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Propose a mechanism for the reaction of pent-1-yne with HBr in the presence of peroxides. Show why anti-Markovnikov orientation results.

Short Answer

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Anti-Markovnikov orientation results to form the more stable20free radical.

Step by step solution

01

Step-by-Step SolutionStep 1: Anti- Markovnikov rule

When the unsymmetrical reagent (HBr, HCl, etc) is added to an unsymmetrical alkene, the positive part of the unsymmetrical reagent goes to that unsaturated bond (double or triple) having a greater number of hydrogen atoms.

The product formed from such a reaction is governed by the anti-Markovnikov rule.

Anti- Markovnikov rule governs the product for many chemical reactions.

02

Mechanism for the reaction of pent-1-yne with HBr in the presence of peroxide 

In the initiation step, radicals are formed.

Initiation step

In the propagation step, the bromide radical adds to triple bond of pent-1-yne to generate alkyl radical on the secondary carbon atom, and then the secondary alkyl radical abstract hydrogen atom from HBr to generate product and bromide radical.

Mechanism showing the reaction of pent-1-yne with HBr in the presence of peroxides

03

Reason for anti-Markovnikov orientation

The anti- Markovnikov orientation result when pent-1-yne react with HBr in the presence of peroxide to form a more stable free radical. The more substituted radical is more stable; thus, 20radical is more stable than 10radical.

Hence, the bromine radical first attack forms a more stable 20radical forming the anti- Markovnikov oriented product.

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Most popular questions from this chapter

Question:Show how you would accomplish the following synthetic transformations. Show all intermediates.

(²¹)2,2-»å¾±²ú°ù´Ç³¾´Ç²ú³Ü³Ù²¹²Ô±ð→b³Ü³Ù-1-²â²Ô±ð

(²ú)2,2-»å¾±²ú°ù´Ç³¾´Ç²ú³Ü³Ù²¹²Ô±ð→b³Ü³Ù-2-²â²Ô±ð

(³¦)²ú³Ü³Ù-1-²â²Ô±ð→o³¦³Ù-3-²â²Ô±ð

(d)trans-³ó±ð³æ-2-±ð²Ô±ð→h±ð³æ-2-²â²Ô±ð

(±ð)2,2-»å¾±²ú°ù´Ç³¾´Ç³ó±ð³æ²¹²Ô±ð→h±ð³æ-1-²â²Ô±ð

(´Ú)³¦²â³¦±ô´Ç»å±ð³¦²â²Ô±ð→cis-cyclodecene

(²µ)³¦²â³¦±ô´Ç»å±ð³¦²â²Ô±ð→trans-cyclodecene

(h)hex-1-yne→hexan-2-one, CH3COCH2CH2CH2CH3

(i)hex-1-yne→hexanal, CH3(CH2)4CHO

(j)trans-³ó±ð³æ-2-±ð²Ô±ð→cis-hex-2-ene

When 2,2-dibromo-1-phenylpropane is heated overnight with sodium amide at 150 °C, the major product (after addition of water) is a different foul-smelling compound of formula C9H8. Propose a structure for this product, and give a mechanism to account for its formation.

Question: Show how you might synthesize the following compounds, using acetylene and any suitable alkyl halides as your starting materials. If the compound given cannot be synthesized by this method, explain why. (a) hex-1-yne (b) hex-2-yne (c) hex-3-yne (d) 4-methylhex-2-yne (e) 5-methylhex-2-yne (f) cyclodecyne.

Question: Show how you would synthesize each compound, beginning with acetylene and any necessary additional reagents.

(a) prop-2-yn-1-ol (propargyl alcohol)

(b) hept-2-yn-4-ol

(c) 2-phenylbut-3-yn-2-ol

(d) 3-methylhex-4-yn-3-ol

Question:Predict the products of the reaction of but-1-yne with the following reagents.

(a)1 equivalent of HCl

(b)2 equivalents of HCl

(c)excess H2 , Ni

(d) H2 , Pd /BaSO4, quinolone

(e)1 equivalent of Br2

(f)2 equivalents of Br2

(g)cold, dilute KMnO4

(h)warm, concd.KMnO4 , NaOH

(i)Na, liquid ammonia

(j) NaNH2

(k) H2SO4/HgSO4, H2O

(l) Sia2BH, then , H2O2,OH-

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