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Ber [7]
3 years ago
12

Comparing the 2-bromobutane + methoxide and 2-bromobutane + t-butoxide reactions, choose the statements that BEST describe the d

ata and mechanism. a. the mechanism for this reaction is E2 b. an increase in 1-butene was observed when t-butoxide was used c. an increase in 1-butene was observed when methoxide was used d. the mechanism for this reaction is E1 e. no significant difference was observed
Chemistry
1 answer:
Elina [12.6K]3 years ago
8 0

Answer:

an increase in 1-butene was observed when t-butoxide was used

Explanation:

When a base reacts with an alkyl halide, an elimination product is formed. This reaction is an E2 reaction.

Here we are to compare the reaction of two different bases with one substrate; 2-bromobutane. Both reactions occur by the E2 mechanism but follow different transition states due to the size of the base.

The Saytzeff product, 2-butene, is obtained when the methoxide is used while the non Saytzeff product, 1-butene, is obtained when t-butoxide is used.

The Saytzeff rule is reliable in predicting the major products of simple elimination reactions of alkyl halides given the fact that a small/strong bases is used for the elimination reaction. Therefore hydroxide, methoxide and ethoxide bases give similar results for the same alkyl halide substrate. Bulky bases such as tert-butoxide tend to yield a higher percentage of the non Saytzeff product and this is usually attributed to steric hindrance.

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In run 1, you mix 7.9 mL of the 43 g/L MO solution (MO molar mass is 327.33 g/mol), 3.13 mL of the 0.040 M SnCl2 in 2.0 M HCl so
blsea [12.9K]

Answer:

Concentration of H3O⁺  [H3O⁺] = 0.864 M

Explanation:

Given that:

The mass concentration of MO = 43 g/L

The volume of MO = 7.9 mL = 7.9 × 10⁻³ L

Recall that

The mass number of MO = Mass concentration of MO × Volume of MO

The mass number of MO = (43 g/L) * (7.9 × 10⁻³ L)

The mass number of MO =  0.3397 g

number of  moles of MO = (mass number of MO) / (molar mass of MO)

number of  moles of MO = (0.3397 g) / (327.33 g/mol)

moles of MO = 0.00104 mol

The total volume = 7.9 mL + 3.13 mL + 5.49 mL + 3.43 mL

The total volume = 19.95 mL = 19.95 × 10⁻³ L

Concentration of MO [MO} =(number of moles of MO) / (total volume)

[MO] = 0.00104 mol  /  19.95 × 10⁻³ L

[MO] = 5.2130 × 10⁻⁸ M

the number of moles of H3O⁺ = molarity of HCl in the solution × the volume of HCl in solution

the number of moles of H3O⁺ = [(2.0 M) * (3.13 mL)] + [(2.0 M) * (5.49 mL)]

the number of moles of H3O⁺ = 17.24 mmol

Concentration of H3O⁺  [H3O⁺] = (the number of moles of H3O⁺) / (total volume)

Concentration of H3O⁺  [H3O⁺] = (17.24 mmol) / (19.95 mL)

Concentration of H3O⁺  [H3O⁺] = 0.864 M

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