1° alcohol
Summary
A primary (1°) alcohol has its OH on a carbon bonded to just one other carbon, written R–CH₂OH. The most testable point is oxidation: PCC stops at the aldehyde, while strong oxidants such as Jones reagent or KMnO₄ carry it on to the carboxylic acid.
Detail
Meaning: A primary (1°) alcohol has its OH on a carbon bonded to only one other carbon, so the group is R–CH₂OH. Ethanol and 1-butanol are examples; methanol, with no carbon neighbors, is usually grouped with them. The carbinol carbon carries two hydrogens, so it can be oxidized in two steps. It is also unhindered, so substitution tends to go by SN2, because a primary carbocation is too unstable for SN1.
On the MCAT: Oxidation is the classic question. PCC stops at the aldehyde, while stronger oxidants such as Jones reagent (CrO₃, H₂SO₄), K₂Cr₂O₇ or KMnO₄ carry it on to the carboxylic acid. In reverse, LiAlH₄ reduces aldehydes, carboxylic acids and esters to primary alcohols, whereas NaBH₄ reduces aldehydes and ketones but generally not acids or esters. Converting the OH into a good leaving group (TsCl, PBr₃, SOCl₂) sets up an SN2 reaction, and the passage may ask you to predict the product or the mechanism.
Common confusion: Students count the length of the chain instead of the carbons attached to the carbinol carbon. 1-Propanol is primary; 2-propanol is secondary. Oxidation to a ketone belongs to secondary alcohols, and tertiary alcohols, which have no H on the carbinol carbon, resist oxidation by these reagents.
Sources
- Organic Chemistry by Klein
- Organic Chemistry by Wade
- Organic Chemistry by Clayden, Greeves and Warren
Written for the MCAT and checked before it was published. For study reference only. Spotted an error? Let us know.