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2 years ago

Draw the alcohol that is the product of the reduction of 3-methylpentanal.

Chemistry

1 answer:

natali 33 [55]2 years ago

7 0

Answer: The product from the reduction reaction is

CH3-CH2-CH(CH3)-CH2-CH2OH

IUPAC name; 3- Methylpentan-1-ol

Explanation:

Since oxidation is simply the addition of oxygen to a compound and reduction is likewise the addition of hydrogen to a compound.

Therefore, hydrogen is added onto the carbon atom adjacent to oxygen in 3- methyl pentanal

CH3 CH2 CHCH3 CH2 CHO thereby -CHO( aldehyde functional group) are reduced to CH2OH ( Primary alcohol) which gives;

3-methylpenta-1-ol .

The structure of the product is:

CH3-CH2-CH(CH3)-CH2-CH2OH

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2 years ago

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Pierce conducts an experiment in which waves collide in a way that the energy increases, which means that has occurred

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It means the waves collides and constructive interference occurred.

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2 years ago

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You have to prepare some 2 M solutions, with 10 g of solute in each. What volume of solution will you prepare, for each solute b

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0.045 L or 45 mL

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Vol in L = Moles / Molarity

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1 year ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$

$E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol$

$E_a=22.87\ kJ/mol$

<u>The activation energy for this reaction = 23 kJ/mol.</u>

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2 years ago