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2 years ago

The electrophilic bromination or chlorination of benzene requires ______ along with the halogen.

Chemistry

1 answer:

Oksanka [162]2 years ago

8 0

The electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.

<h3>What is bromination of benzene?</h3>

The bromination or chlorination of benzene is an example of an electrophilic aromatic substitution reaction.

During the reaction, the bromine forms a sigma bond to the benzene ring, yielding an intermediate. Subsequently a a proton is removed from the intermediate to form a substituted benzene ring.

This reaction is achieved with the help of Lewis acid as catalysts.

Thus, the electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.

Learn more about bromination of benzene here: brainly.com/question/26428023

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I need help with my homework.Calculate the number of formula units in: 41.5 grams CaBr2

lana66690 [7]

Answer:

$\text{ }1.25\times10^{23}\text{ formula units}$

Explanation:

Here, we want to calculate the number of formula units in the given molecule

We start by getting the number of moles

To get the number of moles, we have to divide the mass given by the molar mass

The molar mass is the mass per mole

The molar mass of calcium bromide is 200 g/mol

Thus, we have the number of moles as follows:

$\frac{41.5}{200}\text{ = 0.2075 mol}$

The number of formula units in a mole is:

$1\text{ mole = 6.02 }\times10^{22}\text{ formula units}$

The number of formula units in 0.2075 mole will be:

$0.2075\text{ }\times6.02\times10^{23}\text{ = }1.25\times10^{23}\text{ formula units}$

5 0

2 years ago

Draw the product formed when the compound shown below undergoes a reaction with hbr in ch2cl2. 2-methylbut-2-ene

kenny6666 [7]

Methylbut-2-ene undergoes asymmetric electrophilic addition with hydrogen bromide to produce two products:

$\text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}$ , 2-bromo-2-methylbutane;
$\text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}$ , 2-bromo-1-methylbutane.

It is expected that $\text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}$ would end up being the dominant product.

Explanation

Molecules of methylbut-2-ene contains regions of high electron density at the pi-bonds. Those bonds would attract hydrogen atoms with a partial positive charge in polar hydrogen bromide molecules and could occasionally induce heterolytic fission of the hydrogen-bromide bond to produce positively-charged hydrogen ions $\text{H}^{+}$ and negatively-charged bromide ions $\text{Br}^{-}$ .

$\text{H-Br} \to \text{H}^{+} + \text{Br}^{-}$

The positively-charged hydrogen ion would then attack the methylbut-2-ene to attach itself to one of the two double-bond-forming carbon atoms. It would break the pi bond (but not the sigma bond) to produce a carbocation with the positive charge centered on the carbon atom on the other end of the used-to-be double bond. The presence of the methyl group introduces asymmetry to the molecule, such that the two possible carbocation configurations are structurally distinct:

$\text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3}$ ;
$\text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3}$ .

The carbocations are of different stabilities. Electrons in carbon-carbon bonds connected to the positively-charged carbon atom shift toward the electron-deficient atom and help increase the structural stability of the molecule. The electron-deficient carbon atom in the first carbocation intermediate shown in the list has three carbon-carbon single bonds after the addition of the proton $\text{H}^{+}$ as opposed to two as in the second carbocation. The first carbocation- a "tertiary" carbocation- would thus be more stable, takes less energy to produce, and has a higher chance of appearance than its secondary counterpart. The polar solvent dichloromethane would further contribute to the stability of the carbocations through dipole-dipole interactions.

Both carbocations would then combine with bromide ions to produce a neutral halocarbon.

$\text{H}_3\text{C}-\text{C}^{+}(\text{C}\text{H}_3)-\text{CH}_2 - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{Br}-\text{CH}_2 - \text{CH}_{3}$
$\text{H}_3\text{C}-\text{C}(\text{C}\text{H}_3)\text{H}-\text{C}^{+}\text{H} - \text{CH}_{3} + \text{Br}^{-} \to \text{H}_3\text{C}-\text{C}(\text{CH}_3)\text{H}-\text{CHBr} - \text{CH}_{3}$

The position of bromine ions in the resultant halocarbon would be dependent on the center of the positive charge in the carbocation. One would thus expect 2-bromo-2-methylbutane, stemming from the first carbocation which has the greatest abundance in the solution among the two, to be the dominant product of the overall reaction.

3 0

3 years ago

Galvanic cell runs on the following reaction: Al (s) + Fe3+ (aq) → Al3+ (aq) + Fe (s) Draw a diagram for this Galvanic cell, lab

MAVERICK [17]

See image for details.

5 0

4 years ago

Which metal is used to make gasmasks

Lubov Fominskaja [6]

Explanation:

Gas mask filters include activated carbon,

absorbents that trap toxins in millions of micro-pores. It is the same compound used to filter water and treat ingestion of poisons. The activated carbon traps the toxins, but in gas masks it is further augmented with metal oxides, such as copper and molybdenum, to help break down the toxins.

A lot of research is being done in Berkeley labs

8 0

3 years ago

(a) Calculate the density of a 374.5-g sample of copper if it has a volume of 41.8 cm3. (b) A student needs 15.0 g of ethanol fo

o-na [289]

Answer:

For a: The density of the sample of copper is $8.96g/cm^3$