When tert-pentyl bromide is treated with 80% ethanol, the amounts of alkene products indicated below are detected on analysis:
CH3CH=C(CH3)2 (32%)
CH3CH2C=CH2 (there is also a CH3 single bonded underneath the single C, but I didn't know how to draw it in) (8%)
tert-Pentyl alcohol, tert-Pentyl ethyl ether (60%)
Explain why Compound I is formed in far greater amount than the terminal alkene.