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Rainbow [258]
3 years ago
11

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:2500-3300cm^{-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 3000cm^{-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 3000cm^{-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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The answer is attached below

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\\ \sf\longmapsto 14u+3(1u)

\\ \sf\longmapsto 14u+3u

\\ \sf\longmapsto 17g/mol

We know.

No of moles=Given mass/Molar mass

\\ \sf\longmapsto Given\;Mass=17(5)

\\ \sf\longmapsto Given \:Mass\:of\:NH_3=85g

Now

Lets write the balanced equation

\\ \sf\longmapsto N_2+3H_2=2NH_3

  • There is 2moles of Ammonia
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Now

\boxed{\sf No\:of\:Molecules =No\:of\;moles\times Avagadro\:no}

For Hydrogen

\\ \sf\longmapsto 3\times 6.023\times 10^{23}

\\ \sf\longmapsto 18.069\times 10^{23}

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For Ammonia

\\ \sf\longmapsto 2\times 6.023\times 10^{23}

\\ \sf\longmapsto 12.046\times 10^{23}

\\ \sf\longmapsto 1.2\times 10^{22}molecules

For Nitrogen

\\ \sf\longmapsto 1\times 6.023\times 10^{23}

\\ \sf\longmapsto 6.023\times 10^{23}molecules

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