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stepan [7]
3 years ago
15

Which ketone in each pair is more reactive?

Chemistry
1 answer:
KiRa [710]3 years ago
4 0

Answer:

a. 2-heptanone is more reactive than 4-heptanone

b. chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone

Explanation:

The reactivity of the carbonyl compound (ketone ) is affected by the steric effect. The steric effect is a hindrance that occurs in the structure or reactivity of a molecule, which is affected by the physical size and the proximity of the adjacent parts of the molecule.

Between 2-heptanone or 4-heptanone, 2-heptanone is more reactive than 4-heptanone. This is because 2-heptanone is less affected by the steric hindrance, unlike the 4-heptanone.

Similarly, the reactivity of the carbonyl compound (ketone) is also affected by the polarity on the carbon compound, which is associated with how electronegative the substituent attached is to the carbonyl compound. From the periodic table, the electronegativity of the Halogen family decreases down the group. Therefore chlorine is more electronegative than bromine.

As such, chloromethyl phenyl ketone is more reactive than bromomethyl phenyl ketone.

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3 years ago
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A student titrated 20 ml of 0.410 m hcl with 0.320 m naoh. determine the volume of naoh needed at equivalence point
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Answer:

25.6mL NaOH

Explanation:

We are given the Molarity of the solution (\frac{moles}{liters}) and the volume of the solution (.02L).

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moles = \frac{.410 moles}{L} *.02L

This gives us .0082 moles of HCl.

We then find the moles of NaOH that are needed to react with the HCl using the equation.

HCl + NaOH = NaCl + H_{2} O

As HCl and NaCl have a 1:1 ratio, we need .0082 mol of NaOH.

Dividing this value by the Molarity of the solution

\frac{.0082mol}{.320mol/L}

Gives us the answer, in Liters (.0256), which we can then divide by 100 convert to mL.

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3 years ago
What volume of a 0.3300M solution of sodium hydroxide would br ruired to titrate 15.00 mL of 0.1500 M oalic Acid?
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Answer:

3.41 mL

Explanation:

At equivalence point  from the reaction given,

Moles of Oxalic\ Acid = 2 × Moles of NaOH

Considering

Molarity_{Oxalic\ Acid}\times Volume_{Oxalic\ Acid}=2\times Molarity_{NaOH}\times Volume_{NaOH}

Given  that:

Molarity_{NaOH}=0.3300\ M

Volume_{NaOH}=?

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Molarity_{Oxalic\ Acid}=0.1500\ M

So,  

Molarity_{Oxalic\ Acid}\times Volume_{Oxalic\ Acid}=2\times Molarity_{NaOH}\times Volume_{NaOH}

2\times 0.3300\times Volume_{NaOH}=0.1500\times 15.00

Volume_{NaOH}=3.41\ mL

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3 years ago
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Volume of 1 mole of N2 at 30 c and 1.2 ATM
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We should indicate first and foremost the assumption of ideal gas behavior so that we can use the ideal gas equation written below:

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