Give the product of the reaction of excess benzene (2 equivalents) with dichloromethane and alcl3.

Chemistry

1 answer:

SOVA2 [1]3 years ago

5 0

Hello!

The product of the reaction of excess benzene with dichloromethane is diphenylmethane.

The reaction that benzene undergoes with dichloromethane is a Friedel-Crafts Alkylation, in which firstly Benzene is alkylated to Benzyl Chloride, and after that, the Benzyl Chloride reacts with another equivalent of Benzene to form the compound Diphenylmethane. The Reaction Scheme is shown in the image below:

Have a nice day!

You might be interested in

Solid sodium reacts violently with water, producing heat, hydrogen gas, and sodium hydroxide. How many molecules of hydrogen gas

Zinaida [17]

Answer: Number of molecules of hydrogen gas $6.32\times 10^{32}$

Explanation:

$2Na+2H_2O\rightarrow 2NaOH+H_2$

Number of moles of sodium = $\frac{\text{mass of sodium}}{\text{molar mass of sodium}}=\frac{48.7 g}{23 g/mol}=2.11mol$

According to reaction , 2 moles of sodium produces 1 mole of hydrogen gas , then 2.11 mol of sodium will= $\frac{1}{2}\times 2.11 mol$ of hydrogen gas that is 1.05 moles of hydrogen gas.

Number of molecules = $N_A(\text{Avogadro number})\times$ moles of substance

Moles of hydrogen gas formed = 1.05 moles

Number of molecules of hydrogen gas = $N_A\times$ moles of hydrogen gas

Number of molecules of hydrogen gas $=6.022\times 10^{23} mole^{-1}\times 1.05 mole=6.32\times 10^{32}$

8 0

3 years ago

Why do single cell organisms divide?

mixer [17]

Single-celled organisms which use asexual reproduction can do so very rapidly simply by dividing into two equal halves. This is called binary fission. In yeasts the cell does not divide equally in two halves; instead, there is a large mother cell and a smaller daughter cell. This is called budding.

7 0

3 years ago

Dissolution of KOH, ΔHsoln:

swat32

Using Hess's law we found:

1) By adding reaction 10.2 with the reverse of reaction 10.1 we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) ΔH (10.3)

2) The ΔHsoln must be subtracted from ΔHneut to get the total change in enthalpy (ΔH).

The reactions of dissolution (10.1) and neutralization (10.2) are:

KOH(s) → KOH(aq) ΔHsoln (10.1)

KOH(s) + HCl(aq) → H₂O(l) + KCl(aq) ΔHneut (10.2)

1) According to Hess's law, the total change in enthalpy of a reaction resulting from differents changes in various reactions can be calculated as the sum of all the enthalpies of all those reactions.

Hence, to get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

We need to add reaction 10.2 to the reverse of reaction 10.1

KOH(s) + HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq) + KOH(s)

Canceling the KOH(s) from both sides, we get reaction 10.3:

KOH(aq) + HCl(aq) → H₂O(l) + KCl(aq) (10.3)

2) The change in enthalpy for reaction 10.3 can be calculated as the sum of the enthalpies ΔHsoln and ΔHneut:

$\Delta H = \Delta H_{soln} + \Delta H_{neut}$