Nitrogen and Phosphorus Containing Compounds

hello everybody my name is Iman welcome back to my YouTube channel today we&#39;re going to do a practice problem set that relates to our nitrogen and phosphorus containing compounds lecture let&#39;s go ahead and get started problem one says which of the following amino acids does not have an L in tumor in other words this problem is pretty much asking you which of these following amino acids it doesn&#39;t have a chyal center it&#39;s Alpha carbon is not a chyal center now when we were discussing amino acids we said that every amino acid except one that Alpha carbon had four different groups attached to it the exception was glycine now here you see the structure for glycine if we look at that Alpha carbon we have that amino group we have that carboxy group it has that hydrogen it&#39;s our group is another hydrogen so that Alpha carbon does not have four different groups attached to it therefore that Alpha carbon is not chyal and glycine is not going to have a an antier all right so in short glycine&#39;s R Group is a hydrogen atom and so this amino acid is a chyal because the central carbon Alpha carbon is not bonded to four different substituents the other amino acids are all chyal and therefore have both L and D in anti that means the correct answer for one is D two says which of the following would be formed if methyl bromide were reacted with thide and followed by hydrolysis with an aquous base now this reaction is similar to the Gabriel synthesis we saw all right we have the same nucleophile the electrophile is a little bit different but pretty much the same kind of mechanism theide acts as a nucleophile the methyl carbon acts like an electrophile and then this bromine group this this bromide I should say acts as the leaving group and so this reaction between methyl bromide and thide results in the formation of methyl thide and then if you&#39;re going to take that reaction and then have subsequent hydrolysis that occurs what&#39;s going to happen is you&#39;re going to form the following all right this part of the molecule is going to to leave all right and if you remember from our Gabriel synthesis what happens here near the end is that you&#39;re going to form this part of the molecule it&#39;s going to get split I guess we can say split just to help us visualize what happens when we treat it o this is not letting me erase let&#39;s try to erase this perfect when we&#39;re doing that last step where it&#39;s followed by hydrolysis with an aquous base what we get in that step is now we&#39;re going to have the following molecule all right this is a carbonal all right all right and then this group is essentially after getting proteinated we&#39;re going to have this group right here all right so that subsequent hydrolysis is going to yield this is called a methyl amine all right and so that is going to line up with answer Choice B all right just as a reminder I&#39;m going to go back to our notes cuz we did talk about the Gabriel synthesis all right we we started off with potassium thide depicted right here I&#39;m going to circle it in blue all right and this was the electrophile what you notice is that this nitrogen gets connected to this group here with bromine as the leaving group and then there&#39;s you treat it with base it gets deproteinated then you treat it with um this bromo alane you have an sn2 reaction you get the following R group that attaches to that carbon and then you treat it with base all right and you you allow it to get hydrolized in the presence of base and what you notice is whatever group was attached here all right whatever group was attached here is removed all right and protonated and then the ends of these two carbonal groups in this molecule this what was once thide gets you know ox oygen at each end negatively charged oxygens so that&#39;s exactly what&#39;s happening here in this problem all right that&#39;s how we figured out that what we get as an end product here is methyl amine answer Choice B for two fantastic 3 says which of the following amino acids contains sulfur we have cysteine Serene and methionine all right so of course for for for the MCAT please be familiar with your am you know acid structures their names their full names their three-letter code names and their onlet code names all right and so here we can start to go through each op option all right we have cistin if we look at cysteine what we notice is that it does have a sulfur atom all right so cysteine does contain a sulfur check good what about saring all right saring is right here let&#39;s find it all right does this amino acid have a sulfur atom I don&#39;t see any sulfur atoms so no all right and then methon 9 if we look at methon 9 right here also it has a sulfur atom so out of these three options cysteine and methionine both have and both contain sulfur so 1 and three the correct answer here is going to be B four says nylon a polyamide is produced from hexane diamine and a substance X this substance X is most probably a blank now this requires us to recall what how do you form an amide an amide is formed from an amine plus a carboxy group or it&#39;s AAL derivatives now in this question the first thing that you&#39;re given all right in the production of nylon is you know that you need hexane diamine so you have the amine in the problem right and amine is already given in this question and so the compound that needs to be identified this substance X it has to be an AAL compound and the only ACL compound that&#39;s given in these choices is going to be the carboxilic acid and so just working off that logic the answer here is answer Choice B fantastic five says intermediates in the streker synthesis include all of the following nitrogen containing functional groups except blank now during the ster synthesis what we talked about was that ammonia attacks a carbonal all right ammonia attacks a carbonal forming an amine so we can obviously cancel out answer Choice B this amine is attacked by a cyanide forming an amine so we can go ahead and cancel out answer Choice D and then what you form is also a nitrol all right which cancels out answer Choice a all right what we don&#39;t form is an amide amide bonds are formed between amino acids but they don&#39;t appear during the ster synthesis so if you remember the ster synthesis it had two big steps all right in step one all right you start with an alahh ammonium chloride and potassium cyanide all right the carbonal oxygen is proteinated we can even look at our mechanism here just to remind mind ourselves all right so um the carbonal oxygen is protonated and then what we see is that ammonia can attack the carbonal carbon forming an amine all right and then the amine carbon is susceptible to nucleophilic attack by the cyanide all right by the cyanide CN anion from potassium cyanide it attacks all right and now you get at the end of Step One an amino nital and then in step two the nital nitrogen is protonated and then a water molecule attacks leading to the creation of a molecule with a bond with both an amine and a hydroxy group on the same carbon all right and then this amine um is attacked by another equivalent of water a carbonal is formed kicking off ammonia creating the carboxilic acid functionality all right and then essentially what you end up with at the end of all this is your amino acid all right so in that second step the amino acid is generated from that Amino nital so again what we see is that there is a nital group there&#39;s an amine and there is an amine what there isn&#39;t is an amide so five all right is C all right we see all of the following nitrogen containing functional groups except amides fantastic six says a biochemist is syn sizing veine shown below using the struer synthesis which of the following carbonal containing compounds would be a propriate starting reactant in this synthesis now the ster synthesis creates an amino acid from an alide the carbonal carbon ultimately becomes the alpha carbon of the amino acid any remaining alkal chain becomes the R Group so if we look at this right here right any remaining alkal chain becomes the R Group all right and so if we&#39;re starting with for example um two methyl propanel all right and then we treat it with ammonium chloride and and and and potassium cyanide all right we can get the following amuno nital all right and then it can form the following amino acid which is exactly what looking for all right what we notice is this R Group determines everything all right again again what this needs to be is is two methyl propenol so this R Group if it was two methyl propenol it would be able to with the starting aldah right here all right with this starting aldah you would be able to go through the struer synthesis all right form this am this this amuno nital where the R Group is again two methyl propanel and then in the Second Step form that amino acid where the R Group is that again two methyl propanel and if that is the group that is going to be here in the R Group then what we get is veiling all right and so what really did what what&#39;s really crucial to take away from this ster synthesis is that whatever amino acid you are forming you have to be cautious you have to be informed about what its R Group is because that R Group needs to be present in your starting aldah so if you look at this amino acid this veiling here&#39;s the alpha carbon here&#39;s the amino group here&#39;s the carboxy group there&#39;s a hydrogen and then there&#39;s this group right here this group right here is what&#39;s going to be added right here as an R Group to this aldah and it will be called two methyl propanel all right and so that is our appropriate starting group this is two methyl propanel propanel that is our starting material here is where what we would get the amino nitril all right and then this is the ending amino acid fantastic so then the answer here for six is going to be C 7even says why is the CN Bond of an amide planer all right the answer choices say one it has a partial double bond character two it is sp3 hybridized and three says it has some SP2 character all right why is the CN Bond of an amide planer now one resonance structure of a CN bond in an amide has the double bond between the C and the n and not between the C and an O and so what that means is that the CN Bond of an amide it it has some SP2 character and SP2 hybridized atoms exhibit planer geometry all right and so it has partial double bond character all right and it has some SP2 character both of these play a role in why the CN Bond of an amine is planer and so we&#39;re looking for one and three the correct answer is going to be D now a says which of the primary methods of amino acid synthesis results in an optically active solution all right a says the stuer synthesis B say only B says the Gabriel synthesis only C says both D says neither now we covered a really important point when we were talking about both the ster synthesis and the Gabriel synthesis and we said that they both contain planer intermediates and that&#39;s really important because if they have planer intermediates they can be attacked from either side by a nucleophile and if that&#39;s the case then that&#39;s going to result in a rasmic mixture of an aners if you have an equal rmic mixture of enantiomers the solution will therefore therefore be optically inactive all right so both the streer synthesis and the Gabriel synthesis result in products that are a rasic mixture of inan humors and if you have a racemic mixture all right it&#39;s going to be a solution that&#39;s optically inactive and that means if we&#39;re looking for uh which primary method of amino acid synthesis results in optically active solution the answer is neither all right neither of these result in optically Active Solutions both of these methods result in the production of rasmic mixtures of inan tumors and th that results in the solution being optically inactive so eight is D nine says during the Gabriel synthesis thite serves as the blank all right during the Gabriel synthesis if you remember thite acts a attacks um a secondary carbon in D um ethyl bromomalonate all right that the secondary carbon is the electrophile bromide is the leaving group and thide is the nucleophile so if we go right back here just to refresh our memories this thide all right this is our nucleophile all right this right here this carbon is our electrophile all right bromine right here is our leaving group so thide is our nucleophile nine is a beautiful 10 says each of the following reaction types occurs during the Gabriel synthesis except all right so this is really good because what you&#39;re noticing here in this practice problems is that these are very big picture questions about the struer synthesis and the Gabriel synthesis now if you remember the Gabriel synthesis we can go back and look at it all right it includes two and I&#39;m going to use my orange highlighter here two nucleophilic substitution reactions all right followed by hydrolysis all right hydrolysis and decarbox all right so it has sn2 reactions two nucleophilic substitution reactions hydrolysis and decarbox if we look at our answer choices we can go ahead and cancel out a b and d dehydration the loss of a water molecule is actually not part of this synthesis and so each of the following reaction types occurs except dehydration 10 is C beautiful 11 says at physiological pH which two forms of phosphoric acid have the highest concentrations so the pka2 of phosphoric acid is going to be really close to to physiological uh pH and so what that&#39;s what that means is that we&#39;re going to have a h24 concentration pretty much almost equal to hp42 minus so the pka2 of phosphoric acid is close to physiological pH and so it would have lost one of its hydrogens already since this is at at pka2 and it&#39;s going to be approximately equal to the concentration of um hpo4 all right so h24 and hpo4 are going to be um the highest concentrations are going to be about equal and so the answer for 11 is going to be B 12 cells in aquous solution pyrro phosphate will likely blink form insoluble complexes be stable and inert degrade into inorganic phosphate or decrease the polarity of the solvent now pyrro phosphate is unstable in aquous solution all right and it will degrade to form two equivalents of inorganic phosphate all right now that is going to be answer Choice C it will degrade into an organic phosphate now the solvent is water which should retain its polarity regardless of the presence of solute so that regardless of the presence of solutes that&#39;s going to eliminate answer Choice D and pyrro phosphate and inorganic phosphate they are small charged molecules all right which are are relatively soluble all right so again these answer choices don&#39;t make any sense 12 is C pyrro phosphate will degrade into likely degrade into inorganic phosphate 13 says what would be the charge of aspartic acid at ph7 all right so the amino acid in question is aspartic acid which is an acidic amino acid because it contains an extra carbox group as part of its rchain all right just to remind you I&#39;m going to quickly go back to our amino acid table and I&#39;m going to circle asart uh aspartic acid or aspartate right here all right there we go all right notice how it has an extra carboxy group as part of its rchain at neutral pH both the carboxy groups are ionized so that means there are two negative charges one over here one over here here all right only one of those charges is neutralized by the positive charge that would be at the amino group all right and on that amino group so the molecule is going to have what it&#39;s going to have one negative charge plus one negative charge plus one positive charge for a total of a minus one negative charge so the molecule has an overall negative charge the answer for 13 is going to be 14 says when a bond is created between two nucleotide triphosphates in DNA synthesis the small molecule released from this reaction is blank now as DNA is synthesized if you remember from our discussion it forms phosphodiester bonds and it releases pyrro phosphate now pyrro phosphate is an inorganic phosphate containing molecule but it is not the single phosph pH at group commonly referred to as an organic phosphate all right and so when a bond is created between two nucleotide triphosphates in DNA the small molecule released is going to be pyo phosphate all right so 14 is a now last but not least 15 says the hydrogens of phosphoric acid have PKA values that blank a says allow High buffering capacity over small PH range B say allow moderate buffering capacity over a large PH range C says allow low buffering capacity over a small PH range and D says do not allow buffering now phosphoric acid it has three hydrogens all right H3 P4 all right has three hydrogens with PKA values that are spread across the PH range this allows some degree of buffering over almost the entire standard PH range from 0 to 14 all right now this means that it&#39;s we wouldn&#39;t say high buffering capacity but we would say it allows a moderate buffering capacity over a large PH range so the answer for 15 is going to be B all right with that we&#39;ve completed our practice problem set 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

hello everybody my name is Iman welcome back to my YouTube channel today we&amp;#39;re going to do a practice problem set that relates to our nitrogen and phosphorus containing compounds lecture let&amp;#39;s go ahead and get started problem one says which of the following amino acids does not have an L in tumor in other words this problem is pretty much asking you which of these following amino acids it doesn&amp;#39;t have a chyal center it&amp;#39;s Alpha carbon is not a chyal center now when we were discussing amino acids we said that every amino acid except one that Alpha carbon had four different groups attached to it the exception was glycine now here you see the structure for glycine if we look at that Alpha carbon we have that amino group we have that carboxy group it has that hydrogen it&amp;#39;s our group is another hydrogen so that Alpha carbon does not have four different groups attached to it therefore that Alpha carbon is not chyal and glycine is not going to have a an antier all right so in short glycine&amp;#39;s R Group is a hydrogen atom and so this amino acid is a chyal because the central carbon Alpha carbon is not bonded to four different substituents the other amino acids are all chyal and therefore have both L and D in anti that means the correct answer for one is D two says which of the following would be formed if methyl bromide were reacted with thide and followed by hydrolysis with an aquous base now this reaction is similar to the Gabriel synthesis we saw all right we have the same nucleophile the electrophile is a little bit different but pretty much the same kind of mechanism theide acts as a nucleophile the methyl carbon acts like an electrophile and then this bromine group this this bromide I should say acts as the leaving group and so this reaction between methyl bromide and thide results in the formation of methyl thide and then if you&amp;#39;re going to take that reaction and then have subsequent hydrolysis that occurs what&amp;#39;s going to happen is you&amp;#39;re going to form the following all right this part of the molecule is going to to leave all right and if you remember from our Gabriel synthesis what happens here near the end is that you&amp;#39;re going to form this part of the molecule it&amp;#39;s going to get split I guess we can say split just to help us visualize what happens when we treat it o this is not letting me erase let&amp;#39;s try to erase this perfect when we&amp;#39;re doing that last step where it&amp;#39;s followed by hydrolysis with an aquous base what we get in that step is now we&amp;#39;re going to have the following molecule all right this is a carbonal all right all right and then this group is essentially after getting proteinated we&amp;#39;re going to have this group right here all right so that subsequent hydrolysis is going to yield this is called a methyl amine all right and so that is going to line up with answer Choice B all right just as a reminder I&amp;#39;m going to go back to our notes cuz we did talk about the Gabriel synthesis all right we we started off with potassium thide depicted right here I&amp;#39;m going to circle it in blue all right and this was the electrophile what you notice is that this nitrogen gets connected to this group here with bromine as the leaving group and then there&amp;#39;s you treat it with base it gets deproteinated then you treat it with um this bromo alane you have an sn2 reaction you get the following R group that attaches to that carbon and then you treat it with base all right and you you allow it to get hydrolized in the presence of base and what you notice is whatever group was attached here all right whatever group was attached here is removed all right and protonated and then the ends of these two carbonal groups in this molecule this what was once thide gets you know ox oygen at each end negatively charged oxygens so that&amp;#39;s exactly what&amp;#39;s happening here in this problem all right that&amp;#39;s how we figured out that what we get as an end product here is methyl amine answer Choice B for two fantastic 3 says which of the following amino acids contains sulfur we have cysteine Serene and methionine all right so of course for for for the MCAT please be familiar with your am you know acid structures their names their full names their three-letter code names and their onlet code names all right and so here we can start to go through each op option all right we have cistin if we look at cysteine what we notice is that it does have a sulfur atom all right so cysteine does contain a sulfur check good what about saring all right saring is right here let&amp;#39;s find it all right does this amino acid have a sulfur atom I don&amp;#39;t see any sulfur atoms so no all right and then methon 9 if we look at methon 9 right here also it has a sulfur atom so out of these three options cysteine and methionine both have and both contain sulfur so 1 and three the correct answer here is going to be B four says nylon a polyamide is produced from hexane diamine and a substance X this substance X is most probably a blank now this requires us to recall what how do you form an amide an amide is formed from an amine plus a carboxy group or it&amp;#39;s AAL derivatives now in this question the first thing that you&amp;#39;re given all right in the production of nylon is you know that you need hexane diamine so you have the amine in the problem right and amine is already given in this question and so the compound that needs to be identified this substance X it has to be an AAL compound and the only ACL compound that&amp;#39;s given in these choices is going to be the carboxilic acid and so just working off that logic the answer here is answer Choice B fantastic five says intermediates in the streker synthesis include all of the following nitrogen containing functional groups except blank now during the ster synthesis what we talked about was that ammonia attacks a carbonal all right ammonia attacks a carbonal forming an amine so we can obviously cancel out answer Choice B this amine is attacked by a cyanide forming an amine so we can go ahead and cancel out answer Choice D and then what you form is also a nitrol all right which cancels out answer Choice a all right what we don&amp;#39;t form is an amide amide bonds are formed between amino acids but they don&amp;#39;t appear during the ster synthesis so if you remember the ster synthesis it had two big steps all right in step one all right you start with an alahh ammonium chloride and potassium cyanide all right the carbonal oxygen is proteinated we can even look at our mechanism here just to remind mind ourselves all right so um the carbonal oxygen is protonated and then what we see is that ammonia can attack the carbonal carbon forming an amine all right and then the amine carbon is susceptible to nucleophilic attack by the cyanide all right by the cyanide CN anion from potassium cyanide it attacks all right and now you get at the end of Step One an amino nital and then in step two the nital nitrogen is protonated and then a water molecule attacks leading to the creation of a molecule with a bond with both an amine and a hydroxy group on the same carbon all right and then this amine um is attacked by another equivalent of water a carbonal is formed kicking off ammonia creating the carboxilic acid functionality all right and then essentially what you end up with at the end of all this is your amino acid all right so in that second step the amino acid is generated from that Amino nital so again what we see is that there is a nital group there&amp;#39;s an amine and there is an amine what there isn&amp;#39;t is an amide so five all right is C all right we see all of the following nitrogen containing functional groups except amides fantastic six says a biochemist is syn sizing veine shown below using the struer synthesis which of the following carbonal containing compounds would be a propriate starting reactant in this synthesis now the ster synthesis creates an amino acid from an alide the carbonal carbon ultimately becomes the alpha carbon of the amino acid any remaining alkal chain becomes the R Group so if we look at this right here right any remaining alkal chain becomes the R Group all right and so if we&amp;#39;re starting with for example um two methyl propanel all right and then we treat it with ammonium chloride and and and and potassium cyanide all right we can get the following amuno nital all right and then it can form the following amino acid which is exactly what looking for all right what we notice is this R Group determines everything all right again again what this needs to be is is two methyl propenol so this R Group if it was two methyl propenol it would be able to with the starting aldah right here all right with this starting aldah you would be able to go through the struer synthesis all right form this am this this amuno nital where the R Group is again two methyl propanel and then in the Second Step form that amino acid where the R Group is that again two methyl propanel and if that is the group that is going to be here in the R Group then what we get is veiling all right and so what really did what what&amp;#39;s really crucial to take away from this ster synthesis is that whatever amino acid you are forming you have to be cautious you have to be informed about what its R Group is because that R Group needs to be present in your starting aldah so if you look at this amino acid this veiling here&amp;#39;s the alpha carbon here&amp;#39;s the amino group here&amp;#39;s the carboxy group there&amp;#39;s a hydrogen and then there&amp;#39;s this group right here this group right here is what&amp;#39;s going to be added right here as an R Group to this aldah and it will be called two methyl propanel all right and so that is our appropriate starting group this is two methyl propanel propanel that is our starting material here is where what we would get the amino nitril all right and then this is the ending amino acid fantastic so then the answer here for six is going to be C 7even says why is the CN Bond of an amide planer all right the answer choices say one it has a partial double bond character two it is sp3 hybridized and three says it has some SP2 character all right why is the CN Bond of an amide planer now one resonance structure of a CN bond in an amide has the double bond between the C and the n and not between the C and an O and so what that means is that the CN Bond of an amide it it has some SP2 character and SP2 hybridized atoms exhibit planer geometry all right and so it has partial double bond character all right and it has some SP2 character both of these play a role in why the CN Bond of an amine is planer and so we&amp;#39;re looking for one and three the correct answer is going to be D now a says which of the primary methods of amino acid synthesis results in an optically active solution all right a says the stuer synthesis B say only B says the Gabriel synthesis only C says both D says neither now we covered a really important point when we were talking about both the ster synthesis and the Gabriel synthesis and we said that they both contain planer intermediates and that&amp;#39;s really important because if they have planer intermediates they can be attacked from either side by a nucleophile and if that&amp;#39;s the case then that&amp;#39;s going to result in a rasmic mixture of an aners if you have an equal rmic mixture of enantiomers the solution will therefore therefore be optically inactive all right so both the streer synthesis and the Gabriel synthesis result in products that are a rasic mixture of inan humors and if you have a racemic mixture all right it&amp;#39;s going to be a solution that&amp;#39;s optically inactive and that means if we&amp;#39;re looking for uh which primary method of amino acid synthesis results in optically active solution the answer is neither all right neither of these result in optically Active Solutions both of these methods result in the production of rasmic mixtures of inan tumors and th that results in the solution being optically inactive so eight is D nine says during the Gabriel synthesis thite serves as the blank all right during the Gabriel synthesis if you remember thite acts a attacks um a secondary carbon in D um ethyl bromomalonate all right that the secondary carbon is the electrophile bromide is the leaving group and thide is the nucleophile so if we go right back here just to refresh our memories this thide all right this is our nucleophile all right this right here this carbon is our electrophile all right bromine right here is our leaving group so thide is our nucleophile nine is a beautiful 10 says each of the following reaction types occurs during the Gabriel synthesis except all right so this is really good because what you&amp;#39;re noticing here in this practice problems is that these are very big picture questions about the struer synthesis and the Gabriel synthesis now if you remember the Gabriel synthesis we can go back and look at it all right it includes two and I&amp;#39;m going to use my orange highlighter here two nucleophilic substitution reactions all right followed by hydrolysis all right hydrolysis and decarbox all right so it has sn2 reactions two nucleophilic substitution reactions hydrolysis and decarbox if we look at our answer choices we can go ahead and cancel out a b and d dehydration the loss of a water molecule is actually not part of this synthesis and so each of the following reaction types occurs except dehydration 10 is C beautiful 11 says at physiological pH which two forms of phosphoric acid have the highest concentrations so the pka2 of phosphoric acid is going to be really close to to physiological uh pH and so what that&amp;#39;s what that means is that we&amp;#39;re going to have a h24 concentration pretty much almost equal to hp42 minus so the pka2 of phosphoric acid is close to physiological pH and so it would have lost one of its hydrogens already since this is at at pka2 and it&amp;#39;s going to be approximately equal to the concentration of um hpo4 all right so h24 and hpo4 are going to be um the highest concentrations are going to be about equal and so the answer for 11 is going to be B 12 cells in aquous solution pyrro phosphate will likely blink form insoluble complexes be stable and inert degrade into inorganic phosphate or decrease the polarity of the solvent now pyrro phosphate is unstable in aquous solution all right and it will degrade to form two equivalents of inorganic phosphate all right now that is going to be answer Choice C it will degrade into an organic phosphate now the solvent is water which should retain its polarity regardless of the presence of solute so that regardless of the presence of solutes that&amp;#39;s going to eliminate answer Choice D and pyrro phosphate and inorganic phosphate they are small charged molecules all right which are are relatively soluble all right so again these answer choices don&amp;#39;t make any sense 12 is C pyrro phosphate will degrade into likely degrade into inorganic phosphate 13 says what would be the charge of aspartic acid at ph7 all right so the amino acid in question is aspartic acid which is an acidic amino acid because it contains an extra carbox group as part of its rchain all right just to remind you I&amp;#39;m going to quickly go back to our amino acid table and I&amp;#39;m going to circle asart uh aspartic acid or aspartate right here all right there we go all right notice how it has an extra carboxy group as part of its rchain at neutral pH both the carboxy groups are ionized so that means there are two negative charges one over here one over here here all right only one of those charges is neutralized by the positive charge that would be at the amino group all right and on that amino group so the molecule is going to have what it&amp;#39;s going to have one negative charge plus one negative charge plus one positive charge for a total of a minus one negative charge so the molecule has an overall negative charge the answer for 13 is going to be 14 says when a bond is created between two nucleotide triphosphates in DNA synthesis the small molecule released from this reaction is blank now as DNA is synthesized if you remember from our discussion it forms phosphodiester bonds and it releases pyrro phosphate now pyrro phosphate is an inorganic phosphate containing molecule but it is not the single phosph pH at group commonly referred to as an organic phosphate all right and so when a bond is created between two nucleotide triphosphates in DNA the small molecule released is going to be pyo phosphate all right so 14 is a now last but not least 15 says the hydrogens of phosphoric acid have PKA values that blank a says allow High buffering capacity over small PH range B say allow moderate buffering capacity over a large PH range C says allow low buffering capacity over a small PH range and D says do not allow buffering now phosphoric acid it has three hydrogens all right H3 P4 all right has three hydrogens with PKA values that are spread across the PH range this allows some degree of buffering over almost the entire standard PH range from 0 to 14 all right now this means that it&amp;#39;s we wouldn&amp;#39;t say high buffering capacity but we would say it allows a moderate buffering capacity over a large PH range so the answer for 15 is going to be B all right with that we&amp;#39;ve completed our practice problem set 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

Transcript for:Nitrogen and Phosphorus Containing Compounds

Transcript for:
Nitrogen and Phosphorus Containing Compounds