HNMR - Peak Splitting If you have $\ce{CH3-\color{\red}{CH2}-CH2Br}$

Then the $\ce{CH2}$ (in red) would have a peak splitting of $12$ because $(n+1)(m+1)$
where $n$ for the $\ce{CH3}$ is $(3+1=4)$ and m for $\ce{CH2}$ is $(2+1=3)$ so its $4\times3=12$

source:




  In the general case, a signal will be split into $(n + 1) \times (m + 1)$
  peaks for an $\ce{H}$ atom that is coupled to a set of $n~\ce{H}$ atoms with one
  coupling constant, and to a set of $m~\ce{H}$ atoms with another coupling
  constant.


How can we determine differences in the coupling constant by just looking at the structure? Otherwise how do I know when to apply $n+1$ OR $(n+1)(m+1)$?

But in the other textbook it says:

$\ce{CH3-\color{\red}{CH2}-CH2-COOH}$

The $\ce{CH2}$ (in red) has a peak splitting of $6$ because $(n+1)$ where $n=5$, $(5+1=6)$ and the $n$ is the $3~\ce{H}$ on $\ce{CH3}$ plus the $2~\ce{H}$ on $\ce{CH2}$.

Source:



Which is the correct method? Or are both right? and why? 

Alternatively symmetric adjacent hydrogen groups: 

$\ce{CH3–\color{\red}{CH2}–CH3}$

Is the $\ce{CH2}$ split to 4 peaks due to symmetric adjacent hydrogen environments? Due to $n+1$, where $n=3$ so $(3+1=4)$.  I read that it should be split into 7 peaks - but where are they getting 7 from - is it $n+1$, where $n=6$ so $(6+1=7)$. For symmetric adjacent groups do you allow $n$ to be the sum of all hydrogens adjacent/or not?
Both are correct to some degree.

The first book predicting 12 lines is being very theoretical.  In principle with narrow enough line widths 12 lines would be observed.  However, due to coupling constants (J) being similar from the beta protons to the alpha and gamma protons, in practice with reasonable resolution, only 6 lines are observed (the first book is actually explaining this in fine print).


(source: ucalgary.ca)  

Credit: Univ. Calgary