If the pKa is larger than the pH are most of the protonated amino groups going to be deprotonated? (No. The equilibrium will shift to the acidic side, that is the protonated form. - GS).
This blog is a learning resource for undergraduate students studying organic chemistry. It consists of a database of actual questions and answers about organic chemistry collected by a chemistry professor teaching the subject.
Topics
- Halides-alcohols-ethers (94)
- Conjugation (60)
- Alkenes-alkynes (51)
- Carboxylic acid derivatives (39)
- Structure-bonding (39)
- Aldehydes and ketones (36)
- Alkanes-cycloalkanes (31)
- Stereochemistry (27)
- Spectroscopy (21)
- Acids-bases (19)
- Amines (18)
- Useful links (2)
Monday, November 30, 2015
Dr. Sereda, in order to separate a mixture containing a hydrocarbon and an amine, I am thinking you would first want to dissolve the mixture in diethyl ether since they are both soluble in this, transfer this into a separatory funnel, and then add a strong acid which will react with the amine, converting it into an ammonium salt which is soluble in water. So then in the separatory funnel, the organic layer would include the hydrocarbon component and the aqueous layer would include the amine. (The aqueous layer will contain the ammonium salt. - GS). Is this method correct? (Yes, but not complete. Now you need to get your amine back. - GS).
Saturday, November 28, 2015
Sunday, November 22, 2015
For increasing reactivity of carboxylic acid derivatives in nucleophilic acyl substitution, the order of reactivity goes nitriles<amides<esters and acids<anhydrides< acid chlorides, right? (Almost. Remove the acid from the row, because it forms a salt instead of undergoing acyl substitution. - GS). If so, is that the same reactivity for hydrolysis? (Yes. - GS). Thank you!
Thursday, November 19, 2015
Tuesday, November 17, 2015
Dr. Sereda, Im a little lost on the reaction of anhydrides part, I know you covered it in class today, but could you maybe skim over it again please? (The leaving group is acetate, and the reaction does not produce a strong acid, so does not require an external amine to neutralize it. For cyclic anhydrides, the flavor of the reaction is that the “leaving group” leaves the reaction center, but stays a part of the molecule, to which it is tethered. - GS).
Monday, November 16, 2015
Thursday, November 12, 2015
Could you explain why the carbonyl group of ethyl acetate is more stable than that of ethyl thioacetate? (The carbonyl group is more stabilized in esters, because the lone electron pair on oxygen (period 2) better fits by size to the pi-system of the C=O group ((period 2) than the lone electron pair on sulfur does (period 3).
Friday, November 6, 2015
So an aldehyde is when the C in the carbonyl group is connected to a CH3 group and an H? (H and C or two H are bonded with the carbonyl. - GS). And a ketone is when that H is something else right? (Two carbons are bonded with the carbonyl. - GS). So the difference (in structure) between them would be just that one “R” group? (Presence of H makes the difference. - GS).
Tuesday, November 3, 2015
If you are comparing two compounds with only primary amines, and one compound has more primary amines than the other, would the one with more have a higher boiling point? (True, if there are no groups that make H-bonds stronger than amino-groups (e.g. alcohols, carboxylic acids. - GS). If so, would this have to do not only with hydrogen bonds, but also van der waals forces? (Van der Waals forces are negligible compared with H-bonds,- GS).
Thursday, October 29, 2015
How do you distinguish the basicity levels for amines? Like what makes an amine a weaker/stronger base? And why? (The more the lone electron pair of N is involved in conjugation, the less basic it is. Also, sp2-hybridized nitrogen is less basic than sp3-hybridized nitrogen, because the lone pair has lower energy and, therefore, less reactive. - GS).
Wednesday, October 28, 2015
Resonance structures provide a means to show electron delocalization which in turn stabilizes the structure, right? (Yes. - GS). So for simplicity’s sake this means that structures that can be demonstrated using resonance structures for it are more stable than a similar molecule that cant form resonance, right? (Correct. - GS).
^^Could you explain why electron delocalization would be seen as stabilizing if a carbocation or anion would constantly result from this formation? (Electrons will have more space and more opportunities to lower their energy. Same on the math language: there will be more coefficients to optimize in the linear combination of atomic orbitals - GS). Wouldn't a neutral structure with no charged carbons be more stable than one that would observe resonance characteristics? (The neutral structure will be most contributing. - GS).
How do you know which reaction product is less stable? For example, in the addition of HCl to conjugated dienes, the major product formed is 1,2-addition results in major product which is less stable than 1,4-addition. Is 1,4- just typically more stable? (Yes, because of a more substituted C=C bond - GS).
Are halogens inductive acceptors and resonance donors? (Yes, but Cl, Br, and I are also resonance acceptors because of the empty d-orbitals - GS). So, are halogens resonance donors and resonance acceptors? (No. They are resonance donors and inductive acceptors. Cl, Br, and I are also resonance acceptors because of the empty d-orbitals - GS).
Monday, October 26, 2015
Thursday, October 15, 2015
Why would a compound that mimics a plant growth hormone be used as a commercial lawn fertilizer? Wouldn't that just cause the plant to grow more quickly? (Yes, it is not a nutrient. In larger concentrations, they are used as herbicides, because they make the plant grow so unnaturally fast, that the plant dies quickly. - GS).
Brominated organic molecules are used commercially as flame retardants, why is that? (I thinks that the three following contributors are important: 1. Br-containing molecules act as traps for free radicals, so they speed up free radical chain termination; 2. Non-combustible molecules (HBr and Br2) are produced, and they extinguish fire like carbon dioxide; 3. Bromine does not bond well with oxygen, so the energy spend on breakage of the C-Br and C-H is not compensated enough by a relatively weak H-Br bond. - GS).
Alcohols are a weak acid, which is why we can intake alcohol without experiencing severe irritation, right? (Ethanol causes irritation when its concentration exceeds 40%, which corresponds to the hydrate of ethanol. Above this concentration, ethanol acts as a drying agent. It was a part of M.S. thesis of Dmitri Mendeleev, which is much less known than his periodic table of elements. - GS). But just out of curiosity is the fact that alcohol is indeed a acid (even though its weak) one of the reasons alcoholics experience extreme symptoms such as ulcers (which could be described as an erosion or “break” of intestinal lining, etc.)? Is that alright to observe those effects in that manner, or is that side effect due to a different characteristic altogether? (No. Those adverse reactions are mostly because of oxidation of ethanol to a toxic ethanal. - GS).
So is it wrong to think that more polar molecules often have lower boiling points if the molecular size is similar? (Polar molecules have higher boiling points because of stronger electrostatic intermolecular interaction. - GS). Larger molecules have higher boiling points? (Yes, because of stronger Van-der-Waals intermolecular interaction. - GS).
Wednesday, October 14, 2015
Could you go over how the free radicals one more time? I still don’t fully understand it. Thanks (Ok. Please, also review this: http://ochem.orgfree.com/SR.html - GS).
Hi Dr. Sereda, I was wondering if you could explain a little better the relationships with the boiling points, polarity, molecular size, and molecular shape? (Molecular shape determines polarity. Polarity and molecular size, and again shape, determine intermolecular interactions. Intermolecular interactions determine the boiling points. - GS).
Is a rearrangement reaction going to take place when the conditions are in line with a SN1 reaction? Because a rearrangement reaction is where a carbocation gets reformed into something more stable and in an SN1 reaction is where one of the steps to get to the product is through a carbocation intermediate. (Yes. SN1 and any other reaction involving carbocations can be complicated by carbocation rearrangements. - GS).
Dr. Sereda would you mind explaining how we differentiate between SN1 (http://ochem.orgfree.com/sn1.html - GS) and SN2 (http://ochem.orgfree.com/sn2.html - GS). reactions? (See the links that I inserted to your question. - GS).
Tuesday, October 6, 2015
Different compounds can be expressed in terms of optically pure compounds as (+) or (-), could you explain how we are supposed to differentiate between when its (+) or negative? Is it simply in regards to the rotation it expresses so it has a negative then it would be because it is the enantiomer of the optical compound? (If a compound has positive rotation, its enantiomer will be for sure negative, and vice versa. You cannot determine an absolute value or sign of rotation just by the structure. An experiment is needed. - GS).
^^Sorry this may be naive, but what do we gain by knowing if its (+) and (-)? (It is part of a compound’s identity and purity, like its melting point. - GS).
^So you could use it as another way to identify a compound? (Yes, and evaluate its purity, which also embraces monitoring chemical reactions of that compound. Mutarotation is a great example. - GS).
Monday, October 5, 2015
Friday, October 2, 2015
Tuesday, September 29, 2015
The relative density of a polymer is due to branching factors mainly, right? So would it be correct to assume that thicker materials such as thick plastic sheets (for example) would be due to a polymer that is less branched (aks more “linear” of a structure) (Yes. - GS). whereas the plastic for packaging is more branched making it more flexible? (Flexibility is mostly determined by conformational mobility. - GS). Just out of curiosity, how would you go about manipulating this concept, so that you could produce a compound of desired density? (In industry, radical-induced polymerization of ethylene causes the radical centers hopping along the chain, which causes branching. Polymerization directed by a metal (Ziegler-Natta catalyst), produces mostly linear chains. An extreme example of this type of manipulation is making aerogels in supercritical carbon dioxide. - GS). Meaning if I were to say add heat to the compound/ substance the material would become more flexible? (Yes, which is why under freezing temperatures, some plastics and rubbers become inflexible, fragile, and may fall apart. It is a big issue for aircraft landing gear and spaceships. - GS).
Monday, September 28, 2015
Rate limiting step is that determined by the step with the highest Transition state only or is it the largest difference between the minima (intermediate) and the transition state that determines the rate limiting step? (The difference between the transition state and the preceding intermediate or starting material determines the activation energy and, therefore, the reaction rate. - GS).
Sunday, September 27, 2015
Tuesday, September 22, 2015
Did you say that trans is typically more stable than cis ? (Yes, if there is no attraction between the substituents. - GS). if so would the same apply for Z configuration in terms of it being more stable than E? Thanks (There is no trend. It all depends on the repulsion and attraction between specific substituents. - GS).
Monday, September 21, 2015
Wednesday, September 16, 2015
Monday, September 14, 2015
Sunday, September 13, 2015
Friday, September 11, 2015
Tuesday, September 8, 2015
I know that angular strain decreases stability, and that a larger ring decreases the effects of angular strain, so in general a larger cyclic ring would be more stable but how would trans and cis play an effect? Would you still consider a larger ring to be more stable even if trans were involved in a slightly smaller ring? (Stability of cycles sharply goes up from 3- to 6-membered ring, then slightly decreases and remains virtually flat. Trans- is usually more stable than cis-, however, conformational analysis should be performed on a case-by-case basis. Except for the textbook-easy conformational analysis of cyclohexane, for larger cycles it is really convoluted and requires full-blown computational research. The good news is that overwhelming majority of useful cyclic compounds have 6-membered rings! - GS).
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