[Dynamo22]

Having a tough time figuring out this reaction sequence and the product. It seems obvious but the given molecular formula is throwing me off because I'm not sure how you eliminate the oxygen from the starting material.

Starting material: 4-pentanoic acid
Reagents in order: 1) SOCl2, 2) Piperidine 3)LiAlH4, 4) H3O+
Product: C10H19N

I know there are 2 DU. The only product I could come up with is attached. Not sure if that is correct or not. Any hints would be appreciated.

[discodermolide]

What reactions of carboxylic acids do you know of? especially with SOCl₂

[Dynamo22]

Acylation of the carboxylic acid. Cl replaces the OH group. Which then I figured you would add that group to the piperidine.

[discodermolide]

OK, what's next then. what does the LiAlH₄ do?

[Dynamo22]

Should reduce the carbonyl to a 2° alcohol. But how do you go about removing the hydroxy group with those reagents? That is what I'm not sure of.

[discodermolide]

Well you reduced it to the alcohol, what does the acid do?

[Dynamo22]

Well, it would probably remove it. I thought it protonated the oxygen, thats where I was getting confused. Would it create the double bond also?

[discodermolide]

Yes water is eliminated to give your structure WITHOUT the double bond at the end of the chain.

[Dan]

Starting material: 4-pentanoic acid

I assume you mean 4-pentenoic acid?

Also, the product you propose (C₁₀H₁₇N) does not match the formula for the product given in the question (C₁₀H₁₉N).

Look up LiAlH₄ reduction of amides. It does not generally give alcohols:

Should reduce the carbonyl to a 2° alcohol.

This secondary alcohol you mention - do you mean a hemiaminal? You are half way there - the hemiaminal will react further with LiAlH₄.

[discodermolide]

I assumed the opposite of Dan's observation.
I assume also you were given the reagents, or did you come up with them yourself?

[Dynamo22]

Yes it was 4-pentenoic acid. Thanks for the heads up. The reagents were given. I'm picking up what you're throwing down now. Makes much more sense. I wasn't even taking into account that the carboxylic acid turned into an amide.

[discodermolide]

Never mind.

[Dynamo22]

Alright, I have a new wrinkle in this problem. This is the context of the whole thing:
When 1,5-dibromopentane reacts with ammonia, among several products isolated is a water-soluble compound A that rapidly gives a precipitate of AgBr with acidic AgNO3 solution. Compound A is unchanged when treated with dilute base, but treatment of A with concentrated NaOH and heat gives a new compound B (C10H19N) that decolorizes Br2 in CCl4. Compound B is identical to the product obtained from the reaction sequence shown below. Identify A and B. Hint: First, identify compound B by completing the reaction sequence. The result of the bromine test suggests the presence of a particular functional group in compound B.

The structure I figured out for B is correct, which is the one we hashed out. For structure A, I really don't know what it could be...we've never really emphasized reactions like this. I would imagine you would need a NH with two 5 carbon chains coming off each side. But I am not sure what would be attached to the ends of those chains in order to 1) close the ring and 2) create an alkene. I assume treating with the concentrated NaOH and heat would do a ring closing metathesis, but again..we haven't learned that with amines yet. Any help to get this darn thing figured out would be much appreciated.

[Dan]

I assume treating with the concentrated NaOH and heat would do a ring closing metathesis

Bad assumption.

You start with an alkyl bromide (electrophile) and an amine (nucleophile). Can you suggest some kind of reaction between them?
Perhaps google the reactivity of alkyl halides to get you started.

As the question states, there are a number of potential products from the first reaction, so you need to work backwards from B as well to get A. You know what B is - it is an alkene formed by treatment of A with NaOH. NaOH is a strong base, what kind of reactions can it do with organic compounds?

[orgopete]

I agree with Dan here. If B is obtained from A and it gives a positive test for bromine, what are the possibilities? Also, hint Hoffmann.