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Chapter 4: Problem 58

It is feasible to prepare tertiary butyl ethyl ether by treating ethyl bromide with sodium tertiary butoxide, rather than treating tertiary butyl bromide with sodium ethoxide because (a) sodium ethoxide will not react with t-butyl bromide. (b) the first reaction is faster and gives more yield. (c) t-butyl bromide undergoes dehydrohalogenation to give alkene in presence of sodium ethoxide. (d) t-butyl bromide undergoes rearrangement.

Short Answer

Expert verified
Answer: Tertiary butyl bromide undergoes dehydrohalogenation to give alkene in the presence of sodium ethoxide, which results in a lower yield of the desired ether product.

Step by step solution

01

Option A

Sodium ethoxide will not react with t-butyl bromide. This statement is incorrect. Sodium ethoxide will definitely react with t-butyl bromide to form an ether, namely tertiary butyl ethyl ether. However, we need to consider the reaction conditions and the possibility of side reactions.
02

Option B

The first reaction is faster and gives more yield. It is true that the first reaction, which is between ethyl bromide and sodium tertiary butoxide, is faster and gives more yield compared to the second reaction (between tertiary butyl bromide and sodium ethoxide). The reason is that alkoxide ions are strong nucleophiles, and ethyl bromide is a primary alkyl halide. Primary alkyl halides are more reactive towards nucleophilic substitution reactions due to less steric hindrance. In the second reaction, the tertiary alkyl halide (t-butyl bromide) is highly sterically hindered, making the reaction less favorable. Hence, option (b) seems to be valid.
03

Option C

T-butyl bromide undergoes dehydrohalogenation to give alkene in the presence of sodium ethoxide. This statement is true. Tertiary alkyl halides tend to undergo dehydrohalogenation, which is an E2 elimination reaction, in the presence of strong bases. Sodium ethoxide is a strong base, and E2 elimination is a major competing reaction for the nucleophilic substitution reaction. This results in the formation of an alkene, which means that the desired ether product would be lesser in yield. Hence, this is the main reason tertiary butyl bromide must be avoided.
04

Option D

T-butyl bromide undergoes rearrangement. Rearrangement does occur in some reactions involving alkyl halides, particularly in the case of carbocation formation (such as during an SN1 reaction). However, rearrangement is not a key consideration in this specific exercise because we are discussing an SN2 reaction (nucleophilic substitution), where carbocation formation and rearrangement are not involved. So, considering the analysis of all the given options, we can conclude that: The correct answer is: (c) t-butyl bromide undergoes dehydrohalogenation to give alkene in the presence of sodium ethoxide.

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