Phenol reacts with methyl chloroformate in the presence of NaOH to form product A. A reacts with...

without need get instant video solutions to all your maths physics chemistry and biology doubts just click the image of the question crop the question and get instant video solution download doubt and dab today hey everyone so in this question it is gained that phenol reacts with methyl chloroform it in the presence of nh to form product a now a reacts with bromine to form product b a and b are respectively so first of all we have a phenol molecule in the presence of methylcolor format and nh is there so i'll write down the phenol molecule

that is o h to which a benzene group is going to be attached now it is said that the reaction is happening in the presence of a base that is naoh so what will this nah do it will have a positive charge on nega sodium and negative charge on the o h so here we have a acidic proton from this case that is a hydrogen over here so what will this o h minus will do it will take up the hydrogen and there will be removal of h2o molecule and this becomes a phenoxide and having

o minus charge an na positive present as a counter anion so this is the phenoxide ion which we have now now this phenoxide ion is said to be treated with like the reaction is going to be happened in the presence of a methyl chloroformate so methyl chloroform it has a structure of first of all methyl means ch3 chloroformat means we will have o c double bond o cl okay so it is methyl chloroformate now what is going to happen this o minus which is present over here is going to attack on the carbonyl carbon because

the electron density over here is less and the electron density is more in this case so bond shifting will happen this will come back which will help in the removal of cl group right so there will be removal of nacl from this part and the product which we are going to get if i write it down over below so since there is no space present i'll write it over here the product is going to be formed is o so c double bond o it is okay and to this we have connection of och3 group and

to this o oxygen we have a benzene group attached now it is said this is going to be my product a now this product a is reacted with what we are reacting this with bromine to form product b a bromine is what we are going to do the bromination reaction in the bromination reaction what happens is there is generation of an electrophile which is br positive and benzene are supposed to undergo electrophilic substitution reaction because the electron density in the ring is quite high and over here this group has electrons present on the oxygen so

what will happen in this case there is going to be increase in electron density because these are electron releasing group erg that means electron result releasing group and this makes electron resulting groups releasing groups makes this benzene molecule ortho and paradigm the ring becomes activated drink is activated now here what will happen the electron density is going to increase at the ortho and para positions because what happens is there will be shift and resonance and therefore it accumulation of negative charge at this part and similarly this will undergo resonance again and accumulation of negative charge

on this part so the electron density is maximum and ortho and para position so again resonance this is also auto position now what happens is this is the negative charge which is going to happen after the resonance so we don't need to write this right now because we have shown the three bonds already right so what can happen now the attack of bromine can either happen at the ortho position or the para position but the one which is preferred is the para position why because oco ch3 group is already a very bulky group and br

positive is again a very bulky group so the presence of earth at ortho position of this bromine attack of electrophile on the ortho position will make the groups unstable due to steric hindrance so due to avoiding steric hindrance what will happen para substitution is going to happen in this case now the molecule will have a structure o c double bond o o c h 3 and add the para position to this group we will have a bromine molecule attached to this so this is going to be my product b now we'll have to select a

and b from the options which we have so we see option number a we have oco ch3 so that is correct for now then we have at the para position to this because this becomes ortho this becomes meta and para so at para position we have bromine so this is also correct option a becomes our correct answer if you move forward to the second option here we have a oh group which is not needed so this is incorrect automatically again oh group this becomes incorrect here we have oco ch3 but here at the meta position

the bromine group is added so again this becomes incorrect only correct answer is option a thank you for class 6-12 itj and neet level trusted by more than 5 crore students download doubt and app today