Question

The major product formed in the following reaction sequence is:
Step 1: \(\mathrm{C_6H_{13}{-}C{\equiv}CH}\) is treated with catecholborane, then heated.
Step 2: the product from Step 1 is treated with \(\mathrm{Br_2}\), then \(\mathrm{NaOMe}\).

• A. \(\mathrm{C_6H_{13}{-}CH{=}CH{-}Br}\) (E-configured: the hexyl chain and \(\mathrm{Br}\) on opposite sides of the double bond)
• B. \(\mathrm{C_6H_{13}{-}CH{=}CBr_2}\) (1,1-dibromoalkene: both \(\mathrm{Br}\) atoms on the terminal alkene carbon)
• C. \(\mathrm{C_6H_{13}{-}CH{=}CH{-}Br}\) (Z-configured: the hexyl chain and \(\mathrm{Br}\) on the same side of the double bond)
• D. \(\mathrm{C_6H_{13}{-}CBr{=}CH{-}Br}\) (1,2-dibromoalkene: one \(\mathrm{Br}\) on each alkene carbon)

Hint:

Catecholborane hydroboration of the terminal alkyne gives the (E)-alkenylboronate (syn addition, boron on the terminal carbon); \(\mathrm{Br_2}\) then \(\mathrm{NaOMe}\) swaps boron for bromine with net inversion of the double bond geometry, so the product is the (Z)-vinyl bromide, not the (E)-isomer or a dibromide.

Answer 1

The Correct Option is C

Step 1: Hydroboration of the terminal alkyne with catecholborane.
Catecholborane (\(\mathrm{HBcat}\)) adds across the triple bond of the terminal alkyne \(\mathrm{C_6H_{13}{-}C{\equiv}CH}\) by a concerted, syn addition of the \(\mathrm{H{-}B}\) bond, with boron going to the less hindered terminal carbon (anti-Markovnikov regiochemistry, typical for hydroboration). Because the addition is syn, both new bonds form on the same face of the triple bond, so the resulting alkenylboronate has the hexyl chain and the boron group on opposite faces of the new double bond. This gives the (E)-1-alkenylboronate: \(\mathrm{C_6H_{13}{-}CH{=}CH{-}Bcat}\), with \(\mathrm{C_6H_{13}}\) and \(\mathrm{Bcat}\) trans to each other.

Step 2: Bromination of the alkenylboronate.
Treating the (E)-alkenylboronate with \(\mathrm{Br_2}\) adds bromine across the same carbon-carbon double bond in an anti fashion, placing the two new bromine atoms, and the retained boron group, on defined faces of the resulting \(\mathrm{C{-}C}\) bond.

Step 3: Base-induced anti-periplanar elimination (bromo-deboronation).
Adding \(\mathrm{NaOMe}\) promotes an anti-periplanar elimination that expels the boronate group together with one bromide, regenerating a carbon-carbon double bond in the same position as before. This elimination proceeds with a net inversion of the alkene geometry relative to the original alkenylboronate, the boron is formally replaced by bromine, but on the opposite face. Since the starting alkenylboronate was (E), chain and boron trans, the product vinyl bromide comes out (Z), chain and \(\mathrm{Br}\) cis, that is, on the same side of the double bond.

Step 4: Identify the product.
The overall transformation is a net anti-Markovnikov, stereospecific conversion of the terminal alkyne into a single monobromoalkene, with inversion of the alkene geometry set up in the hydroboration step. The product is \(\mathrm{C_6H_{13}{-}CH{=}CH{-}Br}\) with the hexyl chain and \(\mathrm{Br}\) on the same side of the double bond, that is, the (Z)-isomer.

Why the other options are wrong:
Option (A), the (E)-vinyl bromide, would be the product only if the boron-to-bromine exchange happened with retention of configuration, but the bromination and anti-elimination sequence inverts the geometry, so (E) is not the major product.
Options (B) and (D) both carry two bromine atoms on the alkene. This sequence replaces the single boron substituent with a single bromine, it does not add a second equivalent of \(\mathrm{Br}\) across the final double bond, so a dibromoalkene, whether 1,1- as in B, or 1,2- as in D, is not the product.

Final Answer:
The major product is the (Z)-vinyl bromide, \(\mathrm{C_6H_{13}{-}CH{=}CH{-}Br}\) with the chain and \(\mathrm{Br}\) cis. \[ \boxed{\text{(C)}} \]