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 $\mathrm{NaOMe}$.

Question

Write the major product in the following reaction:
ethyl chloride (CH3CH2Cl)

Answer 1

This sequence is a classic two-step method for turning a terminal alkyne into a single, stereodefined vinyl bromide: hydroboration sets the alkene geometry with boron in place, and the bromination/elimination step swaps that boron for bromine, flipping the geometry in the process.

Step 1, hydroboration: catecholborane delivers $\mathrm{H}$ and $\mathrm{B}$ to the triple bond on the same face, syn addition, with boron landing on the terminal, less hindered carbon. Because both new atoms arrive from one face, the hexyl chain, already on the alkyne, ends up trans to the boron group in the product alkene: this is the (E)-alkenylboronate, $\mathrm{C_6H_{13}CH{=}CHBcat}$.
Step 2, bromination then base: $\mathrm{Br_2}$ first adds across the double bond from one face, anti addition, giving a vicinal bromo compound that still carries the boron group. $\mathrm{NaOMe}$ then pulls out the boron and an adjacent bromide in a single anti-periplanar elimination step, which rebuilds the double bond.
Net stereochemical outcome: an addition followed by an anti-elimination at the same two carbons flips the original alkene geometry. Since boron started trans to the chain (E), bromine ends up cis to the chain in the new double bond, giving the (Z)-vinyl bromide.
Why not the dibromides: options with two bromines on the double bond, $\mathrm{C_6H_{13}CH{=}CBr_2}$ or $\mathrm{C_6H_{13}CBr{=}CHBr}$, would require an extra bromine to stay on the alkene permanently. Here bromine only ever occupies the position boron is leaving from, one bromine net stays, the rest is consumed as bromide leaving with the boron.

Only one candidate is both a monobromide and has the geometry consistent with inversion from the (E)-alkenylboronate: the (Z)-alkene, $\mathrm{C_6H_{13}CH{=}CHBr}$ with the chain and bromine on the same side.

Let's summarize:

Catecholborane hydroboration of a terminal alkyne gives the (E)-alkenylboronate (chain trans to boron).
$\mathrm{Br_2}$ then $\mathrm{NaOMe}$ swaps boron for bromine with net inversion of alkene geometry.
The product therefore has the chain and bromine cis, the (Z)-vinyl bromide.

The major product is option (C), the (Z)-configured vinyl bromide.

Answer 2

This sequence is a classic two-step method for turning a terminal alkyne into a single, stereodefined vinyl bromide: hydroboration sets the alkene geometry with boron in place, and the bromination/elimination step swaps that boron for bromine, flipping the geometry in the process.

Step 1, hydroboration: catecholborane delivers $\mathrm{H}$ and $\mathrm{B}$ to the triple bond on the same face, syn addition, with boron landing on the terminal, less hindered carbon. Because both new atoms arrive from one face, the hexyl chain, already on the alkyne, ends up trans to the boron group in the product alkene: this is the (E)-alkenylboronate, $\mathrm{C_6H_{13}CH{=}CHBcat}$.
Step 2, bromination then base: $\mathrm{Br_2}$ first adds across the double bond from one face, anti addition, giving a vicinal bromo compound that still carries the boron group. $\mathrm{NaOMe}$ then pulls out the boron and an adjacent bromide in a single anti-periplanar elimination step, which rebuilds the double bond.
Net stereochemical outcome: an addition followed by an anti-elimination at the same two carbons flips the original alkene geometry. Since boron started trans to the chain (E), bromine ends up cis to the chain in the new double bond, giving the (Z)-vinyl bromide.
Why not the dibromides: options with two bromines on the double bond, $\mathrm{C_6H_{13}CH{=}CBr_2}$ or $\mathrm{C_6H_{13}CBr{=}CHBr}$, would require an extra bromine to stay on the alkene permanently. Here bromine only ever occupies the position boron is leaving from, one bromine net stays, the rest is consumed as bromide leaving with the boron.

Only one candidate is both a monobromide and has the geometry consistent with inversion from the (E)-alkenylboronate: the (Z)-alkene, $\mathrm{C_6H_{13}CH{=}CHBr}$ with the chain and bromine on the same side.

Let's summarize:

Catecholborane hydroboration of a terminal alkyne gives the (E)-alkenylboronate (chain trans to boron).
$\mathrm{Br_2}$ then $\mathrm{NaOMe}$ swaps boron for bromine with net inversion of alkene geometry.
The product therefore has the chain and bromine cis, the (Z)-vinyl bromide.

The major product is option (C), the (Z)-configured vinyl bromide.

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 \(\mathrm{NaOMe}\).

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Reaction mechanisms

Questions Asked in GATE CY exam

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 \(\mathrm{NaOMe}\).