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Elden [556K]
3 years ago
11

Which of the following alkyl halides is most likely to undergo rearrangement in an SN1 reaction?

Chemistry
1 answer:
konstantin123 [22]3 years ago
5 0

Answer:

(b) 2-chloro-3,3-dimethylpentane

Explanation:

A rearrangement occurs in a SN1 reaction when the carbocation is formed. More substituted carbocations are more stable because of the electron-donating effect of alkyl groups and the hyperconjugation.

An alkyl halide with a carbon more substituted than the carbon where is the C-X bond, is most likely to undergo rearrangement.

(a) 1-bromo-4-methylcyclohexane

In this molecule could occur a rearrangement but the carbocation need to jump three bonds.

(b) 2-chloro-3,3-dimethylpentane

This alkyl halide is most likely to occur a rearrangement because the near carbon is a quaternary carbon.

(c) 3-bromopentane

Can't occur a rearrangement because all carbons are secondary

(d) bromocyclohexane

Also, can't occur a rearrangement because all carbons are secondary.

Thus, the most likely to undergo rearrangement in an SN1 reaction is (b) 2-chloro-3,3-dimethylpentane

I hope it helps!

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<u>Step 1:</u> Data given

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Mass of carbon dioxide = 211.4 kg

Mass of urea produced = 170.3 kg

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<u>Step 2:</u> The balanced equation

2NH3(aq) + CO2(aq) --> CH4N2O(aq) + H2O(l)

<u>Step 3:</u> Calculate moles of NH3

Number of moles = mass / Molar mass

Moles NH3 = 122500 grams / 17.031 g/mol

Moles NH3 = 7192.77 moles

<u>Step 4:</u> Calculate moles of CO2

Moles CO2 = 211400 / 44.01 g/mol

Moles CO2 = 4803.45 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

NH3 is the limiting reactant. It will completely be consumed (7192.77 moles).

CO2 is in excess. There will be consumed 7192.77/2 = 3596.4 moles

There will remain 4803.45 - 3596.4 = 1207.05 moles of CO2

<u>Step 6:</u> Calculate moles of urea produced:

For 2 moles NH3 consumed, we need 1 moles of CO2 to produce 1 mole urea and 1 mole H2O

For 7192.77 moles of NH3, we have 3596.4 moles of urea produced

<u>Step 7: </u>Calculate mass of urea

Mass urea = moles urea * molar mass urea

Mass urea = 3596.4 moles * 60.06 g/mol

Mass urea = 216000 grams = 216 kg = theoretical yield

<u>Step 8</u>: Calculate % yield

% yield = (actual yield / theoretical yield)*100%

% yield = (170.3 / 216) *100% = 78.8%

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