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Chapter 26: Problem 102

A compound with the same hydrogen-to-carbon ratio as cyclobutane is (a) \(\mathrm{C}_{4} \mathrm{H}_{10} ;\) (b) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}\) (c) \(\mathrm{CH}_{3} \mathrm{C}=\mathrm{CCH}_{3} ;\) (d) \(\mathrm{C}_{6} \mathrm{H}_{6}\).

Short Answer

Expert verified
None of the options has the same hydrogen-to-carbon ratio as cyclobutane.

Step by step solution

01

Calculate the Hydrogen-to-Carbon Ratio of Cyclobutane

The molecular formula of cyclobutane is \(C_4H_8\). The hydrogen-to-carbon ratio is therefore \(8/4 = 2\). This means there are twice as many hydrogen atoms as there are carbon atoms in cyclobutane.
02

Check the Hydrogen-to-Carbon Ratio in each Option

Calculate the hydrogen-to-carbon ratio for each compound given in the options and compare it with that of cyclobutane.\n(a) \(\mathrm{C}_{4} \mathrm{H}_{10}\) has a ratio of \(10/4 = 2.5\)\n(b) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{3}\) has a ratio of \(6/4 = 1.5\)\n(c) \(\mathrm{CH}_{3} \mathrm{C}=\mathrm{CCH}_{3}\) has a ratio of \(6/4 = 1.5\)\n(d) \(\mathrm{C}_6H_6\) has a ratio of \(6/6 = 1\)
03

Select the Compound with the Same Hydrogen-to-Carbon Ratio as Cyclobutane

After comparing the hydrogen-to-carbon ratio in each option to that of cyclobutane, it can be concluded that none of the options has the same ratio as cyclobutane. Therefore, none of the given options (a), (b), (c), or (d) has the same hydrogen-to-carbon ratio as cyclobutane.

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Most popular questions from this chapter

With appropriate sketches, represent chemical bonding in terms of the overlap of hybridized and unhybridized atomic orbitals in the following molecules. (a) \(\mathrm{C}_{4} \mathrm{H}_{10} ;\) (b) \(\mathrm{H}_{2} \mathrm{C}=\mathrm{CHCl} ;\) (c) \(\mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CH}\).

Write structural formulas corresponding to these condensed formulas. (a) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCH}_{2} \mathrm{C}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{Cl}_{3} \mathrm{CCH}_{2} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{CH}_{2} \mathrm{Cl}\).

Draw the \(E\) and \(Z\) isomers of (a) 2 -chloro-2-butene; (b) 3 -methyl- 2 -pentene.

Consider the following molecular formulas. How many elements of unsaturation are there in each case? (a) \(\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{NO}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}_{3} ;\) (c) \(\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{ClO}\).

Draw as many structural isomers as you can for cyclic ethers (no - OH groups) having the formula \(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}\) Try to draw at least six. (There are more than six.).
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