Monday, November 30, 2015


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).

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).

Is decarboxylation the loss of a carboxy-group in the form of carbon dioxide? (Yes. - GS).

Saturday, November 28, 2015


Is it correct to think that diethyl ether will not form a carboxylic acid upon acidic hydrolysis since it does not contain a carbonyl group? (Yes. - GS).

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


Esters make good soaps, right? (Not necessarily. - GS). To look for a soap, do you want to find something that has a long hydrocarbon tail with a polar end? (Yes, and the polar end should better be ionized. - GS).

For the “rule” that weak bases make good leaving groups, does that apply to every type of substitution  reaction? Or just certain ones? (For all reactions when the leaving group leaves with accepting an electron pair. - GS).

amides are more acidic than amines due to the conjugation of the electron pair with the carbonyl group, right? (Yes, because this conjugation is stronger in the conjugate base. - GS).
Is hydrolysis of esters the backwards reaction of Fischer esterification? (Only under acidic conditions. - GS).

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).

Why are amides less basic than amines? Is it because of the lone electron pair on nitrogen? (No. Because the lone electron pair on nitrogen is deactivated by conjugation with the carbonyl group. - GS).

Monday, November 16, 2015


I’m confused as to why cyclopentadiene is more reactive than trans,trans-2,4-hexadiene and cis, cis-2,4-hexadiene. Is it because of the s-cis conformation? Thanks. (Because cyclopentadiene is locked in the s-cis conformation. - GS).

How would you decide which molecule is more reactive between ethyl acetate and sodium acetate for nucleophilic acid substitution? Ethyl acetate is much more active than sodium acetate, because ethoxide is a weaker base than O2- - GS).

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


Dr. Sereda, for an ozonolysis reaction, can the reactant have two double bonds in its structure? And if so, how would the reaction proceed? Like which double bond would go through the reaction? (Yes. Multiple C=C bonds react independently, and the compound may split to more than two pieces. - GS).

Can you explain what an enolate ion is? (Conjugate base related to abstraction of the alpha-proton. - GS).

Dr. Sereda, what exactly is an alpha hydrogen? (Hydrogen bonded with the carbon next to carbonyl. - GS). and how do you determine the acidity of it? (Standard approach: mostly look at the stability if the conjugate base. - GS).

Being more sterically hindered makes a molecule more reactive, correct? (Steric hindrance increases potential energy. To connect it with reactivity, this change must be monitored through the whole potential energy diagram. - GS).

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).

How do I figure out what the starting materials for the ozonolysis are? (To figure out the ozonolysis products, completely break C=C bonds, and attach oxygens to the pieces. To go back to the starting materials, do the opposite. - GS). Thank you

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).