Tutorial 6 Solutions Page 1 of 7 CHEMISTRY 1AA3-TUTORIAL chem1aa3/tutorial/answer/tut-6... · Tutorial…

  • Published on
    16-Aug-2018

  • View
    231

  • Download
    6

Embed Size (px)

Transcript

<ul><li><p>Tutorial 6 Solutions Page 1 of 7 </p><p>CHEMISTRY 1AA3-TUTORIAL 6-WEEK E WEEK OF FEBRUARY 26, 2001 </p><p>1. For each of the following molecules: write the Lewis structure, draw the molecule with the correct molecular shape (include bond angles), decide whether the molecule is polar, indicate the hybridization of the central atom. (TAs: choose a couple of these to focus on. The students might suggest the ones that theyre most puzzled about.) Molecule Lewis </p><p>structure Molecular shape Polarity? Hybridization </p><p>of central atom a) OF2 F O F </p><p>OF F</p></li><li><p>Tutorial 6 Solutions Page 2 of 7 </p><p>f) SO2 O S O S</p><p>OO</p></li><li><p>Tutorial 6 Solutions Page 3 of 7 </p><p>N</p><p>CN</p><p>C</p><p>CC</p><p>O</p><p>CH</p><p>O H</p><p>CO</p><p>C</p><p>C C</p><p>H</p><p>H</p><p>H</p><p>C</p><p>H</p><p>N</p><p>N</p><p>N</p><p>H</p><p>H</p><p>H</p><p>H</p><p>H</p><p>H</p><p>OH</p><p> a) How many carbon atoms are sp3 hybridized? </p><p>All of the tetrahedral Cs (all the Cs with 4 single bonds) are sp3 hybridized. There are 6 of them (i.e. all of the Cs EXCEPT the ones in the six-membered ring). </p><p>b) How many carbon atoms are sp2 hybridized? The C atoms in the six-membered ring are all triangular (trigonal) planar (they are all surrounded by three electron-pair domains). Therefore these 4 C atoms are sp2 hybridized. </p><p>c) Which atom is sp hybridized? The middle N of the N3 group has only two electron-pair domains (i.e. linear geometry), and is therefore sp hybridized. </p><p>d) How many bonds are in the molecule? The framework of the molecule is made of bonds. That is, there is a bond between every pair of bonded atoms. Therefore, the total number of bonds is 33. </p><p>e) How many bonds are in the molecule? When you see a multiple bond, the first bond is a bond, and any additional bonds must be bonds. Therefore, there are 5. </p><p>f) What is the N-N-N bond angle in the azide (-N3) group? The central N has two electron-pair domains, so the geometry around it must be linear. That is, the angle is 180. </p><p>g) What is the H-O-C bond angle in the side group attached to the five-membered ring? The O has four electron-pair domains, so the geometry around it is based on a tetrahedron (actually, its bent). The angle is slightly less than 109. Remember that Lewis structures show only which atoms are connected to which, not the correct geometry! </p><p>h) What is the hybridization of the oxygen atom in the CH2OH group? The O has four electron-pair domains, so its sp3 hybridized. </p></li><li><p>Tutorial 6 Solutions Page 4 of 7 </p><p> 3. For the following molecules, draw a complete orbital picture to show the </p><p>bonding involved. Your picture should clearly show the 3-D shape of the molecule. Clearly indicate the types of orbitals involved (s, p, sp, sp2, sp3, etc.) and the types of bonds formed. </p><p>a) H2CO </p><p>C OH</p><p>H Note that both the C and the O are surrounded by three electron-pair domains, and so they are sp2-hybridized. Therefore, the structure can be built by combining two sp2-hybrid building blocks (see p. 389 in your text): </p><p>p</p><p>sp2 sp2sp2</p><p>sp2-hybridized atom </p><p> psp2 sp2</p><p>sp2</p><p>sp2-hybridized atom </p><p>Now bring these puzzle pieces together, and add the H atoms (let the H 1s orbitals interact with the unbonded sp2 orbitals on the C). Note the lone pairs in two of the sp2-hybrid orbitals on the O: </p><p>psp2 sp2s</p><p>p</p><p>sp2 sp2</p><p>ss </p><p>Note that all of the end-to-end overlaps produce bonds; there is also one bond formed by side-to-side overlap of p orbitals. </p><p>b) HCN H C N Note that both the C and the N are surrounded by two electron-pair domains, and so they are sp-hybridized. Therefore, the structure can be built by combining two sp-hybrid building blocks (see p. 389 in your text): </p><p>p</p><p>psp sp</p><p>sp-hybridized atom</p><p>p</p><p>psp sp</p><p>sp-hybridized atom Now bring these puzzle pieces together, and add the H atom (let the H 1s orbital interact with the unbonded sp orbital on the C). Note the lone pair in the sp-hybrid orbital on the N: </p></li><li><p>Tutorial 6 Solutions Page 5 of 7 </p><p>p</p><p>psp sps</p><p>p</p><p>psp sp</p><p> Note that all of the end-to-end overlaps produce bonds; there are also two bonds formed by side-to-side overlap of p orbitals. </p><p>c) allene, CH2=C=CH2 </p><p>C C CH</p><p>H</p><p>H</p><p>H Note that the two Cs on the ends are surrounded by three electron-pair domains, and so they are sp2-hybridized. The central C has only two electron-pair domains, and is sp-hybridized. Therefore, the structure of allene can be built by combining the following building blocks (see p. 389 in your text): </p><p>p</p><p>sp2 sp2sp2</p><p>sp2-hybridized atom </p><p>p</p><p>psp sp</p><p>sp-hybridized atom </p><p>psp2 sp2</p><p>sp2</p><p>sp2-hybridized atom </p><p>Now lets start building the carbon chain, one pair at a time: </p><p>p</p><p>sp2 sp2sp2</p><p>p</p><p>psp sp</p><p> p</p><p>sp2 sp2sp2</p><p>Note that the two atoms were brought together such that the p orbitals were lined up side by side, allowing a bond to form. </p><p>This is important to note before we bring in the remaining C. We see that the unbonded p orbital remaining on the sp-hybridized atom is perpendicular to the page. Therefore, if we want to make another bond, well have to rotate the other atom so that its p orbital is also pointing out of the page (otherwise, the two p orbitals couldnt line up with each other): </p><p>p</p><p>sp2 sp2sp2</p><p>p</p><p>psp sp</p><p>sp2p</p><p>sp2</p><p>sp2 </p><p> Now bring these puzzle pieces together, and add the H atoms (let the H 1s orbitals interact with the unbonded sp2 orbitals): </p></li><li><p>Tutorial 6 Solutions Page 6 of 7 </p><p>sp2p</p><p>sp2</p><p>sp2ss</p><p>ssp</p><p>sp2 sp2</p><p>p</p><p>psp sp</p><p> Note that all of the end-to-end overlaps produce bonds; there are also two bonds formed by side-to-side overlap of p orbitals. </p><p>Finally, for allene, use your picture to decide whether the four hydrogen atoms are in the same plane. If not, what is their spatial relationship? From this picture, we can see that the Hs are NOT all in the same plane. The plane that contains the Hs on one end is perpendicular to the plane that contains the Hs on the other end. </p><p>TAs: Ill bring a molecular model to the tutorial rooms; it should really help the students see the solution. </p><p> 4. Draw as many structural isomers of C5H10 as you can. Can you find more than I </p><p>could? Be careful not to include any duplicates! Note that this formula CANNOT represent an acyclic alkane: there arent enough Hs to satisfy the general alkane formula, CnH2n+2. This is CnH2n: two Hs are missing! So all isomers of this formula must have a double bond or a ring. </p><p>Its a good idea to be systematic, so that you dont miss any isomers! Start with chains of 5 Cs, then try 4 Cs, etc. Dont forget the rings! </p><p>And finally, be careful to show the correct number of Hs on each C to bring the total number of bonds to 4. </p><p>CH2=CHCH2CH2CH3</p><p>CH3CH=CHCH2CH3</p><p>CH2=CCH2CH3</p><p>CH3</p><p>CH3C=CHCH3</p><p>CH3</p><p>CH3CHCH=CH2</p><p>CH3</p><p>CH3</p><p>CH3 CH3</p><p>CH3</p><p>CH3CH2CH3</p><p> So I found a total of 10! </p></li><li><p>Tutorial 6 Solutions Page 7 of 7 </p><p>5. Arrange the following sets of compounds in order of increasing boiling point (and, as always, be prepared to justify your answer!). a) CH3CH2CH2CH2CH3 &lt; CH3CH2CH2CH2Cl &lt; CH3CH2CH2CH2OH </p><p>The compound with the weakest intermolecular forces will have the lowest bp, because for that compound, the least energy will be required to overcome the IMFs. </p><p>Pentane will have the lowest bp. Because it is nonpolar (C-H bonds have a very small difference in electronegativities), the only forces attracting one molecule to another are dispersion forces (the weakest type of IMF). </p><p>1-Chlorobutane will have the intermediate bp. The C-Cl bond is slightly more polar than C-H bonds, so dipole-dipole interactions will be greater than in pentane. And probably more importantly, since Cl is a larger, more polarizable atom than C, the dispersion forces in 1-chlorobutane will be stronger than the dispersion forces in butane. </p><p>Finally, 1-butanol will have the highest bp; because of the OH group, it is capable of hydrogen bonding. (Remember that hydrogen bonds are NOT covalent bonds; they are simply very strong dipole-dipole interactions between different molecules.) </p><p>b) </p></li></ul>