All categories

2 years ago

What are the conditions needed for reaction to happen in aldehydes and ketones?

Chemistry

1 answer:

miv72 [106K]2 years ago

7 0

Explanation:

You may not realise it, but you come across aldehydes and ketones many times a day. Take cakes and biscuits, for example. Their golden, caramelised crust is formed thanks to the Mailliard reaction. This is a process that occurs at temperatures above 140° C, when sugars with the carbonyl group in foods react with nucleophilic amino acids to create new and complex flavours and aromas.

Another example is formaldehyde. Correctly known as methanal, it is the most common aldehyde in industry. It has multiple uses, such as in tanning and embalming, or as a fungicide. However, we can also react it with different molecules to make a variety of more useful compounds. These include polymers, adhesives and precursors to explosives. But how do aldehydes and ketones react, and why?You should remember from Aldehydes and Ketones that they both contain the carbonyl functional group , . This is a carbon atom joined to an oxygen atom by a double bond. Let's take a closer look at it.

If we compare the electronegativities of carbon and oxygen, we can see that oxygen is a lot more electronegative than carbon.

You might be interested in

A runner is traveling around the track below at a constant speed. Is she accelerating?

Black_prince [1.1K]

<h2>ANSWER:</h2>

D.No, she is not accelerating because her speed is constant.

<h2>EXPLANATION:</h2>

The rate of change of velocity with respect to time is known as acceleration.

acceleration =change in velocity/time

As she is running at constant speed and there is no change in velocity so she is not accelerating.

4 0

3 years ago

Read 2 more answers

Determine the AMOUNT OF NO2, LIMITING REACTANT, AND THE AMOUNT AND NAME OF EXCESS REACTANT.

zepelin [54]

Answer:

balanced equation mole ratio 5 2 mol NO/1 mol O2

10.00 g O2 3 1 mol O2/32.00 g O2 5 0.3125 mol O2

20.00 g NO 3 1 mol NO/30.01 g NO 5 0.6664 mol NO

actual mole ratio 5 0.6664 mol NO/0.3125 mol O2 5 2.132 mol NO/1.000 mol O2

Because the actual mole ratio of NO:O2 is larger than the balanced equation mole

ratio of NO:O2, there is an excess of NO; O2 is the limiting reactant.

Mass of NO used 5 0.3125 mol O2 3 2 mol NO/1 mol O2 5 0.6250 mol NO

0.6250 mol NO 3 30.01 g NO/1 mol NO 5 18.76 g NO

Mass of NO2 produced 5 0.6250 mol NO2 3 46.01 g NO2/1 mol NO2 5 28.76 g NO2

Excess NO 5 20.00 g NO 2 18.76 g NO 5 1.24 g N

Explanation:

3 0

2 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

You are given an unknown gaseous binary compound (that is, a compound consisting of two different elements). when 10.0 g of the

Luda [366]

Answer is: a possible identity for the unknown compound is C₃H₈.

m(binary compound) = 10.0 g.

m(H₂O) = 16.3 g; mass of water.

M(O₂) = 32 g/mol; molar mass of oxygen.

M(binary compound) = 1.38 · 32 g/mol.

M(binary compound) = 44.16 g/mol.

n(binary compound) = 10 g ÷ 44.16 g/mol.

n(binary compound) = 0.225 mol; amount of substance.

n(H₂O) = 16.3 g ÷ 18 g/mol.

n(H₂O) = 0.9 mol; amount of water.

m(H₂O) : n(binary compound) = 0.9 mol ÷ 0.225 mol.

m(H₂O) : n(binary compound) = 4 : 1.

Unknown cpmpound has 4 times more hydrogen than water, it has 8 hydrogen atoms.

Second element in compound is carbon:

M(X) = 44.16 g/mol - 8 · 1.01 g/mol.

M(X) = 36.08 g/mol ÷ 3.

M(C) = 12.01 g/mol.

8 0

3 years ago

Describe where the "Reference Carbon" (the one that determines if the structure is D or L) is located in a carbohydrate with mor

tresset_1 [31]

Answer:

Farthest from the carbonyl carbon.

Explanation:

Reference carbon that determined the absolute D and L configuration is located farthest from the carbonyl carbon.

In other words, reference carbon is that assymentric carbon which is located farthest from the carbolyl carbon and has configuration similar to D- or L-glyceraldehyde isomers.

D and L configuration is decided by the direction of -OH group attached to the reference carbon.

In L-isomer, -OH group is attached to the left side of the reference carbon and in D-isomer, -OH group is attached to the right side of the reference carbon.

5 0

3 years ago