In a basic solution, why are ketones more reactive than ethers? I know you said that ethers cannot be deprotonated by bases because they do not have an -OH bond, but neither does the ketone. (Most ketones have more acidic alpha-hydrogens. Also, a base can add to the C=O group. - 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)
Tuesday, November 29, 2016
In the lecture notes, it says, “Reactivity with nucleophiles is decreased by the stabilizing conjugation, which depends on the electron donating ability of X.” This confuses me because I thought that weak bases were good leaving groups and, therefore, more reactive to nucleophilic substitution? That is correct. (Weak bases usually hold an electron pair better, so stabilization of the carbonyl group decreases, and, therefore, reactivity increases. It is another part of the explanation of the basicity trend. - GS). So does that mean that strong acids are bad leaving groups and are more reactive to nucleophilic substitution because they want to donate their electrons?
During class when you were talking about Histones and the ability of the amines in them to form spikes, what did you want us to know for that? I understand if there’s a spike in the wound structure of the protein and DNA that it won’t unravel easily but what else is there to know about the importance of Histones and how they relate to the chemistry we are learning now? (Just know that the amino-groups of histones can act as nucleophiles in the nucleophilic acyl substitution with a thioester AcetylCoa. You do not need to know anything from biochemistry for the exam. - GS).
Tuesday, November 8, 2016
I know we went over the mechanism of alkylation of amines in class but could you explain it again? (It is SN2 with amine as a nucleophile: http://ochem.orgfree.com/sn2.html - GS). Could you also explain how alkylation of amines is different from acylation of amines? (Acylation of amines proceeds by the nucleophilic acyl substitution that will be discussed in unit 10. - GS).
Monday, November 7, 2016
Sunday, November 6, 2016
Monday, October 31, 2016
I know you touched on this at the end of class, but i am still confused. When trying to determine whether or not hydrogen bonds will form between a molecule and water we look for lone pairs, correct?? (Yes. - GS). What about when trying to determine if a molecule will form hydrogen bonds with itself?? (It happens when the molecule contains a both H-bond donor, and a H-bond acceptor. - GS).
What is the difference between alkylation and acylation of amines? Could you please go through an example of each? (Alkylation is introduction of an alkyl group to the amine’s nitrogen. Example - preparation of lecithin from methyl iodide. Acylation is introduction of an acyl group to the amine’s nitrogen. Example - synthesis of acetaminophen from para-aminophenol. - GS).
Wednesday, October 26, 2016
What is n in 4n+2? I understand that it is an integer but what number does it represent? Principal Quantum number? (No, it comes from the mathematical definition of an odd number of orbitals (2n+1). - GS). It is still not clear. (Each orbitals holds 2 electron, so the number of electrons = 2X(2n+1) - 4n+2. - GS).
Tuesday, October 25, 2016
Can you please review how to determine if a compound is aromatic or not depending on the number of conjugated electrons? I understand how to count the number of conjugated electrons but am unsure of how that relates to aromaticity. Thanks (An aromatic system must contain 4n+2 electrons (n=0,1,2,....) in a cyclic conjugation (uninterrupted by sp3-hybridized atoms. - GS).
Why is having a donor in the Meta position so unfavorable? (Since the sigma-complex has the formal positive charge distributed between ortho- and para-positions, the groups at those positions have the strongest effect on the reaction, so the donor speeds up ortho- and para-substitution much more than meta-substitution. - GS).
To clarify- if you have an incoming substituents in the metha or ortho position, there is a strong influence and if you have an incoming substituent in the ortho position there is little influence? (No. Since the sigma-complex has the formal positive charge distributed between ortho- and para-positions, the groups at those positions have the strongest effect on the reaction.The incoming substituent has almost no influence, because it attaches to an sp3-carbon with no formal charge - GS).
Monday, October 24, 2016
So far we learned about ortho and para positions but do compounds ever go to the meta position and if so when does that happen? (Electrophiles always enter the meta-position,, but preferably - when the existing group has no donating effects. - GS). Also can you go over why meta isn’t as favorable as ortho or para please, I’m still confused why it isn’t? (The sigma-complex has the positive charge distributed over the ortho- and para-positions, so it is stabilized better by ortho- and para- electron-donating substituents, which makes the ortho- and para-orientation preferred. Ortho- and para-substituents with no electron-donating effects are are having trouble withdrawing electrons from the positively-charged ortho- and para-positions, which makes ortho- and para-orientation unfavorable, so the meta-substitution is preferred. - GS).
Can you go over another example of the pi-complex and sigma-complex mechanisms? I don’t really know the distinction between the two quite yet. Thanks!! (They are steps of the same mechanism of aromatic electrophilic substitution: http://ochem.orgfree.com/SEAr.html - GS).
Friday, October 21, 2016
How do we know if lone pairs participate in regards to aromatic and antiaromatic compounds? (It participates if it can be perpendicular to the plane of the ring and, therefore, align with the pi-system. It will always happen if that space is not already occupied by a p-orbital of the pi-system. - GS).
Wednesday, October 19, 2016
Tuesday, October 11, 2016
I understand that hydroboration followed by oxidation is used when we need to perform hydration of a double bond against markovnikov’s rule, however, at what point is the markovnikov’s rule broken? According to markovnikov, the halogen of the hydrogen halide attaches to the carbon of alkene bearing the fewer number of hydrogen and greater number of carbons, so does this mean the rule is broken when the conditions are changed as to where it attaches? (That was the original rule formulation of Markovnikov’s himself. Now, we understand the rule broader: instead of halogen, it can be any negative part of the reagent. The rule usually works if addition proceeds through a carbocation, which is not the case for hydroboration. - GS).
Sunday, October 9, 2016
Does the SP3 bond break when its fragments are rotated? (There are no sp3 bonds. - GS). Also, can you please give an explanation on this concept. Thanks! (Sigma-bonds do not break, because the bond-forming orbitals (may involve sp3-orbitals), overlap in only one area. Pi-bonds break, because the bond-forming p-orbitals overlap in two areas, which have to disconnect during the rotation.. - GS).
Thursday, October 6, 2016
When selecting R or S configuration, I know that we 1. Rank substituents by highest priority (highest atomic mass) 2. Find out which way priority is decreasing 3. If priority is decreasing clockwise, then R configuration… If priority is decreasing counterclockwise, then S configuration 4. Change R configuration to S configuration, or vice-versa, if lowest priority group is pointing to the front….. I understand all of this, however, I always get confused when the lowest priority group is neither pointing towards you or away from you…. Do we still flip the configuration if the lowest priority group is in the plane of the paper? (Flip the lowest in priority group, which is in the plane of paper with the group behind paper, determine configuration as usual, and flip your answer. - GS). Or do we instead have to rotate the molecule in our minds in order to get the lowest priority facing away, and then go ahead and figure out the configuration? Spasibo. (This will also work. Choose either method, depending on your individual strengths. - GS). Your advice to flip the lowest priority, which is in the plane of paper, with the group behind the paper, determine configuration, and flip the answer sounds much better. That really cleared things up. Thank you!!
Actually, please explain the entire concept of Solvent Effects on Nucleophilicity. Please be specific on what we need to know. Thanks. (As opposed to protic solvents, aprotic solvents selectively solvate cations, leaving anions “untamed”. This reverses the nucleophilicity trend of anions in groups, such is from F to I - GS).
Wednesday, October 5, 2016
Tuesday, October 4, 2016
Tuesday, September 20, 2016
How do you determine the major product? I have been going over questions from the database and just when I think I’m doing it right I get the question wrong. Could you please clarify on that. (The major product is either lowest in energy (for equilibrium conditions), or the fastest forming product - GS).
Wednesday, August 31, 2016
Could you please explain how to tell the strength of an acid from looking at its structure? (For neutral acids, remove H+ and evaluate stability of the remaining anion (= conjugate base), taking into consideration electronegativity of atoms holding the negative charge, and number of resonance structures. - GS).
Monday, August 29, 2016
Thursday, August 25, 2016
Tuesday, April 26, 2016
Wednesday, April 6, 2016
Wednesday, March 23, 2016
How does a polar, protic solvent reduce the free energy of activation and stabilize an intermediate in an SN1 reaction? (By solvation of the very polar transition state - GS). And, why does a stabilized intermediate reduce the energy barrier? (Because the less polar starting compound is not stabilized as efficiently as the transition state - GS).
Tuesday, March 15, 2016
Dr. Sereda, I was looking over some of the previous year’s quizzes and in the preparation of 5-methyl-2-hexyne from propyne, what would be the nucleophile? Would it be the same as the alkylating agent? Thanks. (The nucleophile is deprotonated alkyne, which attacks the alkylating agent and kicks out the leaving group from it. Overall, it results in the introduction of the alkyl group to the alkyne. - GS).
Monday, February 22, 2016
Tuesday, February 16, 2016
How do you prove that the Beckmann’s rearrangement pathway where the leaving group and alkyl migration occur simultaneously is prefered over a pathway with a nitrenium ion intermediate? (For the simultaneous pathway, the group in the anti-position with respect to the oxime hydroxyl will migrate. If the nitrenium ion forms, both group will have a good chance to migrate. - GS).
Subscribe to:
Posts (Atom)