Question

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

Answer 1

Hint:Solve each reaction stepwise as given in the question, firstly there is a nucleophilic attack of $C{H_3}$ group on carbonyl compound of each option, $conc.\,{H_2}S{O_4}$ 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 $conc.\,{H_2}S{O_4}$ , in this step hydroxide group will convert into hydronium ion ${}^ + O{H_2}$ , which is a very good leaving group. It leaves and to form an olefin there is a hydrogen just next to that carbon atom.
${C_6}{H_5}COO{C_2}{H_5}\, + \,C{H_3}MgBr \to {C_6}{H_5} - C(OH)(C{H_3}) - O - {C_2}{H_5}$
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 $conc.\,{H_2}S{O_4}$ 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.
$CH{}_3OC{H_2}CO{C_6}{H_5}\, + \,C{H_3}MgBr \to C{H_3} - O - C{H_2} - C(C{H_3})(OH) - {C_6}{H_5}$
$C{H_3} - O - C{H_2} - C(C{H_3})(OH) - {C_6}{H_5}\xrightarrow{{conc.\,{H_2}S{O_4}}}C{H_3} - O - CH = C(C{H_3}) - {C_6}{H_5}$
$C{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 $C{H_3}CO - $ group which on reacting with $NaOH$ and ${I_2}$ gives iodoform $CH{I_3}$ . This must be the right answer.
In option (C) we have $C{H_3}CO - {C_6}{H_5} - COC{H_3}$ this compound reacts in the same way as all others did, but atlast the ketone form is having larger number of carbon atoms ${C_9}{H_{11}}{O_2}$ , therefore this can’t be our right option.
In the last part option (D) we have ${C_6}{H_5}COO{C_6}{H_5}$ , there are two phenyl groups in this compound. So after all the steps we get compound ${C_6}{H_5} - C(OH)(C{H_3}) - O - {C_6}{H_5}$ .
As we see here also there is no hydrogen just next to carbon having $OH$ . 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 ${C_8}{H_8}O$ , 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 $C{H_3}CO - $ group or compounds which give secondary alcohol on reduction.