Jump to content


Redox


2 replies to this topic

#1 Andy-1872

    Newbie

  • Members
  • Pip
  • 4 posts
  • Gender:Male

Posted 24 April 2011 - 06:34 PM

Was wondering how to answer questions like these:

21. Which of the following reactions can be
classified as reduction?
A CH3CH2OH → CH3COOH
B CH3CH(OH)CH3 → CH3COCH3
C CH3CH2COCH3 → CH3CH2CH(OH)CH3
D CH3CH2CHO → CH3CH2COOH

Any help would be appreciated.

#2 Marcus

    Site Swot

  • Members
  • PipPipPipPip
  • 147 posts
  • Location:Aberdeen
  • Interests:Guitar - Passed my grade 6 :D :D<br />Write computer programs - won Heriot-Watt programming challenge 2006 and 2007 :D<br />
  • Gender:Male

Posted 24 April 2011 - 07:24 PM

View PostAndy-1872, on 24 April 2011 - 06:34 PM, said:

Was wondering how to answer questions like these:

21. Which of the following reactions can be
classified as reduction?
A CH3CH2OH → CH3COOH
B CH3CH(OH)CH3 → CH3COCH3
C CH3CH2COCH3 → CH3CH2CH(OH)CH3
D CH3CH2CHO → CH3CH2COOH

Any help would be appreciated.

Hi

If you notice all the choices are related to alcohols, ketones, aldehydes, and carboxylic acids where in higher you mainly look at the oxidation reactions of these, so you are look for the reaction that is 'backwards' hence indicating reduction.

Let's look at each of the options:

A primary alcohol -> carboxylic acid
B secondary alcohol -> ketone
C ketone -> secondary alcohol
D aldehyde -> carboxylic acid

Well looking at those, it must be either B or C, as they are opposites of each other, now you should know that secondary alcohols are oxidised to ketones, the answer must be C.

You could also use rule (for organic molecules): reduction reduces the oxygen/hydrogen ratio, whereas oxidation increases this.

A 1/6 -> 1/2 (increase)
B 1/8 -> 1/6 (increase)
C 1/8 -> 1/10 (reduction)
D 1/6 -> 1/3 (increase)

again answer must be C

hope this helps show a couple of ways of going about this problem
=-=-=Marcus=-=-=

#3 Andy-1872

    Newbie

  • Members
  • Pip
  • 4 posts
  • Gender:Male

Posted 28 April 2011 - 03:29 PM

Ah that makes clear sense now, thanks :)





1 user(s) are reading this topic

0 members, 1 guests, 0 anonymous users