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Question: An ester (A) with molecular formula \({C_9}{H_{10}}{O_2}\) was treated with excess of \(C{H_3}MgBr\)...

An ester (A) with molecular formula C9H10O2{C_9}{H_{10}}{O_2} was treated with excess of CH3MgBrC{H_3}MgBr and the compound so formed was treated with conc. H2SO4conc.\,{H_2}S{O_4} to form olefin (B). Ozonolysis of (B) gave ketone with formula C8H8O{C_8}{H_8}O which shows positive iodoform test. The structure of (A) is?
A) C6H5COOC2H5{C_6}{H_5}COO{C_2}{H_5}
B) CH3OCH2COC6H5CH{}_3OC{H_2}CO{C_6}{H_5}
C) CH3CO−C6H5−COCH3C{H_3}CO - {C_6}{H_5} - COC{H_3}
D) C6H5COOC6H5{C_6}{H_5}COO{C_6}{H_5}

Explanation

Solution

Solve each reaction stepwise as given in the question, firstly there is a nucleophilic attack of CH3C{H_3} group on carbonyl compound of each option, conc. H2SO4conc.\,{H_2}S{O_4} is given to make hydroxide ion as hydronium ion, which is a better leaving group and makes olefin. After ozonolysis, don’t forget to check the formula of the compound formed because by this we can easily check on the formation of the right product.

Complete step by step solution:
In the first option (A), methyl group attack as a nucleophile on the carbonyl group we get an alcoholic group after hydrolysis. Now our next step is the reaction of our formed compound with conc. H2SO4conc.\,{H_2}S{O_4} , in this step hydroxide group will convert into hydronium ion +OH2{}^ + O{H_2} , which is a very good leaving group. It leaves and to form an olefin there is a hydrogen just next to that carbon atom.
C6H5COOC2H5 + CH3MgBr→C6H5−C(OH)(CH3)−O−C2H5{C_6}{H_5}COO{C_2}{H_5}\, + \,C{H_3}MgBr \to {C_6}{H_5} - C(OH)(C{H_3}) - O - {C_2}{H_5}
In this reaction we see that there is an oxygen atom and not a hydrogen atom forming an olefin. So this option is wrong.
In option (B), a similar process takes place, firstly attack by the methyl group on carbonyl followed by hydrolysis. The hydroxyl group formed in the compound then reacts in the presence of conc. H2SO4conc.\,{H_2}S{O_4} forming hydronium ion. The hydronium ion leaves from one carbon atom and in this option there is a hydrogen atom just next to carbon so formation of olefin takes place.
CH3OCH2COC6H5 + CH3MgBr→CH3−O−CH2−C(CH3)(OH)−C6H5CH{}_3OC{H_2}CO{C_6}{H_5}\, + \,C{H_3}MgBr \to C{H_3} - O - C{H_2} - C(C{H_3})(OH) - {C_6}{H_5}
CH3−O−CH2−C(CH3)(OH)−C6H5→conc. H2SO4CH3−O−CH=C(CH3)−C6H5C{H_3} - O - C{H_2} - C(C{H_3})(OH) - {C_6}{H_5}\xrightarrow{{conc.\,{H_2}S{O_4}}}C{H_3} - O - CH = C(C{H_3}) - {C_6}{H_5}
CH3−O−CH=C(CH3)−C6H5→Ozonolysis CH3−O−CHO+ C6H5COCH3final productC{H_3} - O - CH = C(C{H_3}) - {C_6}{H_5}\xrightarrow{{Ozonolysis}}\,\mathop {C{H_3} - O - CHO + \,{C_6}{H_5}COC{H_3}}\limits_{final\,product}
The ketone form here is having CH3CO−C{H_3}CO - group which on reacting with NaOHNaOH and I2{I_2} gives iodoform CHI3CH{I_3} . This must be the right answer.
In option (C) we have CH3CO−C6H5−COCH3C{H_3}CO - {C_6}{H_5} - COC{H_3} this compound reacts in the same way as all others did, but atlast the ketone form is having larger number of carbon atoms C9H11O2{C_9}{H_{11}}{O_2} , therefore this can’t be our right option.
In the last part option (D) we have C6H5COOC6H5{C_6}{H_5}COO{C_6}{H_5} , there are two phenyl groups in this compound. So after all the steps we get compound C6H5−C(OH)(CH3)−O−C6H5{C_6}{H_5} - C(OH)(C{H_3}) - O - {C_6}{H_5} .
As we see here also there is no hydrogen just next to carbon having OHOH . Therefore here also olefin cannot form.

So the correct option would be option B.

Note: Always try to write the compound in short formula as suggested in question like C8H8O{C_8}{H_8}O , by this we can check that after each reaction we are getting the desired product. Write reaction to reaction by applying reagents. Keep in mind that the iodoform test is given by CH3CO−C{H_3}CO - group or compounds which give secondary alcohol on reduction.