Question

Draw the major, neutral organic product(s) for each reaction below.

Incorrect. When reacting with a strong base, tertiary alkyl halides will undergo an elimination reaction instead of a substitution reaction.

Answer 1

General guidance

Concepts and reason

A reaction is given in which an organic reactant reacts with sodamide and tertiary butyl bromide, whose correct product needs to be formed. Sodamide (${\\rm{NaN}}{{\\rm{H}}_2}$) act as a base which can abstract an acidic proton and can generate a nucleophile which may attack at tertiary haloalkane to form the suitable elimination product.

Thus, identify the role of the given reagent correctly and write the correct product of the reaction.

Fundamentals

Sodamide is a very strong base which result in the abstraction of an acidic proton and can generate a carbanion which can either act as a nucleophile or a base.

E2 elimination reaction takes place in one step through the formation of a transition state which result in the simultaneous removal of beta-hydrogen and halogen atom. In haloalkane, the order of reactivity for the E2 elimination is, $3^\\circ < 2^\\circ < 1^\\circ < {\\rm{C}}{{\\rm{H}}_3}$.

But in the presence of sterically bulky base E2 is favored even with the tertiary alkyl halide no substitution product is formed.

Step-by-step

Step 1 of 2

Write the product of the reaction in which the given substrate or organic reactant reacts with the reagent 1, which is sodamide as shown below:

NaNH2 prop-1-yn-1-ide Propyne

Explanation | Hint for next step

The H attached to the ${\\rm{sp}}$ hybridized carbon atom is highly acidic in nature as the corresponding conjugate base is contains the negative charge on ${\\rm{sp}}$ hybridized carbon atom, thus, this can be easily abstracted by the base.

In the given reaction propyne reacts with a base, like sodamide to generate a negative charge on the carbon atom as shown below:

Step 2 of 2

Consider the reaction between the carbanion generated in step 1 with the given tertiary alkyl halide as shown below:

The products are as follows:

The carbanion generated in step 1 can act as nucleophile as well as base, therefore, it can abstract the beta hydrogen from tertiary alkyl bromide to form the elimination product as shown below:

Sterically hindered base generally results in the formation of elimination product with tertiary alkyl halide that follows E2 elimination mechanism as compared to the substitution product.

Thus, beta hydrogen will be abstracted from the alkyl halide by the strong base and result in removal of halogen. Hence, alkene will be formed as a product.

Answer

The products are as follows:

Answer only

The products are as follows: