Question

3. If sodium methoxide in methanol was used in place of potassium t butanol, your GC analysis would look much more like the 2-butanol/sulfuric acid reaction products. In addition, a new, non-alkene product would be present in the GC analysis from a competing, non-elimination mechanism. Explain why the alkene distribution would change, what the new product would be, and why this new product would form. 4. If 2-bromobutane was reacted with trimethylamine to produce sec-butyl trimethyl- ammonium bromide (see below), and then reacted with sodium ethoxide in ethanol, how would you expect the relative distribution of the alkene products to change? Do some research on "Hofmann elimination" in the textbook to help you answer this question. (CH3)3N NaOCH CHy Br

Answer 1

SOLUTION:

3.

If sodium methoxide in methanol was used , the GC analysis would look much more like the 2-butanol/sulfuric acid reaction products. In addition, a new, non-alkene product would be present in the GC analysis.

When potassium t-butoxide in t-butanolis used the product is different due to the strong basicity, A new non alkene product in the GC system indicte a possible anion generation

which is stabilized by the solvent leading to addition or clevage product. rhe sodium methoxide or 2-butanol in sullfuric acid behave in a similar way due to the soft nature of the reagent and th eliminationn occurs rather than a clevage.

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