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2 years ago

Provide only the major alkene product that results when n,n-dimethylhexan-2-amine undergoes cope elimination?

1 answer:

5 0

The major alkene product that results when n,n-dimethylhexan-2-amine undergoes cope elimination is hexene or hex-1-ene.

The reaction in which an amine is oxidize to an intermediate called an N-oxide which , when heated , acts as base in an intramolecular elimination reaction. The oxidation of tertiary amine into N-oxide is called cope reaction.

This elimination gives the less substituted alkene along with more substituted alkene which is Zaitsev product.

Example: Cope elimination of n,n-dimethylhexan-2-amine form hexene.

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A scientist measures the standard enthalpy change for the following reaction to be -17.2 kJ : Ca(OH)2(aq) 2 HCl(aq)CaCl2(s)

elixir [45]

Answer: $\Delta H^{0}=-173.72$ kJ/mol

Explanation: Enthalpy Change is the amount of energy in a reaction - absorption or release - at a constant pressure. So, Standard Enthalpy of Formation is how much energy is necessary to form a substance.

The standard enthalpy of formation of HCl is calculated as:

$\Delta ^{0}=\Sigma H_{products}-\Sigma H_{reactants}$

$Ca(OH)_{2}_{(aq)}+2HCl_{(aq)}$ → $CaCl_{2}_{(s)}+2H_{2}O_{(l)}$

Standard Enthalpy of formation for the other compounds are:

Calcium Hydroxide: $\Delta H^{0}=$ -1002.82 kJ/mol

Calcium chloride: $\Delta H^{0}=$ -795.8 kJ/mol

Water: $\Delta H^{0}=$ -285.83 kJ/mol

Enthalpy is given per mol, which means we have to multiply by the mols in the balanced equation.

Calculating:

$-17.2=[-795.8+2(285.85)]-[-1002.82+2\Delta H]$

$-17.2=-1367.46+1002.82-2\Delta H$

$2\Delta H=17.2-364.64$

$\Delta H=-173.72$

So, the standard enthalpy of formation of HCl is -173.72 kJ/mol

8 0

2 years ago

24 POINTS!!!!!!!!!!

Oliga [24]

Your answer is going to be A, because it was shoved harder, it will go faster

8 0

3 years ago

Read 2 more answers

Hexane and Butane are alkanes. Describe the structure of alkanes

dem82 [27]

Alkanes do not have a double bond, thus they are saturated. This makes them the simplest form of hydrocarbon, containing only hydrogen and carbon atoms. They are not very reactive as they don't have a double bond which can open up and join on to other molecules.

8 0

3 years ago

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Element abundant in the body are not abundant on Earth. True or False?

Lena [83]

False. The element abundant in our body is oxygen and on earth, it is still oxygen. The body is more than 50% water. A water molecule is 11% hydrogen and the rest is oxygen. That is why you can expect oxygen to be abundant in the body. Same is true with the earth. Majority of the oxygen is from the bodies of water the earth is rich of.

6 0

3 years ago

Calculate each of the following: (b) Mass % of Mn in potassium permanganate

Brilliant_brown [7]

Mass percent is the estimation of the mass of the element in a compound in percentage form. The mass % of manganese (Mn) in potassium permanganate is 34.76 %.

<h3>What is the mass percentage?</h3>

The mass percentage is defined by dividing the mass of the element by the molecular mass of the compound followed by multiplying it by 100%. It is given as,

Mass % = (Atomic Mass of element ÷ Molecular Mass) × 100

As it is known that,

Atomic Mass of manganese = 54.94 g/mol

Molecular Mass of potassium permanganate = 158.034 g/mol

Substituting values to calculate mass % as:

% Mn = (Atomic Mass of Mn ÷ Molecular Mass of KMnO₄) × 100

= (54.94 g/mol ÷ 158.034 g/mol) × 100

= 34.76 %

Therefore, 34.76% Mn is present in KMnO₄.

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3 0

2 years ago