Outline the synthesis from lauric acid \(\left(\mathrm{n}-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COOH}\right.\), dodec- anoic acid) of the following compounds: (a) 1-bromododecane; (b) tridecanoic acid (C \(_{13}\) acid); (c) 1 -tetradecanol; (d) 1 -dodecene; (e) dodecane; (f) 1-dodecyne; (g) methyl n-decy1 ketone; (h) 2-dodecanol; (i) undecanoic acid; (j) 2 -tetra- decanol; (k) 2 -methy \(1-2\) -tetradecanol.

(a) Synthesis of 1-bromododecane from lauric acid: 1. Reduce lauric acid to dodecanol with LiAlH4: \(\mathrm{n}-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COOH + LiAlH}_{4} \rightarrow \mathrm{n}-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{CH}_{2} \mathrm{OH}\). 2. Convert dodecanol to 1-bromododecane with HBr: \(n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{CH}_{2} \mathrm{OH} + \mathrm{HBr} \rightarrow n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{CH}_{2} \mathrm{Br} + \mathrm{H}_{2}\mathrm{O}\). (b) Synthesis of tridecanoic acid from lauric acid: 1. Convert lauric acid to dodecanoyl chloride with SOCl2: \(n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COOH} + \mathrm{SOCl}_{2} \rightarrow n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COCl} + \mathrm{SO}_{2} + \mathrm{HCl}\). 2. Friedel-Crafts acylation with benzene: \(n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COCl} + \mathrm{C}_{6} \mathrm{H}_{6} \rightarrow n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCl} + \mathrm{HCl}\). 3. Basic hydrolysis: \(n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCl} + \mathrm{OH}^{-} \rightarrow n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\). (c) Synthesis of 1-tetradecanol from lauric acid: 1. Convert lauric acid to dodecanoyl chloride with SOCl2: \(n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COOH} + \mathrm{SOCl}_{2} \rightarrow n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COCl} + \mathrm{SO}_{2} + \mathrm{HCl}\). 2. Grignard reaction with EtMgBr: \(n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COCl} + CH_{3}CH_{2}\mathrm{MgBr} \rightarrow n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COCH}_{2} \mathrm{CH}_{3} + \mathrm{MgBrCl}\). 3. Reduce the ketone to 1-tetradecanol with NaBH₄: \(n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{COCH}_{2} \mathrm{CH}_{3} + \mathrm{NaBH}_{4} \rightarrow n-\mathrm{C}_{11} \mathrm{H}_{23} \mathrm{CH}_{2}\mathrm{OH} \mathrm{CH}_{2} \mathrm{CH}_{3}\).