A) \[{{(C{{H}_{3}})}_{2}}CHC{{H}_{2}}COOCH{{(C{{H}_{3}})}_{2}}\]
B) \[{{(C{{H}_{3}})}_{2}}CHCOOC{{H}_{2}}CH{{(C{{H}_{3}})}_{2}}\]
C) \[{{(C{{H}_{3}})}_{2}}CHCOO{{C}_{2}}{{H}_{5}}\]
D) None of the above
Correct Answer: B
Solution :
Aq KOH converts -X (where, X = halogens) into \[OH,\]so the structure of C must be\[{{(C{{H}_{3}})}_{2}}CHC{{H}_{2}}X\]. Moreover, compound B on reduction with\[Pd/BaS{{O}_{4}}\]and after that with \[LiAl{{H}_{4}}\]gives an alcohol, so it must be an acyl halide and A must be an ester. By using the structures of B and C, structure of A is found. The reactions are as follows \[\underset{'A'}{\mathop{{{(C{{H}_{3}})}_{2}}CHCOOC{{H}_{2}}CH{{(C{{H}_{3}})}_{2}}}}\,\xrightarrow[-POC{{l}_{3}}]{PC{{l}_{5}}}\] \[\underset{'A'}{\mathop{{{(C{{H}_{3}})}_{2}}CHCOCl}}\,+\underset{'C'}{\mathop{{{(C{{H}_{3}})}_{2}}CHC{{H}_{2}}Cl}}\,\] \[\underset{'B'}{\mathop{{{(C{{H}_{3}})}_{2}}CHCOCl}}\,\xrightarrow[{}]{(i)\,Pb/BaS{{O}_{4}}}{{(C{{H}_{3}})}_{2}}CHCHO\] \[\xrightarrow[{}]{LiAl{{H}_{4}}}{{(C{{H}_{3}})}_{2}}CHCHO\] \[{{(C{{H}_{3}})}_{2}}CHC{{H}_{2}}Cl\xrightarrow{Aq.\,KOH}{{(C{{H}_{3}})}_{2}}CHC{{H}_{2}}OH\]
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