Question

Oct-1-yne (hc≡cch2ch2ch2ch2ch2ch3) reacts rapidly with nah, forming a gas that bubbles out of the reaction mixture, as one product. oct-1-yne also reacts rapidly with ch3mgbr, and a different gas is produced. write balanced equations for both reactions and identify the gases formed.

Answer 1

- Addition of NaH releases hydrogen gas.

- Addition of
`CH_3MgBr` releases methane.

Step-by-step explanation:

Hello,

On the attached picture, you will find both reactions as shown below:

- Addition of NaH: in this case, due to the fact that the alkynes are slightly acid, they are able to donate one proton, thus, the hydrogen bonded with the triple-bonded carbon will be released as hydrogen gas.

- Addition of
`CH_3MgBr`: in this case, a Grignard reactant, magnesium methyl bromide, is added so the methyl from such reactant will accept the hydrogen bonded with the triple-bonded carbon, forming a grignard intermediate and methane gas.

Best regards.

Answer 2

As we know that terminal Alkyne (alkyne with triple bond at the end) are acidic in nature, they can donate proton. When Sodium Hydride is reacted with 1-Octyne , the hydride abstract the terminal proton and forms Hydrogen Gas, while Methyl Magnesium Bromide when reacted with 1-Octyne eliminates Methane gas. Both reactions are given below,

Result:
Gas evolved in reaction of NaH is H₂.

Gas evolved in reaction of CH₃MgBr is CH₄.