Differences between ¹³C-decoupled NMR experiment vs a ¹³C-coupled

How does the acquisition time changes between $$\ce{^{13}C}$$–decoupled NMR experiment vs a $$\ce{^{13}C}$$–coupled?

Would the acquisition time be longer if a decoupling of the protons (H) is carried out?

• Are you talking about 1H-decoupled 13C NMR? When you say 13C-decoupled that means that you are removing couplings to 13C nuclei. There is usually no real need for that since 13C has very low abundance... – orthocresol Feb 21 '19 at 0:34

The answer is a bit complicated because there are a number of ways to apply decoupling (see eg this site which describes different decoupler gating modes), depending on the goal. For simplicity I limit the discussion to acquisition of a 1D $$\ce{^{13}C}$$ spectrum:
1. $$\ce{^1H}~$$ decoupling during the entire experiment: this provides a signal enhancement due to the NOE from $$\ce{^1H}$$ to $$\ce{^{13}C}$$, and suppresses couplings, which reduces spectral complexity and also increases carbon s/n due to collapsed multiplets.
Applying decoupling to obtain a signal enhancement from NOEs allows you to reduce the overall acquisition time (reduce the number of FIDs you need to acquire) relative to no NOE. This typically works for all but quaternary carbons. Decoupling during acquisition also can allow you to reduce the overall acquisition time, even if you do not obtain an NOE enhancement, by collapsing signals from multiplets into single peaks. The longest experiment would be a truly quantitative one, and particularly if you want full coupling information and s/n equivalent to that from a decoupled experiment. Then you would not apply decoupling at all. In that case you would have to wait on the order of $$\mathrm{5 \times } \ce{^{13}C} ~\mathrm{T_1}$$ and acquire enough scans to get the desired s/n.