Integrative Problems. These problems require the integration of multiple concepts to find the solutions. An organometallic compound is one containing at least one metal-carbon bond. An example of an organometallic species is \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right) \mathrm{MBr},\) which contains a metal-cthyl bond. How does the hybridization of the starred carbon atom change, if at all, in going from reactants to products? c. What is the systematic name of the product? (Hint: In this shorthand notation, all the \(\mathrm{C}-\mathrm{H}\) bonds have been eliminated and the lines represent \(\mathrm{C}-\mathrm{C}\) bonds, unless shown differently. As is typical of most organic compounds, cach carbon atom has four bonds to it and the oxygen atoms have only two bonds.) a. If \(\mathrm{M}^{2+}\) has the electron configuration \([\mathrm{Ar}] 3 d^{10},\) what is the percent by mass of \(\mathrm{M}\) in \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right) \mathrm{MBr} ?\) b. A reaction involving \(\left(\mathrm{CH}, \mathrm{CH}_{2}\right) \mathrm{MBr}\) is the conversion of a ketone to an alcohol as illustrated here:

Step by step solution

01

1. Hybridization change of carbon atom

To determine the hybridization change, we need to look at the number of electron groups (bonds + lone pairs) around the carbon atom before and after the reaction. In the reactant, \(\left(\mathrm{CH}_{3}\mathrm{CH}_{2}\right) \mathrm{MBr}\), the starred carbon atom is bonded to one hydrogen atom, one carbon atom, and the metal (M). So, it has three bonding electron groups. This corresponds to sp2 hybridization. In the product, the alcohol as it's illustrated here, the starred carbon atom is bonded to one hydrogen atom, one carbon atom, and one oxygen atom (within the OH group). It still has three bonding electron groups. Thus, the hybridization remains as sp2. So, there is no change in hybridization.

02

2. Systematic name of the product

The IUPAC name for the product can be found by identifying the functional group and the carbon chain. The product is an alcohol, which has an -OH functional group. The alcohol's carbon chain contains two carbon atoms, so it is an ethyl group. Therefore, the systematic name of the product is Ethanol.

03

3. Percent by mass of \(\mathrm{M}\) in \(\left(\mathrm{CH}_{3}\mathrm{CH}_{2}\right) \mathrm{MBr}\)

Given that the electron configuration of \(\mathrm{M}^{2+}\) is \([\mathrm{Ar}] 3 d^{10}\), this indicates that the metal is \(Zn\) with an atomic mass of 65.38 u. Now, let's calculate the molecular weight of the compound \(\left(\mathrm{CH}_{3}\mathrm{CH}_{2}\right) \mathrm{MBr}\): Weight of \(\mathrm{CH}_{3}\mathrm{CH}_{2}\) = (2 * 12.01) + (5 * 1.008) = 24.02 + 5.04 = 29.06 u Weight of \(\mathrm{MBr}\) = 65.38 u (Zn) + 79.904 u (Br) = 145.284 u Total molecular weight = 29.06 + 145.284 = 174.344 u Now, we can find the percent by mass of the metal (\(\mathrm{M}\)) in the compound: Percent by mass = \(\frac{\text{Weight of M}}{\text{Total molecular weight}} \times 100\) Percent by mass = \(\frac{65.38}{174.344} \times 100 \approx 37.50\%\) The percent by mass of \(\mathrm{M}\) in \(\left(\mathrm{CH}_{3}\mathrm{CH}_{2}\right) \mathrm{MBr}\) is approximately 37.50%.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!