Question

Predict the Product. Provide the stable organic product(s) for each reaction below.

11:46 PM Tue Oct 29 42% cdn.inst-fs-iad - prod.inscloudgate.net CONCEPT QUESTION Predict the Product. Provide the stable organic product(s) for each reaction below NaBH, CH3ОH 1 2. HСI, H,о NaBHa (1 eq.), CH,он 1 2. НС, Н,0 NaBH(ОCH3)з. (1 еq.), CH,Он 1. 2. HСI, Н,о LIAIH4 (1 eq.), THF 1. 2. HCI, H,о

Answer 1

Answer:

1. NaBH4, MOH 2. HCI, H20 1-p-tolylethanone 1-p-tolylethanol NaBH4 sodium borohydride is reducing agent selectively for carbonyl compounds aldehyde and ketone (in priority order if 1 eqv. only). SBH reduces ketone to secondary alcohol. NaBH4 in methanol solvent first forms alkyl borate which on HCI,H20 i.e. acid wash gives corresponding alcohol. U 1. NaBH4 1 eqv., MeOH 2. HCI, H20 OH 4-oxocyclohexanecarbaldehyde нон 4-(hydroxymethyl)cyclohexanone SBH reduces aldehydic functional group preferentially over ketone. The very reason is the more electrophilicity of Carbonyl C in aldehyde than that of ketone. In aldehyde there is only one alkyl group which have electron donating inductive effect at carbonyl C in aldehyde but in ketone there are 2 alkyl group which low ers its electrophilicity than aldehyde case. OH NABH(OME), is far more selective than NaBH, and reduces aldehyde selectively over acid chloride. 1. NaBH (OME), MeOH 2. HCI, H2O LiAlH4 can reduce both aldehyde and acid chloride but aldehyde prior to acid chloride when only 1 eqv. of BrLiAlH4 used. Br 1. LIAIH4 (1 eqv.), THF 2. HCI, H20 HO 6

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