Answer
Verified
35.1k+ views
Hint: First when the Grignard reagent acts as catalyst the ketone group will be attacked changing into a complex. Then with the help of aqueous acid, the chlorine group will leave with the Grignard group.
Complete step by step answer:
First, let us see what is the reactant?
This compound is 5-chloropentan-2-one. This is the compound that comes under the ketonic group with chlorine as the substituent at the 5th place.
Now, the catalyst mentioned are (i)- \[\text{C}{{\text{H}}_{3}}\text{MgBr}\]
This is called the Grignard reagent. They react with aldehyde, ketones, and ester to form alcohol in two steps. The first involves the nucleophilic attack of the Grignard reagent to the carbonyl group to form an adduct. The second step involves hydrolysis of the adduct with water or preferably with dilute hydrochloric acid or dilute sulphuric acid to form an alcohol.
(ii)- aqueous acid
This catalyst helps to remove the negative part of the chain-forming specific product. because acid has hydrogen ions which combine with the negative part to form the by-product.
Now when the reaction takes place,
The product formed is
Let us understand the mechanism:
First, the Grignard reagent attacks the ketonic group, this leads to the breaking of double bond and formation of OMgBr complex. The other part of the Grignard reagent i.e. methyl group will attach to the carbon atom having the double bond.
In the next step, the chlorine bond will break which leads to acquire a positive charge on the carbon atom.
Now, the negative charge of the oxygen atom and positive charge of carbon atom will combine to form a cyclo structure.
Therefore, Option D is correct
Note: You may get confused between option (b) and (d) because in the catalyst Grignard reagent is given which mainly forms the alcohol and in all the options, only option (b) contains the alcohol group. But due to the presence of aqueous acid, it will form a cyclic structure.
Complete step by step answer:
First, let us see what is the reactant?
This compound is 5-chloropentan-2-one. This is the compound that comes under the ketonic group with chlorine as the substituent at the 5th place.
Now, the catalyst mentioned are (i)- \[\text{C}{{\text{H}}_{3}}\text{MgBr}\]
This is called the Grignard reagent. They react with aldehyde, ketones, and ester to form alcohol in two steps. The first involves the nucleophilic attack of the Grignard reagent to the carbonyl group to form an adduct. The second step involves hydrolysis of the adduct with water or preferably with dilute hydrochloric acid or dilute sulphuric acid to form an alcohol.
(ii)- aqueous acid
This catalyst helps to remove the negative part of the chain-forming specific product. because acid has hydrogen ions which combine with the negative part to form the by-product.
Now when the reaction takes place,
The product formed is
Let us understand the mechanism:
First, the Grignard reagent attacks the ketonic group, this leads to the breaking of double bond and formation of OMgBr complex. The other part of the Grignard reagent i.e. methyl group will attach to the carbon atom having the double bond.
In the next step, the chlorine bond will break which leads to acquire a positive charge on the carbon atom.
Now, the negative charge of the oxygen atom and positive charge of carbon atom will combine to form a cyclo structure.
Therefore, Option D is correct
Note: You may get confused between option (b) and (d) because in the catalyst Grignard reagent is given which mainly forms the alcohol and in all the options, only option (b) contains the alcohol group. But due to the presence of aqueous acid, it will form a cyclic structure.
Recently Updated Pages
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
The path difference between two waves for constructive class 11 physics JEE_MAIN
What is the difference between solvation and hydra class 11 chemistry JEE_Main
IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main
Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main
Consider the following oxyanions PO43P2O62SO42MnO4CrO4S2O52S2O72 class 11 chemistry JEE_Main
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
The mole fraction of the solute in a 1 molal aqueous class 11 chemistry JEE_Main
The strongest oxidising agent among the following is class 11 chemistry JEE_Main
Assertion Internal forces cannot change linear momentum class 11 physics JEE_Main
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
Oxidation state of S in H2S2O8 is A 6 B 7 C +8 D 0 class 12 chemistry JEE_Main