Predict the products from reaction of 1 -hexyne with the...

Predict the products from reaction of 1 -hexyne with the following reagents: (a) 1 equiv HBr (b) 1 equiv $\mathrm{Cl}_{2}$ (c) $\mathrm{H}_{2},$ Lindlar catalyst (d) $\mathrm{NaNH}_{2}$ in $\mathrm{NH}_{3},$ then $\mathrm{CH}_{3} \mathrm{Br}$ (e) $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HgSO}_{4}$ (f) 2 equiv HCl

Predict the products from reaction of 1 -hexyne with the following reagents:
(a) 1 equiv HBr
(b) 1 equiv $\mathrm{Cl}_{2}$
(c) $\mathrm{H}_{2},$ Lindlar catalyst
(d) $\mathrm{NaNH}_{2}$ in $\mathrm{NH}_{3},$ then $\mathrm{CH}_{3} \mathrm{Br}$
(e) $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HgSO}_{4}$
(f) 2 equiv HCl

Key Concepts

Electrophilic Addition to Alkynes and Markovnikov's Rule

This concept explains how electrophiles add to the triple bond of an alkyne. In these reactions, the distribution of the added elements (such as hydrogen and a halogen) follows Markovnikov's rule whereby the proton (from HX) attaches to the carbon with more hydrogen atoms already, resulting in a more stable, substituted carbocation intermediate before the nucleophile attacks.

Stereospecific Anti Addition in Halogenation Reactions

In halogen addition reactions to alkynes, such as with Cl2 or HBr, the mechanism often involves formation of a cyclic halonium ion intermediate. This leads to anti addition, meaning that the two halogen substituents add to opposite faces of the triple bond, resulting in a trans configuration in the resulting dihalogenated alkene when two equivalents are used.

Partial Hydrogenation Using Lindlar Catalyst

This concept covers the catalytic hydrogenation of alkynes to cis-alkenes. The Lindlar catalyst, a poisoned palladium catalyst, facilitates the addition of hydrogen in a controlled, syn (same side) fashion. The result is the selective formation of a cis-alkene from the alkyne without further reduction to an alkane.

Acetylide Anion Formation and Alkylation

Terminal alkynes possess an acidic hydrogen that can be deprotonated by a strong base to form an acetylide anion. This anion is a good nucleophile and can participate in SN2 reactions with alkyl halides, allowing for the formation of new carbon–carbon bonds and the extension of the carbon chain.

Hydration of Alkynes and Keto-Enol Tautomerism

Under acidic conditions with a mercuric salt catalyst, alkynes undergo hydration to form enol intermediates. These unstable enols quickly tautomerize to the more stable keto form, typically yielding methyl ketones. This transformation is important both synthetically and mechanistically in organic chemistry.

Sequential Addition of Electrophiles to Alkynes

When alkynes react with an excess of electrophilic reagents such as HCl, multiple additions occur. The first equivalent forms a vinyl halide, and the subsequent addition can lead to geminal dihalide formation. This sequential addition illustrates how reaction conditions and the molar ratio of reagents can influence the final product structure in alkyne chemistry.

Transcript

00:02 Problem 9 .22 tells us to start with one hexine and use the reagents given to protect the products.

00:13 In part a, we are reacting one hexine with one equivalent of hbr.

00:20 And the electrophilic or nucleophilic triple bond is going to attack the electrophilic hydrogen, forming a new bond, and then we're left with the negative bromide.

00:36 This proceeds via a carbocadion intermediate.

00:49 So we're going to have a carbokadion in the markovnikov position, and this is going to be electrophilic, and the nucleophilic bromide will attach there.

01:34 In part b, we are reacting one hexine with one equivalent of cl2.

01:42 This is going to proceed in a similar fashion to the reaction above, that we're going to get a trans molecule and this is our product for part b.

02:15 In part c we should already know that hydrogen and a lindlar catalyst will convert a triple bond and reduce it to a double bond.

02:25 It will give the cis form of a double bond but since we have a terminal triple bond here we do not have cis or trans orientation.

02:47 So here we have one hexene.

02:52 Now let's look at part c.

02:54 We have a strong base which will remove the terminal hydrogen from the triple bond, and this is in liquid pneumonia.

03:15 So we will have a negative charge on our terminal carbon, and then we'll react it with ch3br, which will add another carbon to the end of the triple bond...

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