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2 years ago

Addition of water to an alkyne gives a keto-enol tautomer product. draw an enol that is in equilibrium with the given ketone.

1 answer:

8 0

If water were added to an alkyne and the product was 2-pentanone, then the alkyne must have been 1-pentyne. 2-pentyne would have given a mixture of 2- and 3-pentanone.

The enol that would give 2-pentanone would have been pent-1-en-2-ol. Because an equilibrium favors the ketone so greatly, equilibrium is not a good description. If the ketone were treated with bromine, little reaction would be seen as the enol content would be too low. If a catalyst were added, NaOH for example, then formation of the enolate of pent-1-en-2-ol would form and react with bromine. This would eventually give a bromoform product. Under acidic conditions, the enol would favor formation of the more substituted enol consistent with alkene stability.

You might be interested in

How many grams of hydrogen are produced if 30.0 g of zinc reacts?

alekssr [168]

0.925 grams if using hydrochloric acid in the reaction. 0.462 grams if using sulfuric acid in the reaction. 0.000 grams if using nitric acid in the reaction. Assuming you're using HCl or a similar acid for this reaction, the equation for the reaction is: Zn + 2 HCl ==> ZnCl2 + H2 So each mole of zinc used, produces 1 mole of hydrogen gas, or 2 moles of hydrogen atoms. So we need to look up the atomic weights of both zinc and hydrogen. Atomic weight zinc = 65.38 Atomic weight hydrogen = 1.00794 Moles zinc = 30.0 g / 65.38 g/mol = 0.458855919 mol Since we produce 2 moles of hydrogen atoms per mole of zinc, multiply by 2 and the atomic weight of hydrogen to get the mass of hydrogen produced. So 0.458855919 * 2 * 1.00794 = 0.92499847 grams. Rounding to 3 significant figures gives 0.925 grams. To show the assumption of the acid used, the balanced equation for sulfuric acid would be Zn2 + H2SO4 ==> Zn(SO4)2 + H2 Which means that for every mole of zinc used, 1 mole of hydrogen gas is generated (half that produced via hydrochloric acid). If nitric acid were used, the reaction is 4Zn + 10HNO3 ==> 4Zn(NO3)2 + N2O + 5H2O Which means that NO hydrogen gas is generated. The only justification for assuming hydrochloric acid is used is that it's a fairly common acid that's easy to obtain. But as shown above with 2 alternative acids, the amount of hydrogen gas generated is very dependent upon the exact chemical reaction occurring and asking "How many grams of hydrogen are produced if 30.0 g of zinc reacts?" is a rather silly question unless you specify EXACTLY what the reaction is.

3 0

2 years ago

Find the enthalpy of neutralization of HCl and NaOH. 87 cm3 of 1.6 mol dm-3 hydrochloric acid was neutralized by 87 cm3 of 1.6 m

Vesna [10]

Answer : The correct option is, (A) -101.37 KJ

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.6mole/L\times 0.087L=0.1392mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.1392 mole of HCl neutralizes by 0.1392 mole of NaOH

Thus, the number of neutralized moles = 0.1392 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $87ml+87ml=174ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 174ml=174g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 174 g

$T_{final}$ = final temperature of water = 317.4 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=174g\times 4.18J/g^oC\times (317.4-298)K$

$q=14110.008J=14.11KJ$

Thus, the heat released during the neutralization = -14.11 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -14.11 KJ

n = number of moles used in neutralization = 0.1392 mole

$\Delta H=\frac{-14.11KJ}{0.1392mole}=-101.37KJ/mole$

Therefore, the enthalpy of neutralization is, -101.37 KJ

3 0

2 years ago

A solution of 20.0 g of which hydrated salt dissolved in 200 g H2O will have the lowest freezing point? (A) CuSO4 • 5 H2O (M = 2

Andrews [41]

Answer:

(D) Na₂SO₄•10H₂O (M = 286).

Explanation:

The depression in freezing point of water by adding a solute is determined using the relation:

ΔTf = i.Kf.m,

Where, ΔTf is the depression in freezing point of water.

i is van't Hoff factor.

Kf is the molal depression constant.

m is the molality of the solute.

Since, Kf and m is constant for all the mentioned salts. So, the depression in freezing point depends strongly on the van't Hoff factor (i).

van't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

(A) CuSO₄•5H₂O:

CuSO₄ is dissociated to Cu⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(B) NiSO₄•6H₂O:

NiSO₄ is dissociated to Ni⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(C) MgSO₄•7H₂O:

MgSO₄ is dissociated to Mg⁺² and SO₄²⁻.

So, i = dissociated ions/no. of particles = 2/1 = 2.

(D) Na₂SO₄•10H₂O:

Na₂SO₄ is dissociated to 2 Na⁺ and SO₄²⁻.

So, i = dissociated ions/no. of particles = 3/1 = 3.

∴ The salt with the high (i) value is Na₂SO₄•10H₂O.

So, the highest ΔTf resulted by adding Na₂SO₄•10H₂O salt.

4 0

2 years ago

Determine ΔH for the reaction CaCO3 → CaO + CO2 given these data: 2 Ca + 2 C + 3 O2 → 2 CaCO3 ΔH = −2,414 kJ C + O2 → CO2 ΔH = −

kicyunya [14]

Answer:

The ΔH for the reaction is -456.5 KJ

Explanation:

Here we want to determine ΔH for the reaction;

Mathematically;

ΔH = ΔH(product) - ΔH(reactant)

In the case of the first reaction;

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3) ...........................(*)

From the other reactions, we can get the respective ΔH for the individual molecule in the reaction

In second reaction;

Kindly note that for elements, molecule of gases, ΔH = 0

What this means is that throughout the solution;

ΔH(Ca) = 0 KJ

ΔH(O2) = 0 KJ

ΔH(C) = 0 KJ

Thus, in writing the equation for the subsequent chemical reactions, we shall need to write and equate the overall ΔH for the reaction to that of the product alone

So in the second reaction

ΔH = 2ΔH(CaCO3)

Thus;

-2414/2 = ΔH(CaCO3)

ΔH(CaCO3) = -1,207 KJ

Moving to the third reaction, we have;

ΔH = ΔH(CO2)

Hence ΔH(CO2) = -393.5 KJ

For the last reaction;

ΔH = ΔH(CaO)

Hence ΔH(CaO) = -1270 KJ

Going back to equation *

ΔH = ΔH(CaO) + ΔH(CO2) - ΔH(CaCO3)

Using the values of the ΔH of the respective molecules given above,

ΔH = -1270 + (-393.5) - (-1207)

ΔH = -456.5 KJ

8 0

2 years ago

The volume of hcl gas required to react with excess ca to produce 11.4 l of hydrogen gas at 1.62 atm and 62.0 °c is ________ l.

seropon [69]

Answer:

22.8 L

Step-by-step explanation:

We can use Gay-Lussac's Law of Combining Volumes to solve this problem:

Gases at the same temperature and pressure react in simple whole-number ratios.

1. Write the chemical equation.

Ratio: 2 L 1 L

Ca(s) + 2HCl(g) ⟶ CaCl₂(s) + H₂(g)

V/L: 11.4

2. Calculate the volume of HCl.

According to the law, 2 L of HCl form 1 L of H₂.

Then, the conversion factor is (2 L HCl/1 L H₂).

Volume of HCl = 11.4 L H₂ × (2 L HCl/1 L H₂)

= 22.8 L HCl

3 0

2 years ago