Compound A has the molecular formula $C_{14} H_{25} B r$ and was obtained by reaction of sodium acetylide with 1,12-dibromododecane.On treatment of compound A with sodium amide,it was converted to compound B $(C_{14} H_{24})$ Ozonolysis of compound B gave the diacid $H O_{2} C (C H_{2})_{12} C O_{2} H$ Catalytic hydrogenation of compound B over Lindlar palladium gave compound $C (C_{14} H_{26}$ ,and hydrogenation over platinum gave compound $D (C_{14} H_{28})$ Sodium-ammonia reduction of compound B gave compound $E (C_{14} H_{26})$ Both C and E yielded $O = C H (C H_{2})_{12} C H = O$ on ozonolysis Assign strutures to compounds A through E so as to be consistent with the observed transformations.

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Solution

The structures of compounds A to E are as shown below.
Actylide ion displaces one bromine atom of 1,12-dibromododecane to form compound A. Soda amide abstracts acidic hydrogen from compound A and the carbanion formed displaces the bromine atom to form cyclic compound B. Ozonolysis of compound B breaks carbon carbon double bond to form the diacid $H O_{2} C (C H_{2})_{12} C O_{2} H$ . Catalytic hydrogenation of Lindlar catalyst over palladium gives compound C. In this reaction, carbon carbon triple bond is selectively reduced to double bond. Hydrogenation in presence of Pt gives compound D in which carbon carbon triple bond is reduced to single bond. Compound B on reduction with sodium in ammonia gives compound E in which the triple bond is reduced to trans double bond.

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