Question

The reagents used for the following reactions is/are:
$C{{H}_{3}}-CH=CH-{{C}_{2}}{{H}_{5}}\to C{{H}_{3}}CHO+{{C}_{2}}{{H}_{5}}CHO$
This question has multiple correct options.
(A) ${{O}_{3}}$ and Zn/steam
(B) Baeyer's reagent
(C) $KMn{{O}_{4}}/{{H}_{2}}S{{O}_{4}}$
(D) Lemieux reagent

Answer 1

In this question, we can see that pent-2-ene is converted into ethanal (or acetaldehyde) and propanal. By determining the product when the reagents given interact with pent-e-ene, we can find which reagent satisfies the given reaction.

Complete answer:
When pent-2-ene is reacted with reagent (A) ${{O}_{3}}$ and Zn/steam, it undergoes the ozonolysis process to form ethanal and propanal.
When pent-2-ene is reacted with reagent (B) Baeyer's reagent i.e. alkaline solution of cold potassium permanganate, it will lose its color to form pentane-1,2-diol.
When pent-2-ene is reacted with reagent (C) $KMn{{O}_{4}}/{{H}_{2}}S{{O}_{4}}$, the alkene undergoes oxidation to form carbonyl compounds carboxylic acid (or ethanoic acid) and propanoic acid.
When pent-2-ene is reacted with reagent (D) Lemieux reagent, the alkene undergoes oxidative cleavage to form ethanal and propanal.
Lemieux reagent is prepared by making an aqueous solution of sodium periodate $NaI{{O}_{4}}$ and a trace of potassium permanganate $KMn{{O}_{4}}$ and can be used to locate the position of unsaturation in an alkene.
The product formed from the Lemieux oxidation of alkene is the same as the product formed in the last stage of ozonolysis of alkene. Aldehydes are formed in both processes.

Since, reagents (A) and (D) form the products required, the answer is option (A) and option (D).

Note:
Since the Lemieux oxidation also produces various side products, the yield of required aldehyde is low and hence usually ozonolysis is used to prepare aldehydes from alkenes. It should be noted that during preparation of aldehydes from an alkene by ozonolysis method, an ozonide intermediate cation is formed.