Question

Why do alkyl groups have low electronegativity versus hydrogen? How does this affect the alkalinity of alcohols?

Answer 1

Hint: When we talk of hyperconjugation, and wonders about how $C - H$ bonds can stabilize alkyl carbocations. It really isn't about electronegativity, but about the p orbital on the central C atom being vacant (and thus electropositive). In general, delocalized negative charge stabilizes a molecule.

Complete answer:
We know that alkanes have a high pKa, on the order of \[50\~60\] generally. That physically means it is difficult to deprotonate them. The equilibrium can be shown like this:
\[RCH3\;\underset {} \leftrightarrows RC{H^{2 - }} + {H^ + }\]
If one manages to deprotonate an alkane, then what would result is an exceptionally great (Lewis) base, whose electron density also happens to be extremely localized onto the frontal carbon atom, allowing for easy electron donation. Greater the nucleophilicity which is the ability to gather electron density easily and donate it quickly, but great basicity is the thermodynamic instability of the compound in question. Looking for alkanes, alcohols have hydroxyl groups (−OH), which are electron-donating as well, back onto the alkyl group's frontal carbon, since their σ bonding molecular orbitals are doubly-occupied. That means the electron density is localized more in the $C - O$ bond (towards $O$ , since it is more electronegative than C) than the $O - H$ bond. This weakens the $O - H$ bond and makes that hydrogen acidic, instead of an alkyl hydrogen.

Note:
We need to know that alcohols are (much) more acidic than alkanes. This is reflected in the pKa of alcohols being approximately $16$, and not \[50\~60\]. Note that since this is a log scale, actually, alcohols are about $1034$ times more capable of singly-ionizing than alkanes.