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ivann1987 [24]
3 years ago
9

γ−Butyrolactone (C4H6O2, GBL) is a biologically inactive compound that is converted to the biologically active recreational drug

GHB by a lactonase enzyme in the body. Since γ−butyrolactone is more fat soluble than GHB, it is more readily absorbed by tissues and thus produces a faster onset of physiological symptoms. γ−Butyrolactone shows an absorption in its IR spectrum at 1770 cm−1 and the following 1H NMR spectral data: 2.28 (multiplet, 2 H), 2.48 (triplet, 2 H), and 4.35 (triplet, 2 H) ppm. What is the structure of γ−butyrolactone?

Chemistry
1 answer:
Mashutka [201]3 years ago
6 0

Answer:

Here's what I get.  

Explanation:

The name tells me the compound is a lactone (a cyclic ester).

1. IR spectrum

1770 cm⁻¹

Esters and unstrained lactones normally absorb at 1740 cm⁻¹.

This peak is shifted to a higher frequency by ring strain.

A five-membered lactone absorbs at 1765 cm⁻¹, and a four-membered lactone at 1840 cm⁻¹.

The compound is probably a five-membered lactone.

2. NMR spectrum

2.28 m (2H)  

2.48 t (2H)

4.35 t (2H)

This indicates three CH₂ groups arranged as X-CH₂-CH₂-CH₂-Y.

The X-CH₂-  and -CH₂-Y signals would each be triplets, being split by the central -CH₂- group.

The central -CH₂- signal would be a multiplet, split by the non-equivalent hydrogens on either side.

The peak at 4.35 ppm indicates that the group is adjacent to an oxygen atom ( -CH₂- = 1.3; -CH₂-O- = 3.3 - 4.5).

The peak at 2.42 ppm indicates that the group is adjacent to a carbonyl group (-CH₂-C=O = 1.8 - 2.5.

The only way to fit these pieces together is if γ-butyrolactone has the structure shown below.

Confirmation:

(a) The IR spectrum shows a carbonyl peak at 1770 cm⁻¹.

(b) The NMR spectrum matches that given in the problem.

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Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

\text{calculating the initial HI}= \frac{mol}{V}

                                       =\frac{9.3 \times 10 ^ -3}{2}

                                      =0.00465 \ Mol

\text{Similarly}\ \  I_2 \ \  \text{follows} \ \  H_2 = 0 }

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

HI  = 0.00465 - 2x\\\\ I_{2}  \ eq = H_2 \ eq = 0 + x \\\\

It is defined that:

I_2 = 6.29 \times 10^{-4}  \ M \\\\x = I_2 \\\\

HI \ eq= 0.00465 - 2x \\

          =0.00465 -2 \times 6.29 \times 10^{-4} \\\\ =  0.00465 -\frac{25.16 }{10^4}  \\\\   = 0.003392\  M

Now, we calculate the position:  

For the reaction H 2(g) + I 2(g)\rightleftharpoons  2HI(g), you can calculate the value of Kc at 1000 K.  

data expression for Kc

2HI \rightleftharpoons  H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}

         = \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386

calculating the reverse reaction

H_2(g) + I_2(g)\rightleftharpoons  2HI(g)

Kc = \frac{1}{Kc} \\\\

     = \frac{1}{0.034386}\\ \\= 29.081\\

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