The electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.
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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
Answer:
A whale bleeding from a cut.
Explanation:
Shivering in the cold, breathing hard during exercise, and sweating because it's hot are ways the body is working to maintain good functioning and to keep the body alive. Bleeding from a cut is not an example of this.
Answer:
A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)
B - Increase(P), Increase(q), Decrease (R)
C - Triple (P) and reduce (q) to one third
Explanation:
<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>
P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.
In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.
Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.
Answer:
The particles in the water are always moving
Explanation:
