Give the major product that forms when (Z)-3-methyl2-pentene reacts with each of the following reagents: (a) \(\mathrm{HI} ;\) (b) \(\mathrm{H}_{2}\) in the presence of a platinum catalyst; (c) \(\mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{H}_{2} \mathrm{SO}_{4} ;\) (d) \(\mathrm{Br}_{2}\) in \(\mathrm{CCl}_{4}\).

Short Answer

Expert verified
The major products of the reactions are as follows: (a) 2-iodo-3-methylpentane, (b) 3-methylpentane, (c) 2-methyl-3-pentanol, and (d) (2R, 3R)-2,3-dibromo-3-methylpentane and its enantiomer (2S, 3S)-2,3-dibromo-3-methylpentane.

Step by step solution

01

Reaction with Hydroiodic Acid (\(\mathrm{HI}\))

The reaction with \(HI\) will follow Markovnikov's rule, with the halogen attaching to the most substituted carbon of the alkene. The iodine will attach to the tertiary carbon, resulting in 2-iodo-3-methylpentane as the major product.
02

Reaction with Hydrogen (\(\mathrm{H}_{2}\)) in the Presence of a Platinum Catalyst

The reaction with hydrogen gas in the presence of a platinum catalyst will result in the hydrogenation of the alkene, leading to the addition of hydrogen across the double bond. This will yield 3-methylpentane.
03

Reaction with Water (\(\mathrm{H}_{2}\mathrm{O}\)) in \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

The reaction with water in the presence of sulphuric acid will lead to the hydration of the alkene, following Markovnikov's rule. This will result in the formation of 2-methyl-3-pentanol.
04

Reaction with Bromine (\(\mathrm{Br}_{2}\)) in \(\mathrm{CCl}_{4}\)

The reaction with bromine in carbon tetrachloride will lead to the addition of bromine across the double bond in an anti fashion, resulting in the formation of (2R, 3R)-2,3-dibromo-3-methylpentane and its enantiomer (2S, 3S)-2,3-dibromo-3-methylpentane.

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