Methylene Group using Aldol reaction I What a beautiful JEE Advanced problem ! 2024 Q-10 I

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This problem is related to the concept of aldol condensation and Caninaro reaction so problem says complete reaction of acetal deide with excess formal deide upon heating with concentrated NaOH solution gives p and

Q so acetal deide is completely reacted because it says complet compl reaction and it also says excess formal deide this means more than one molecule of formal deide has reacted with acetal deide furthermore question also says compound P does not gives tolerance test whereas Q on acidification gives positive tolerance test this means P does not have an alhy group but Q has an alhy alhy group that is it contains CH group treatment of p with excess cyclohexanone in the presence of catalytic amount of power to sulphonic acid gives product R so when p is treated with cyclohexanone it gives R and it also says excess of cyclohexanone this means more than one molecule of cyclohexanone has reacted with p furthermore it says catalytic amount of PTSA this means this is not participating in the reaction as reactant or as product this is a catalyst so you can safely ignore while writing the reaction product or reaction mechanism you have to find sum of the number of methylene groups that is ch2 groups and oxygen atom in R so how many ch2 groups you have and how many oxygen atoms you have in compound

R so let us try to solve this

problem so what are the hints that we have the first hint is p is not an alahh this means it does not have an alhy group CH group but Q gives positive Toler test this means Q has a CH group now further hint I have that is we have no no can act as a base or it can act as a nucleophile it is reacting with acetal deide that is ch3 CH and ch3 CH has acidic hydrogen so this means another hint I have that is acid based reaction of course na NaOH can act as a base or it can act as a nucleophile if there is a chance of acid base reaction first acid base reaction will take

place so let us try to write the mechanism so we have estal deide so this molecule is acal deide ch3 CH now it is reacting with base oh minus and this o minus is coming from NaOH now this hydrogen is acidic because of this Carbonic group this Alpha hydrogen becomes acidic now o minus is a base and it will take this H+ and it will form H2 molecule and at the same time this Bond pair will come to this carbon and it will form a carb an so now I have a carbanion and a water molecule

now this carbon will react with formal deide so this is the formal deide it also has a positively charged carbon this oxygen basically pulling all the electron towards itself toward its side then this will have Delta plus charge and this will have Delta minus charge so this is this carbon can be attacked so now this molecule that is the carbon can act as a nucleophile and this carbon will attack on this carbon so basically there is a carbon carbon Bond formation at the same time this Bond pair will come to the oxygen and a negative charge will appear on this oxygen atom so now I have this molecule now the next step this o minus can take H+ from water so this o minus can take H+ from here and this Bond pair will go at oxygen atom and it will form o minus so basically o minus is used in the first step and it was returned in the last step and what I have now here this side I have an alhy group and this side I have alcohol group now what will happen

next now see what has happened initially if you see this what has happened if you see this molecule was initially here and on this molecule at at this carbon one hydrogen is removed and at this hydrogen what we have we have ch2 so this carbon and this carbon is same so here I have hydrogen and here I have hydrogen so two hydrogen here also I have two hydrogen but one hydrogen is missing and in place of one hydrogen what I have

ch2 so what I have done in this whole process one hydrogen has been removed and it has been attached by a ch2 oh group now the same thing can happens once again and I can remove one hydrogen and I can add ch2 H group so basically this process can happen two times so I will remove one hydrogen here from here and I will add ch2 group and another hydrogen and I will add ch2 group so I have removed one hydrogen added ch2 oh group and here also one ch2 oh group now we don't have any acidic hydrogen on this carbon so what will happen next so now oh minus cannot get one proton now oh minus will act as a nucleophile and as it nucleophile it it will attack on formal deide molecule it will attack on this carbon and this Bond will open so oxygen will have a negative charge now if you remember at this place you will have canaro