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vekshin1
3 years ago
7

Predict the two most likely mechanisms for the reaction of 1-iodohexane with potassium tert-butoxide.

Chemistry
1 answer:
Rudiy273 years ago
6 0

Answer:

C) SN2 and E2

Explanation:

For this question, we have analyzed the <u>substrate</u> and the <u>base/nucleophile</u>. The substrate, in this case, is 1-iodohexane and the base/nucleophile is potassium tert-butoxide.

<u>Substrate</u>

<u />

In the 1-iodohexane the iodide "I" is bonded to a primary carbon (carbon 1). Therefore we will have a <u>primary substrate</u>. If we have a primary substrate an Sn1 can not take place. We can not have a <u>primary carbocation</u> due to this instability. So, we can disccard options A) and B).

<u>Base/nucleophile</u>

<u />

In the potassium tert-butoxide we have an ionic compound. A positive charge is placed in the potassium atom a negative charge is placed in the oxygen of the ter-butoxide ion. So, we will have a <u>strong base</u> (a molecule with the ability to remove electrons) and a <u>strong nucleophile</u> (a molecule with ability to bond with an electrophile). With all this in mind, w<u>e can not have an E1 reaction</u>.

With both analyses, the answer is C).

See figure 1

I hope it helps!

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Given that the molar mass of NaCl is 58.44 g/mol, what is the molarity of a solution that contains 87.75 g of NaCl in 500. mL of
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<u>Given information:</u>

Mass of NaCl (m) = 87.75 g

Volume of solution (V) = 500 ml = 0.5 L

Molar mass of NaCl (M) = 58.44 g/mol

<u>To determine:</u>

The molarity of NaCl solution

<u>Explanation:</u>

Molarity is defined as the number of moles of solute(n) dissolved per liter of solution (V)

i.e. M = moles of solute/liters of solution = n/V

Moles of solute (n) = mass of solute (m)/molar mass (M)

moles of NaCl = 87.75 g/58.55 g.mol-1 = 1.499 moles

Therefore,

Molarity of NaCl = 1.499 moles/0.5 L = 2.998 moles/lit ≅ 3 M

<u>Ans: (D)</u>

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Which statements regarding the Henderson-Hasselbalch equation are true? If the pH of the solution is known as is the pKa for the
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Answer:

1, 2, and 3 are true.

Explanation:

The Henderson-Hasselbalch equation is:

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

  • If the pH of the solution is known as is the pKa for the acid, the ratio of conjugate base to acid can be determined. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If you know pH and pka:

10^(pH-pka) = \frac{[A^-]}{[HA]}

The ratio will be: 10^(pH-pka)

  • At pH = pKa for an acid, [conjugate base] = [acid] in solution. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

0 = log₁₀ \frac{[A^-]}{[HA]}

10^0 = \frac{[A^-]}{[HA]}

1 = \frac{[A^-]}{[HA]}

As ratio is 1,  [conjugate base] = [acid] in solution.

  • At pH >> pKa for an acid, the acid will be mostly ionized. <em>TRUE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH >> pKa,  10^(pH-pka) will be >> 1, that means that you have more [A⁻] than [HA]

  • At pH << pKa for an acid, the acid will be mostly ionized. <em>FALSE</em>

pH = pka + log₁₀ \frac{[A^-]}{[HA]}

If pH << pKa,  10^(pH-pka) will be << 1, that means that you have more [HA] than [A⁻]

I hope it helps!

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