exam qs

• Heating with acidified sodium dichromate (VI)


  the orange solution turns green, indicates the compound is being oxidised

  typically oxidises alcohols and aldehydes
• addition of 2,4 DNPH 


  bright orange precipitate is formed, indicates presence of carbonyl group (C=O) which could be aldehyde or ketone
• A no. alcohol so although it can be oxidised, it doesn't react with 2,4 DNPH to form precipitate

  B. no. ketone so although it can form a bright orange precipitate with 2,4 DNPH, it does not oxidise further under mild conditions

  C. no. secondary alcohol, it can be oxidised to form ketone which will react with 2,4 DNPH to form precipitate but not carbonyl compound initially

  D. yes. aldehyde so can be oxidised to form carboxylic acid, can form bright orange precipitate with 2,4 DNPH due to carbonyl

  
• step 1: the smell of peppermint remains because menthone is a ketone and does not oxidise easily under given conditions

  step 2: NaBH4 is a reducing agent, this reduces menthone (ketone) to menthol (secondary alcohol), indicates loss of smell during reduction

  ANSWER: A

  
• write the structural formula of the organic product formed when hydroxyethanal reacts with tollens' reagent

  HOCH2CHO -> HOCH2COOH

  tollens reagent is a mild oxidising agent so can only oxidise aldehyde (-CHO) group
• the biochemist also reacted hydroxyethanal with acidified dichromate by heating under reflux. write an equation for this oxidation.

  HOCH2CHO + 3[O] -> HOOCCOOH + H2O

  powerful oxidising agent used so

  aldehyde group oxidised once (no water produced)

  primary alcohol oxidised twice due to reflux

  water formed from oxidation of primary alcohol -> aldehyde
• the biochemist then reduced hydroxyethanal using aqueous NaBH4. write the structural formula of the organic product.

  HOCH2CHO -> HOCH2CH2OH

  This happens because NaBH₄ is a strong reducing agent that adds hydrogen to the carbonyl group (C=O) of the aldehyde, turning it into an alcohol.