry of this transformation – the precise molar relationship between the reactants and products.
The balanced chemical equation for this reaction is deceptively simple: CnH2n + Br2 → CnH2nBr2. For every mole of alkene that participates in this dance, one mole of bromine joins in, and one mole of dibromoalkane emerges as the final result. This one-to-one stoichiometry is our guiding principle as we navigate the world of grams and moles.
The problem tells us that 64.8 grams of dibromoalkane are formed. To translate this mass into the language of moles, we need the molar mass of the dibromoalkane. The general formula for a dibromoalkane is CnH2nBr2, and its molar mass can be expressed as: M(CnH2nBr2) = 12n + 2n + 2 * 79.9 = 14n + 159.8 g/mol. Here, n represents the number of carbon atoms in the alkene, a crucial piece of information that we are seeking. To proceed, we calculate the moles of dibromoalkane formed:
Moles of CnH2nBr2 = Mass / Molar Mass = 64.8 g / (14n + 159.8 g/mol)
Since one-third of the original alkene reacted with bromine, the moles of alkene involved in this reaction are equal to the moles of dibromoalkane formed. Let's denote the moles of alkene as moles(alkene). Then, we have:
moles(alkene) = 64.8 / (14n + 159.8) moles
This equation is a cornerstone of our investigation. It links the known mass of dibromoalkane to the unknown number of carbon atoms in the alkene. But it's only one piece of the puzzle. To fully unveil the alkene's identity, we must now turn our attention to the reaction with hydrochloric acid.
The remaining two-thirds of the unknown alkene engaged in a separate reaction, this time with hydrochloric acid (HCl). This reaction, another example of electrophilic addition, results in the formation of a monochloroalkane. The balanced chemical equation for this transformation is: CnH2n + HCl → CnH2n+1Cl. Again, the stoichiometry is beautifully simple: one mole of alkene reacts with one mole of HCl to produce one mole of monochloroalkane.
This time, we're told that 55.5 grams of monochloroalkane are formed. Just as we did with the dibromoalkane, we need to convert this mass into moles. The molar mass of the monochloroalkane (CnH2n+1Cl) is: M(CnH2n+1Cl) = 12n + 2n + 1 + 35.5 = 14n + 36.5 g/mol.
Now, we can calculate the moles of monochloroalkane: Moles of CnH2n+1Cl = Mass / Molar Mass = 55.5 g / (14n + 36.5 g/mol)
Remember that two-thirds of the original alkene reacted with HCl. This means that the moles of alkene involved in this reaction are equal to the moles of monochloroalkane formed. We can express this as:
(2/3) * (total moles of alkene) = 55.5 / (14n + 36.5) moles
This equation provides us with a second crucial piece of information. It connects the mass of monochloroalkane to the same unknown variable, n, the number of carbon atoms in the alkene. Now, we have two equations, each whispering a clue about the alkene's identity. The next step is to bring these equations together, to see how they can illuminate the path to our solution.
We now stand at a pivotal point in our investigation. We have two equations, each born from the stoichiometry of the reactions involving our mystery alkene:
- Moles of alkene (from Br2 reaction): moles(alkene) = 64.8 / (14n + 159.8)
- Moles of alkene (from HCl reaction): (2/3) * (total moles of alkene) = 55.5 / (14n + 36.5)
These equations are our Rosetta Stone, capable of translating the quantitative data into the molecular identity of the alkene. But to decipher them, we need to recognize a crucial relationship: the total moles of alkene are constant, regardless of which reaction we consider. The amount of alkene that reacted with bromine plus the amount that reacted with hydrochloric acid must equal the total amount of alkene present at the beginning.
Let's represent the total moles of alkene as 'T'. Then, the moles of alkene that reacted with bromine (equation 1) represent one-third of T, and the moles of alkene that reacted with HCl (equation 2) represent two-thirds of T. Therefore, we can write:
(1/3) * T = 64.8 / (14n + 159.8)
(2/3) * T = 55.5 / (14n + 36.5)
Now, we have a system of two equations with two unknowns (T and n). The most elegant way to solve this system is to divide the second equation by the first. This maneuver will eliminate 'T', leaving us with a single equation in terms of 'n':
[(2/3) * T] / [(1/3) * T] = [55.5 / (14n + 36.5)] / [64.8 / (14n + 159.8)]
Simplifying this equation, we get:
2 = [55.5 * (14n + 159.8)] / [64.8 * (14n + 36.5)]
This equation, though seemingly complex, is our key to unlocking the mystery. It's a mathematical statement that encapsulates the stoichiometric relationships of the reactions, a precise expression of the dance between the alkene and its halogen partners. To solve for 'n', we need to carefully navigate the algebraic landscape, expanding, rearranging, and isolating the variable that holds the alkene's secret.
Our journey to identify the unknown alkene has led us to a crucial equation:
2 = [55.5 * (14n + 159.8)] / [64.8 * (14n + 36.5)]
To solve for 'n', the number of carbon atoms in the alkene, we must embark on an algebraic ascent. First, we clear the fraction by multiplying both sides of the equation by the denominator:
2 * [64.8 * (14n + 36.5)] = 55.5 * (14n + 159.8)
Next, we distribute the constants on both sides:
129.6 * (14n + 36.5) = 55.5 * (14n + 159.8)
1814.4n + 4730.4 = 777n + 8869
Now, we gather the 'n' terms on one side and the constant terms on the other. Subtract 777n from both sides:
1814.4n - 777n + 4730.4 = 8869
1037.4n + 4730.4 = 8869
Subtract 4730.4 from both sides:
1037.4n = 8869 - 4730.4
1037.4n = 4138.6
Finally, we isolate 'n' by dividing both sides by 1037.4:
n = 4138.6 / 1037.4
n ≈ 3.99
This result is tantalizingly close to a whole number. Since 'n' represents the number of carbon atoms in the alkene, it must be an integer. The value of 3.99 strongly suggests that n = 4. This is a crucial breakthrough in our quest to identify the alkene. But before we declare victory, let's consider the implications of this result and confirm our findings.
Our algebraic journey has led us to the value of n ≈ 4. This strongly suggests that our unknown alkene has four carbon atoms. In the realm of organic chemistry, alkenes with four carbon atoms have the general formula C4H8. But there's a twist: the formula C4H8 encompasses several isomers, molecules with the same chemical formula but different structural arrangements. The most common isomers are but-1-ene and but-2-ene.
To definitively identify our alkene, we need to consider the context of the problem. We know that the alkene reacts with both bromine and hydrochloric acid. While both but-1-ene and but-2-ene would react with these reagents, our stoichiometric calculations have pointed us towards a specific value of 'n'. This strongly suggests that we are dealing with a single alkene, not a mixture of isomers.
Given our result of n ≈ 4, we can confidently conclude that our unknown alkene is likely but-2-ene. But-2-ene is a symmetrical alkene, which means that the addition of HCl would yield a single major product, 2-chlorobutane. But-1-ene, on the other hand, would yield a mixture of 1-chlorobutane and 2-chlorobutane.
To be absolutely certain, we could, in theory, perform additional experiments to distinguish between the isomers. However, within the scope of this problem and the information provided, but-2-ene stands out as the most plausible answer. Our stoichiometric analysis has successfully guided us through the maze of possibilities, revealing the identity of the mystery alkene.
In this chemical detective story, we embarked on a quest to identify an unknown alkene. Guided by the principles of stoichiometry and armed with the quantitative data from its reactions with bromine and hydrochloric acid, we successfully unmasked the molecule. Through careful calculations and insightful interpretations, we navigated the world of grams and moles, ultimately arriving at the solution: the unknown alkene is likely but-2-ene.
This journey highlights the power of stoichiometry in unraveling chemical mysteries. By understanding the quantitative relationships between reactants and products, we can decipher the hidden language of chemical reactions and gain profound insights into the nature of molecules. The story of the unknown alkene serves as a testament to the elegance and precision of chemistry, where even the most elusive identities can be revealed through the careful application of scientific principles.