Based on your answer to Problem 11-68, predict the...

Based on your answer to Problem $11-68,$ predict the product(s) and show the mechanism for each reaction below. $$ \begin{array}{l} \text { (a) } \mathrm{CH}_{3} \mathrm{CH}_{2} \\ \mathrm{C}_{\mathrm{H}}{ }^{\prime}=\mathrm{C}^{\prime} \end{array} $$ (b) $$ \sum_{\mathrm{Ph}}^{\mathrm{Br}} \mathrm{C}=\mathrm{C}^{\mathrm{CH}_{3}} \frac{1 . \mathrm{Na}^{+} \mathrm{NH}_{2}}{2 . \mathrm{H}_{3} \mathrm{O}^{+}} $$ (c) $$ \sum_{\mathrm{Cl}}^{\mathrm{H}} \mathrm{C}=\mathrm{C}^{\mathrm{H}} \quad \frac{1 . \mathrm{Na}^{+} \mathrm{NH}_{2}}{2 . \mathrm{H}_{3} \mathrm{O}^{+}} $$

Based on your answer to Problem $11-68,$ predict the product(s) and show the mechanism for each reaction below.
$$
\begin{array}{l}
\text { (a) } \mathrm{CH}_{3} \mathrm{CH}_{2} \\
\mathrm{C}_{\mathrm{H}}{ }^{\prime}=\mathrm{C}^{\prime}
\end{array}
$$
(b)
$$
\sum_{\mathrm{Ph}}^{\mathrm{Br}} \mathrm{C}=\mathrm{C}^{\mathrm{CH}_{3}} \frac{1 . \mathrm{Na}^{+} \mathrm{NH}_{2}}{2 . \mathrm{H}_{3} \mathrm{O}^{+}}
$$
(c)
$$
\sum_{\mathrm{Cl}}^{\mathrm{H}} \mathrm{C}=\mathrm{C}^{\mathrm{H}} \quad \frac{1 . \mathrm{Na}^{+} \mathrm{NH}_{2}}{2 . \mathrm{H}_{3} \mathrm{O}^{+}}
$$

Key Concepts

Role of Strong Bases in Dehydrohalogenation

NaNH2 is a very strong, non?nucleophilic base that is commonly used to promote double dehydrohalogenation reactions. Its strength is sufficient to remove relatively less acidic hydrogens adjacent to halogens, driving the formation of unsaturated bonds such as alkynes. The use of NaNH2 thus facilitates successive elimination events, which are key to transforming dihalides into alkynes.

E2 Elimination Mechanism

This mechanism involves the concerted removal of a proton by a base and elimination of a leaving group, leading to the formation of a ? bond. In the reactions in question, the strong base (NaNH2) abstracts a hydrogen atom while the leaving group (halide) departs simultaneously, often resulting in the formation of an alkene or alkyne depending on the substrate structure and number of eliminations.

Alkyne Synthesis via Double Elimination

The synthesis of alkynes from vicinal or geminal dihalides involves two successive elimination reactions. The first elimination typically forms an alkene intermediate, and the second elimination, with the aid of a strong base, converts the alkene into an alkyne. This strategy is a cornerstone in organic synthesis for constructing carbon?carbon triple bonds from pre?functionalized saturated or unsaturated precursors.

Leaving Group Ability in Elimination Reactions

The efficiency of the elimination process is significantly impacted by the quality of the leaving group. Halides like bromide and chloride exhibit different leaving group abilities, with bromide generally being better than chloride. This difference can affect the rate of elimination and the extent to which the reaction proceeds under a given set of conditions.

Reaction Mechanisms Involving Halogenated Alkenes and Alkynes

Understanding the stepwise mechanism, including sequential deprotonation and elimination from halogenated alkenes, is crucial. In these reactions, careful consideration is required to determine which hydrogen is abstracted by the base and how the electronic effects induced by substituents (like phenyl, methyl, etc.) influence the formation of the ? system. This deeper mechanistic insight allows chemists to predict the regiochemistry and stereochemistry of the reaction products.

Transcript

Please give Ace some feedback