Q. 25 - An Introduction to General, Organic, and Biological Chemistry

Question

 Butyraldehyde is a clear liquid solvent with an unpleasant odor. It has a low boiling point and is highly flammable. Butyraldehyde has a density of $0.802 gm/mL$. The unpleasant odor of old gym socks is due to butyraldehyde.

1. . Draw the condensed structural formula for butyraldehyde.
2. Draw the line-angle formula for butyraldehyde
3.  What is the IUPAC name of butyraldehyde?
4.  Draw the condensed structural formula for the alcohol that is produced when butyraldehyde is reduced.
5.  Write the balanced chemical equation for the complete combustion of butyraldehyde.
6.  How many grams of oxygen gas are needed to completely react with 15.0 mL of butyraldehyde?
7.  How many liters of carbon dioxide gas are produced at STP from the reaction in part f? 

Step-by-Step Solution

Answer

Our required answers are:

1. Condensed structure of butyraldehyde:
   
2. Line angle formula of butyraldehyde:
   
3. Butanal is IUPAC name of butyraldehyde
4. Structural formula of product:
   
5. $2{\mathrm{C}}_4{\mathrm{H}}_8\mathrm{O}+ 11{\mathrm{O}}_2\to 8{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}+\mathrm{energy}$ is balanced equation of combustion of butyraldehyde.
6. $28.512 g$ of oxygen required for complete reaction of$15.00 mL$ of butyraldehyde.
7. ${\mathrm{CO}}_2$ produced in reaction of part f is $14.512 L$ at STP.

Part(a) Step 1: Given Information

We need to draw the condensed structural formula for butyraldehyde 

Part(a) Step 2: Explanation

Butyraldehyde in compound of 4 carbon atoms and an aldehyde group present in it at any terminal carbon. 

So below is condensed structure of butyraldehyde:

 

Part(b) Step 1: Given Information

We need to draw the line-angle formula for butyraldehyde

Part(b) Step 2: Explanation

As specified about condensed form of butyraldehyde in part(b) step $2$.

We get below as line angle structure:

Part(c) Step 1: Given Information

We need to specify IUPAC name of butyraldehyde.

Part(c) Step 2: Explanation

We use -al as suffix for aldehydic groups as specified under rules of IUPAC. So, IUPAC name of  butyraldehyde is butanal. 

Part(d) Step 1: Given Information

We need to draw the condensed structural formula for the alcohol that is produced when butyraldehyde is reduced.

Part(d) Step 2: Explanation

When aldehyde is reduced it form an primary alcohol. Below is reaction mechanism that shows reduction of an aldehyde 

So, according to above common reduction method butyraldehyde gets reduced to butanol. Below is condensed structure of butanol:

Part(e) Step 1: Given Information

We need to write the balanced chemical equation for the complete combustion of butyraldehyde.

Part(e) Step 2: Explanation

On combustion of butyraldehyde we get following equation:

$2{\mathrm{C}}_4{\mathrm{H}}_8\mathrm{O}+11{\mathrm{O}}_2\to 8{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}+ \mathrm{energy}$

Part(f) Step 1: Given Information

We need to find grams of oxygen gas are needed to completely react with $15.00 mL$ of butyraldehyde 

Part(f) Step 2: Explanation

$11$moles of data-custom-editor="chemistry" ${\mathrm{O}}_2$ required to react completely with $2$mole of butyraldehyde

Mass of  butyraldehyde= Volume*Density= $15\times 0.802=12.03 g$

Moles of butyraldehyde = $\frac{12.03}{74}=0.162$

Moles of oxygen required= $\frac{11}{2}\times 0.162=0.891$

Hence, $28.512$ grams of oxygen gas are needed to completely react with $15.00 mL$of butyraldehyde.

Part(g) Step 1: Given Information

 We need to how many liters of carbon dioxide gas are produced at STP from the reaction in part f

moles of oxygen consumed= $0.891$(from step 2 part(f))

Part(g) Step 2: Explanation

On combustion of $11$ moles of oxygen it leads to formation of $8$ moles of carbon dioxide.

Using law of conservation of mass we get,

Moles of carbon dioxide formed = $\frac{8}{11}\times$moles of oxygen

$=\frac{8\times 0.891}{11}=0.648$

So, $0.648$ moles of carbon dioxide formed

Carbon dioxide formed in liters at STP=$0.648\times 22.4=14.51 L$

Hence, $14.51 L$ of carbon dioxide gas is produced at STP from reaction in part f.