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9701 MCQ Solution 79

Question 79: [Organic > Hydroxyl Compounds]

When the apparatus below was used with compound Z, a brick-red precipitate formed in the right-hand tube.
Which compound could be Z?
1 CH3CH(OH)CH3
2 CH3CH2CH2OH
3 CH3OH
Reference: Past Exam Paper –June 2003 Paper 1 Q39



Solution 119:
Answer: C.
1, 2 and 3 are all alcohols. When an alcohol is presented to heated acidified potassium dichromate they will oxidise. In 1, the alcohol has 1 H atom bonded to the alcohol-bearing C atom and is secondary which will form a ketone. In 2 and 3 the alcohols have 2 or more H atoms bonded to the alcohol-bearing C atom and are primary which will form aldehydes. Aldehydes turn into a brick red precipitate when presented to Fehling’s reagent. Thus 2 and 3 are the only possible compounds that Z can be. Thus the correct answer is option C

9701 MCQ Solution 69

Question 69: [Organic > Hydroxyl Compounds]
In a preparation of ethene, ethanol is added a drop at a time to a heated reagent Y. To purify the ethane it is bubbled through a solution Z and then collected.
What could reagent reagent Y and solution Z be?

reagent Y
solution Z
A
acidified K2Cr2O7
dilute NaOH
B
concentrated H2SO4
dilute H2SO4
C
concentrated H2SO4
dilute NaOH
D
ethanolic NaOH
concentrated H2SO4

Reference: Past Exam Paper –June 2003 Paper 1 Q29



Solution 69:
Answer: C.

Alcohol undergoes dehydration to form an alkene. Concentrated H2SO4 is among the possible conditions for this reaction. This eliminates options A and D as possible answers. After dehydration occurs there may be remaining H2SO4 mixed with the alkene. This remainder must be neutralised. It is acidic and in option B, the addition of more H2SO4 will contaminate the alkene more, thus is an incorrect answer. A basic solution will neutralise the acid, such as the dilute NaOH in option C, which is the correct answer.