Explain how the following changes would affect the rate of the reaction of 2-bromo-3-methylbutane with methanol: Part A The alkyl halide is changed to 2-chloro-3-methylbutane. The alkyl halide is changed to 2-chloro-3-methylbutane. The reaction will be slower because the leaving group will be poorer. The reaction will be faster because the leaving group will be better. The reaction will be slower because the leaving group will be better. The rate of the reaction does not change.

The reaction will be slower because the leaving group will be poorer.

Explanation:

The given reaction is an example of reaction.

In the first step, bromide ion leaves the molecule to produce a carbocation. This is the rate determining step.

In the second step, methanol attacks the carbocation to form product of the reaction. This is a fast step and does not have influence on rate of reaction.

Therefore, if we replace Br with Cl then formation of carbocation will require more activation energy as is a better leaving group than due to higher polarizability of

Hence, the reaction will be slower when the alkyl halide is changed to 2-chloro-3-methylbutane.

indium, and thallium have three electrons i think

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.