When 3-methylpentan-3-ol undergoes an elimination reaction in the presence of
hot concentrated H2SO4, the product which is MOST likely to form is:
A. 3-methylpent-3-ene
B. 2-ethylbut-2-ene
C. 3-methylpentane
D. 3-pentanone
E. 3-methylpent-2-ene
There will be several products formed. However, it is known that this is an elimination reaction, and water can only be eliminated in such case. Therefore, possible products are alkenes.
Options A) and B) are violating the naming rules (these are wrong names of the same compound from the option E);
Option C) is incorrect because this is an alkane, not alkene;
Option D) is incorrect because this is a ketone, not alkene;
The only remaining option is E), and it correctly represents one of the products of this reaction. The basic scheme is below.
Answer: E
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