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3 years ago

Which region in the IR spectrum could be used to distinguish between butanoic acid and 2-butanone?

Chemistry

1 answer:

Tems11 [23]3 years ago

4 0

Answer:2500-3300 $cm^{-1}$

Explanation:We can distinguish between 2-butanone and butanoic acid by analyzing the spectra .

The structure of the two compounds are slightly different as butanoic acid has an OH group whereas for 2-butanone there is no OH group present.

Please refer the attachments for structure of compounds.

Both the compounds will show carbonyl stretching frequency at around 1700-1750cm^{-1}[/tex] due to the presence of carbonyl group in both the compounds.

A broad intense peak at around 3000 $cm^{-1}$ will be observed only for butanoic acid due to the presence of OH group and 2-butanone would not show any broad intense peak at around 3000 $cm^{-1}$ .

So by analyzing the IR spectra and identifying the intense broad peak of OH we can easily distinguish between butanoic acid and 2-butanone.

Due to hydrogen bonding in between two carboxylic acid molecules ,peak broadening for OH group in butanoic acid takes place.

Generally the ketones show IR stretching in around 1700-1730cm^{-1}[/tex] due to the presence of carbonyl group.

Generally carboxylic acids show IR stretching for carbonyl group at around 1700-1760cm^{-1}[/tex] and a broad intense peak for OH at around 2500-3000cm^{-1}[/tex].

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How many molecules of H2O are there in 1.0 g of H2O?

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Ten pints of 15 % salt solution are mixed with 15 pints of 10 % salt solution. What is the

sergij07 [2.7K]

$Concentration\ rate=\frac{mass\ of\ product}{mass\ of\ substance}*100\%\\\\
15\%=\frac{m_1}{ms_1}*100\%\\ms_1=10pints\\
15\%=\frac{m_1}{10}*100\%\\
\frac{15}{100}=\frac{m_1}{10}\ \ |*10\\
m_1=\frac{150}{100}=1,5pints\\\\
10\%=\frac{m_2}{15}*100\%\\
0,1=\frac{m_2}{15}\ \ |*15\\
1,5=m_2\\
Resulting\ solution:\\
ms=ms_1+ms_2=15+10=25pints\\m=m_1+m_2=1,5+1,5=3pints
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