Malonic ester contains an active methylene group and two hydrogen atoms of active methylene group which are easily replaced by alkyl and acyl groups in the presence of a strong base. The carbonation intermediate formed makes $S_{N} 2$ attack on alkyl or acyl groups. In the sequence of reactions given above, the final product [ $S$ ] is :

C H_{2} = C H - C H_{2} - C H_{2} - C O - C O O H

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C H_{2} = C H - C H_{2} - C O - C H_{2} - C O O H

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C H_{2} = C H - C H_{2} - C H_{2} - C O O H

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C H_{3} - C H = C H - C H_{2} - C O O H

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Solution

The correct option is C $C H_{2} = C H - C H_{2} - C H_{2} - C O O H$
Sodium hydride abstracts the proton from the active methylene group of diethyl malonate to obtain the carbanion [ $P$ ]. This acts as a nucleophile and attacks 1-chloro pop-1-ene. Chloro group is replaced in $S_{N} 2$ mechanism to obtain [ $Q$ ]. The ester groups are hydrolyzed in presence of acid to obtain [ $R$ ].
This is followed by thermal decarboxylation to obtain [ $S$ ]. Thus [ $S$ ] is $C H_{2} = C H - C H_{2} - C H_{2} - C O O H$

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Similar questions

(A) C H_{3} - C H_{2} - C | C O O H H - - C H - C H_{3}

(B) C H_{3} - C | C O O H H - C H_{2} - - C H - C H_{3}

(C) C | C O O H H_{2} - C H_{2} - C H_{2} - - C H - C H_{3}

C H_{3} - C H_{2} - C H | C O O H - C H_{2} - C l | C H - C H_{2} - C H_{3}

C H_{3} - C O O H

C H_{2} = C H - C O O H

C_{6} H_{5} - C O O H

C_{6} H_{5} - C H_{2} - C O O H

C H_{3} - C H_{2} - C O O H

C H_{3} - C H | O H - C O O H

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