[orgo814]

2 questions: 1) Why when 1-butyne is reacted with H2(in excess)/Pt, the product is butane while when it is reacted with H2, Ni2B(P-2) the product is 1-butene? Why would there be more hydrogenation with the excess hydrogen/Pt catalyst?

2) When 1 butyne is reacted with NaNH2 in liquid  NH3, then (CH3)3CBr, we yield products  based off of an E2 reaction. However, when 1-butyne is reacted with NaNH2 in liquid NH3 then CH3I we get 2-pentyne? Why did the first scenario go by elimination?

This is Organic Chemistry 1 so please do not exceed that level of knowledge

[discodermolide]

Question 2 first: Can MeI undergo an E2 reaction?

Question 1; I'm not sure what Ni₂B(P-2) is can you explain this.
Anyway what happens when alkynes are hydrogenated with a deactivated catalyst, Hint: This reaction has a name.

[orgo814]

I understand my first question. Apparently H2(excess) means it is hydrogenated twice (at least in these problems). Sometimes my textbook is a little unclear in what it's asking us.

However, I still don't get WHY the alkyne can only react with the tert-butyl bromide by elimination (E2). Why can't it just the bromine simply leave the tert butyl group and the lone pair on the alkyne react with the tert butyl forming that molecule. Does the tertiary carbon have anything to do with preventing this sort of reaction? The problem after it is the same thing but with CH3-I and it reacts the way I described above. Confused..

[discodermolide]

I ask again, can methyl iodide undergo an E2 elimination?
In the other case why should the bromine just leave, you can't have atoms just upping and walking away otherwise we would all just fall apart.
What is it about t-Butylbromide that seems to favour an E2 elimination? Make a model of it.