Can you clarify on the nomenclature? For example in the class notes, why wouldn't it be 4-ethyl-3,5,7-methyloctane? (Instead it's 5-ethyl-2,4,6-methyloctane, pg 5)

Alexis,

Page 7? The numbering always starts at the end closest to the first side group, so that means carbon number 2 will be the one with a methyl group if numbering from the left as the molecule is drawn. Starting on the other end, there is not side group until carbon # 3.

Hi, I have a few questions over the book homework problems.
1.) Why does 37b exhibit optical isomerism? Doesn't there have to be a carbon bound to 4 different groups for optical isomerism?

2.) In 63a, is there a reason for the numbering of the hydrocarbon base chain (I see that the longest chain is 9 carbons but there is a side group/substituent at the same number of C's in when counting from either side)?

3.) In 99, what makes the molecules able to be stereoisomers? Is it because if any of the bonds were flipped/written differently the tetrahedral shapes around carbons would prevent the exact same overall 3D shape?

Sawyer_D.....#37b: Carbon #3 from the left does have 4 different groups attached: H, methyl, ethyl, and butyl. #63s: We're splitting the nomenclature hairs really fine here but methyl come before phenyl alphabetically so it wins when they have the same carbon number in a chain. #99: The term stereoisomers includes both optical isomers (enantiomers) and geometric isomers (e.g., cis-trans isomers). So, look for carbon atoms with 4 different groups attached and/or C=C double bonds where cis-trans isomers may exist depending on the substituent pattern

Question #35: Why couldn't:
H3C—CH—CH2—CH2—CH3
...........|
...........CH2
...........|
...........CH3

be one of the nine structural isomers?
Also, some of the isomers lead with CH3, and others lead with H3C. There is not a difference right?

Question #47: It asks to list the products, but the answers only have the alkanes listed. Wouldn't HBr be a product for parts a&b? And HCl for parts c&d?

Also another question on 47, the substitution will only replace a hydrogen connected to a carbon on the base chain, right? Not a hydrogen connected to a carbon on one of the substituents?

Question #55: There is more than one way to draw a correct structure, right? For part b, I drew the structure as one straight line, still with the double bond and everything. Would that still be acceptable? The book kinda staggered it.

Question #99: Is part a an optical isomer because the 3rd carbon has four different groups attached to it? Is part b a cis-trans isomer because there is a methyl group on the second carbon and a methyl group on the 3rd carbon? Is part c a cis-trans isomer because there are two propyl groups on the 3rd carbon?

Allison_R.....#35: Your structure is the same as the 3rd one in the book's answer. Either way, it's 3-methylhexane, a 6-carbon chain with a CH3 group attached to the 3rd C of a 6C chain. #47: Indeed, HBr and HCl are products of those substitution reactions. Generally, you'll find that organic chemists tend to focus only on the organic reactants and products. Important "details" like writing balanced equations, etc. are often omitted. In Gen Chem, we know better, right? #55b: The kink in the chain makes the structure fit the column space better. Don't you see that? #99: Yes, you're correct on all accounts. Good job!

For #63a, along with 3,5-dimethyl-7-phenylnonane, could the compound be also named 5,7-dimethyl-3-phenylnonane?

Davis,

The 3, 5 name gives lower numbers to two of the groups. If we start on the end nearest one side-group in simple cases, then when there are two groups on the same carbon, that end become the end to begin counting.

On Question #51d, why does it say that there are six carbon groups instead of five? I thought the side groups didn't count? And isn't it ethyl, not dimethyl?

Alexis,

That"side group" is the main chain, bent out of what you like to see as the "line". See 2c and 2f on Dr. Patty's Additional homework for some more examples. Organic compounds are not all drawn with the main chain parallel to the edge of the page.

Hello,

I am confused by 2c on the additional problems. I don't see the fourth methyl group (on the 6 carbon, according to the name).

Thank you

For 2c,

I think I figured it out. Is the CH2 attached to the CH3 on carbon 6 a part of the skeleton? I confused it for an ethyl group.

For 3e, I am confused why carbons 3 and 6 have methyl groups and not double bonds.

When you have a chain with a double bond on one end and a group on the other end, do you start at the group carbon or double bond carbon (if they are an equal number of carbons from the end)? Referring to additionals 2d,e.

Emme,

Carbon #6 in 2c has no hydrogen atoms. It is bonded to two methyl groups (CH3)and the the rest of the chain. There is no ethyl group. Those carbons than bend down and to the right are part of the longest chain of 8 atoms. This is the to show how the longest chain can be anywhere.

2c

The longest chain starts on the left. there is a methyl group on C2 C 3 and there are two methyl groups on C6.

Hello,
On #11 of Dr.Patty's Additional Problems, I see that there are 2 blanks when asked about the hybridization of the expanded compound.
I put that the hybridization is sp with 180* degrees apart. Is this correct? If so, what else am I missing for this question?
Thanks!

Hello,

How do you tell if there is an ethyl group on a carbon? Looking at 37b, I didn't see the ethyl group because I thought the two left carbons were a part of the chain.

Jadan # 11

Yes, it is correct as shown in the answers at the end.

37 B.

There is not ethyl group for the purposes of naming. In this problem, you are not naming the compound. I is simply 3-methylheptane. What you are supposed to do is determine if there is and OPTICAL isomer. Going back to the lecture, where this is addressed, you need to see if there is an asymmetric carbon which makes the molecule chiral and give the possibility that there are optical isomers, (2 mirror image compounds).

A chiral atom has 4 DIFFERENT GROUPS. Stand at C# 3. Looking one way there is a H, another way is a CH3 group, another direction is a CH2CH3 group, and the fourth direction of the tetrahedral C has a CH2CH2CH2CH3 groups. So C3 is asymmetric, the molecule is chiral, and there are optical mirror image isomers called enantiomers.

Not sure if it's been asked - On Dr. Pattys additional problems ch.21. Problem #2, Part C, it is named ....octane, however the chain has only 7 carbons (if I'm doing this right?). Why would it be octa?

^^ I was think the same think as the longest chain is in the center

^^^Nevermind, I figured it out, it's like a "U" shape that gives you octa.

Yep, There ain't no ethyl group in 2c. Each one of those problems has incredible teaching value. So does the recorded lecture!

Also, for everyone (and there are lots of you), please see the snaking yellow line for the longest chain on page 7 of the annotated notes (not the typed problem, the one handwritten and explained in the recorded lecture) Oh are there recorded lectures? Hmm! Maybe I'd better see if they are still up on YouTube.

On Dr. Pattys additional problems, #3c, why is it "....cyclohexadiene"? Why are you writing cyclo? Looking at the Youtube lecture for Part C (around time 15:10), you did a similar example, but just wrote "....hexene".

How many double bonds in that ring. One, two, three, .....di, tri........

Book Problem Chapter 21 #51d

Why is the answer 2,4-dimethyl-3-hexene? I get that answer when I start from the left, however the double bond is closer to the right, therefore I started from the right, and got 2-ethyl,4-methyl-2-pentene.

Book Problem Chapter 21 #51d

Hexene is the longest chain. Chains don't have to be written in a straight line. Sometimes they bend up, down, back, forward just to fit into the space allotted. There is no ethyl group, See annotated outlines page 7, handwritten example done in the recorded lecture.

Book Problem Chapter 21 #51d

I understand chains don't have to be in a straight line, as I noted when I listened to the recorded lecture and saw page 7. However, on page 10 of the annotated outlines, under the alkene section, it says "number from the end closest to the double bond," so I guess I'm confused since 51d is an alkene and when I start from the side closest to the double bond I get pentene

Chapter 21 #51d

You can include the double bond in the bent chain. Start with C1 which is the first one on the left in this case. C4 is on the right side of the double bond. C# 5 is the C drawn below C# 4 and C#^ is the end of what you think is an ethyl group. That makes a chain of 6 C atoms.

Chapter 21 #51d

Also see pages 994 and 995 in the book

Chapter 21 #51d

So for alkenes we dont just number from the end closest to the double bond like page 10 says?... we find the longest carbon chain that includes the double bond and then number so the double bond has the lowest number possible?

Chapter 21 #51d
yes

For alkane free radical substitution reactions, will there only be one atom that can be substituted for a hydrogen? Like, if adding Cl2, would only one Cl be incorporated into the alkane?

Sawyer_D....Free radical substitution reactions are generally not selective, meaning that they occur unpredictably, often with multiple H substitutions leading to mixtures of products. Absent any other information, it's essentially impossible to predict the products. As a result, such reactions have relatively little practical utility, especially in laboratory scale reactions, in synthetic organic chemistry.

I need help comprehensively with #99. I understand that part a is a optical isomer because the third carbon is attached to 4 different groups. My question comes down to the next two parts. On part b the methyl groups are on the third and fifth carbon, not the second and third. Is this cis-trans because the methyl group on the third carbon is not able to rotate? Is there no way to say if it’s cis or trans because the other methyl group can still rotate? It’s just not very clear to me. On part c I do not understand how it could be a cis-trans isomer because there isn’t two propyl, there is only one.

For cis-trans isomerization to occur, you need a C=C double bond with two of the same side groups (e.g., 2 CH3 groups), one on each C of the double bond. A simple example is 2-butene (Tro, p 996), where the 2 CH3 groups around the C=C bond can be on the same (cis) or opposite (trans) sides. There is restricted rotation around the C=C double bond due to the overlapping p orbitals that form the pi component of the double bond. This occurs in #99b in a similar manner with the CH3 groups on carbons 2 and 3 (draw the structure out to see that there are actually 2 CH3 groups as in 2-butene). Thus, there are cis-trans isomers for 99b. The confusion, I think, occurs in 99c because the textbook's answer is wrong! When this structure is drawn properly, C-2 has a H and a CH3 attached while C-3 has 2 propyl groups attached. Since one carbon of the double bond has two identical groups, there is no possibility of cis-trans isomerization because it doesn't matter how the groups on the other carbon are arranged. Such differences are easier to see with simpler molecules. Compare 2-butene that has cis-trans isomers with 1-butene that does not. Once that makes sense, take another look at 99b and 99c.

Chapter 21 #73 D
I am confused on if this is a oxidation of an alcohol or an elimination? is the NaCr2O7 the Ox agent? And is the H2SO4 the concentrated acid.. ?
can you walk me through the steps to get to the products?

-Is it important to know about types of reactions not in the notes but in the book problems such as 73c (alcohol reacting with metal)? Also, should we be able to identify types of catalysts needed for certain reactions?
-On 91a, shouldn't the substituent other than methyl be sec-butyl not isobutyl?
-On 107b, why is the ratio 11:3?

Bella.....#73d: Yes, this is an oxidation reaction of an alcohol. Na2Cr2O7 is the oxidizing agent and H2AO4 keeps the solution acidic which is required for the oxidation reaction. Since this example is a primary alcohol, the initial product is the aldehyde. Aldehydes also will readily oxidize as well, especially with strong oxidants like dichromate, to form the carboxylic acid as shown. You should be familar with both steps of the oxidation overall process.

Sawyer....In general, emphasize the reactions covered in the notes but, as usual, "if it's in the homework, it's fair game."

I have a question about #2 on quiz 9. Could hydrocarbon A not also be an alkane undergoing substitution? If not, why? Also, when you ask for complete structural formulas do you only mean the fully condensed skeletal-type structure?

Sawyer_D.....(1) In general, organic reactions involve the conversion of one functional to another. Alkanes have no functional groups, hence very few useful reaction. While they do undergo free radical substitution, those reactions are not selective, meaning that you could never expect only 2 Br atoms to end up exactly where they need to be in the desired product. Also, those reactions require yhe molecular halogens (Cl2, Br2, etc.) not the hydrogen halides (HCl, HBr, etc.). On the other hand, alkenes readily undergo a wide variety of addition reactions with small molecules (H2O, HX, etc.), leading to alcohols, haloalkanes, etc. The orientation of addition is often directed by Markovnikov's rule, as in this HBr/diene reaction, to locate the Br groups specifically on C-2 and C-3 as needed. Review PowerPoint slide 13 again and also the textbook for addition reactions of alkenes. (2) "Complete structural formula" may be either fully-condensed (stick figure) or partially condensed (e.g., CH3CH2OH) as long as the C-C connectivity and location of functional groups are clearly shown. Other types of structures in the problem can be a guide as well.

When arranging parts of a name alphabetically, do we refer to the "b" in t-butyl not the t? (I saw an answer that seemed to do this...). Also, is sec-butyl the same as 2-butyl?

Sawyer.....Yes and Yes

For reactions that can involve multiple steps with the same type of reactant (such as alkynes undergoing additions to become alkenes then alkanes, or primary alcohols being oxidized to aldehydes which are then oxidized to carboxylic acids), if that reactant is in excess, should we just expect the final product of all of those steps? (like if we are asked for one product, should we go with the initial product of the first reaction or the final one, such as the alkene vs the alkane or the aldehyde vs the carboxylic acid?)

Sawyer.....It's important to know that such things are possible. However, unless the problem indicates otherwise, it's probably safe to assume that only the first step occurs.