Reaction of CH3CH2MgBr with ethylene epoxide gives:

Explanation

The reaction of CH3CH2MgBr (ethylmagnesium bromide) with ethylene oxide (ethylene epoxide) typically results in the formation of 1,2-propanediol. This reaction is an example of a nucleophilic substitution reaction in which the nucleophilic carbon of the organometallic reagent attacks the electrophilic carbon of the epoxide, leading to the formation of an alcohol.The reaction can be represented by the following equation:CH3CH2MgBr + C2H4O-> CH3CH2CH2OHIn this reaction, the ethylmagnesium bromide acts as a nucleophile, attacking the epoxide ring and displacing the bromide ion. The resulting intermediate is an alkoxide ion, which quickly protonates to form the corresponding alcohol, 1,2-propanediol.It's important to note that the reaction conditions, such as temperature, solvent, and presence of any catalysts, can influence the outcome of the reaction. However, under typical conditions, the reaction described above is the most common result.Ethylene epoxide reacts with Grignard's Reagent to produce an additive product, which upon hydrolysis yields a primary alcohol. The type of primary alcohol depends upon the number of C atoms in the grignard reagent's 'R' group (RMgX).Since we have two C in Grignard's Reagent and two in ethylene epoxide, the total number of carbon atoms in primary alcohol will be four, so the name of the alcohol formed is 1-butanol.