[slimlingsaucer]

CH3CH2CH2Br + LiAlH4 ---) ? e2??

[organic help]

Well, LAH (LiAlH₄) is a strong base, and your substrate is a primary alkyl halide and definitely capable of E₂, so i would say that you are seeing E₂ out of this reaction.  Any thoughts from anyone else?

[choongaloonga]

But what comes from this reaction?Maybe LiAlH4 gives H and takes Br so it produces alkane and LiAlH3Br or?
iF IT'S e2 THEN ALKENE SHOULD BE THE PRODUCT..

[movies]

This is a reduction reaction, not an elimination.  The product would be propane.

[Yggdrasil]

Could this be thought of something like an S_N2 reaction?  The LAH donates a hydride that acts as the nucleophile and the Br is the leaving group.

[slimlingsaucer]

I don't see how this could be elimination because the base should take Hydrogen and how is LiAlH4 capable of that?
SN2 seems to be the better solution..H reacts like Nucleophile and bonds with C and Br is the leaving group..

Well, LAH (LiAlH₄) is a strong base, and your substrate is a primary alkyl halide and definitely capable of E₂, so i would say that you are seeing E₂ out of this reaction.  Any thoughts from anyone else?

How?could you explain..what products could it give if it is e2?

[macman104]

It's not E2.  Like Movies said, LiAlH4 reduces primary alkyl halides to alkanes

Hey, look publications!

[choongaloonga]

Could this be thought of something like an S_N2 reaction?  The LAH donates a hydride that acts as the nucleophile and the Br is the leaving group.

Yes I think it can.

[movies]

Could this be thought of something like an S_N2 reaction?  The LAH donates a hydride that acts as the nucleophile and the Br is the leaving group.

Yes I think it can.

The results of the paper linked by macman suggest that the mechanism may in fact proceed through radical intermediates.  It's a good read!