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Lady bird [3.3K]
2 years ago
9

Compounds X has the formula C7H15Cl; Y is C7H15Br. X undergoes base-promoted E2 elimination to give a single alkene product Z. Y

likewise reacts under similar conditions to give a single alkene product that is isomeric with Z Catalytic hydrogenation of Z affords 3-ethylpentane. X readily reacts in SN2 fashion with sodium iodide in acetone. Y does not undergo a similar SN2 reaction. Propose structures for X and Y.
Chemistry
1 answer:
djverab [1.8K]2 years ago
3 0

Answer:

See explanation and image attached

Explanation:

Let us examine the statements in the question carefully. First of all, we will discover that the products of the E2 elimination of the both compounds are isomeric. However Y does not undergo SN2 reaction as  X does.

The fact that SN2 reaction does not occur in Y confirms that the bromine atom is attached to a tertiary carbon atom and SN2 reaction does not occur due to steric hinderance. Since X undergoes SN2 reaction in aprotic solvent, the chlorine atom must be attached to a secondary carbon atom.

However, E2 reactions does occur with tertiary alkyl halides when strong bases such as OH^- or RO^- are used.

The question also stated that the catalytic hydrogenation of Z affords 3-ethylpentane.

Putting all these together, the structures of X and Y have been suggested in the image attached to this answer.

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When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for
balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-

Whereas the half reactions are:

Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O

Next, we exchange the transferred electrons:

1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow  4N^{4+}O^{2-}_2+4H_2O

Afterwards, we add them to obtain:

Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O

By adding and subtracting common terms we obtain:

Sn^0+4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O

Finally, by removing the oxidation states we have:

Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

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.........

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