HURRY PLEASE HELP 25PTS

Predict the aldol product when the following ketone undergoes self-condensation in the presence of NaOH. Do not show the dehydra

Anit [1.1K]

Answer:

β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal.

Explanation:

When acetaldehyde is treated with dil.NaOH it undergoes self condensation as it contains alpha-hydrogen atom in its compound forming β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal. This compound upon further heating will eliminate a molecule of water forming aldol condensation product namely Crotonaldehyde Or But-2-en-al. see the diagram attached.

5 0

3 years ago

If the rate law for a reaction A → P is rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122 M, calcula

DedPeter [7]

Rate = 3.37x10-3 M^-1 min-1 [A]^2 and the initial concentration of a is 0.122M.

A rate law indicates the rate of a chemical response depends on reactant concentration. For a response inclusive of the price regulation commonly has the form rate = ok[A]ⁿ, in which okay is a proportionality constant known as the fee regular and n is the order.

The charge of a chemical response is, perhaps, its maximum crucial asset because it dictates whether or not a reaction can arise all throughout an entire life. knowing the charge regulation, an expression concerning the price to the concentrations of reactants can assist a chemist to modify the response conditions to get an extra suitable rate.

half-life is the time taken for the radioactivity of a substance to fall to 1/2 its authentic cost whereas implies existence is the common life of all the nuclei of a particular risky atomic species.

Learn more about rate law here:-brainly.com/question/7694417

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3 0

2 years ago

Write a net ionic equation to show how codeine, c18h21o3n, behaves as a base in water.

Lera25 [3.4K]

Actually, the ionic equation for this is a reversible equation since codeine is a weak base. Any weak base or weak acids do not completely dissociate which makes them a reversible process. The ionic equation for this case is:

<span>C18H21O3N + H3O+ </span><=> C18H21O3NH+ + H2O

7 0

3 years ago

A mysterious white powder could be powdered sugar (C12H22O11), cocaine (C17H21NO4), codeine (C18H21NO3), norfenefrine (C8H11NO2)

rodikova [14]

Norfenefrine (C₈H₁₁NO₂).

<h3>Further explanation</h3>

We will solve a case related to one of the colligative properties, namely freezing point depression.

The freezing point of the solution is the temperature at which the solution begins to freeze. The difference between the freezing point of the solvent and the freezing point of the solution is called freezing point depression.

$\boxed{ \ \Delta T_f = T_f(solvent) - T_f(solution) \ } \rightarrow \boxed{ \ \Delta T_f = K_f \times molality \ }$

<u>Given:</u>

A mysterious white powder could be,

powdered sugar (C₁₂H₂₂O₁₁) with a molar mass of 342.30 g/moles,
cocaine (C₁₇H₂₁NO₄) with a molar mass of 303.35 g/moles,
codeine (C₁₈H₂₁NO₃) with a molar mass of 299.36 g/moles,
norfenefrine (C₈H₁₁NO₂) with a molar mass of 153.18 g/moles, or
fructose (C₆H₁₂O₆) with a molar mass of 180.16 g/moles.

When 82 mg of the powder is dissolved in 1.50 mL of ethanol (density = 0.789 g/cm³, normal freezing point −114.6°C, Kf = 1.99°C/m), the freezing point is lowered to −115.5°C.

<u>Question: </u>What is the identity of the white powder?

<u>The Process:</u>

Let us identify the solute, the solvent, initial, and final temperatures.

The solute = the powder
The solvent = ethanol
The freezing point of the solvent = −114.6°C
The freezing point of the solution = −115.5°C

Prepare masses of solutes and solvents.

Mass of solute = 82 mg = 0.082 g
Mass of solvent = density x volume, i.e., $\boxed{ \ 0.789 \ \frac{g}{cm^3} \times 1.50 \ cm^3 = 1.1835 \ g = 0.00118 \ kg \ }$

We must prepare the solvent mass unit in kg because the unit of molality is the mole of the solute divided by the mass of the solvent in kg.

The molality formula is as follows:

$\boxed{ \ m = \frac{moles \ of \ solute}{kg \ of \ solvent} \ } \rightarrow \boxed{ \ m = \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

Now we combine it with the formula of freezing point depression.

$\boxed{ \ \Delta T_f = K_f \times \frac{mass \ of \ solute \ (g)}{molar \ mass \ of \ solute \times kg \ of \ solvent} \ }$

It is clear that we will determine the molar mass of the solute (denoted by Mr).

We enter all data into the formula.

$\boxed{ \ -114.6^0C - (-115.5^0C) = 1.99 \frac{^0C}{m} \times \frac{0.082 \ g}{Mr \times 0.00118 \ kg} \ }$

$\boxed{ \ 0.9 = \frac{1.99 \times 0.082}{Mr \times 0.00118} \ }$

$\boxed{ \ Mr = \frac{0.16318}{0.9 \times 0.00118} \ }$

We get $\boxed{ \ Mr = 153.65 \ }$

These results are very close to the molar mass of norfenefrine which is 153.18 g/mol. Thus the white powder is norfenefrine.

<h3>Learn more</h3>

The molality and mole fraction of water brainly.com/question/10861444
About the mass and density of ethylene glycol as an antifreeze brainly.com/question/4053884
About the solution as a homogeneous mixture brainly.com/question/637791

Keywords: a mysterious white powder, sugar, cocaine, codeine, norfenefrine, fructose, the solute, the solvent, dissolved, ethanol, normal freezing point, the freezing point depression, the identity

7 0

3 years ago

Read 2 more answers

In the periodic table how many elements could be classified as metals in group IV?

lord [1]

4 elements are classified in group 4 like titanium, zirconium, hafnium and ruthorfordium

8 0

3 years ago