Ask question

Ask question

All categories

3 years ago

13

Question 044 Which of the following sequences converts 2-methylpropene and sodium acetylide into 3-methylbutanal? 1) HBr; 2) NaC

CH; 3) O3; 4) DMS 1) NaCCH; 2) H2/Ni2B; 3) O3; 4) DMS 1) HBr, ROOR; 2) NaCCH; 3) H2/Ni2B 4) O3; 5) DMS 1) HBr, ROOR; 2) NaCCH; 3) O3; 4) H2O 1) HBr; 2) NaCCH; 3) O3; 4) H2O

2 answers:

Minchanka [31]3 years ago

8 0

Answer:

1) HBr; 2) NaCCH; 3) O3; 4) H2O

Explanation:

The first step is formation of alkyl halide followed by reaction with sodium acetylide, to form 3-methylbutene, which is then followed by oxidation reaction with O3& H2O to 3-methylbutanal

Kruka [31]3 years ago

6 0

Answer:

The sequence of reagents' incorporation for the conversion of 2-methylpropene into 3-methylbutanal is <u>option (C.):</u>

1. $HBr,ROOR$ ; 2. $NaCCH,H_{2}/Ni_{2}B$ ; 3. $O_{3}, DMS$

Explanation:

The summary of this reaction's mechanism is as follows:

<em>Radical hydro-halogenation</em> of 2-methylpropene through a technique known as <em>"peroxide effect"</em> in which the HBr addition causes free radical formation of the alkene
<em>Nucleophilic SN2 substitution </em>reaction (acetylide)
<em>Hydrogenation</em>
<em>Ozonolysis of the alkene</em>: after the addition reaction, the intermediate ozonide is converted to aldehyde using dimethyl sulfide (DMS)

You might be interested in

Picric acid has been used in the leather industry and in etching copper. However, its laboratory use has been restricted because

olchik [2.2K]

Answer:

0.3023 M

Explanation:

Let Picric acid = $H_{picric}$

So, $H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

The ICE table can be given as:

$H_{picric}$ + $H_2}O$ ⇄ $H_3}O^+$ + $Picric^-$

Initial: 0.52 0 0

Change: - x + x + x

Equilibrium: 0.52 - x + x + x

Given that;

acid dissociation constant ( $K_a$ ) = 0.42

$K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}$

$0.42 = \frac{[x][x]}{0.52-x}}$

$0.42 = \frac{[x]^2}{0.52-x}}$

0.42(0.52-x) = x²

0.2184 - 0.42x = x²

x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)

Using the quadratic formula;

$\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ ; ( where +/- represent ± )

= $\frac{-0.42+/-\sqrt{(0.42)^2-4(1)(-0.2184)} }{2*1}$

= $\frac{-0.42+/-\sqrt {0.1764+0.8736} }{2}$

= $\frac{-0.42+\sqrt {1.0496} }{2}$ OR $\frac{-0.42-\sqrt {1.0496} }{2}$

= $\frac{-0.42+1.0245}{2}$ OR $\frac{-0.42-1.0245}{2}$

= $\frac{0.6045}{2}$ OR $-\frac{1.4445}{2}$

= 0.30225 OR - 0.72225

So, we go by the +ve integer that says:

x = 0.30225

x = [ $H_3}O^+$ ] = [ $Picric^-$ ] = 0.3023 M

∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).

6 0

3 years ago

What is the vapor pressure of an aqueous solution made up of 60.0g of urea (CH4N2O) in 180g of water? the vapor pressure of wate

djyliett [7]

To get the answer you use the Law of Raoult.

Raoult's law states that the decrease of the vapor pressure of a liquid is proportional to the molar fraction of the solute.

ΔP = Pa * Xa

Here Pa = 0.038 atm

And Xa = N a / (Na + Nb), where Na is number of moles of A and Nb is number of moles of b

Na = mass of urea / molar mass of urea = 60 g / (molar mass of CH4N2O)

molar mass of CH4N2O = 12 g/mol + 4*1g/mol + 2*14 g/mol + 16 g/mol = 60 g/mol

Na = 60 g / 60 g/mol = 1 mol

Nb = mass of water / molar mass of water = 180g / 18g/mol = 10 mol

Xa = 1 mol / (10 mol + 1 mol) = 1/11 =0.09091

ΔP = Pb * Xa = 0.038 atm * 0.09091 = 0.0035 atm

Then, the final vapor pressure of water is Pb - ΔP = 0.038atm - 0.0035atm = 0.035 atm.

Answer: 0.035 atm

7 0

3 years ago

Solids we need to live

gogolik [260]

Foods are a solid we need to survive

6 0

3 years ago

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of pho

wlad13 [49]

<h3>Answer:</h3>

18.58 liters of hydrogen gas

<h3>Explanation:</h3>

We are given;

The equation;

3Mg + 2H₃(PO₄) → Mg₃(PO₄)₂ + 3H₂

Atoms of Magnesium = 7.179 x 10^23 atoms
Mass of phosphoric acid as 54.21 g

We are required to determine the volume of hydrogen gas produced;

Step 1; moles of Magnesium

1 mole of an element contains 6.02 × 10^23 atoms

therefore;

Moles of Mg = (7.179 x 10^23 ) ÷ (6.02 × 10^23)

= 1.193 moles

Step 2: Moles of phosphoric acid

moles = Mass ÷ Molar mass

Molar mass of phosphoric acid = 97.994 g/mol

Therefore;

Moles of Phosphoric acid = 54.21 g ÷ 97.994 g/mol

= 0.553 moles

Step 3: Determine the rate limiting reagent

From the mole ratio of Mg to Phosphoric acid (3 : 2);

1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

Therefore, phosphoric acid is the rate limiting reagent

step 4: Determine the moles of hydrogen produced

From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

Therefore; moles of Hydrogen = moles of phosphoric acid × 3/2

= 0.553 moles × 3/2

= 0.8295 moles

Step 5: Volume of hydrogen gas

1 mole of a gas occupies a volume of 22.4 liters at STP

Therefore;

Volume of Hydrogen = 0.8295 moles × 22.4 L/mol

= 18.58 Liters

Therefore; 18.58 liters of hydrogen gas will be produced

4 0

3 years ago

How many potassium atoms are in 250.0 grams of potassium?

loris [4]

Hi!

The correct option would be 3.85x10^(24)

To find the number of atoms in 250g of potassium, we need to first calculate the number of atoms in

1 mole of Potassium = 39g which contains 6.022x10^(23) atoms of K

<em>(Avogadro's constant value for the amount of molecules/atoms in one mole of any substance)</em>

<em>Solution</em>

So as 39g of Potassium contains 6.022x10^(23) K atoms

1g of Potassium would contain 6.022x10^(23) / 39 = 1.544 x10^(22) atoms

So 250g of Potassium would contain 1.544x10^(22) x 250 = 3.86x10^(24) atoms

3 0

3 years ago