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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.

To learn more about alkene ,

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3 years ago

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The compound Xe(CF3)2 decomposes in afirst-order reaction to elemental Xe with a half-life of 30. min.If you place 7.50 mg of Xe

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Answer:

$t=147.24\ min$

Explanation:

Given that:

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Where, k is rate constant

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Using integrated rate law for first order kinetics as:

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Where,

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Given that:

The rate constant, k = 0.0231 min⁻¹

Initial concentration $[A_0]$ = 7.50 mg

Final concentration $[A_t]$ = 0.25 mg

Time = ?

Applying in the above equation, we get that:-

$0.25=7.50e^{-0.0231\times t}$

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3 years ago

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How many moles of xenon gas, Xe, would occopy 37.8L at stp

Charra [1.4K]

Answer:

The number of moles of xenon are 1.69 mol.

Explanation:

Given data:

Number of moles of xenon = ?

Volume of gas = 37.8 L

Temperature = 273 K

Pressure = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

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R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Now we will put the values in formula.

1 atm × 37.8 L = n × 0.0821 atm.L/ mol.K ×273 K

37.8 atm.L = n × 22.413 atm.L/ mol.

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3 years ago