hello everybody my name is Iman welcome back to my YouTube channel today we're going to do a practice problem set that relates to chapter 7 for MCAT organic chemistry this was alahh and ketones part two let's go ahead and get started this first problem says what is the product of the reaction below so if we look at these two molecules we're given we're given an alahh and an alcohol so we have one mole of alahh that's going to react with one mole of alcohol and they're going to react through a nucle iilc addition reaction to form a product that's called a Hemi aetl all right so what's going to happen here is this oxygen is going to attack right here at this carbon all right it's going to add itself right there this double bond if we're going to make a bond we got a break a bond this double bond is going to break and it's going to dump its electrons on the oxygen and the oxygen will become will uh be proteinated with a hydrogen here and what we're going to get is a Hemi aetl so in a Hemi aetl an alcohol group an O group so a a ether group a hydrogen atom and an R Group are attached to the same carbon atom so this carbon atom right here is going to have that R Group this methyl group ch3 it's going to have that oxygen that will get proteinated all right it's going to still have that Hy it's going to still have that hydrogen that's right there and then in addition this group that was just added n o and this is 1 2 3 C3 H7 all right so the molecule that we're going to form here is going to be molecule D that's exactly what we drew right here we have this carbon has an alcohol group has a hydrogen has this methyl and then it has this oc3 H7 group so problem number one is D two says the reaction below is an example of BL blank all right so what we see here all right this is an alahh an alahh that has a carbonal group this is the keto form all right and then what we see here is a double bond in an alcohol group this is the enol form and so this is going to be to toiz and toiz is the interconversion of two isomers in which a hydrogen and a double bond are moved so the keto and ters of alahh and ketones are common examples of toomer that you're going to see on the MC on teste all right note that the equilibrium lies to the left because the keto form is more stable just like we covered in lecture so the correct answer here for two is going to be B this reaction is an example of toiz fantastic 3 says which of the following reactions would produce the compound below so the reactions okay the reactions that are listed in the answer choices all right if we look at them are examples of aldol condensation now in the presence of a base the alpha hydrogen all right the alpha hydrogen is abstracted from an aldah forming an enate ion and then this enolate ion then attacks the carbonal group of the other alahh all right and it's going to form the Alo that we want so what what are those molecules going to look like so that we can follow that step by step all right so let's draw this out all right so we have the following molecule that we can start off with all right here is the alpha carbon so there's an alpha hydrogen here all right if this reacts with some sort of Base that base is going to take the hydrogen hydrogen's going to dump its electrons here to form a double bond and then this double Bond's going to break and dump its electrons on the oxygen all right so we essentially get a molecule then that has a resonance that happens in this area right here all right with with when those electrons are dumped now this all right in the presence of a base that Alpha hydrogen abstracted from an aldah forms this enolate all right this enolate ion that we see here okay c3h ch ch H all right and it has that negative charge because of those dumped electrons from the hydrogen now this enolate ion then attacks the carbonal carbon of an of another alahh molecule so if we draw the alahh that we started off with right here okay this enolate can attack the carbonal carbon all right right there at itself this double bond breaks off all right fantastic now it's going to form it's going to form the following molecule when it does that all right we're going to still have our carbonal Hy our alahh part right here all right and it's going to add right here our o group all right so we've added these two forms to form an alol all right so again let's go over this in the presence of a base the alpha hydrogen is stolen from an aldah and you form an enolate ion this is the enolate ion it can then attack the carbonal group of another alahh all right it can attack the carbonal group of another aldhy and so then what you can form when it does that is the Alo all right it can form an alol which is what we wanted ultimately so what we're going to need to start off with all right is ch3 ch2 CH this is the the the alahh that we started off with all right two of those two of the same alahh all right and so it's going to be answer Choice D fantastic all right so just to also help us picture where each carbon is going and also if these electrons are dumped on the oxygen it will get protonated from from from the uh uh it'll it'll still get proteinated so that's why you have an alcohol group what I also want to do before we move on is number this so this aldah is going to be 1 2 3 all right and the enolate ion we're going to call this carbon 4 five and six all right so the Enola ion here at position five attacks the carbon of the carbonal at position three all right and then the double bond in the carbonal dumps its electrons on the oxygen now you have a oxygen with extra set of electrons and a negative charge it will get protonated to form that alcohol group all right so here are and I didn't draw this properly this should have right here I think I accidentally erased it so what are our numbering which carbon is which in this aldol all right this is going to be 1 2 3 all right three is that carbonal carbon that we attacked that the Ana ion attacked we dumped the electrons on the oxygen it was O minus it got proteinated hence the alcohol right here all right and then right here is four five and six all right this is that methyl four right here and then 5 six is the alahh part all right of the enolate Fantastic so three is going to be answer Choice D four says why does the equilibrium between keto and enal toomer lie far to the keto side all right and one says the keto form is more thermodynamically stable two says the Eno form is lower energy and three says the enal form is more thermodynamically stable all right so let's review the points that we made in lecture the keto enal equilibrium lies far to the keto side because the keto form is significantly more thermodynamically stable than the Eno form and this thermodynamic stability stems from the fact that the oxygen is more electronegative than the carbon and so then the ketoers puts more electron density around the oxygen than the Eno toomer if the Eno toomer is is is less thermodynamic thermodynamically stable then it is the higher energy form all right it is higher energy than the keto toomer all right so considering all that then the only statement that's true is one the keto form is more thermodynamically stable it is the keto form that's lower energy not the en form so this is untrue and then three says the Eno form is more thermodynamically stable also untrue the keto form is more thermodynamically stable and it is the lower energy form all right the correct answer for four is going to be a only statement one was true all right five says the aldol condensation is an example of which type of reaction um is an example of which reaction types one says dehydration two says cleavage and three says nucleophilic addition the aldol condensation is both a dehydration reaction because a molecule of water is lost like we covered in the two-step process for aldol condensation in the lecture all right and it's also nucleophilic addition reaction because the nucleophilic enolate attacks and bonds to the carbonal carbon and so it is both a dehydration reaction and a nucleophilic addition reaction so both one statement one and three are true which means the correct answer for five is going to be B beautiful six says which of the hydrogens in the following molecule is the most acidic all right so let's look we have hydrogen a right here near a double bond all right we have hydrogen B right between these two carbonal we have hydrogen c um that's at this Alpha carbon we have hydrogen D over here now the the most acidic car uh uh the most acidic hydrogen is going to be hydrogen B this hydrogen is on the carbon that's between two carbonal which means that it's particularly acidic and this is due to both the inductive effects of the two oxygen atoms in the carbonal and the resonance stabilization of the annion between the carbonal groups just like we talked about in our lecture so the correct answer here for six is going to be B fantastic seven says when reacted with ammonia at 200° C which enolate of a carbonal containing compound would predominate a says kinetic enolate B says thermodynamic enolate C says neither enolate all right they'd be present in roughly equal proportions D says neither enolate this reaction condition would not form either inate okay cool all right so we have at 200° C all right this is pretty high temperature and we're reacting with ammonia ammonia is a weak base so at high temperatures and with a weak base like ammonia all right these are the exact conditions that would favor the thermodynamic enolate and so the reaction proceeds slowly with the weak base all right that giving the K kinetic inate time to interconvert to the more stable thermodynamic enolate all right so high temperatures weak base those are the right conditions for the thermodynamic inate to predominate so the correct answer for seven here is going to be B eight says which of the following compounds would be most reactive toward a nucleophile all right we have Pentel we have three pentan we have pentane and then we have two non noin my God is that hard to pronounce all right let's break this down alahh are generally more reactive than ketones all right and we said in lecture this is because the additional alkal group of a ketone is sterically hindering all right this alkal group is also electron donating so destabilizing that carbon carbo anion intermediate so right off the bat we can go ahead and eliminate the ketones here which is B and D all right now we have a alahh and we have an alkan the carbonal carbon is highly electrophilic alkanes they lack any significant electrophilicity so we're going to go ahead and cross out that alkane the answer here is going to be Pentel all right this alahh is going to be the most reactive from our answer choices towards a nucleophile nine says Alpha hydrogens of a ketone are acidic due to Blink once says resonance stabilization two says the electron withdrawing properties of the alkal group and three says the electr negative car carbonal oxygen so let's review when Alpha carbons are deprotonated that negative charge is resonance stabilized in part by the electro uh Electro negative carbonal oxygen all right and that Electro negative carbonal oxygen is electron withdrawing and also something to keep in mind is alkal groups they're actually donating which destabilizes carbo anion intermediates all right so if we look at our answer choices again Alpha hydrogens of a ketone are acidic due to Resonance stabilization this is true two says the electron withdrawing properties of the alkal groups this is false alkal groups are electron donating all right and they actually destabilize carbon annion intermediates so one is true two is false false three says the electronegative carbonal oxygen this is a true statement and so the correct answer for nine is going to be B both statements one and three are true statements fantastic 10 says which of the following is considered a toer of the amine functional group all okay fantastic we have S cyanohydrin hydrone enamine and semic carbo carosone all of the answer choices are nitrogen containing functional groups but like we discussed in lecture only enamines are ters of imines all right ammines contain a double bond between a carbon and a nitrogen and enamines contain a double bond between two carbons as well as an amine all right so the correct answer for 10 is going to be C 11 says when suin alahh is treated with lithium diisopropylamide LDA it blank does it become more nucleophilic does it become less nucleophilic does it generate a carboanion all right when suxin alahh or any alahh or Ketone with Alpha hydrogens is treated with a strong base like LDA it's going to form the more nucleophilic enolate carboanion all right it's going to form the more nucleophilic enolate carboanion so it does become more nucleophilic not less nucleophilic and it does indeed generate a carboanion so statement one and three are true and that means the correct answer for 11 is going to be C 12 says which of the following best describes the final product of an aldol condensation all right is it 13 dicarbon is it one two DIC carbonal is it Alpha Beta unsaturated carbonal or is it beta gamma UNS saturated carbonal so like we talked about in lecture alol condensations they it contains two main steps the first step the alpha carbon of an alahh or Ketone is deprotonated and when it's deprotonated it's going to generate the enolate carboanion all right so first step all right our Alpha carbon is deprotonated deprotonated and that forms an enal carbo anion that's that first step then in the Second Step all right well actually still part of the first step this carbon anion that's formed what does it do it can then attack another alahh or Ketone all right so this goes on to attack another alahh or Ketone and what does that form that gives us our alal all right that gives us our alal all right now in the Second Step all right that alol is dehydrated all right it's dehydrated to form a double bond all right and this double bond where does it form it's between the alpha and beta carbons so the molecule is now an alpha beta unsaturated carbonal that means the correct answer to 12 is is going to be C 13 says when benzal ahid is reacted with acetone which will act as a nucleophile now because benzal ahid lacks an alpha proton all right it cannot be reacted with base to form the nucleophilic enolate carbon anion and so what that means is the acetone will act as our nucleophile all right acetone will act as our nucleophile so off the B we can eliminate answer Choice A and B where they try to claim that benzal dhide is the nucleophile now in order to perform this reaction which is an aldol condensation acetone is going to is going to be reacted with a strong base not a strong acid in order to extract the alpha hydrogen and form the enal an ion which will act as a nucleophile all right so acetone will act as a nucleophile but only after reaction with a strong base the correct answer for 13 is going to be answer Choice D 14 says all right last but not least three hydroxy buttin L can be formed by the reaction of blank again this is an example of an alol condensation but stopped after aldol formation before the dehydration step so after the aldol is formed using a strong base the reaction can be halted by the addition of acid all right now let's look at these these answer choices we have methanol and Dy dthy ether ether we have ethanol in base then in acid and then we have C butanol in strong acid and then D methanol and ethanol in catalytic base okay buttin L is a strong acid all right butl in strong acid like we see in answer Choice C would actually likely be likely to deprotonate without gaining the hydroxy group all right and then uh methanal in DL ether it would not be reactive all right it would it would not be reactive because diethyl ether is not a strong enough base to abstract that Alpha hydrogen all right so right off the bat A and C are not good choices all right and then and then the reaction of the two alahh methanal and ethanal all right in catalytic base would actually form um three hydroxy propanel which would dehydrate which would dehydrate to form propenol not three hydroxy buttin l so that would also not be a correct answer the correct answer answer here if we want to form three hydroxy butl is going to be ethanol in base then in acid all right so 14 is going to be B all right with that we've ended our problem set for this chapter let me know if you have any questions comments concerns down below other than that good luck happy studying and have a beautiful beautiful day future doctors