# Why do t-butyl bromide and silver(I) nitrite form the nitroalkane, not the nitrite ester?

We know that reaction of $\ce{AgNO2}$ with alkyl halides is SN1 type reaction and follows HSAB principle. So in the reaction of $\ce{(CH3)3C Br}$ we would expect $\ce{AgBr}$ to form and also a carbocation $\ce{(CH3)3C+}$.

According to HSAB, since the electrophile is hard, we expect the hard nucleophile to attack. So attack is expected from oxygen, and expected product is $\ce{(CH3)3C-ONO}$

But my book states that reaction product is $\ce{(CH3)3C-NO2}$. Why is this product formed?

• Even though it is widely taught, HSAB is quite inappropriate in this context. I don't have the time now to explain that thoroughly, but the key reference is doi.org/10.1002/anie.201007100. See also Jan's answer about cyanide alkylation. I also somewhat question the claim that the nitroalkane RNO2 is formed. According to doi.org/10.1021/ja01626a028 the nitrite ester RONO is obtained in ~65% yield and the nitroalkane RNO2 cannot be observed. – orthocresol Jan 7 '19 at 18:39