A compound containing only carbon and hydrogen is 85.63\(\%\) C by mass. Reaction of this compound with \(\mathrm{H}_{2} \mathrm{O}\) produces a secondary alcohol as the major product and a primary alcohol as the minor product. (See Exercise \(68 . )\) If the molar mass of the hydrocarbon is between 50 and \(60 \mathrm{g} / \mathrm{mol},\) name the compound.

Let's first calculate the mass percentages of carbon and hydrogen in the compound: Mass percentage of C = 85.63% Therefore, mass percentage of H = 100% - 85.63% = 14.37% Since percentages are mass ratios, we can use arbitrary mass for the compound. Let's assume we have 100 g of the compound. Then we would have 85.63 g of carbon and 14.37 g of hydrogen. Now, convert these masses to moles using the molar masses of C and H: Moles of carbon (C) = \(\frac{85.63 \,\text{g}}{{12.01\, \text{g/mol}}}\) = 7.13 moles Moles of hydrogen (H) = \(\frac{14.37\, \text{g}}{{1.01\, \text{g/mol}}}\) = 14.23 moles Next, let's find the mole ratio: Mole ratio (C : H) = \(\frac{7.13}{7.13}\) : \(\frac{14.23}{7.13}\) = 1 : 2 Hence, the empirical formula of the compound is CH2.

To find the molecular formula of the compound, we will use the given molar mass range of 50 to 60 g/mol. The empirical formula mass of CH2 is: \(12.01\, \text{g/mol}\) (C) + \(2 * 1.01\, \text{g/mol}\) (H) = 14.03 g/mol Now, we need to find an integer (n) that when multiplied by the empirical formula mass, falls within the given molar mass range: Lowest molar mass: \(\frac{50}{14.03} \approx\) 3.6 Highest molar mass: \(\frac{60}{14.03} \approx\) 4.3 Since n must be an integer, the molecular formula is 4 times the empirical formula, i.e., C4H8.

We know that the compound reacts with water to form a secondary alcohol as the major product and a primary alcohol as the minor product. This implies that the compound contains an alkene group, which is undergoing hydration reactions. There are several structural isomers of C4H8, and we need to find the one which undergoes hydration such that a secondary alcohol is formed as the major product and a primary alcohol as the minor product. Based on the Markovnikov's rule, the major product should have the hydroxyl group on the more substituted carbon atom. We quickly notice that this behavior is observed with 2-butene (CH3CH=CHCH3), reacting to give 2-butanol (CH3CH(OH)CH(CH3)2) as the major product and 1-butanol (CH3(CH2)3OH) as the minor product. So, the name of the given compound is 2-butene.