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10 days ago

Mechanism of 1-iodobutane reacts with pyridine

Chemistry

1 answer:

Anarel [2.5K]10 days ago

8 0

This is indicative of an elimination reaction via the E2 mechanism. The reaction between 1-iodobutane and pyridine exemplifies an E2 (bimolecular elimination) process, where pyridine serves primarily as a base and the alkyl halide is a primary one, both factors enhancing the E2 mechanism. In this single-step reaction, H and Cl are removed to yield 1-butene.

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Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in

eduard [2509]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, $T_{1} = 100^{o}C$ = (100 + 273) K = 373 K

Starting temperature of block 2, $T_{2} = 0^{o}C$ = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

$mC \Delta T = mC \times \Delta T$

$10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C$

$T_{f} - 100^{o}C = 0^{o}C - T_{f}$

$2T_{f} = 100^{o}C$

$T_{f} = 50^{o}C$

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

$\Delta H = mC \Delta T$

= $10000 g \times 0.385 J/K g \times 323 K$

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

$\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{373}$

= $10000 g \times 0.385 J/K g \times -0.143$

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

$\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}$

= $10000 g \times 0.385 J/K g \times ln \frac{323}{273}$

= $10000 g \times 0.385 J/K g \times 0.168$

= 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

= -554.12 J/K + 647.49 J/K $\Delta S_{total} = \Delta S_{1} + \Delta S_{2}$

= 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and $\Delta S_{total}$ is 93.37 J/K.

6 0

11 days ago

A compound composed of only carbon and chlorine is 85.5% chlorine by mass. propose a lewis structure for the lightest of the pos

lions [2633]

The visual representation is displayed in the following image.

For calculations, consider 100 grams of the compound:

ω(Cl) = 85.5% ÷ 100%.

ω(Cl) = 0.855; signifying the mass percentage of chlorine in the compound.

m(Cl) = 0.855 · 100 g.

m(Cl) = 85.5 g; this represents the mass of chlorine.

m(C) = 100 g - 85.5 g.

m(C) = 14.5 g; indicating the mass of carbon.

n(Cl) = m(Cl) ÷ M(Cl).

n(Cl) = 85.5 g ÷ 35.45 g/mol.

n(Cl) = 2.41 mol; this is the quantity of chlorine.

n(C) = 14.5 g ÷ 12 g/mol.

n(C) = 1.21 mol; this is the quantity of carbon.

n(Cl): n(C) = 2.41 mol: 1.21 mol = 2: 1.

The compound in question is identified as dichlorocarbene CCl₂.

4 0

1 month ago

Read 2 more answers

A sample of helium gas has a volume of 0.180L, a pressure of 0.800a and a temperature of 29 C. What is the new temp or gas of vo

Anarel [2593]

Solution:

The gas's new temperature is 604K

Justification:

Assuming standard temperature and pressure, we can determine the gas's temperature using the ideal gas law;

Step 1: Formulate the general gas law equation

P1V1/T1 = P2V2/T2

Step 2: Insert the values, converting as needed to standard units.

P1 = 0.800 atm

V1 = 0.180 L

T1 = 29°C = 273 + 29 = 302K

P2 = 3.20 atm

V2 = 90 mL = 90 * 10^-3 L = 0.09 L

Step 3: Solve for T2

The new gas temperature T2 is calculated as:

T2 = P2V2T1/(P1V1)

T2 = 3.20 * 0.09 * 302 / (0.800 * 0.180)

T2 = 86.976 / 0.144

T2 = 604K

The gas's new temperature is 604K.

7 0

16 days ago

How many micrograms of iron were in the 8.0 mL sample of Greg's blood?

castortr0y [2720]

The result is: 3.36 micrograms of iron in<span> Greg's blood sample.
</span>m(Fe) = 42 mcg(micrograms).
V(Fe) = 1 dL = 1 dL · 100 mL/1dL.
V(Fe) = 100 mL.
Using proportions: m(Fe): 8 mL = 42 mcg: 100 mL.
Thus, 100 mL · m(Fe) = 8 mL · 42 mcg.
m(Fe) = 336 mL·mcg ÷ 100 mL.
m(Fe) = 3.36 mcg.

4 0

27 days ago

An oxygen atom has a mass of and a glass of water has a mass of . Use this information to answer the question below. Be sure you

VMariaS [2667]

Answer:

Number of moles of oxygen atoms that weigh the same as a glass of water = 3.12 moles

Note: This question lacks certain figures. Below is a complete similar question.

An oxygen atom weighs 2.66*10^-23 g and a glass of water weighs 0.050 kg. What is the weight of one mole of oxygen atoms? Round your result to three significant figures. How many moles of oxygen atoms have a weight equal to the weight of a glass of water? Round your answer to two significant figures.

Explanation:

One mole of a substance comprises the Avogadro number of particles, which is 6.02 * 10²³.

Hence, one mole of oxygen atoms contains 6.02*10²³ atoms.

Weight of a single oxygen atom = 2.66*10⁻²³ g

Weight of one mole of oxygen atoms = weight of a single atom multiplied by the number of atoms in one mole.

Weight of one mole of oxygen atoms = 2.66*10⁻²³ g * 6.02*10²³ = 16.01 g

A glass of water weighs = 0.050 kg or 50 g.

To calculate how many moles of oxygen atoms weigh the same as a glass of water (i.e., 50 g), the following formula is applied;

number of moles = mass/molar mass

mass of oxygen atoms = 50 g, molar mass or weight of one mole of oxygen atoms = 16.01 g

Thus, the number of moles of oxygen atoms = 50 g / 16.01 g = 3.12 moles

4 0

1 month ago