For massless fermions, their chirality (determining under which representation of the Lorentz group they transform) and their helicity (projection of spin onto three-momentum) eigenvalues are the same. For massless antifermions, it's exactly opposite (see e.g. Itzykson & Zuber, Eq. (2-103) and the text below).
When we talk about which particles couple to $W$ bosons in the Standard Model, many sources say "left-handed fermions and right-handed antifermions". Knowing that many references use "..-handed" in terms of "..-chiral", this is not a problem for fermions (since chirality and helicity is the same in the massless limit), but for antifermions, this distinction is important.
Question: Is an antineutrino that participates in the interaction with a $W$ boson right-handed or right-chiral?
My thoughts are as follows: The fermions in the SM are implemented via left-chiral doublets and right-chiral singlets. The $W$ bosons only interact with the doublets. This interaction can be written like (e.g. Cottingham & Greenwood's "Introduction to the SM of particle physics", Eq.(12.15)): $$ \mathcal L = -\frac{g_2}{\sqrt 2} \nu_{eL}^\dagger \tilde\sigma^\mu e_L W_\mu^+ + ... $$ where $\psi$ is the left-chiral doublet. Since a Hermitian conjugation swaps the (0,1/2) and (1/2,0) reps of the Lorentz group, this means that e.g. a left-chiral $e_L$ and a right-chiral $\nu_L^\dagger$ come together. Conclusion: the $W$ bosons interact with left-chiral fermions and right-chiral antifermions. This would imply that (massless) neutrinos and antineutrinos in the SM are both left-handed, since a right-chiral antineutrino is left-handed.