This is related to Weibel Homological Algebra Exercise 3.1.1
All modules considered are left modules. Let $M$ be a left $R-$module. Set $Ann^M(r)=\{m\in M\vert rm=0\}$ for any module $M$. I will treat $R$ as bimodule over $R$ itself but $Ann^R(r)$ treats $R$ as left module.
Consider the following exact sequence for $r$ being zero divisor. $0\to Ann^R(r)\to R\xrightarrow{r\cdot} R\to R/(r)\to 0$ where $(r)=rR$. Let $B$ be a left $R-$module.
It can be split into 2 short exact sequence as the followings.
(1) $0\to Ann^R(r)\to R\xrightarrow{r\cdot}(r)\to 0$
(2) $0\to (r)\to R\to R/(r)\to 0$
Apply $-\otimes_R B$ functor to 1, I have the following exact sequence.
$0\to Tor^1((r),B)\to Ann^R(r)\otimes B\to R\otimes B\xrightarrow{r\cdot\otimes 1_B}(r)\otimes B\to 0$
Now $R\otimes_R B\cong B$ and $(r)\otimes B\cong rB$. Then above sequence is rewritten in the following form.
$0\to Tor^1((r),B)\to Ann^R(r)\otimes B\xrightarrow{multiplication}B\xrightarrow{r\cdot}rB\to 0$
Then I am going to describe kernel of the $r\cdot$(last epi map). $Ker(r\cdot)=Ann^B(r)$ whereas image of "multiplication" map is $Ann^R(r)B$. Clearly $Ann^R(r)B\subset Ann^B(r)$.
$\textbf{Q:}$ From exactness, I deduced $Ann^R(r)B=Ann^B(r)$. The book gives the short exact sequence $0\to Tor^1(R/(r),B)\to Ann^R(r)\otimes B\xrightarrow{multiplication}Ann^B(r)\to Tor^1(R/(r),B)\to 0$. This indicates that I should always get $Tor^1(R/(r),B)=0$ which is not possible in general. Where is my mistake above?