Alcohols
Alcohols are formed when a hydrogen atom in aliphatic hydrocarbon is replaced by –OH group. Alcohols are classified as primary, secondary and tertiary alcohols depending on the attachment of –OH group to primary, secondary and tertiary carbon atom, respectively. In alcohols, the oxygen of the –OH group is attached to carbon by a sigma (σ) bond formed by the overlap of a sp3 hybridised orbital of carbon with a sp3 hybridised orbital of oxygen. Alcohols consist of two parts, an alkyl group and a hydroxyl group. The properties of alcohols are chiefly due to the hydroxyl group. The nature of alkyl groups simply modify these properties. The boiling points of alcohols increase with increase in the number of carbon atoms (increase in van der Waals forces). In alcohols, the boiling points decrease with increase of branching in carbon chain (because of decrease in van der Waals forces with decrease in surface area). Alcohols undergo dehydration (removal of a molecule of water) to form alkenes on treating with a protic acid e.g., concentrated H2SO4 or H3PO4, or catalysts such as anhydrous zinc chloride or alumina. Alcohols are versatile compounds. They act both as nucleophiles and electrophiles. An alcohol with molecular formula C4H10O (A) is prepared by the reduction of corresponding ketone. A is also prepared by the reaction of propanal with methylmagnesium bromide followed by hydrolysis. Compound A on dehydration gives alkene (B).
The IUPAC name of the major alkene (Compound B) formed by the dehydration of compound “A” is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → But-2-ene
Alcohols are dehydrated by heating with concentrated \( \ce{H2SO4} \) or \( \ce{H3PO4} \). The –OH is protonated and leaves as water, giving a carbocation; a proton is then lost from a neighbouring (β) carbon to form the double bond. When there is a choice of β-hydrogens, the Saytzeff (Zaitsev) rule applies: the more substituted alkene (the one with fewer hydrogens on the double-bond carbons) is the major product, because it is more stable.
A is butan-2-ol. Protonation and loss of water give the secondary carbocation, which can lose \( \ce{H^{+}} \) from C-1 or from C-3:
\( \displaystyle \ce{CH3-CH(OH)-CH2-CH3} \xrightarrow{\ce{H^+}} \ce{CH3-C^+H-CH2-CH3 + H2O} \)
\( \displaystyle \text{loss of H from C-3: }\ce{CH3-CH=CH-CH3}\ \text{(but-2-ene, major)} \)
\( \displaystyle \text{loss of H from C-1: }\ce{CH2=CH-CH2-CH3}\ \text{(but-1-ene, minor)} \)
The net reaction is:
\( \displaystyle \ce{CH3-CH(OH)-CH2-CH3} \xrightarrow[\Delta]{\text{conc. }\ce{H2SO4}} \ce{CH3-CH=CH-CH3 + H2O} \)
But-2-ene has two alkyl groups on its double bond, while but-1-ene has only one, so but-2-ene is the major alkene B.
Why the other options are incorrect
Option (1) But-1-ene — Incorrect. It forms too, but as the minor product, because it is the less substituted alkene.
Option (3) 2-methylpropene — Incorrect. It has a branched skeleton and comes from dehydrating 2-methylpropan-2-ol or 2-methylpropan-1-ol; A has an unbranched chain.
Option (4) Propene — Incorrect. It has only three carbon atoms; dehydration removes water but no carbon, so a four-carbon alcohol gives a four-carbon alkene.