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adoni [48]
1 year ago
5

A compound with molecular formula C4H6O4 produces a broad signal between 2500 and 3600 cm-1 in its IR spectrum and produces two

signals in its 1H NMR spectrum (a singlet at 12.1 ppm with a relative integration of 1 and a singlet at 2.4 ppm with a relative integration of 2). Draw the structure of the compound.

Chemistry
1 answer:
tiny-mole [99]1 year ago
3 0

Answer:

Succinic Acid

Explanation:

We have to start using the info given by the IR spectrum. In this case, we have a <u>broad signal in 3600</u>. This indicates the presence of OH in the structure. Therefore we can have an <u>alcohol or an acid</u> in the structure.  

Now, the NMR info tells us that we only have 2 signals, this indicates that we have a <u>very symmetric structure</u>. Also, we have a signal in 12 ppm therefore we can affirm that we have an O-H bond with <u>high polarity</u> (a downfield signal) this behavior is given in the <u>acid groups</u>.

The structure that fulfill these requirements it the succinic acid. In which we only have 2 signals in the 1H- NMR. We have an acid group and we have a formula of  C_4H_6O_4.

<u>For further explanations see the attached images.</u>

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M_{Na^+}=1.36M

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Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+

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M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M

Now, we perform the same procedure but now for the bromide ions:

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Finally, its molarity results:

M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M

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