MC test: Reduction reactions

Multiple choice test on 20.1(4) Reduction reactions

Use the following 'quiz' to test your knowledge and understanding of this sub-topic. You will need access to a periodic table (Section 6 of the IB data booklet). Note that this quiz covers both electrophilic addition and electrophilic substitution reactions.

If you get an answer wrong, read through the explanation carefully to learn from your mistakes.

Which can be reduced to primary alcohols?

I. aldehydes

II. ketones

III. carboxylic acids

Ketones are reduced to secondary alcohols.

 

Which reagent can be used to convert propanoic acid into propan-1-ol?

Lithium aluminium hydride is reacted using ether (an aprotic solvent) as a solvent then the mixture is acidified to obtain the alcohol.

 

Which reagent can be used to obtain propan-2-ol from propanone?

Although sodium borohydide cannot be used to reduce carboxylic acids it is a good reducing agent for aldehydes and ketones to form  primary and secondary alcohols respectively.

 

Sodium borohydride and lithium aluminium hydride are inorganic reducing agents. Which species do they provide that are effectively responsible for the reduction process?

They both produce hydride ions, H− which act as a nucleophile reacting with the δ+ carbon atom of the carbonyl group. 

 

Nitrobenzene can be reduced to phenylamine in two stages. In the first stage nitrobenzene is refluxed with a mixture of tin metal and concentrated hydrochloric acid to form the phenylammonium ion as an intermediate.

The half-equation for the formation of the phenylammonium ion is:

Which row contains the correct values for x, y and z?

Row x y z
1 3 2 1
2 7 6 2
3 5 4 2
4 5 5 2

There are two O atoms on the LHS so put 2H2O on the RHL. Hydrogen atoms are obtained from H+ so three for the phenylammonium ion and four for the water are required making 7 in total. Then balance up the charges on both sides by using 6 electrons.

 

Nitrobenzene can be reduce to phenylamine in two stages. In the first stage nitrobenzene is refluxed with a mixture of tin metal and concentrated hydrochloric acid to form the phenylammonium ion, C6H5NH3+ as an intermediate. In the second stage the phenylammonium ion is converted into phenylamine, C6H5NH2.

Which reagent can be used to bring about the second stage of this conversion?

This is a classic reaction of the conjugate acid of a weak base reacting with a strong base to release the weak base and water.
C6H5NH3+ + OH− → C6H5NH2 + H2O

 

Starting with propanoic acid as the only organic reagent a route to make propyl propanoate consists of two separate steps. 

Step 1: Convert some of the propanoic acid into propan-1-ol.

Step 2: React the propan-1-ol formed in step 1 with the remainder of the propanoic acid to make the ester.

Which row gives the correct reagents and general conditions required?

Row Step 1 Step 2
1 LiAlH4 in ether Reflux in an aqueous solution
2 NaBH4 in ethanol Add concentrated H2SO4 and reflux
3 LiAlH4 in ether followed by acidification Add concentrated H2SO4 and warm
4 NaBH4 in ethanol followed by acidification Warm in an aqueous solution

Only lithium aluminium hydride is a strong enough reducing agent to reduce carboxylic acids to primary alcohols. Esters are made by warming the requisite carboxylic acid and alcohol in the presence of a small amount of concentrated sulfuric acid to act as a catalyst.

 

Butyl butanoate can be prepared from butanal in three separate steps.

Which reagents and conditions could be used for Step 1?

Aldehydes are best reduced to primary alcohols using sodium borohydride in either water or ethanol as the solvent.

 

Butyl butanoate can be prepared from butanal in three separate steps.

Which reagents and conditions could be used for Step 2?

Aldehydes can be oxidized to carboxylic acids using acidified solutions of either potassium dichromate(VI) or potassium manganate(VII)

 

Which route will produce prop-2-yl ethanoate CH3COOCH(CH3)2 using only ethanoic acid and propanone as the organic reagents?

Propanone should be reduced to propan-2-ol which can then form the ester with ethanoic acid.

 

 

Total Score:
All materials on this website are for the exclusive use of teachers and students at subscribing schools for the period of their subscription. Any unauthorised copying or posting of materials on other websites is an infringement of our copyright and could result in your account being blocked and legal action being taken against you.