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alukav5142 [94]
3 years ago
12

Organic compounds undergo a variety of different reactions, including substitution, addition, elimination, and rearrangement. An

atom or a group of atoms in a molecule is replaced by another atom or a group of atoms in a substitution reaction. In an addition reaction, two molecules combine to yield a single molecule. Addition reactions occur at double or triple bonds. An elimination reaction can be thought of as the reverse of an addition reaction. It involves the removal of two atoms or groups from a molecule. A rearrangement reaction occurs when bonds in the molecule are broken and new bonds are formed, converting it to its isomer. Classify the following characteristics of the organic reactions according to the type of organic reaction.
a. Reactions involving the replacement of one atom or group of atoms.
b. Reactions involving removal of two atoms or groups from a molecule.
c. Products show increased bond order between two adjacent atoms.
d. Reactant requires presence of a π bond.
e. Product is the structural isomer of the reactant.

1. Substitution reaction
2. Addition reaction
3. Elimination reaction
4. Rearrangement reaction
Chemistry
1 answer:
riadik2000 [5.3K]3 years ago
5 0

Answer:

Reactions involving the replacement of one atom or group of atoms. - Substitution reaction

Reactions involving removal of two atoms or groups from a molecule - Elimination reaction

Products show increased bond order between two adjacent atoms - Elimination reaction

Reactant requires presence of a π bond - Addition reaction

Product is the structural isomer of the reactant - Rearrangement reaction

Explanation:

When an atom or a group of atoms is replaced by another in a reaction, then such is a substitution reaction. A typical example is the halogenation of alkanes.

A reaction involving the removal of two atoms or groups from a molecule resulting in increased bond order of products is called an elimination reaction. A typical example of such is dehydrohalogenation of alkyl halides.

Any reaction that involves a pi bond is an addition reaction because a molecule is added across the pi bond. A typical example is hydrogenation of alkenes.

Rearrangement reactions yield isomers of a molecule. Rearrangement may involve alkyl or hydride shifts in molecules.

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Choose the correct answer.
ahrayia [7]

Explanation:

Covalent compounds (solid, liquid, solution) do not conduct electricity. Metal elements and carbon (graphite) are conductors of electricity but non-metal elements are insulators of electricity. Ionic bonds are the electrostatic attraction between positive and negative ions.

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Brainliest plz

4 0
3 years ago
A 15.00-mL sample of an NaOH solution of unknown concentration requires 17.88 mL of a 0.1053 M H2SO4 M solution to reach the equ
Anastasy [175]

Answer:

The concentration of the NaOH solution CB = 0.251 M

Explanation:

The balanced equation of reaction is:

H2SO4 + 2NaOH ===> Na2SO4 + 2H2O

Using titration equation of formula

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 1 (from the balanced equation of reaction)

NB is the number of mole of base = 2 (from the balanced equation of reaction)

CA is the concentration of acid = 0.1053 M

CB is the concentration of base = to be calculated

VA is the volume of acid = 17.88 mL

VB is the volume of base = 15.00mL

Substituting

0.1053×17.88/CB×15 = 1/2

Therefore CB =0.1053×17.88×2/15×1

CB= 0.251 M

6 0
4 years ago
HELPPP!!! What is the volume of a 1.31 moles sample of gas if the pressure is 904 mmHg and the temperature is 37. degrees Celsiu
Tju [1.3M]

Answer:

27.99 dm³

Explanation:

Applying

PV = nRT................ Equation 1

Where P = Pressure, V = Volume, n = number of mole, R = molar gas constant, T = Temperature.

From the question, we were aksed to find V.

Therefore we make V the subject of the equation

V = nRT/P................ Equation 2

Given: n = 1.31 moles, T = 37°C = 310K, P = 904 mmHg = (904×0.001316) = 1.1897 atm

Constant: R = 0.082 atm.dm³/K.mol

Substitute these values into equation 2

V = (1.31×310×0.082)/(1.1897)

V = 27.99 dm³

4 0
3 years ago
A 4.0 g sample of iron was heated from 0°C to 20.°C. It absorbed 35.2 J of energy as heat. What is the specific heat of this pie
Crazy boy [7]

Answer:

Explanation:i joule is equal to 0.238902957619 calories so 1251 joules is equal to 298.87 calories divided by 25.0 degrees centigrade is equal to 11.95 calories divided by the 35.2 gram sample weight to get the calories per gram per degree centigrade would come to 0.3396 calories/gram degree centigrade. Presumably this, if correct, could be used to obtain the metal in question by consulting a chart or table with specific heats of various metals because they should always be the same specific heat for each metal.

5 0
3 years ago
Calculate the molarity of sodium ion in a solution made
Arada [10]

Answer:

0.1035 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Moles =Molarity \times {Volume\ of\ the\ solution}

Sodium chloride will furnish Sodium ions as:

NaCl\rightarrow Na^{+}+Cl^-

Given :

For Sodium chloride :

Molarity = 0.288 M

Volume = 3.58 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 3.58×10⁻³ L

Thus, moles of Sodium furnished by Sodium chloride is same the moles of Sodium chloride as shown below:

Moles =0.288 \times {3.58\times 10^{-3}}\ moles

Moles of sodium ions by sodium chloride = 0.00103104 moles

Sodium sulfate will furnish Sodium ions as:

Na_2SO_4\rightarrow 2Na^{+}+SO_4^{2-}

Given :

For Sodium sulfate :

Molarity = 0.001 M

Volume = 6.51 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.51 ×10⁻³ L

Thus, moles of Sodium furnished by Sodium sulfate is twice the moles of Sodium sulfate as shown below:

Moles =2\times 0.001 \times {6.51\times 10^{-3}}\ moles

Moles of sodium ions by Sodium sulfate = 0.00001302 moles

Total moles = 0.00103104 moles + 0.00001302 moles = 0.00104406 moles

Total volume = 3.58 ×10⁻³ L + 6.51 ×10⁻³ L = 10.09 ×10⁻³ L

Concentration of sodium ions is:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity_{Na^+}=\frac{0.00104406}{10.09\times 10^{-3}}

<u> The final concentration of sodium anion = 0.1035 M</u>

4 0
4 years ago
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