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We discussed some factors (electronegativity, resonance and ion size) which influence the acidity of an acid today in class. You can look for any of these features to help you estimate the strength (in general terms) of an acid. Organic acids are, for the most part, weak, but some are weaker than others. The weaker the acid, the stronger its conjugate base. Look, for example, at these two acids: acetic acid and chloroacetic acid - CH3COOH and ClCH2COOH. Which one do you think is more acidic? Both form resonance-stabilized resonance anions, so that can't help you to decide. However, Cl is electron-withdrawing (high electronegativity) and that usually increases the strength of an acid. The electron-withdrawing ability of the Cl polarizes the bonding electrons in the unionized acid. That effect is transmitted to the OH. This makes the O-H bond even more polarized than it is without the Cl in the molecule. This makes loss of the proton (H+) easier than it would be if the Cl were not in the molecule. What you need to do is examine the structure of the acid (or base) and try to analyze the effect of the structure on the acidity, using all the concepts you already know. If you wish further clarification, or more examples, I would be more than happy to talk with you.
2-chloro-2-methylpentane is achiral. There is no carbon which is bonded to 4 different atoms or different groups of atom. Check your models again. They should be superimposable.
I'm not sure which question or answer you are looking at. The trans-1-chloro-3-methylcyclohexane has to have one substituent axial, the other equatorial. There will be an equilibrium between the two conformers of the ring: one with the -Cl (e) and the -CH3 (a) and the other with -Cl (a) and -CH3 (e). The (energetically) favored conformer will be the one with the larger substituent in the (e) position. More of the molecules at any one time will have the -CH3 (e) and -Cl (a.) For the cis-1-chloro-3-methylcyclohexane, there are two chair conformers, one with both substitutents (a) and one with both substituents (e). Clearly it is the latter conformation that is highly favored in the equilibrium. Far more molecules at any one time will be "e,e" than "a,a".
I'd be glad to set up a time to see you. I'll check with you in class today. EKT
(1) Don't worry about the subscripts; I'm not sure you can do that when you submit a question. The ether compound (the one with the two oxygens) can only ACCEPT a hydrogen bond from another molecule. It can "accept" hydrogen bonds at its oxygens. It cannot, however, be a hydrogen bond DONOR. A DONOR must have a hydrogen covalently bonded to a HIGHLY ELECTRONEGATIVE ATOM such as O, N or F. The ether has its hydrogens bonded to carbons only. In order for a molecule to H-bond to other IDENTICAL molecules, which is what the question implied, it must be BOTH a donor and an acceptor. Dimethyl amine (the molecule with the N) can be both a hydrogen bond donor (it has a hydrogen covalently bonded to the "highly" electronegative atom N) and acceptor (it can accept a hydrogen bond from another dimethyl amine molecule at its N atom). (2) An example of an acid is H3O+. An example of a base is CH3NH2 (it's an organic base). An example of a base is CH3NH2; an example of ITS CONJUGATE ACID is CH3NH3+. "Conjugate" refers to a relationship between an acid and a base: it's an acid/base pair that differ from one another by a proton (H+). "A pair of compounds or ions related by gain or loss of one H+ is called a conjugate acid-base pair". (p. 556, "Chemistry and Chemical Reactivity", Kotz and Purcell, Saunders Publishing, 1987.
Allyl and vinyl are the "common" names for these groups. The IUPAC name for allyl is 2-propenyl and the IUPAC name for vinyl is ethenyl. According to Streitwieser, Heathcock and Kosower, "Introduction to Organic Chemistry", 4th ed., "the name allyl is derived from the Latin name for onion and garlic, plants that yield pungent and unforgettable aromas". The allyl group is present in a key amino acid isolated from garlic. I do not have information at hand about the "origin" of the common name vinyl. Perhaps you could find out more about it!??
You'll have to check out http://www.bris.ac.uk/Depts/Chemistry/MOTM/motm.htm to find out!
Check out the Web! Look at http://www.nobel.se/announcement-96/index.html
They continue to interconvert because the energy required to go from one to the other is small enough that it is available from the surroundings. If you make the activation energy (to go from one chair conformation to the other), high enough, you can get interconversion between the two to be very, very slow, such that, at any one time, almost all of the molecules are in the more stable conformation. This was pointed out in class with the t-butyl group. This group greatly favors the equatorial position. You can also make the activation energy "inaccessible". For example, "conversion of one chair conformation to the other of methylcyclohexane has an activation energy of about 10 kcal/mol and is very rapid at room temperature. However, by working at very low temperatures, workers have been able to obtain the pure equatorial conformers of chlorocyclohexane and trideuteriomethoxycyclohexane as solids and in solution at -160C". (Ref: March, Jerry, "Advanced Organic Chemistry", 3r ed., John Wiley & Sons, p.125)
By "larger" we mean "in size", sterically. Fluorine is the most electronegative atom on the Periodic Table, yet it is quite small, smaller than bromine or iodine, whose electronegativities are much smaller. The methyl group is bigger (sterically) than fluorine, but contains only C and H, neither one very electronegative. According to March, Jerry in "Advanced Organic Chemistry", 3rd ed., John Wiley & Sons, "In monosubstituted cyclohexanes, the substituent normally prefers the equatorial position because in the axial position there is interaction between the substituent and the axial hydrogens in the 3 and 5 positions [1,3-diaxial interactions], but the extent of this preference depends greatly on the nature of the group. Alkyl groups have a greater preference than polar groups and for alkyl groups, the preference increases with size. For polar groups, size seems to be unimportant. .....In disubstituted compounds, the rule for alkyl groups is that the conformation is such that as many groups as possible adopt the equatorial position."
Look at the reactants. Look at the products. Find bonds present in the products that aren't in the reactants. Determine by looking at the various reactants and products which bonds had to be broken and which had to be formed to go from reactants to products. You then need the BDE of the bonds in question (those that have changed). The energy required to break a bond is +; the energy released on bond formation is -. The number you multiply by is determined by how many of each of the bonds is broken and/or formed during the reaction. In the reaction between methane, CH4 and chlorine, Cl2, to give CH3Cl and HCl, we had to break only 1 C-H bond (of the methane) and 1 Cl-Cl bond and we formed only 1 new bond, a C-Cl bond in the product hydrocarbon (the other 3 C-H bonds remained unchanged from the reactant). We also formed a H-Cl bond. In the combustion of propane, CH3CH2CH3, all the C-C bonds (2 of them) and all the C-H bonds (8 of them) had to be broken to get to the products (CO2 and H2O. Therefore the BDE of the C-C bond was multiplied by 2 and of the C-H bonds, by 8 (except the BDE of the terminal C-H bonds is slightly different from the two internal C-H bonds so we differentiated between the two). The difference between the energy we put in (BDE) when we have to break bonds (+) and the energy we get out when we form bonds (-) is the overall delta H for the reaction.
Yes. In fact, all saturated hydrocarbons are classified as ALIPHATIC.
It has significance in that it determines the molecular formula, and isomers (structural and others) by definition have the same molecular formula but are not identical molecules (not superimposable). Structural isomers (also called constitutional isomers) do not, however, have to have the same type of bonds. That is, I can write 3 structural isomers with the molecular formula C2H4O. Two of them have double bonds and one has only single bonds (but it has a RING). See the quizzes that are posted on the 4th floor. You cannot actually "calculate" how many structural isomers can be written for a given molecular formula.
In order to be a hydrogen bond donor, a molecule must contain an H covalently bonded to a highly electronegative atom like O or N. In order to be a hydrogen bond acceptor a molecule must contain an O or N. If a molecule has neither or these features, it cannot act as a hydrogen bond donor or acceptor. CH3OH is both an acceptor and a donor; CH3OCH3 is an acceptor only. CH3Cl is neither.
One way is to name them (completely and correctly). If they have the same (correct) name, they are identical. However, if you cannot name them, you must find other ways. Models are the best. However, if you do not have models available, you have to work with the 2-D structures. Try changing the bond ANGLES (do not break bonds and move atoms from one place to another) and comparing. If you can get them to be superimposable, atom for atom, they are identical. I think it is easier to work with mostly condensed structures (don't draw out the C-H bonds) to compare structures.
Dr. Thornton has distributed a sheet which may be of help here. Remember that angles on paper do not necessarily coincide with the angles for the "real" molecule. Bending a "line" of carbon atoms so that there is a 90 degree "kink" does NOT change the bonding of the atoms, and therefore, does not make a different (that is, nonsuperimposable) molecule. For example, for heptane (see sheet on writing structural isomers), there is only 1 way you can bond all 7 carbons "in a line", no other isomers (for the "linear" array of C atoms). Bending it does not produce an isomer.