So, what have we learned from the NaOH/HBr reaction? Hydroxide is reactive because the extra lone pair (i.e. the negative charge) is on a fairly small atom (O). In the presence of HBr, the hydroxide has a potential vehicle for becoming more stable through reaction. The electron-rich O is attracted to the electron-poor H and an O-H bond starts forming. As the introduction of more electrons breaks the “octet” rule at H, the H-Br bond must begin breaking, which makes sense as it’s weak. The H-Br bond pair of electrons becomes a lone pair on Br, which is able to handle that extra lone pair (i.e. the negative charge) because Br is reasonably electronegative but moreso because Br is large and can handle the four lone pairs. Overall, a weak H-Br bond is sacrificed for a stronger O-H bond and the negative charge ends up in a better place on Br.