I need to know what the product of this Sn2 reaction is

(CH3)3N reacting with bromoethane

I thought the product would be CH3CH2N(CH3)3

but my book says it's CH3CH2N(CH3)3Br-

 

is it a mistake or am i just not understanding

The product you have written is almost correct, but remember that you have made a quaternary nitrogen. A nitrogen with four bonds has a positive charge, so it becomes a salt. The Br- they include is the counter ion to that positive charge.

The product you have written is almost correct, but remember that you have made a quaternary nitrogen. A nitrogen with four bonds has a positive charge, so it becomes a salt. The Br- they include is the counter ion to that positive charge.

How would I know it's a quaternary N?

 

I just don't understand why the Br is reatined if it's a substitution reaction. In the other exercises I did where that exercise was (like for example CH3CH2CH2O- reacting with bromoethane, the product is ch3ch2och2ch2ch3)

You'd know because there are four bonds going to the N. The three methyl groups and the ethyl group. The formal charge on that nitrogen is then +1. The Br isn't retained in the same way as it is in the original structure. It is no longer covalently bonded to the ethane, it's ionically bonded to the positive nitrogen. You could draw your product without the Br-; so long as you remembered the positive charge on the nitrogen, it is still correct.

 

The reason that you don't see any Br- in your other example is that you aren't making a salt and it's therefore not needed as a counter ion.

 

When I get home, I'll draw out what I am talking about. Hopefully that will make it clearer.

Here we go.

 

 

In the first example, the nitrogen formally "loses" an electron when it bonds to the carbon in bromoethane In a two electron covalent bond, we generally say that each atom "owns" one electron each. Accordingly, when the neutral nitrogen uses its lone pair of electrons to attack the carbon and form a bond, one of the electrons will still belong to the nitrogen, but the other to the carbon that it's bonded to. The loss of the single electron makes the nitrogen positively charged. The Br- ions will then be attracted to this charge and you end up with the salt, as shown.

 

Compare that to your other example. The alkoxide has a negatively charged oxygen. It attacks in a similar way to the first example and just as the nitrogen formally loses an electron, so does the oxygen. The difference is that it started out negatively charged and so the formal loss of an electron makes it neutral - no need for a counter ion. The bromide will still form a salt, just not with your compound (the alkoxide would have been present as a metal salt, so the Br- will probably make a salt with whatever that metal is after the alkoxide reacts).