3, 3-Dimethyl-2-butanol cannot be prepared by:
Choose the correct answer from the options given below:

Solution

Chemistry ma’am · 3 steps

Draw the target first, (CH₃)₃C-CH(OH)-CH₃, then check what each reaction really ends with.

1. The target. 3,3-Dimethyl-2-butanol is (CH₃)₃C-CH(OH)CH₃. It has 6 carbons, a tert-butyl group, and an -OH on the next carbon. A route works only if it gives this exact skeleton and stops at the alcohol.

   Route	What happens	Final product	Target?
   A	CH₃⁻ from MeMgBr adds to the C=O of (CH₃)₃C-CHO	(CH₃)₃C-CH(OH)CH₃	yes
   B	Acid hydration of CH₂=CH-C(CH₃)₂-CH₂CH₃, a 7-carbon alkene	a 7-carbon alcohol (the 2° cation also rearranges)	no
   C	Ozonolysis splits the symmetric alkene into 2 molecules of (CH₃)₃C-CO-CH₃, then NaBH₄ reduces the ketone	(CH₃)₃C-CH(OH)CH₃	yes
   D	LiAlH₄ reduces (CH₃)₃C-CO-CH₃	(CH₃)₃C-CH(OH)CH₃	yes
   E	Hg²⁺-catalysed hydration of (CH₃)₃C-C≡CH: the enol tautomerises	the ketone (CH₃)₃C-CO-CH₃	no

2. The two failures. B has one carbon too many. Its secondary carbocation also sits next to a quaternary carbon, so an alkyl shift gives a tertiary alcohol instead. E stops at the ketone, because the enol from the alkyne tautomerises to C=O and nothing in the reagents reduces it.
3. So. Only B and E cannot give 3,3-dimethyl-2-butanol. The answer is (a).

The idea to keep. For a 'cannot be prepared' question, first count the carbons in the product. Then ask whether the reaction stops at the right functional group: alkyne hydration gives a carbonyl, not an alcohol.

Answer from the NTA final answer key. Solution by PrepWiser.