Why is ceasium more reactive than sodium

Draw the structure of the product for the reaction of propylmagnesium bromide with cyclohexanone. Assume that the reactions are

Nadya [2.5K]

Answer:

See explanation below

Explanation:

This reaction is known as the Grignard reaction. If's often used to add an alkyl to a molecule, in this case, a ketone. This also helps to reduce the carbonile to an alcohol.

The mechanism is taking place in 3 steps.

The first step involves the attack of the pair of electrons of the oxygen from the cyclohexanone to the MgBr, and the other pair takes the propyl.

The second step involves the breaking of this bond to have the OMgBr and the propyl attached to the carbon of the chain.

The last steps involves an hydration of the molecule, to turn the OMgBr into OH, and that would be the final product.

In this case, the final product will be 1-propyl-cyclohexanol.

The picture below, shows the mechanism and the drawing of the final product.

5 0

3 years ago

Methods: Part A: Preparation of Buffers Make two buffers starting with solid material, which is the most common way to make buff

Alecsey [184]

Answer:

0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.

0,856 g of Tris-HCl and 0,553 g of Tris-base

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀ $\frac{A^{-} }{HA}$

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for phosphate buffer is:

pH = 7,21 + log₁₀ $\frac{HPO4^{2-} }{H2PO4^{-} }$

If desire pH is 7,0 you will obtain:

0,617 = $\frac{HPO4^{2-} }{H2PO4^{-} }$ (1)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] (2)

Replacing (1) in (2) you will obtain:

[H₂PO₄⁻] = 0,0618 M

And with this value:

[HPO₄²⁻] = 0,0382 M

As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:

Na₂HPO₄ = 0,1 L× $\frac{0,0382mol}{1L}$ × $\frac{141,96g}{1mol}$ = 0,542 g of Na₂HPO₄

NaH₂PO₄ = 0,1 L× $\frac{0,0618mol}{1L}$ × $\frac{119,96g}{1mol}$ = 0,741 g of NaH₂PO₄

For tris buffer the equilibrium is:

Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075

Henderson–Hasselbalch equation for tris buffer is:

pH = 8,075 + log₁₀ $\frac{Tris-base }{Tris-H^{+} }$

If desire pH is 8,0 you will obtain:

0,841 = $\frac{Tris-base }{TrisH^{+} }$ (3)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [Tris-base] + [Tris-H⁺] (4)

Replacing (3) in (4) you will obtain:

[Tris-HCl] = 0,0543 M

[Tris-base] = 0,0457 M

As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:

Tris-base = 0,1 L× $\frac{0,0457mol}{1L}$ × $\frac{121,1g}{1mol}$ = 0,553 g of Tris-base

Tris-HCl = 0,1 L× $\frac{0,0543mol}{1L}$ × $\frac{157,6g}{1mol}$ = 0,856 g of Tris-HCl

I hope it helps!

8 0

3 years ago

How does an atom become a positively charged ion

sladkih [1.3K]

When, it donates electons.

as for example take
NaCl ( sodium chloride)

it's an ionic compund,
that means it is formed by donating or gaining electrons

Na is writen first than, it must be electropositive i.e it has donated electons which made it positive and the clorine gains electron so it's electronegative.

Na is positive because
as we know it's atomic number is 11 that means it has 11 protons and 11 electrons
now, when it donate electon it has, greater number of protons whose change is +ve so the atom becomes overall positively charged ion or cation.

and something same happens in clorine and because it gains one electron and the number of electrons increase in it by 1 whise charge is -ve so, the atom becomes negatively charged ion or anion which has a -1 charge.

3 0

3 years ago

Observe the concentrations of the hydronium ions(H3O+), hydroxide ions(OH-), and water (H2O) in the “Water Components” chart on

tia_tia [17]

FOLLOWING!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

6 0

3 years ago

If you know the number of protons what else do you know the number of?

pav-90 [236]

You know the atomic number

5 0

4 years ago

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