1,1-Dibromoheptane is treated first with excess sodium amide (NaNH2) and then treated with bromoethane. Hydrogenation of this product with Lindlar's catalyst produces what major product?Select answer from the options below(E)-2-octene(E)-3-nonene(E)-2-nonene(Z)-3-nonene(Z)-2-nonene(Z)-2-octene
Question
1,1-Dibromoheptane is treated first with excess sodium amide (NaNH2) and then treated with bromoethane. Hydrogenation of this product with Lindlar's catalyst produces what major product?Select answer from the options below(E)-2-octene(E)-3-nonene(E)-2-nonene(Z)-3-nonene(Z)-2-nonene(Z)-2-octene
Solution
Primero, tratamos 1,1-dibromoheptano con exceso de sodio amida (NaNH2). Este paso genera un alquino terminal debido a la eliminación de los átomos de bromo y la formación de un triple enlace. La reacción se puede representar así:
1,1-dibromoheptano + NaNH2 → hept-1-ino
Luego, tratamos el hept-1-ino con bromoetano. Este paso implica una reacción de alquilación, donde el bromoetano agrega un grupo etilo al carbono terminal del alquino, formando un alquino interno:
hept-1-ino + bromoetano → non-2-ino
Finalmente, hidrogenamos el non-2-ino usando el catalizador de Lindlar. Este catalizador es conocido por su capacidad de hidrogenar alquinos a alquenos cis (Z):
non-2-ino + H2 (Lindlar) → (Z)-2-noneno
Por lo tanto, el producto mayoritario de esta serie de reacciones es (Z)-2-noneno.
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