Questions and Answers - Chem 227

General

• When are the review sessions?
Every Sunday at 8PM in Room 2102.

Chapter 1

• Question 1.4c. CH?Cl₂ I know that since C has 4 ve and since you already have 2 on Cl you put just 2 more on the hydrogen but why do you look at the C and at Cl on some of the problems? On some problems you look at C and on others you don't .
The problem is to place the correct subscript where there is a ?. Carbon has to have four bonds to it. Therefore CCl₄ and CH₂Cl₂. Silicon is in group 4 so it is SiF₄, while aluminum is in group 3 so AlH₃.
• 1.4e. CH3NH? Again how come you look at the N and not the carbon?
Look where the ? is. In the case of CH₃NH?, carbon has its four bonds, three to the hydrogens and one to the nitrogen. (We read molecular formula like we read words - from left to right.) Nitrogen in group 5 forms 3 bonds (with a pair of non-bonded electrons). In this case it is bonded to the carbon (remember that's how we got the fourth bond to C) and the ?H. Since we need two more bonds, it must be 2H: CH₃NH₂.
• 1.18. NI?
The number of valence electrons is always equal to the number of the group to which the atom belongs! Nitrogen atoms in a molecular formula will always have three bonds and a pair of non-bonded electrons UNLESS there is a formal charge (+ or -) on the nitrogen.
• 1.30c P(CH₃)₃-how do you know the entire shape? Won't the P be tetrahedral since it is sp³ hybridized and same with the CH₃?? How do you predict the whole shape of trimethylphosphine???
If you consider just the atoms of PY₃ (in this case Y = CH₃), the shape defined is a 3-sided pyramid. IF you consider the electrons as a group and draw your lines to the 'end' of the orbital containing the non-bonded pair, you will get a tetrahedral shape. See the nitrogen example on p. 29.
• 1.31. I know that it will be tetrahedral since the C is sp³ hybridized, but how do you know which atom comes out of the page and which one goes in and which one lies on the plane?
GOOD QUESTION. It doesn't make any difference!! You can put the CH₃, for example, at any of the locations you cite and it will represent THE SAME MOLECULE! Make a molecular model with your model kit and see how you can look at the molecule (and thus draw a representation) for several angles. In fact there are more than four different ways this molecule can be drawn that makes a different PICTURE, but we are interested in what molecule it represents!
• 1.33b. Could you draw the OH as 0-H instead of putting them together as -0H ?
Yes. The line between O and H would be needed for a completely line-bond representation. Chemists often want to use the condensed structures.
• 1.42b. Does the carbon have 6v.e b/c of the positive charge that is incurred on it giving it 2 extra v.e.?
It does not have 2 extra valence electrons. This is covered better in Section 2.3 where formal charge is discussed. One carbon and three hydrogen atoms would have a total of seven valence electrons. In the structure shown in 1.42, there are only 6 valence electrons present (two for each line bond); therefore the structure has one less electron than a neutral system which makes it have a charge of +1.
• 1.43a. How do you get 8 v.e.? Is it b/c of the 3 bonds (=6 v.e) + 1 lone pr(2 v.e.) which give a total of 8v.e.?
YES. Eight valence electrons is one greater than the seven a neutral system would have; therefore the charge of -1.
• 1.46-1.48 Is it just by trial and error to draw and get 2 of each formulas or is there a particluar way to do it. On some of the problems i got 3 diff.ways to draw the bonds but wasn't sure if my cmpnd actually exists. Is there a variety of answers??
You'll see more examples in Chapter 2. Remember that there can be more than one way to represent many molecules. As long as the representation can be interpreted (following normal conventions) as the same structure as shown in the Study Guide, it will be correct.

Chapter 2

• 2.2. H₃C-MgBr; isnt Br more electronegative than C giving it H3C(d+)-MgBr(d-). The electronegativity value of Br is even greater than that of C from the chart given in ch 2.
The Carbon is attached to magnesium - not to the bromine. C-Mg-
• 2.4. how do you do this one. i dont see how the other dipole moments cancel out to give you the net dipole moment. the lone pairs confuse me about the direction of their dipole moment.could you do this one in class if possible.
You have to do a mental vector analysis. The C-O dipole and the O-: dipoles are the most important. Therefore there will be a net vector toward the oxygen, but displaced from the center of the oxygen toward the electron pairs. The oxygen is sp³-hybridized and tetrahedral if you consider the electron pairs as groups .
• 2.6c. why does Cl go in that manner-if you rearranged the placement of the atoms in the tetrahedral shape around the C-you could get a 0 dipole moment with the Cl cancling each other out and and H canceling each other out. Also-how do you know the direction of the net dipole moment-do you take the the x vector and the y vector and get a net vector median in the middle?
MAKE A MODEL OF THIS ONE. You cannot get the Cl cancelling each other. The answer to question 2 is yes.
• 2.19c Can you please explain how I would go about counting the number of Carbon atoms in question 2.19 part (c). I am not sure if you count it as one carbon atom when the two lines from the different structures intersect.
A carbon atom is at the intersection of lines (either single lines for double bonds or double lines for double bonds). In 2.19c there are two carbons which could be considered part of both the 5-membered ring and the 6-membered ring. Thus, there are a total of 8 carbons present.
• 2.42. why do the pka have to be smaller that water?what does water have to do with it?
OH^- + H-Y ----> <---- HOH + Y^-
Water is the conjugate acid of OH^-. If the pKa of H-Y and HOH were the same, the Keq would be = 1 and we would have equal amounts of OH^- and Y^-. To get more than half of the OH^- consumed, the pKa of H-Y must be lower than that of water (a stronger acid reacting to give a weaker acid as the product).
• 2.35b. In the answer how come in CH₃ONa the O and Na have formal charges? why does O have 3 lone pairs. I am getting that the cmpd should have a total of 14v.e. therefore only 2 lone prs on O and not 3 lone prs.
Because sodium does not form covalent bonds. The two ions should be written farther apart; they are not sharing an electron pair. There are 14 valence electrons (8 around the carbon with four bonds (including a C-O bond) and 6 in the 3 unshared pairs around O.
• For the first exam do we need to be able to calculate dipole moments or just understand the calculation. I am also a little confused about the dipole moments. Can only a polar molecule have a dipole moment?
You will not need to do numerical calculations of dipole moment. You must have polar bonds for the molecule to have a dipole moment. Molecules with more than more than one polar bond will have a molecular dipole moment unless the bond dipoles are oriented in such a way as to cancel each other exactly. All molecules that we classify as polar molecules will have a dipole moment.

Chapter 3

• I'm having a hard time understanding how the prefix "iso" affects a structure. In the book I see what isobutane and etc. look like, but maybe if you tell me; the words could help me see it and understand it better.
An 'iso' structure will always have the following group, (CH3)₂CH--,
(CH₃)₂CH- = isopropyl
isopropylcyclopentane = 5-membered ring with an isopropyl attached.
(CH₃)₂CH-CH₂ = isobutyl
(CH₃)₂CH-CH₃ = isobutane
(CH₃)₂CH-CH₂CH₂-- = isopentyl
(CH₃)₂CH-CH₂CH₃ = isopentane

• Problem 3.6: The answer to the book says there are two isomers for C₃H₈O, but I drew out three. One with hydroxyl group on the outside, one with an ether, and another one with a hydroxyl group on the central carbon. Am I wrong anywhere?
The question has an additional restriction that makes your second structure, the ether, incorrect. The problem asks for _alcohols_ with formula C₃H₈O.

Chapter 4

• When would twist-boat conformations be used? Do we need to know how to draw them? If so, could you show us an example?
You will not need to draw this conformation. Information is provided to help understand how one chair conformation can get to the alternative chair conformation. How can we determine if a cis or trans disubstituted isomer is more stable? Is it determined by the equatorial and axial position? I believe that anything in the equatorial position would be the most stable. Is this a correct statement? As shown in Table 4.2, all substituents shown here have a positive axial steric strain, which means that the conformation with the substiuent in the equatorial position is more favorable. For disubstitued cases, See Problem 4.18, we compare the values from the table to see which groups most strongly favors being equatorial. On page 107 of the text it says that "different conformers can't usually be isolated because they interconvert too rapidly." Is this saying that molecules are constantly converting from one conformation to the next? (Are the atoms in constant motion, rotating around the carbon?) Most changes between conformations have energy barriers (going from staggered through eclipsed or from chair to boat to chair) that are 2 - 15 kcal/mol. At room temperature, there is enough energy being imparted to molecules through collisions to overcome such barriers easily (more on this concept in Chap. 5). Therefore, molecules are in constant motion (moving through space and colliding, vibrating (bond lengths and angles), and rotating about single bonds. There are conformations of lower energy, but the molecule keeps being knocked out of these.
• I have a question for problem number 4.18. In (b), the solution says that besides the 1,3-diaxial interaction, there's also interaction between CH₃ and CH₂CH₃ and in the ring-flipped form, between CH₂CH₃ and H. But later on in (d), what seem to me similar kinds of interaction are not mentioned. Why?
This is a little hard to do without structures. Look at the structures in the Answer Guide while considering my reply. In part b, the methyl group and the ethyl group are on adjacent carbons with one axial and one equatorial IN BOTH CONFORMATIONS. Therefore, the dihedral angle between them is 60 degrees - a gauche butane-type interaction! In one conformation there are two axial interaction between a CH₃ and axial hydrogens while in the other there are two axial interactions between a -CH₂CH₃ and axial hydrogens. Since the intereaction between CH₃ and CH₂CH₃ is the same in both conformations, it could be ignored (it cancels) in determining the energy difference. The energy difference between the two conformations is thus the difference between the steric strain of two axial interactions with a methyl (2 x 3.8 kJ/mol) and of two axial interactions with an ethyl group (2 x 4.0 kJ/mol) = 0.4 kJ/mol. In part d, the two substituents are not adjacent so there are no interactions between the substituents as in part b. Review the 1,2-dimethylcyclohexane case shown in Figure 4.3.

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This page was last modified on September 15, 1998