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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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Consider the reaction of Mg3N2 with H2O to form Mg(OH)2 and NH3. If 4.33 g H2O is reacted with excess Mg3N2 and 6.26 g of Mg(OH)
Len [333]

Answer:

89.34%

Explanation:

First, write a balanced reaction.

Mg3N2 + <u>6</u>H2O --> <u>3</u>Mg (OH)2 + <u>2</u>NH3

Next determine the moles of the known substance, or limiting reagent ( H2O)

n= m/MM

n ( H2O) = 4.33/(1.008×2)+16

n(H2O)= 0.2403

Use the mole ratio to find the moles of Mg(OH)2

0.2403 ÷2

n (Mg (OH)2) = 0.1202

Next, find the theoretical mass of Mg (OH)2 that should have been produced

m= n × MM

m= 0.1202 × (24.305 + (16×2) +(1.008 ×2))

=7.007g

To find percentage yield, divide the experimental amount by the theoretical amount and multiply by 100.

6.26/ 7.007 × 100

=89.34%

7 0
3 years ago
What is the atomic mass/mass number of the atom in the diagram above?
Lelechka [254]

Answer:

the atomic number is 5

the atomic mass is 11

Explanation:

The atomic number is the amount of protons inside the nucleus, and this number also equals the amount of electrons. Since it shows you the nucleus and the electrons, all you need to do is count the protons (positive charge inside the nucleus) or count all the electrons (negative charge outside the nucleus, in the rings) and you should have your atomic number.

As for mass, all you need to do is count all the protons and neutrons inside the nucleus and add them up. Protons = 5, Neutrons = 6. (you add them since the equation for atomic mass is Atomic Mass = Protons + neutrons. This works every time)

5+6= 11, so your atomic mass is 11

I hope this helps :)

3 0
2 years ago
If the value of kc at 25oc is 3.7108, and the equilibrium concentrations for n2 and h2 are 0.000105 m and 0.0000542 m, respectiv
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Euilibrium constant:

Kc=(NH3)^2/((N2)((H2)^3))

As concentration of N2=0.000105, H2=0.0000542

so equation will become:

3.7=(NH3)^2/(0.000105)*(0.0000542)^3

NH3=√(3.7*0.000105*(0.0000542)^3)

NH3=7.8×10⁻⁹

So concentration of ammonia will be 7.8×10⁻⁹.

5 0
3 years ago
If we place the like poles of two magnets facing each other what will happen?
Leviafan [203]
The answer to that is D
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3 years ago
Calculate the frequency (Hz) and wavelength (nm)
sergey [27]

Answer:

wavelength, λ =  486.6 nm

frequency, f = 6.16 * 10¹⁴ Hz

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Using the wavelength equation; 1/λ = (1/hc) * 2.18 * 10⁻¹⁸ J * (1/nf² - 1/ni²)

Where nf is the final energy level; ni is the initial energy level; h is Planck's constant = 6.63 * 10⁻³⁴ J.s; c is velocity of light = 3 * 10⁸ m/s

1/λ = 1/(6.63 * 10⁻³⁴ J.s * 3 * 10⁸ m/s) * 2.18 * 10⁻¹⁸ J * (1/2² - 1/4²)

1/λ = 2.055 * 10⁶ m

λ = 4.866 * 10⁻⁷ m

wavelength, λ =  486.6 nm

b.  Frequency

Using f = c/λ

f = (3 * 10⁸ m/s) / 4.866 * 10⁻⁷ m

frequency, f = 6.16 * 10¹⁴ Hz

7 0
3 years ago
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