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This question appeared on an exam and I translate it into English.

In the NMR $\ce{H^1}$ spectrum of a pure substance, with molecular formula $\ce{C4H8Br2}$, the following chemical shifts (in ppm) are observed: 1.95 (s), 3.95 (s): the proportion between the areas of integration of these signals is 3:1. This spectrum corresponds to the molecule: (A) 2,3-dibromobutane. (B) 1,2-dibromobutane. (C) 1,2-dibromomethylpropane. (D) 1,1-dibromomethylpropane. (E) 1,3-dibromomethylpropane.

Note that no access to chemical shift data or graphs etc was given. The answer provided is (C). I can see that 1,2-dibromomethylpropane has 6 equivalent hydrogens (in the methyl groups) and 2 equivalent ones in $\ce{-CH_2{}-}$ group, corresponding to a 3:1 ratio as in the question.

Answer: 1,2-dibromomethylpropane dibromomethylpropane

However, I cannot see why (A) cannot be an answer. It has a ratio of 6:2 equivalent hydrogens.

Doubt: 2,3-dibromobutane dibromobutane

I know I am not taking into account the chemical shifts. So I am not sure I could do that, therefore asking for help.

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    $\begingroup$ In it's present form, I would prefer 1,2-dibromo-2-methylpropane for the first structure shown over the currently used caption of 1,2-dibromomethylpropane. $\endgroup$
    – Buttonwood
    Nov 12, 2021 at 10:59
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    $\begingroup$ Note that the names (C) 1,2-dibromomethylpropane, (D) 1,1-dibromomethylpropane, and (E) 1,3-dibromomethylpropane are all not valid. $\endgroup$
    – Loong
    Nov 13, 2021 at 16:07

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It is a matter of coupling distance if singlet signals split into multiplets to be resolved as such in the NMR experiment. (Other factors, beyond your question, e.g., the difference between a C(sp³), or C(sp²), too.)

One of the general rules here, the greater the distance between to nuclei which in principle may couple with each other (i.e., a matter of their nuclear spin), the lesser the coupling, and the lesser the split of the signals. So if you compare two cases

enter image description here

you count the number of bonds between the two types of hydrogen atoms in question. Assuming coupling by bonding,* you have four bonds separating the hydrogens (case 1, where coupling is denoted as $^4\!J_\text{HH}$), or only three (case 2, $^3\!J_\text{HH}$).

For illustration, compare the 1H NMR spectra of ethyl acetate and propyl acetate. A $^3\!J_\text{HH} = 6...\pu{8 Hz}$ (reference), is resolved today routinely, which includes spectrometers of low magnetic field strength:

enter image description here

($\ce{^1H}$-NMR spectrum of EtOAc recorded in $\ce{CDCl3}$, $\pu{90 MHz}$; published by 1999-03-31 by SDBS)

Now compare an analogue spectrum of propyl acetate, recorded in the same solvent and with the same field strength, where a $^4\!J_\text{HH}$ coupling occurs, with particular attention to the triplet around $\pu{1 ppm}$ about the terminal $\ce{-CH3}$ group:

enter image description here

($\ce{^1H}$ NMR spectrum of propyl acetate recorded in $\ce{CDCl3}$, $\pu{90 MHz}$, published by 1999-03-31 by SDBS)

While the lines of the triplet at least got broader, the split of the signal by $^3\!J_\text{HH}$ coupling is much stronger than the simultaneous one by $^4\!J_\text{HH}$ coupling.

*) A second mechanism is coupling across space, although molecular rotation causes this to be averaged to zero and thus not visible in solution-state NMR.

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