A compound with molecular formula C4H8O2, shows absorptions at 2500-3300 cm-1 (broad), 1720 cm-1 and at 1200 cm-1 on the IR spec
1 answer:
Answer:
Butanoic Acid:
I have attached the structure of Butanoic Acid. Please refer to the attachment.
Explanation:
Solution:
First of all, we need to calculate the double bond equivalent (DBE) for the compound given:
Compound = C4H8O2
DBE = (C +1) - (H + X - N)/2
C = no. of carbons
H = no. of Hydrogen
X = no. of Halogens
N = no. of Nitrogen
DBE = (4+1) - (8/2)
DBE = 5-4
DBE = 1
It indicates the presence of 1 double bond in the possible structure.
Further structure can be identified by looking the peak values of the spectrum. So, we have,
2500 - 3300 = it indicates the presence of -COOH group
1720 = it indicates the (C=0) stretching
1200 = It indicates the (C-O) stretching
SO, the possible structure is
Butanoic Acid:
I have attached the structure of Butanoic Acid. Please refer to the attachment.
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Answer:
See explanation below (Brainlist please)
Explanation:
First of all, we need to understand what is an acidic hydrogen.
An acidic hydrogen, is the atom of hydrogen which is more propense to undergo an acid base reaction, and form a stable ion or molecule in the process.
In other words, is the hydrogen that is more vulnerable to get substracted in an acid base reaction to form another compound.
Knowing this information, gives us an idea of how a molecule can be formed and which kind of compound is formed.
Now, in this question, we have 5 molecules. Each of them is either a ketone or aldehyde, so this mean that we have the carbonile group (C = O), which means that is easier to identify the acidic hydrogen. This is because the Carbonile group is an attractor group, so, it will attract the charges by inductive effect (in some cases by resonance), and the molecule is more stable.
This can be shown by drawing the enolate ion that is formed once the molecule undergo the acid base reaction. As it's an enolate form that we are looking for, then it means that the ketone or aldehyde is undergoing an electrofilic attack with a base. This base will substract the most acidic hydrogen to form a better and stable enolate. The acidic hydrogen and the enolate form can be seen in the attached picture.
a) In the case of acetaldehyde, the most acidic will be the hydrogen of carbon 2, because the hydrogen from the carbonile, once it's substracted, the charge of the carbon cannot be stabilized by resonance. Carbon 2 hydrogens, can do this job easily.
b) Propanal happens something similar to acetaldehyde, the terminal hydrogen cannot be substracted, and carbon 3, once the hydrogen is gone, the negative charge cannot be stabilized by resonance, so hydrogens of carbon 2 can do this.
c) in the case of acetone, is easier to look because we only have the C = O between two methyl group, so you can use either carbon 1 or 3 to do the job.
d) 4 heptanone the most acidic hydrogen would be carbon 3 or 5, because they are closer to the C=O and the ion can be stabilized by resonance.
e) Finally in ciclopentanone, the most acidic hydrogen would be carbon 2 or 5.
See picture for a better understanding.
Hope it helps.
