[rmjmu507]

I'm trying to determine how bromobenzene and benzaldehyde would react in the presence of n-BuLi...

I've come up with two options, with justifications for each, but cannot narrow it down anymore:

1- n-BuLi will react with bromobenzene, forming phenyllithium. This will then react with benzaldehyde to form biphenylmethanol.
      Doubts: I know Li addition to benzene is immeasurably slow without using a strong sigma donor - is this also the case for alkylbenzenes? Because if so, then I would argue that the organolithium reagent would react with the aldehyde as that would be the kinetically-favored product.
2- n-BuLi will add to the aldehyde to make 1-phenyl-1-pentanol.
      Doubts: I'm not sure what would happen with the lithium atom...

Any advice?

[snorkack]

I'm trying to determine how bromobenzene and benzaldehyde would react in the presence of n-BuLi...

I've come up with two options, with justifications for each, but cannot narrow it down anymore:

1- n-BuLi will react with bromobenzene, forming phenyllithium. This will then react with benzaldehyde to form biphenylmethanol.
      Doubts: I know Li addition to benzene is immeasurably slow without using a strong sigma donor - is this also the case for alkylbenzenes? Because if so, then I would argue that the organolithium reagent would react with the aldehyde as that would be the kinetically-favored product.

No way. The reaction would just eliminate LiBr. So
Ph-Br + Bu-Li -> Ph-Bu + Li-Br
and not
Ph-Br + Bu-Li -x> Ph-Li + Bu-Br

2- n-BuLi will add to the aldehyde to make 1-phenyl-1-pentanol.
      Doubts: I'm not sure what would happen with the lithium atom...

Simple. Cation for the alcoholate, so:
Bu-Li + Ph-CH=O -> Ph-CHOLi-Bu

[rmjmu507]

This is not two different reactions...both PhBr and benzaldehyde are present, along with nBuLi, and I'm trying to figure out where the organolithium reagent would react.

[discodermolide]

As the second poster noted n-Butyllithium is quite nucleophillic. In the presence of a mixture, as you have, the aldehyde is by far the most reactive. The Li will end up in an ionic system with the C-O^- which originated from the aldehyde. Some of the bromobenzene will also react, how much will probably depend on the temperature.

[rmjmu507]

is temperature dependence the only determining factor?

[snorkack]

As the second poster noted n-Butyllithium is quite nucleophillic. In the presence of a mixture, as you have, the aldehyde is by far the most reactive.

So the first step would be overwhelmingly
BuLi + PhCHO -> Ph(Bu)CHOLi
and a small side reaction of
BuLi + PhBr -> PhBu + LiBr

The Li will end up in an ionic system with the C-O^- which originated from the aldehyde. Some of the bromobenzene will also react, how much will probably depend on the temperature.

React with what?
React with BuLi as a side reaction?
React with BuLi excess after PhCHO has been consumed?
Or react with alcoholate?
Production of ether is conceivable:
Ph(Bu)CH-O-Li + Br-Ph -> Ph(Bu)Ch-O-Ph + LiBr
But does it take place significantly, or is (secondary) alcoholate too weak nucleophile and aromathic bromide too inert for nucleophilic attack to form a phenyl ether?

[discodermolide]

React with BuLi.
I don't think you can form the ether by direct substitution of bromobenzene with the alcoholate.

[orgopete]

This looks very hypothetical. I think the result will depend upon how the reaction is done. I cannot imagine introducing all of these reagent instantaneously. What is added to what?

[rmjmu507]

it's all mixed together.

i'm pretty sure the butyl carbanion would attack the carbonyl.

thanks for all your feedback/discussion!

[orgopete]

it's all mixed together.

How is that possible? If benzaldehyde is added to bromobenzene, initially the reaction will contain a much larger concentration of bromobenzene than benzaldehyde. If an equimolar amount were added, they will only be equal at the end of the addition. It should be apparent that order of addition will affect the result. If BuLi is first added to benzaldehyde, it will react first. If benzaldehyde is added last, it will react with the product of a reaction of bromobenzene and BuLi. Given that BuLi is the most reactive reagent, when it is added will determine the outcome. My understanding of organic chemistry suggests that bromobenzene would not be added last as it would not react and it's addition would be unnecessary.

[snorkack]

it's all mixed together.

If benzaldehyde is added last, it will react with the product of a reaction of bromobenzene and BuLi.

They wouldn´t. Because the product, a simple hydrocarbon (butylbenzene) is quite inert to benzaldehyde.

Given that BuLi is the most reactive reagent, when it is added will determine the outcome. My understanding of organic chemistry suggests that bromobenzene would not be added last as it would not react and it's addition would be unnecessary.

Benzaldehyde is much less liable to react with a hydrocarbon than bromobenzene with an alcoholate.

[orgopete]

it's all mixed together.

If benzaldehyde is added last, it will react with the product of a reaction of bromobenzene and BuLi.

They wouldn´t. Because the product, a simple hydrocarbon (butylbenzene) is quite inert to benzaldehyde.

Given that BuLi is the most reactive reagent, when it is added will determine the outcome. My understanding of organic chemistry suggests that bromobenzene would not be added last as it would not react and it's addition would be unnecessary.

Benzaldehyde is much less liable to react with a hydrocarbon than bromobenzene with an alcoholate.

I didn't think this problem required a lot of thinking, but I was wrong. I thought the only reasonable reaction was a transmetallation reaction followed by addition to benzaldehyde. I didn't think any other possibility seemed plausible. Snorkack virtually affirms this, though in a contrary fashion. What seems missing is to do a Google search. Okay, I did. Can I find an example? Sure, Wittig in 1940. See slide 5 or so here, http://www.princeton.edu/chemistry/macmillan/group-meetings/SL-LiExchange.pdf.