Answer:
(i) \[\underset{\text{Benzaldehyde}}{\mathop{{{C}_{6}}{{H}_{5}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-H}}\,+\underset{\text{Propanal}}{\mathop{C{{H}_{3}}C{{H}_{2}}CHO}}\,\xrightarrow[\begin{smallmatrix} \,\,\text{(Claisen}\,\text{Schmidt} \\ \,\,\,\,\,\text{condensation)} \end{smallmatrix}]{NaOH}\]\[\underset{\text{Aldol}}{\mathop{\left[ {{C}_{6}}{{H}_{5}}-\overset{\begin{smallmatrix} \,\,\,\,\,OH \\ | \end{smallmatrix}}{\mathop{C}}\,H-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,H-CHO \right]}}\,\xrightarrow[-{{H}_{2}}O]{{}}\] \[\underset{\begin{smallmatrix} \\ \text{2-Methyl-3-phenylprop-2-en-1-al}\,(X) \end{smallmatrix}}{\mathop{{{C}_{6}}{{H}_{5}}-\overset{3}{\mathop{C}}\,H=\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{2C\,}}\,-CHO}}\,\]
(ii)\[\underset{\text{Butan-1-ol}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OH}}\,\,\xrightarrow[(ii)\,{{H}^{+}},\,{{H}_{2}}O]{(i)\,Alk.\,KMn{{O}_{4}}}\,\underset{\text{Butanoic}\,\,\text{acid}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COOH}}\,\]
(iii) \[\underset{\text{Ethyl}\,\text{undecanoate}}{\mathop{C{{H}_{3}}{{(C{{H}_{2}})}_{9}}COO{{C}_{2}}{{H}_{5}}}}\,\xrightarrow[(ii)\,{{H}_{2}}O]{(i)\,DIBAL-H}\,\underset{\text{Undecanal}}{\mathop{C{{H}_{3}}{{(C{{H}_{2}})}_{9}}CHO}}\,\]
Thus, Y = (i) Alk. \[KMn{{O}_{4}},\](ii)\[{{H}^{+}},\,{{H}_{2}}O\] and \[Z=(i)\]\[DiBAL-H,\](ii)\[{{H}_{2}}O\]
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