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2 years ago

Design a synthesis of 2-ethyl-2-hexenoic acid from alcohols of four carbons or fewer.

Chemistry

1 answer:

Studentka2010 [4]2 years ago

4 0

Answer:

Enolate Alkylation

The anions from ketones, called enolates, can act as a nucleophile in SN2 type reactions. Overall an α hydrogen is replaced with an alkyl group and a new carbon-carbon bond is formed. These alkylations are affected by the same limitations as SN2 reactions previously discussed. A good leaving group, chloride, bromide, iodide or tosylate, should be used. Also, secondary and tertiary leaving groups should not be used because of poor reactivity and possible competition with elimination reactions. Lastly, it is important to use a strong base, such as LDA or sodium amide, for preparing the enolate from the ketone. Using a weaker base such as hydroxide or an alkoxide leaves the possibility of multiple alkylations occurring, and competing SN2 reactions with the base.

Explanation:

Design is illustrated in the attached document

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The complex ion, [ni(nh3)6] 2+, has a maximum absorption near 580 nm. calculate the crystal field splitting energy (in kj/mol) f

Wewaii [24]

In given data:
maximum absorption wavelength λ = 580 nm = 580 x 10⁻⁹ m
write the equation to find the crystal field splitting energy:
E = hC / λ
Here, E is the crystal field splitting energy, h = 6.63 x 10⁻³⁴ J.sec is Planck's constant and C = 3 x 10⁸ m/sec is speed of light.
substitute in the equation above:
E = (6.64 x 10⁻³⁴ x 3 x 10⁸) / (580 x 10⁻⁹) = 3.43 x 10⁻¹⁹J
This crystal field splitting energy is for 1 ion.
Number of atoms in one mole, NA = 6.023 x 10²³
to calculate the crystal field splitting energy for one mole:
E(total) = E x NA
= (3.43 x 10⁻¹⁹) x (6.023 x 10²³) = 206 kJ/ mole

5 0

2 years ago

PLS HELP ASAP, WILL GIVE 100 POINTS.

Ivan

Answer: values of the quantum numbers: -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

location of the electron: In the 7th energy level away from the nucleus.

Explanation:

From the description of the problem, the magnetic number is given is as -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6 and the electron is located in the 7th energy level away from the nucleus. Basically, the problem is testing for the understanding of the principal quantum numbers which gives the location of electrons and the magnetic quantum number that shows the spatial orientation of the orbitals.

The orbital designation of the describe electron is 7d

Magnetic quantum number is limited by the azimuthal quantum number which is the quantum number describing the possible shapes. The azimuthal is given as L= n-1. "n" is the principal quantum number which is 7. Therefore L is 6 and the magnetic quantum numbers are -6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6

The position of the electron is given by the principal quantum number which represents the main energy level in which the orbital is located or the average distance from the nucleus.

Got it from quasarJose

Hope it helps

5 0

1 year ago

Identify the following redox reactions by type. Check all that apply. (a) Fe + H2SO4 → FeSO4 + H2 combination decomposition disp

EastWind [94]

Answer :

(a) displacement reaction

(b) combination reaction

(c) disproportionation reaction

(d) displacement reaction

Explanation :

(a) The given balanced chemical reaction is,

$Fe+H_2SO_4\rightarrow FeSO_4+H_2$

This reaction is a single replacement reaction or displacement in which the the more reactive element (Fe) replace the less reactive element (H).

(b) The given balanced chemical reaction is,

$S+3F_2\rightarrow SF_6$

This reaction is a combination reaction in which the two reactants molecule combine to form a large molecule or single product.

(c) The given balanced chemical reaction is,

$2CuCl_2\rightarrow Cu+CuCl_2$

This reaction is a disproportionation reaction in which the chemical species gets oxidized and reduced simultaneously. It is also considered as a redox reaction.

(d) The given balanced chemical reaction is,

$2Ag+PtCl_2\rightarrow 2AgCl+Pt$

This reaction is a single replacement reaction or displacement in which the the more reactive element (Ag) replace the less reactive element (Pt).

7 0

2 years ago

Larisa pumps up a soccer ball until it has a gauge pressure of 61 kilopascals. The volume of the ball is 5.2 liters. The air tem

Nuetrik [128]

<h3>Answer:</h3>

B. 0.33 mol

<h3>Explanation:</h3>

We are given;

Gauge pressure, P = 61 kPa (but 1 atm = 101.325 kPa)

= 0.602 atm

Volume, V = 5.2 liters

Temperature, T = 32°C, but K = °C + 273.15

thus, T = 305.15 K

We are required to determine the number of moles of air.

We are going to use the concept of ideal gas equation.

According to the ideal gas equation, PV = nRT, where P is the pressure, V is the volume, R is the ideal gas constant, (0.082057 L.atm mol.K, n is the number of moles and T is the absolute temperature.

Therefore, to find the number of moles we replace the variables in the equation.
Note that the total ball pressure will be given by the sum of atmospheric pressure and the gauge
Therefore;
Total pressure = Atmospheric pressure + Gauge pressure

We know atmospheric pressure is 101.325 kPa or 1 atm

Total ball pressure = 1 atm + 0.602 atm

= 1.602 atm

That is;

PV = nRT

n = PV ÷ RT

therefore;

n = (1.602 atm× 5.2 L) ÷ (0.082057 × 305.15 K)

= 0.3326 moles

= 0.33 moles

Therefore, there are 0.33 moles of air in the ball.

4 0

2 years ago

Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 <em> initial partial pressure</em>

-x -x +x +x <em> react</em>

105.1-x 7.96-x x x <em>equilibrium</em>

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0

2 years ago

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