Alcohol Etherification: Finding The Right Molecule

Hey guys, let's dive into a cool chemistry problem today! We're talking about etherification, specifically the direct etherification of a saturated monohydric alcohol. You know, those simple alcohols like methanol, ethanol, and the like. The kicker is that the ether product we get has a molecular mass that's a whopping 43.75% higher than the original alcohol. Our mission, should we choose to accept it, is to figure out which alcohol was the starting material for this reaction. Get ready to flex those chemistry muscles!

Understanding Etherification and Molecular Mass

So, what exactly is this direct etherification we're talking about? In simple terms, it's a reaction where two alcohol molecules come together, with the help of an acid catalyst (usually sulfuric acid), to form an ether and a water molecule. Imagine two identical alcohol buddies holding hands, and then they decide to pair up, kicking out a water molecule in the process. The general reaction looks something like this: $2 ext{ROH} ightarrow ext{ROR} + ext{H}_2 ext{O}$. Here, 'R' represents an alkyl group, which is basically a chain of carbon and hydrogen atoms. Since we're dealing with a saturated monohydric alcohol, we know that 'R' will be a simple alkane chain without any double or triple bonds, and each alcohol molecule has just one hydroxyl (-OH) group. Think of methanol ($ ext{CH}_3 ext{OH}$), ethanol ($ ext{CH}_3 ext{CH}_2 ext{OH}$), propanol ($ ext{CH}_3 ext{CH}_2 ext{CH}_2 ext{OH}$), and so on. These are our building blocks.

Now, let's talk about molecular mass. This is simply the sum of the atomic masses of all the atoms in a molecule. For example, water ($ ext{H}_2 ext{O}$) has a molecular mass of about 18 g/mol (2 * 1.01 for hydrogen + 16.00 for oxygen). When we say the ether's molecular mass is 43.75% greater than the alcohol's, it means we need to do some math. If the alcohol's molecular mass is 'M', then the ether's molecular mass is $M + 0.4375M = 1.4375M$. This percentage increase is the key to unlocking the mystery alcohol.

To make this even clearer, let's break down the molecular mass difference. In the etherification reaction ($2 ext{ROH} ightarrow ext{ROR} + ext{H}_2 ext{O}$), we start with two molecules of alcohol (each with molecular mass $M$) and end up with one molecule of ether (ROR) and one molecule of water ($ extH}_2 ext{O}$). The molecular mass of the ether (ROR) is essentially the molecular mass of two 'R' groups plus the mass of the oxygen atom that bridges them. The molecular mass of the alcohol (ROH) is the mass of the 'R' group plus the mass of the hydroxyl group (-OH). When two alcohols combine to form an ether, we lose a water molecule ($ ext{H}_2 ext{O}$). The mass of the ether (ROR) is therefore the mass of two alcohols minus the mass of water $M_{ ext{ether} = 2M_{ ext{alcohol}} - M_{ ext{water}}$.

We know the molecular mass of water is approximately 18 g/mol. So, $M_{ ext{ether}} = 2M_{ ext{alcohol}} - 18$. We are also given that $M_{ ext{ether}} = 1.4375 imes M_{ ext{alcohol}}$. Now we have two equations and two unknowns ($M_{ ext{ether}}$ and $M_{ ext{alcohol}}$), which means we can solve for the molecular mass of the alcohol. This is where the real detective work begins, guys!

Setting Up the Chemical Equation and Solving for the Alcohol's Mass

Alright, let's get down to the nitty-gritty of setting up the equations. We've established the basic reaction for direct etherification: two molecules of a saturated monohydric alcohol react to form an ether and a molecule of water. Let's denote the molecular formula of the saturated monohydric alcohol as $ ext{C}n ext{H}{2n+1} ext{OH}$.

The general reaction is: $2 ext{C}_n ext{H}_{2n+1} ext{OH} ightarrow ( ext{C}_n ext{H}_{2n+1})_2 ext{O} + ext{H}_2 ext{O}$.

Let $M_{ ext{alcohol}}$ be the molecular mass of the alcohol $ ext{C}n ext{H}{2n+1} ext{OH}$. The molecular mass of the ether, $( ext{C}_n ext{H}_{2n+1})_2 ext{O}$, is given as being 43.75% greater than $M_{ ext{alcohol}}$.

Mathematically, this can be expressed as:

$M_{ ext{ether}} = M_{ ext{alcohol}} + 0.4375 imes M_{ ext{alcohol}}$

$M_{ ext{ether}} = 1.4375 imes M_{ ext{alcohol}}$

Now, let's think about the relationship between the molecular masses of the alcohol, the ether, and water. When two alcohol molecules combine to form an ether, a water molecule is eliminated. Therefore, the mass of the ether formed is equal to the mass of two alcohol molecules minus the mass of one water molecule.

$M_{ ext{ether}} = 2 imes M_{ ext{alcohol}} - M_{ ext{water}}$

We know the molecular mass of water ($ ext{H}_2 ext{O}$) is approximately $2 imes 1.008 ( ext{H}) + 16.00 ( ext{O}) imes 1 ext{ (O)} ext{ g/mol} hickapprox 18.016 ext{ g/mol}$. For simplicity in these types of problems, we often use 18 g/mol.

So, our equation becomes:

$M_{ ext{ether}} = 2 imes M_{ ext{alcohol}} - 18$

Now we have a system of two equations with two variables ($M_{ ext{ether}}$ and $M_{ ext{alcohol}}$):

1. $M_{ ext{ether}} = 1.4375 imes M_{ ext{alcohol}}$
2. $M_{ ext{ether}} = 2 imes M_{ ext{alcohol}} - 18$

We can substitute the first equation into the second one:

$1.4375 imes M_{ ext{alcohol}} = 2 imes M_{ ext{alcohol}} - 18$

Now, let's solve for $M_{ ext{alcohol}}$. We need to gather the $M_{ ext{alcohol}}$ terms on one side:

$18 = 2 imes M_{ ext{alcohol}} - 1.4375 imes M_{ ext{alcohol}}$

$18 = (2 - 1.4375) imes M_{ ext{alcohol}}$

$18 = 0.5625 imes M_{ ext{alcohol}}$

Finally, divide by 0.5625 to find $M_{ ext{alcohol}}$:

M_{ ext{alcohol}} = rac{18}{0.5625}

Let's perform this division. $18 / 0.5625 = 32$. Yes, that's right! The molecular mass of the alcohol is 32 g/mol.

This is a crucial step, guys. We've used the given percentage increase and the fundamental chemistry of ether formation to pinpoint the exact molecular mass of our mystery alcohol. Now, the next step is to identify which common saturated monohydric alcohol fits this molecular mass. It's like finding a suspect based on their fingerprint!

Identifying the Alcohol Using its Molecular Formula

We've cracked the code, everyone! We've determined that the molecular mass of the saturated monohydric alcohol involved in this etherification reaction is 32 g/mol. Now, the mission is to identify which alcohol has this specific mass. Remember, we're dealing with saturated monohydric alcohols, which have the general formula $ ext{C}n ext{H}{2n+1} ext{OH}$. Let's break down the calculation of the molecular mass based on this formula.

The molecular mass ($M_{ ext{alcohol}}$) can be calculated as follows:

$M_{ ext{alcohol}} = (n imes ext{Atomic Mass of C}) + ((2n+1) imes ext{Atomic Mass of H}) + ( ext{Atomic Mass of O}) + ( ext{Atomic Mass of H})$

Using the approximate atomic masses: Carbon (C) $ hickapprox 12$ g/mol, Hydrogen (H) $ hickapprox 1$ g/mol, and Oxygen (O) $ hickapprox 16$ g/mol.

So, the formula for the molecular mass becomes:

$M_{ ext{alcohol}} = (n imes 12) + ((2n+1) imes 1) + 16 + 1$

$M_{ ext{alcohol}} = 12n + 2n + 1 + 16 + 1$

$M_{ ext{alcohol}} = 14n + 18$

Now, we set this expression equal to the molecular mass we found: 32 g/mol.

$14n + 18 = 32$

Let's solve for 'n', the number of carbon atoms in the alcohol's alkyl group:

$14n = 32 - 18$

$14n = 14$

n = rac{14}{14}

$n = 1$

Bingo! We found that $n=1$. This means our saturated monohydric alcohol has one carbon atom in its alkyl group. Plugging $n=1$ back into the general formula $ ext{C}n ext{H}{2n+1} ext{OH}$, we get:

$ ext{C}1 ext{H}{(2 imes 1)+1} ext{OH}$

$ ext{C}_1 ext{H}_3 ext{OH}$

This is Methanol! The chemical formula for methanol is $ ext{CH}_3 ext{OH}$.

Let's quickly verify. The molecular mass of methanol is $12.01 ( ext{C}) + 4 imes 1.008 ( ext{H}) + 16.00 ( ext{O}) hickapprox 32.04 ext{ g/mol}$. This matches our calculated molecular mass of 32 g/mol perfectly.

So, the alcohol subjected to the etherification reaction is Methanol.

Checking the Options and Final Answer

We've done the hard work, guys, and arrived at our answer: Methanol. Now, let's look at the options provided to make sure we're on the right track and to finalize our response.

The options given are:

A. Etanolul (Ethanol) B. Metanolul (Methanol) C. Discussion category : chimie

Option C isn't actually an alcohol; it's the category of the discussion. So, we can disregard that. We are left with Ethanol and Methanol.

Let's quickly calculate the molecular mass of Ethanol to see if it fits the criteria. Ethanol has the formula $ ext{C}_2 ext{H}_5 ext{OH}$. Its molecular mass is $(2 imes 12) + (5 imes 1) + 16 + 1 = 24 + 5 + 16 + 1 = 46$ g/mol. If we had started with Ethanol, the ether formed would be diethyl ether, $( ext{C}_2 ext{H}_5)_2 ext{O}$. Its molecular mass would be $2 imes 46 - 18 = 92 - 18 = 74$ g/mol. Let's check the percentage increase: $(74 - 46) / 46 imes 100 = 28 / 46 imes 100 hickapprox 60.87$. This is not the 43.75% given in the problem.

Our calculation clearly showed that an alcohol with a molecular mass of 32 g/mol is required, and this corresponds to Methanol ($ ext{CH}_3 ext{OH}$).

Therefore, the correct answer is B. Metanolul (Methanol).

It's always a good idea to double-check your work, especially when dealing with percentages and molecular masses. The steps involved were: understanding the etherification reaction, relating the molecular masses of the alcohol and ether, setting up and solving equations to find the alcohol's molecular mass, and finally identifying the alcohol based on its general formula and calculated mass. Chemistry problems can sometimes feel like solving a puzzle, and this one was a fun one!

Keep practicing, and you'll get even better at these kinds of problems. Cheers!