In each reaction box, place the best reagent and conditions from the list below.
1) Br2
2) excess NaNH2
3) CH3CH2Br
4) H2, Lindlard's catalyst
Explanation: Reaction of alkene with Br2 produced dibromo compound as product. NaNH2 is a strong base which gives dehydrohalogenation and produces acetylide ion which on further reaction with alkyl halide produces an extended carbon chain as product. Reduction of alkyne with H2,Lindlard's catalyst produces cis-alkene as product.
In each reaction box, place the best reagent and conditions from the list below. In each...
In each reaction box, place the best reagent and conditions from the list below. BH3/THF PBr3 excess NaNH2 HBr H20, H2SO4, HgSO4 NaBH4 CH3BT H2O2, NaOH, H20 Hz, Lindlar catalyst CH3CH2Br Br2 Na, NH3 (1) 2 equiv. NaNH2
In each reaction box, place the best reagent and conditions from the list below. You have the wrong reagent for the final step. An oxidizing agent is necessary to convert the boron functional group into an enol, which will tautomerize to the aldehyde. 1) Br2 DEDY 2) excess NaNH2, then H20 3) BH3/THF 2 equiv. NaNH2 H2, Lindlar catalyst bromocyclohexane HBr H20, H2SO4, HgSO4 NaBH4 CH3Br CH3CH2Br H2O2, NaOH, H20 SOCI2
In each reaction box, place the best reagent and conditions from the list below 1) 2) 3) 4) HBr PВrg Вн/THF excess NaNH2 H20, H2SO4, HgSO4 NaBH4 CH3Br H2, Lindlar catalyst H2O2, NAOH, H20 СH-CH2Br Br2 Na, NH3 () 2 equiv. NaNH2
In each reaction box, place the best reagent and Conditions from the list below. H C C H Br Br H2, Lindlar catalyst H20, H2SO4, Hgso4 CH3CH2Br HBr 1-equiv BH3/THF HBr 2-equiv CH3CH2CH2Br CH3Br H202/NaOH Br2 1-equiv NaOH NaNH2 Br2 2-equiv
In each reaction box, place the best reagent and conditions from the list below. O OY NaBHA SOCI CH3CH2Br 2 equiv. NaNH2 BH/THF HBr Brz H2O3, NaOH, H2O bromocyclohexane H2. Lindlar catalyst H20, H2SO4, HgSO4 CH3Br excess NaNHz, then H20
In each reaction box, place the best reagent and conditions from the list below. In each reaction box, place the best reagent and conditions from the list below 1) 2) 3) 4) PBrs H2, Lindlar catalyst Вн-THF CH5CH2Br NaBH4 CH3Br H2O, H2SO4, HgSO4 Na, NHa () НаОг, NaOH, Hzо excess NaNH2 HBr Br2 2 equiv. NaNH2
In each reaction box, place the best reagent and conditions from the list below. CH3 2) 3) H2, Lindlar catalyst H2O, H2SO4 CH3CH2Br Br2 bromocyclohexane BH3/THF SOCl2 excess NaNH2 HBr CH3Br H2O2, NaOH, H2O
In each reaction box, place the best reagent and conditions from the list below. mu have the right idea for generating an alkyne in the first two steps. Remember that a primary alkyne also reacts with NaNH2 to form the alkylide ion, so with only two equivalents of NaNH2 you will d up with a mixture of some alkylide anion, some unreacted vinyl bromide and probably some unreacted vicinal dibromide. Br2 O 2 equiv. NaNH2 OY BH3/THF 3) 4) H2O2,...
In each reaction box, place the best reagent and conditions from the list below. In each reaction box, place the best reagent and conditions from the list below. CH3 BH3 THF H20, H2SO4 excess NaNH2 HBr SOCI2 CH3CH2Br H2, Lindlar catalyst H202, NaoH, H20 Br2 bromocyclohexane CH3Br
In each reaction box, place the best reagent and conditions from the list below. OOY NaBHA CH3CH2Br 2 equiv. NaNH2 Brz BHY/THF H2O2. NaOH, H20 SOCI HBR H2. Lindlar catalyst H20, H2SO4, HgSO4 bromocyclohexane excess NaNH2, then H2O CH3Br