Note 1: *You can put more than one item into a bin, so if you need two reactants

Question

Note 1: *You can put more than one item into a bin, so if you need two reactants in the same mix, put them both in the same box.** Note 2: You don't need to necessarily fill all the boxes. Note 3: Try to work these on paper, the same way you would on a test, and put in your own correct answers. Then once you've got your answer, look for the chemicals in the bins. NH2 NH ammonia (perhaps excess) PCC methyl amine ethyl amine PhCH2NH2 LiAIH4 (and workup) PBr3 CH3COC CH3CH2COC CH3CH2CH2COC Bromoethane cyclopentanone 2CrO4 PhCOCH3 PhCOCH2CH3 NaOH propanalNaBH3CN, cat. HBromopropane butanalPhCOCI 3-pentanone Nothing more needed

Explanation / Answer

At first you need to take PhCOCl. The lone pair over nitrogen atom attack on the carbonyl carbon of PhCOCl and via addition elimination process You will get PhCONHCH2CH(CH3)2. In this step you must use a base like NaOH or NaHCO3 to absorb the excess amount of HCl. Next you have to choose a reducing agent which seletively reduce carbonyl moiety (ie CO gr) to get the desired product. Here you can choose LiAlH4 (followedby work up) which selectively reduces carbonyl carbon.(amide to amine reduction)

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