draw the alkane formed when 4,5,5-trimethyl-1-hexyne is treated with two equivalents of hbr.

When an alkyne is reacted with 2 moles of HBr, the Markovnikov rule applies. The H-atoms of the 2HBr attaches to the C atom with more H atoms, while the 2-Br atoms, attaches to the C atom with less H-atoms. The product would be 2,2, dibromo - 4,5,5 - trimethylhexane.

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transitionstate transitionstate · Answer 2 · 2018-11-04T05:43:06+01:00

Answer:

The Alkane formed is 5,5-dibromo-2,2,3-trimethylhexane. as shown below in attached scheme (Green Color).

Explanation:

Alkynes like Alkenes undergo Electrophillic Addition Reactions. The reaction given is a two step reaction. In step 1, the Alkyne adds first equivalent of HBr obeying Markovnikov's rule (i.e. Bromine will add to carbon containing less number of hydrogen atoms) and forms 2-bromo-4,5,5-trimethylhex-1-ene. In step 2, the alkene formed in first step (2-bromo-4,5,5-trimethylhex-1-ene) undergoes addition reaction with the second equivalent of HBr via Markovnikov's rule to produce 5,5-dibromo-2,2,3-trimethylhexane.

The scheme is attached below, Blue color is assigned to starting Alkyne, Red color is assigned to intermediate Alkene and Green color is assigned to product Alkane respectively.