[s3lf]

Hi,

First, I'm a super orgo noob .

I've been trying to think through some alkene synthesis basics and I'm hoping for some insight since the reaction mechanism in McMurry's text isn't present for this.

How would I go about synthesizing a terminal alkene from a simple hydrocarbon chain? For example, from pentane. Or even an alkyl halide like 1-bromopentane.

It seems to me that you'd always have a hydride shift by using KOH/ethanol, or with a dehydration reaction, eliminating OH. Is there something I'm missing? This seems like it should be pretty simple.

Thanks!

[Hunter2]

You can use the alcohol here n-Pentanol and eliminate water by using conc. sulfuric acid. By product can be the Ether and the Sulfate ester.

[easyorganic]

From pentane.  Halogenate (which ain't gonna be very selective) and then strong base (E2).

From alkylhalide.  strong base (E2).

Can you suggest a strong base that might achieve this? 

[s3lf]

Thanks for the replies.

Isn't KOH in ethanol already a strong base?

1-bromopentane + KOH (in ethanol) --> 1-pentene

Is this correct? Or would 2-pentene be formed since the intermediate would be more stable?

[discodermolide]

1-Pentene would be formed because the reaction is an elimination process. In basic solution it will not proceed via a carbonium ion intermediate.

[orgopete]

Thanks for the replies.

Isn't KOH in ethanol already a strong base?

1-bromopentane + KOH (in ethanol) --> 1-pentene

Is this correct? Or would 2-pentene be formed since the intermediate would be more stable?

Neither. I think the major product(s) would be pentane-1-ol (and dipentyl ether). Because the halide is primary, you will not get as much elimination. If you wanted to increase the amount of elimination, you should use KOtBu as the base. Even then, you will still get about 10% ether formation.

The original question is too broad to generalize. A more specific question would be helpful.