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Lady bird [3.3K]
2 years ago
9

Compounds X has the formula C7H15Cl; Y is C7H15Br. X undergoes base-promoted E2 elimination to give a single alkene product Z. Y

likewise reacts under similar conditions to give a single alkene product that is isomeric with Z Catalytic hydrogenation of Z affords 3-ethylpentane. X readily reacts in SN2 fashion with sodium iodide in acetone. Y does not undergo a similar SN2 reaction. Propose structures for X and Y.
Chemistry
1 answer:
djverab [1.8K]2 years ago
3 0

Answer:

See explanation and image attached

Explanation:

Let us examine the statements in the question carefully. First of all, we will discover that the products of the E2 elimination of the both compounds are isomeric. However Y does not undergo SN2 reaction as  X does.

The fact that SN2 reaction does not occur in Y confirms that the bromine atom is attached to a tertiary carbon atom and SN2 reaction does not occur due to steric hinderance. Since X undergoes SN2 reaction in aprotic solvent, the chlorine atom must be attached to a secondary carbon atom.

However, E2 reactions does occur with tertiary alkyl halides when strong bases such as OH^- or RO^- are used.

The question also stated that the catalytic hydrogenation of Z affords 3-ethylpentane.

Putting all these together, the structures of X and Y have been suggested in the image attached to this answer.

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What is the half-life of an isotope that decays to 6.25% of its original activity in 18.9 hours?
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Radioactive material obeys 1st order decay kinetics,
For 1st order reaction, we have 
k = \frac{2.303}{t}Xlog \frac{\text{initial conc.}}{\text{final conc.}}
where, k = rate constant of reaction

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8 0
3 years ago
A diver has a lung capacity of 2.4 L when the pressure is 0.8 atm. What is the volume of the diver’s lungs when the pressure cha
tangare [24]

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The final volume is 1.6 L.

Explanation:

It is given that,

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It is based on Boyle's law. According to this law,

PV=K

K is constant

P_1V_1=P_2V_2

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7 0
3 years ago
How many grams of solute are needed to order to prepare 100.00 mL of a 0.1000 M solution of a compound with a molecular weight o
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The grams of solute are required.

The mass of solute is 3.5 g

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M = Molar mass = 350 g/mol

V = Volume of solution = 100 mL = 0.1 L

n = Number of moles

m = Mass of solute

Molarity is given by

c=\dfrac{n}{V}\\\Rightarrow n=cV\\\Rightarrow n=0.1\times 0.1\\\Rightarrow n=0.01\ \text{moles}

Molar mass is given by

M=\dfrac{m}{n}\\\Rightarrow m=Mn\\\Rightarrow m=350\times 0.01\\\Rightarrow m=3.5\ \text{g}

The mass of solute is 3.5 g

Learn more:

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7 0
2 years ago
Read 2 more answers
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