The structure of the carbonyl compound P, which gives positive iodoform test undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.
The structure of the carbonyl compound P is:
The structure of the carbonyl compound P, which gives positive iodoform test undergoes reaction with MeMgBr followed by dehydration to give an olefin Q. Ozonolysis of Q leads to a dicarbonyl compound R, which undergoes intramolecular aldol reaction to give predominantly S.
The structure of the carbonyl compound P is:
The correct option is B
It has to be a methyl ketone to undergo iodoform test. Without looking at the options, we have no information to begin solving - so we try and look at the options, figure out how things can go further.
Apart from the options - there are two other clues! From P to Q, the third step, sulphuric acid + heat makes things interesting. Also, pay attention to the final step - it mentions “intramolecular aldol”.
From the options only a. and b. are methyl ketones. Since there shall be only one correct answer, let us start with the first methyl ketone as P and find out if it can go through the whole sequence. So what happens in the first step?
So far so good, it is straightforward!
But the last step from P to Q is indicative of Intramolecular Friedel-Crafts alkylation. It doesn’t need a catalyst because the acid protonates the OH group and makes it a very capable leaving group. Once water leaves, there is a stable tertiary carbocation, which begets the cyclized product through the sigma complex or wheland complex intermediate.
Once we are here, we will have to just complete the Ozonolysis -which strangely produces a non-enolizable dicarbonyl. This clearly means we have chosen the wrong option.
So P is:
Now that we know what P is, let us methodically put down the actual reaction and products until the last step:
Look at how deceptively the problem has been worded – they just say “Olefin Q” just to tempt you to consider other options for candidates of P.
Once we get to Q, from there onwards we get to R effortlessly
From here on, we just have to complete the final intramolecular aldol and thus we get S.
It has to be a methyl ketone to undergo iodoform test. Without looking at the options, we have no information to begin solving - so we try and look at the options, figure out how things can go further.
Apart from the options - there are two other clues! From P to Q, the third step, sulphuric acid + heat makes things interesting. Also, pay attention to the final step - it mentions “intramolecular aldol”.
From the options only a. and b. are methyl ketones. Since there shall be only one correct answer, let us start with the first methyl ketone as P and find out if it can go through the whole sequence. So what happens in the first step?
So far so good, it is straightforward!
But the last step from P to Q is indicative of Intramolecular Friedel-Crafts alkylation. It doesn’t need a catalyst because the acid protonates the OH group and makes it a very capable leaving group. Once water leaves, there is a stable tertiary carbocation, which begets the cyclized product through the sigma complex or wheland complex intermediate.
Once we are here, we will have to just complete the Ozonolysis -which strangely produces a non-enolizable dicarbonyl. This clearly means we have chosen the wrong option.
So P is:
Now that we know what P is, let us methodically put down the actual reaction and products until the last step:
Look at how deceptively the problem has been worded – they just say “Olefin Q” just to tempt you to consider other options for candidates of P.
Once we get to Q, from there onwards we get to R effortlessly
From here on, we just have to complete the final intramolecular aldol and thus we get S.
