Q.Explain by writing a reaction, the main product formed on heating 2-methylbutan-2-ol with concentrated sulphuric acid. Will the main product in the above reaction show geometrical isomerism?
Step 1. Identify the substrate and reaction type. 2-Methylbutan-2-ol is (CH3)2C(OH)-CH2-CH3, a tertiary alcohol. Heating with concentrated H2SO4 is the acid-catalysed dehydration of section 15.2.2, a beta-elimination removing -OH from the alpha-carbon (C2) and an H from a beta-carbon.
Step 2. Apply Saytzeff's rule. C2 has two possible beta-carbons to lose H from: either of its two methyl groups (giving the LESS substituted alkene, 2-methylbut-1-ene, CH2=C(CH3)-CH2-CH3), or C3 of the ethyl chain (giving the MORE substituted alkene, 2-methylbut-2-ene, CH3-C(CH3)=CH-CH3). Saytzeff's rule states the more-substituted alkene is the major product, so 2-methylbut-2-ene is the main product formed.
Step 3. Check for geometrical isomerism. In 2-methylbut-2-ene, the double bond is between C2 and C3. C2 carries two substituents: the original C1 methyl and the branch methyl -- both are plain CH3 groups, i.e. IDENTICAL to each other. Since one carbon of the double bond has two identical groups, the molecule cannot exist as distinct cis and trans forms (the same rule that rules out geometrical isomerism for any 1,1-disubstituted-type alkene carbon, section 15.2.1).
Main product: 2-methylbut-2-ene, CH3-C(CH3)=CH-CH3. It does NOT show geometrical isomerism, because the C2 carbon of the double bond carries two identical methyl groups.
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