L Question 17 14 pts Br Cl A) H30+ (work up) T) CH3CH2CH2Li B) NaOH, H20 U) CH3C

Question

L Question 17 14 pts Br Cl A) H30+ (work up) T) CH3CH2CH2Li B) NaOH, H20 U) CH3CH2CH2CH2Li C) CH3CH20Na, CH3CH2OH V) (CH3CH2)2CuL D) (CH3) 3CONa, (CH3) 3COH W) (CH3CH2CH2)2CuLi X) (CH3CH2CH2CH22CuLi E) H2SO4, H2O, heat Y) C6H11Cocl F) O3 Z) C6H10 PPh G) mCPBA AA) 11 H) HSCH2CH2SH, ZnCl2 AB) Cro3, H2SO4, H2O I) CH3CH2CCNa AC) PCC J) CH CH CH CCNa AD) LiAlH4 K) (CH3)2S AE) Raney-Ni, H2 L) Zn(Hg), HCl, heat AF) HgCl 2, H2O, CH30H CaCO3 M) Na, NH3 AG) Cl2 N) OsO4, H202 AH) Br O) CH3CH 2Br Al) Pd/C P) CH3CH2CH2Br AJ) Zn, CH3COOH Q) CH3CH2CH2CH2Br AK) Lindlars Catalyst R) AIC AL) H2 S) CH3CH2Li

Explanation / Answer

First we protect the carbonyl functional group with reagent H to give the functional group thiol. Then we add reagent J to give the alkyne functional group via Sn2 mechanism. Reagent M will give the trans alkene via anti reduction. This alkene can react with reagent AG in CCl4 solvent to give the haloalkane with anti regioselectivity. Then finally the protecting group can be removed with reagent AE. If we wanted to return to the ketone instead of the alkane we could use reagent A to remove the protecting group.

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