Rank the following alkyl halides from most reactive to least reactive in an SN1 reaction: 2-bromo-2-methylpentane, 2-chloro-2-me
thylpentane, 3-chloropentane, and 2-iodo-2-methylpentane. Rank the alkyl halides from most reactive to least reactive. To rank items as equivalent, overlap them. ResetHelp Least reactiveMost reactive
1 answer:
Answer:
2-iodo-2-methylpentane > 2-bromo-2-methylpentane > 2-chloro-2-methylpentane > 2-chloro-2-methylpentane
Explanation:
In this case, the<u> Sn1 reaction</u> would form a <u>carbocation</u>. So, the molecule that can generate a <u>very stable carbocation</u> will be more reactive. We have to remember that <u>tertiary carbocations</u> are the more stable ones. With this in mind, 3-chloropentane would be the <u>least reactive</u> molecules of all.
Then to decide which one is more reactive between the other ones, we have to check the <u>leaving group</u>. In this case, all the atoms are <u>halogens</u>, so if we have a <u>larger atom</u> the leaving group would leave <u>more easily</u>.
With his in mind the larger atom would be I, then Br and finally Cl, therefore the 2-iodo-2-methylpentane would be the <u>more reactive one.</u>
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At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.
The melting point is the temperature at which a substance passes from solid to liquid. Below the melting point, a substance is in the solid state. Above the melting point, a substance is in the liquid or gas state.
The boiling point is the temperature at which a substance passes from liquid to gas. Below the boiling point, a substance is solid or liquid. Above the boiling point, a substance is in the gas state.
At -25 °C, methanol is above the melting point (-97.6 °C) and below the boiling point (64.7 °C). Thus, it is in the liquid state.
At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.
You can learn more about the melting and boiling points here: brainly.com/question/5753603?referrer=searchResults
Answer:
1. pH = 1.23.
2.
Explanation:
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1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
Which is also shown in net ionic notation.
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