Which one of the following compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide?
a. butanal
b. 2-methylbutanal
c. 3-methylpentanal
d. 2, 2-dimethylbutanal

Answers

Answer 1
It has to be b. 2-methylbutanal
Answer 2

The compound does not undergo an aldol addition reaction in presence of aqueous sodium hydroxide is 2, 2-dimethylbutanal.

What is aldol reaction?

The Aldol Reaction occurs when the enolate of an aldehyde or ketone combines with the carbonyl of another molecule at the aplha-carbon under basic or acidic circumstances to produce beta-hydroxy aldehyde or ketone.

For the formation of enolate ion, compound should contain alpha hydrogen in it and among the given compound only 2, 2-dimethylbutanal will not have alpha hydrogens.

Butanal, 2-methylbutanal and 3-methylpentanal will have aplha hydrogens in it so that it takes part in the aldol reaction.

Hence 2, 2-dimethylbutanal does not undergo an aldol addition reaction.

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Related Questions

How many grams of sodium chloride are required to make 2.00 L of a solution with a concentration of 0.100 M?

Answers

Answer:

Mass = 11.688g

Explanation:

Volume = 2.00L

Molar concentration = 0.100M

Mass = ?

These quantities are relatted by the following equation;

Conc = Number of moles / volume

Number of moles = Conc * Volume = 2 * 0.100 = 0.2 mol

Number of moles = Mass / Molar mass

Mass = Number of moles * Molar mass

Mass = 0.2mol * 58.44g/mol

Mass = 11.688g

13. Arrange each group of units from smallest to
largest
a)km, mm, cm, m
b) mg, kg, g
C) L, mL
d) s, ms, min, h

Answers

Answer:

A. mm,cm,m,Km

B. mg, g, Kg

C. mL,L

D. ms, s, min, h

Explanation:

suppose you make lemonade with one can lemonade concentrate mixed with four cans of water. What is the fraction of the final product that is water

Answers

Answer:

0.8 part of the product is water

Explanation:

Volume (or parts) of water = 4

Volume (or parts) of lemonade = 1

Total volume = 4 + 1 = 5

Fraction of water = Volume of water / Total volume = 4 / 5 = 0.8

If there are a 1000 ml per 1 L and a 1000g per kilogram
a. How many ml are there in 5.0 L?
b. How many kg are there in 230g?

Answers

Answer:

hbchbjH j jas a aa  a s ds d as das

Explanation:

If an individual proton has mass 1.007825 amu, and an individual neutron has mass 1.008665 amu, what's the calculated mass of a neptunium-236 nucleus? options: A) 237.92482 amu B) 236.99873 amu C) 237.96682 amu D) 237.04817 amu

Answers

Answer:

C) 237.96682 amu

Explanation:

The symbol for neptunium-236 is given as;

²³⁶₉₃Np

This element has 93 protons and (236 - 93 = 143) neutrons.

Mass Number =Total mass of Protons + Total mass of neutrons

Total Mass pf protons = 93 * 1.007825 amu, = 93.727725 amu

Total mass of Neutrons = 143 * 1.008665 amu = 144.239095 amu

Mass = 144.239095 + 93.727725  = 237.96682 amu

Correct option is option C.

what is the oxidation state of the oxygen atoms in co2,h2o and o2 and what does this information tell you about photosynthesis and respiration

Answers

Answer:

-2, -2 and 0.

- Respiration is a process in which energy is produced and photosynthesis is a process in which energy is used.

Explanation:

Hello,

In this case, oxygen is a substance that is used for animals and us to acquire the energy necessary for several functions by the cellular respiration (we also need glucose), besides, it is a product of the photosynthesis carried out by vegetable cells (plants). Moreover, carbon dioxide and water are used by the plants to produce oxygen we need as well as glucose via the aforementioned photosynthesis, thus, both chemical reactions are shown below:

[tex]C_6H_{12}O_6+6O_2\rightarrow 6CO_2+6H_2O\ \ \ respiration\\\\6CO_2+6H_2O\rightarrow C_6H_{12}O_6+6O_2\ \ \ photosynthesis[/tex]

In such a way, since the oxygen in carbon dioxide and water has an oxidation state of -2 (reduced form) we can say that the respiration is a process in which energy is produced and since the oxygen yielded during the photosynthesis has an oxydation state of 0, we can say that photosynthesis is a process in which energy is used.

Best regards.

Determine the molarity for each of the following solutions A. 1.8×10^4mg of HCL in 0.075L of solutions

Answers

The answer would have to be a

If 11.2 g of naphthalene, C10H8, is dissolved in 107.8 g of chloroform, CHCl3, what is the molality of the solution

Answers

Answer:

CHC12

Explanation:

i am not really sure i am onna do a quick research 4 u tho

What is the concentration in ppm of 4 g of NaCl dissolved in 100 mL of water?

Answers

1mL= 1g

That means that 100mL= 100g

Solute: 4g (smaller) + Solvent: 100g (larger) = Solution : 104g

PPM= solute/solution * 106
PPM= 4/104 * 106
PPM= 4.0769ppm
PPM is approximately 4.08ppm

What happens when an atom of sulfur combines with two atoms of chlorine to produce SCl2?

Answers

Answer:

Each chlorine atom shares a pair of electrons with the sulfur atom. ... Each chlorine atom shares all its valence electrons with the sulfur atom

Why can long chain fatty acids can form micelles in solutions with pH > 7 but are insoluble in pH < 5

Answers

Answer:

In basic conditions that is ( pH > 7 ), the equilibrium shifts towards right and produces a lot of (-ve) negatively charged fatty acids which are polar, In water, since they have long hydrophobic hydrocarbon part, this form micelles where the hydrocarbon part  remain inside the  sphere and -coo- group remain outside the sphere due to H-bonding interaction with water.

At ( pH < 5 ) I.e acidic conditions, the equilibrium shift to the left giving neutral molecules which can not have stronger H-bonding interaction .

So micelles cant form as they become insoluble.

What is the value of the equilibrium constant, K, for a reaction for which ∆G° is equal to –5.20 kJ at 50°C?

Answers

Answer:

6.93

Explanation:

Step 1: Given data

Standard Gibbs free energy (∆G°): -5.20 kJTemperature (T): 50°CEquilibrium constant (K): ?

Step 2: Convert the temperature to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 50°C + 273.15

K = 323 K

Step 3: Calculate K

We will use the following expression.

∆G° = -R × T × ln K

-5.20 × 10³ J = -(8.314 J/mol.K) × 323 K × ln K

K = 6.93

Compound has a molar mass of and the following composition: elementmass % carbon47.09% hydrogen6.59% chlorine46.33% Write the molecular formula of .

Answers

The given question is incomplete. The complete question is:

Compound X has a molar mass of 153.05 g/mol and the following composition:

element mass %

carbon 47.09%

hydrogen 6.59%

chlorine 46.33%

Write the molecular formula of X.

Answer: The molecular formula of X is [tex]C_6H_{10}Cl_2[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 47.09 g

Mass of H = 6.59 g

Mass of Cl = 46.33 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{47.09g}{12g/mole}=3.92moles[/tex]

Moles of H =[tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.59g}{1g/mole}=6.59moles[/tex]

Moles of Cl =[tex]\frac{\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac{46.33g}{35.5g/mole}=1.30moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = [tex]\frac{3.92}{1.30}=3[/tex]

For H = [tex]\frac{6.59}{1.30}=5[/tex]

For Cl =[tex]\frac{1.30}{1.30}=1[/tex]

The ratio of C : H: Cl= 3: 5 :1

Hence the empirical formula is [tex]C_3H_5Cl[/tex]

The empirical weight of [tex]C_3H_5Cl[/tex] = 3(12)+5(1)+1(35.5)= 76.5g.

The molecular weight = 153.05 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{153.05}{76.5}=2[/tex]

The molecular formula will be=[tex]2\times C_3H_5Cl=C_6H_{10}Cl_2[/tex]

For the following reaction, 61.6 grams of bromine are allowed to react with 25.5 grams of chlorine gas. bromine (g) + chlorine (g) bromine monochloride (g) What is the maximum amount of bromine monochloride that can be formed? grams

Answers

Answer:

I need great answers

Explanation:

please rate my answer as great

a. Name a chemical or product that was once considered safe but is now considered
harmful. (1 point)
-

Answers

Answer:

Bisphenol A (BPA)

Explanation:

Bisphenol A (BPA) is a chemical additive commonly found in resins and plastics, such as water bottles or food containers. It can also be found in household electronics, medical devices, dental fillings and sales receipts, just to name a few other applications.

Draw two constitutional isomers that share the molecular formula C3H8S. Your structures will have the same molecular formula but will have different connectivities.

Answers

Answer:

Two constitutional isomers for the compound C3H8S are shown in the attachment below

Explanation:

Constitutional isomers are isomers that have the same molecular formula but different connectivity. Two constitutional isomers for the compound C3H8S are shown in the attachment below.

As per the concept of structural isomers, the  two constitutional isomers that share the molecular formula C₃H₈S are attached in attachment below.

Structural isomers are defined as the isomers  in which atoms are completely  arranged  in a different order but the molecular formula remains the same.

They are the molecules which have same molecular formula but different  connectivities  of atoms  which depend on the order they are put together. An increase in the number of carbon atoms leads to an increase  in the structural isomers.

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The concentration of glucose, C6H12O6, in normal spinal fluid is 75 mg/100g. What is the molality of the solution

Answers

Answer:

4.16x10⁻³m

Explanation:

Molality is defined as the ratio between moles of a solute, in this case glucose, and kg of solvent.

As there are 100g of solvent, the kg are 0.1. Thus, we only need to calculate from the mass of glucose its moles to solve the molality of the solution.

Moles glucose:

There are 75mg = 0.075g of glucose. To conver mass to moles it is necessary molar mass.

Molar mass glucose:

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

72.06 + 12.10 + 96 = 180.16g/mol

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Molality of the solution:

4.16x10⁻⁴ moles of glucose / 0.1kg of solvent =

4.16x10⁻³m

The molarity of the solution is 4.16x10⁻³m

Calculation of the molarity:

We know that the molarity refers to the ratio that arise between the moles of a solute.

Since there are 100 g of solvent so here the kg should be 0.1.

Likewise there is 75 mg so it should be 0.075g

Now the Molar mass glucose should be

6C = 12.01g/mol*6 = 72.06g/mol

12H = 12*1.008g/mol = 12.10g/mol

6O = 6*16g/mol = 96g/mol

So,

= 72.06 + 12.10 + 96

= 180.16g/mol

Now

Moles of 0.075g of glucose:

0.075g * (1 mol / 180.16g) =

4.16x10⁻⁴ moles of glucose

Now finally

Molality of the solution:

= 4.16x10⁻⁴ moles of glucose / 0.1kg of solvent

=4.16x10⁻³m

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How many molecules are there in 3.5 moles of carbon dioxide? A. 63.21 x 10^23 B. 21.07 x 10^23 C. 42.14 x 10^23 D. 6.02 x 10^23

Answers

Answer:

B. 21.07 x 10²³ molecules

Explanation:

Avogadro's Number: 6.022 x 10²³

Step 1: Set up equation

[tex]3.5 mols CO_2(\frac{6.022(10^{23}) moleculesCO_2}{1 mol CO_2})[/tex]

Step 2: Multiply and cancel out units

3.5(6.022 x 10²³) = 21.07 x 10²³ molecules CO₂

Step 3: Convert to proper scientific notation

≈ 2.11 x 10²³ molecules CO₂

The pH of an acid solution is 5.82. Calculate the Ka for the monoprotic acid. The initial acid concentration is 0.010 M.

Answers

Answer:

The answer is

[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]

Explanation:

The Ka of an acid when given the pH and concentration can be found by

[tex]pH = - \frac{1}{2} log(Ka) - \frac{1}{2} log(c) [/tex]

where

c is the concentration of the acid

From the question

pH = 5.82

c = 0.010 M

Substitute the values into the above formula and solve for Ka

We have

[tex]5.82 = - \frac{1}{2} log(Ka) - \frac{1}{2} log(0.010) [/tex][tex] - \frac{1}{2} log(Ka) = 5.82 + 1[/tex][tex] - \frac{1}{2} log(Ka) = 6.82[/tex]

Multiply through by - 2

[tex] log(Ka) = - 13.64[/tex]

Find antilog of both sides

We have the final answer as

[tex]Ka = 2.29 \times {10}^{ - 14} moldm^{ - 3} [/tex]

Hope this helps you

A solution of LiCl in water has XLiCl = 0.0800. What is the molality? A solution of LiCl in water has XLiCl = 0.0800. What is the molality? 4.44 m LiCl 8.70 m LiCl 4.83 m LiCl 4.01 m LiCl

Answers

Answer:

mol LiCl = 4.83 m

Explanation:

GIven:

Solution of LiCl in water XLiCl = 0.0800

Mol of water in kg = 55.55 mole

Find:

Molality

Computation:

mole fraction = mol LiCl / (mol water + mol LiCl)

0.0800 = mol LiCl / (55.55 mol + mol LiCl)

0.0800 mol LiCl + 4.444 mol = mol LiCl

mol LiCl - 0.0800 mol LiCl = 4.444 mol

0.92 mol LiCl = 4.444 mol

mol LiCl = 4.83 m

Amino acids can be synthesized by reductive amination. Draw the structure of the organic compound that you would use to synthesize serine.

Answers

Answer:

Following are the answer to this question:

Explanation:

For the reductive amination of its carbonyl group, amino acids could be synthesized by reducing ammunition, which can be synthesized in the given attachment file:

please find the attachment:

1.78 L of gas is at a pressure of 735 torr. What is the volume in liters (L) when the pressure decreases to 700.0 torr

Answers

Explanation:

using boyles law

p1v1=p2v2

735 x 1.76 = 700 x V2

1293.6 = 700 x V2

V2 = 1293.6/700

V2 = 1.85L

It would require ? Liters of water to dissolve 36 grams of the substance.

Answers

The correct answer is 3 liters

Explanation:

If a substance has a solubility of [tex]12 \frac{grams}{liter}[/tex], this means in 1 liter, the grams that can be dissolved are 12 grams. Now, considering Justin and Ellie need to dissolve 36 grams to calculate the number of liters just divide the total of grams into 12 as each liter dissolves only 12 grams. The process is shown below:

36 grams (the amount that will be dissolved) ÷ 12 (grames dissolved per liter) = 3 liters (liters to dissolved 36 grams)

Answer:

It would be 3 liters

Explanation:

g If the titration of a 10.0-mL sample of sulfuric acid requires 28.15 mL of 0.100 M sodium hydroxide, what is the molarity of the acid

Answers

Answer:

[tex]M_{acid}=0.141M[/tex]

Explanation:

Hello,

In this case, the reaction between sulfuric acid and hydroxide is:

[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]

We can notice a 1:2 molar ratio between the acid and the base respectively, therefore, at the equivalence point we have:

[tex]2*n_{acid}=n_{base}[/tex]

And in terms of volumes and concentrations:

[tex]2*M_{acid}V_{acid}=M_{base}V_{base}[/tex]

So we compute the molarity of sulfuric acid as shown below:

[tex]M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}} =\frac{0.100M*28.15mL}{2*10.0mL}\\ \\M_{acid}=0.141M[/tex]

Best regards.

Advantages of using a resource person in handling the first aid lesson

Answers

The advantage of a resource person would be that it will provide a hands-on activity that will allow the students to experience spacing between organs and on the body of the person.

It will also allow them to identify challenges when doing this and will engage them more in the activity and lesson.

Answer:A resource person add knowledge to the course

Explanation:

GIVING 100 POINTS PLEASE ANSWER SOON When ponds freeze, they freeze at the surface first before freezing at the bottom. Which property of water explains why this might happen? Density Adhesion High boiling point High specific heat

Answers

Answer:

[tex]\huge\boxed{Density}[/tex]

Explanation:

This is because of density. Since Ice is less dense than water, the ice even formed inside the pond then starts floating on the surface of water. Because of this density, Ice floats on water and thus the water freezes at the surface.

Answer:

Density

Explanation:

Water is less dense as a solid than as a liquid, so as ponds freeze, the ice floats at the top and the pond freezes from the top-down.

Click on ALL of the following that use sound waves to communicate with their surroundings and find their way! bats cars dolphins whales birds submarines buses school

Answers

Bats, dolphins, whales, ( some ) birds and submarines all use echolocation ( or sound waves ) to navigate through their surroundings.!

When 5.58g H2 react by the following balanced equation, 32.8g H2O are formed. What is the percent yield of the reaction? 2H2(g)+O2(g)⟶2H2O(l)
A) 11.7%
B) 17.0%
C) 38.9%
D) 65.7%

Answers

Answer:

D) 65.7%

Explanation:

Based on the reaction:

2H2(g)+O2(g)⟶2H2O(l)

2 moles of hydrogen produce 2 moles of water assuming an excess of oxygen.

To find percent yield of the reaction we need to find theoretical yield (The yield assuming all hydrogen reacts producing water). With theoretical yield and actual yield (32.8g H₂O) we can determine percent yield as 100 times the ratio between actual yield and theoretical yield.

Theoretical yield:

Moles of 5.58g H₂:

5.58g H₂ ₓ (1 mol / 2.016g) = 2.768 moles H₂

As 2 moles of H₂ produce 2 moles of H₂O, if all hydrogen reacts will produce 2.768 moles H₂O. In grams:

2.768 moles H₂O ₓ (18.015g / mol) =

49.86g H₂O is theoretical yield

Percent yield:

Percent yield = Actual yield / Theoretical yield ₓ 100

32.8g H₂O / 49.86g ₓ 100 =

65.7% is percent yield of the reaction

D) 65.7%

g The electronic structure of which ONE of the following species cannot be adequately described by a single Lewis formula? (In other words, the electronic structure of which one can only be described by drawing two or more resonance structures?) A) C2H4 B) SO3 2– C) SO3 D) C3H8 E) HCN

Answers

Answer:

C) SO3

Explanation:

Lewis formula shows the bonding between atoms of a molecule and expresses the lone pair present in the atoms.

SO3 or Sulfur trioxide cannot be adequately described by a single Lewis formula because it has majorly 3 resonance structures because Sulfur does not follow the octet rule and can expand electrons in its outer shell.

Hence, the correct answer is C) SO3

When the equation MnO₄⁻ + I⁻ + H₂O → MnO₂ + IO₃⁻ is balanced in basic solution, what is the smallest whole-number coefficient for OH⁻?

Answers

Answer:

The smallest whole-number coefficient for OH⁻ is 2

Explanation:

Step 1: The equation redox reaction is divided into two half equations

Reduction half equation: MnO₄⁻  ----> MnO₂

Oxidation half-equation: I⁻  ---> IO₃⁻

Step 2: Next the atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation;

MnO₄⁻ +  2H₂O ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O

Step 3 : The charges are then balanced by adding electrons to the appropriate sides of each half equation

MnO₄⁻ +  2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

Step 4: Oxidation half equation is multiplied by 2 while reduction half equation is multiplied by 1 to balance the number of electrons gained and lost for the reaction

2MnO₄⁻ +  4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

Step 5 : addition of the two half equations to yield a net ionic equation

2MnO₄⁻  + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻

The smallest whole number coefficient for OH⁻ is 2

A redox reaction is divided into two half equations which are shown below:

Reduction half equation: MnO₄⁻  ----> MnO₂

Oxidation half-equation: I⁻  ---> IO₃⁻

Atoms are balanced by adding OH⁻ ions and H₂O molecules to the appropriate side of each half equation to make the equation complete ;

MnO₄⁻ +  2H₂O ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O

The charges needs to be balanced and this is done by adding electrons to the appropriate sides of each half equation

MnO₄⁻ +  2H₂O + 3e⁻ ----> MnO₂ + 4OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

The equation needs to be balanced by multiplying the oxidation half equation by 2 while reduction half equation is multiplied by 1 to balance the number of electrons on both sides of the equations.

2MnO₄⁻ +  4H₂O + 6e⁻ ----> 2MnO₂ + 8OH⁻

I⁻ + 6OH⁻ ---> IO₃⁻ + 3H₂O + 6e⁻

The two half equations are then added and written together to form a net ionic equation

2MnO₄⁻  + I⁻ + H₂O ----> 2MnO₂ + IO₃⁻ + 2OH⁻

The smallest whole-number coefficient for OH⁻ is therefore 2.

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