|-. Scantron Line 1: A reagent that converts l-pentanol to pentanoic acid (CH3CH

Question

|-. Scantron Line 1: A reagent that converts l-pentanol to pentanoic acid (CH3CH2CH2CH2C(=O)(H) is NaBH4. PCC(pyridinium chlorochromate) [B. Acidic Na2Cr207C DMSO (dimethyl sulfoxide). (CH3)2S-o D Thionyl chloride (SoCl2) E 2-. Scantron Line 2: Which of the following carbocations would NOT be likely to undergo rearrangement? Scantron Line 3: Which step in the acid-catalyzed dehydration of 3,3-dimethyl-2-butanol to (CH3)2C-C(CH3)2 is rate limiting? #1 . Protonation of OH A #2. Loss of H20 #3. Hydride shift C . #3. Alkyl shift [D] . #4. Deprotonation .

Explanation / Answer

1) option C) Acidic Na2Cr2O7 is correct

Explanation: Acidic Na2Cr2O7 is oxidising agent which oxidises primary alcohol to carboxylic acid.

2) option C will not undergo rearrangement

Explanation: option C is a tertiary carbocation and it is more stable.

3) option 2) Loss of H2O is the rate limiting step.

Explanation: dehydration of 3,3-dimethyl-2-butanol involves E1 mechanism and the rate limiting step in E1 mechanism is loss of leaving group and formation of carbocation.

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